THE UNIVERSITY OF CALGARY

A Cubic Extension of the Lucas Functions

by

Eric L. F. Roettger

A THESIS

SUBMITTED TO THE FACULTY OF GRADUATE STUDIES

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE

DEGREE OF DOCTOR OF PHILOSOPHY

DEPARTMENT OF MATHEMATICS AND STATISTICS

CALGARY, ALBERTA

January, 2009

c© Eric L. F. Roettger 2009

THE UNIVERSITY OF CALGARY

FACULTY OF GRADUATE STUDIES

The undersigned certify that they have read, and recommend to the Faculty of

Graduate Studies for acceptance, a thesis entitled “A Cubic Extension of the Lucas

Functions” submitted by Eric L. F. Roettger in partial fulfillment of the requirements

for the degree of DOCTOR OF PHILOSOPHY.

Supervisor, Dr. H. C. WilliamsDept. of Mathematics and Statistics

Co–Supervisor, Dr. S. MullerDept. of Mathematics and StatisticsUniversity of Wyoming

Dr. R. ScheidlerDept. of Mathematics and Statistics

Dr. M. BauerDept. of Mathematics and Statistics

Dr. W. TittelDept. of Physics and Astronomy

Dr. C. BallotLaboratoire Nicolas OresmeUniversite de Caen

Date

ii

Abstract

From 1876 to 1878 Lucas developed his theory of the functions Vn and Un, which

now bear his name. He was particularly interested in how these functions could

be employed in proving the primality of certain large integers, and as part of his

investigations succeeded in demonstrating that the Mersenne number 2127 − 1 is

a prime. Vn and Un can be expressed in terms of the nth powers of the zeros of

a quadratic polynomial, and throughout his writings Lucas speculated about the

possible extension of these functions to those which could be expressed in terms of

the nth powers of the zeros of a cubic polynomial. Indeed, at the end of his life he

stated that “by searching for the addition formulas of the numerical functions which

originate from recurrence sequences of the third or fourth degree, and by studying in

a general way the laws of residues of these functions for prime moduli. . . we would

arrive at important new properties of prime numbers.”

In this thesis we discuss a pair of functions that are easily expressed as certain

symmetric polynomials of the zeros of a cubic polynomial and were undoubtedly

known to Lucas. We show how their properties seem to underlie the theory that

Lucas was seeking. We do this by deriving a number of results which show how

the combinatorial and arithmetic aspects of these functions provide an extension

of Lucas’ theory. Furthermore, we develop many new results, which illustrate the

striking analogy between our functions and those of Lucas. We also argue that, while

Lucas very likely never developed this theory, it was certainly within his abilities to

do so.

iii

Acknowledgments

I am endlessly indebted to my supervisor Dr. H. C. Williams. Without him I have

no doubt this thesis would never have been completed. Beyond his valuable criticism

throughout the writing of this thesis (and the endless rounds of corrections), Hugh

provided me with much-needed guidance and counsel in all parts of my life. He also

provided me with a considerable amount of money through the years, for which I

am also properly grateful. I can sincerely say that no student has ever had a better

supervisor than I.

Thanks to my co-supervisor Dr. Siguna Muller, for her encouragement and for

providing me with such a remarkable thesis topic. Many thanks to my committee

members, Dr. Christian Ballot, Dr. Mark Bauer, Dr. Renate Scheidler, and Dr. Wolf-

gang Tittel for the time they invested in reading my thesis and suggesting changes.

I would like to thank and acknowledge the Natural Sciences and Engineering

Research Council of Canada for funding received. Thanks also to the Faculty of

Graduate Studies and the entire Department of Mathematics at the University of

Calgary.

Many thanks to my colleagues at the University of Calgary Mathematics Depart-

ment: Aaron Christie, for his proofreading; to Pieter Rozenhart, my office mate who

always helped me solve elementary number theory problems; Alan Silvester, who is a

LATEXwizard; and finally Kjell Wooding, who I must thank for all the conversations

we had over beer or, to a lesser extent, over coffee.

Finally, I owe thanks to my family. My father Joe, mother Shirley, sister Jennelle,

and brother David. I thank them for their support throughout this entire venture.

iv

Table of Contents

Approval Page ii

Abstract iii

Acknowledgments iv

Table of Contents v

1 The Problem 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Commentary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4 Previous Extensions of the Lucas Functions . . . . . . . . . . . . . . 181.5 Our Objective . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2 Lucas Sequences 252.1 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.2 Computation of Un, Vn . . . . . . . . . . . . . . . . . . . . . . . . . . 272.3 Arithmetic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.4 Primality Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3 A New Attempt to Generalize the Lucas Sequences 463.1 De Longchamps’ Method . . . . . . . . . . . . . . . . . . . . . . . . . 463.2 Another Cubic Generalization . . . . . . . . . . . . . . . . . . . . . . 473.3 Our Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.4 Addition Formulas for Wn and Cn . . . . . . . . . . . . . . . . . . . . 563.5 Multiplication Formulas for Wn and Cn . . . . . . . . . . . . . . . . . 653.6 Calculating Generalized Lucas Sequences . . . . . . . . . . . . . . . . 70

4 Arithmetic Properties of {Cn} and {Wn} 794.1 Introductory Arithmetic Results . . . . . . . . . . . . . . . . . . . . . 794.2 Preliminary Results for the Law of Repetition for {Cn} . . . . . . . . 944.3 The Polynomial Km(x) . . . . . . . . . . . . . . . . . . . . . . . . . . 964.4 The Law of Repetition for {Cn} . . . . . . . . . . . . . . . . . . . . . 1044.5 The Law of Apparition for {Cn} . . . . . . . . . . . . . . . . . . . . . 1134.6 Solutions of the Cubic . . . . . . . . . . . . . . . . . . . . . . . . . . 115

v

5 Arithmetic Properties of {Dn} 1275.1 Preliminary Results for the Law of Repetition for {Dn} . . . . . . . . 1275.2 The Law of Repetition for {Dn} . . . . . . . . . . . . . . . . . . . . . 1305.3 The Law of Apparition for {Dn} . . . . . . . . . . . . . . . . . . . . . 142

6 Arithmetic Properties of {En} 1546.1 Preliminary Results for {En} . . . . . . . . . . . . . . . . . . . . . . 1546.2 A Law of Apparition for {En} . . . . . . . . . . . . . . . . . . . . . . 1646.3 Further Observations on the Law of Apparition for {En} . . . . . . . 167

7 Primality Testing 1827.1 An Analogue of Lucas’ Fundamental Theorem . . . . . . . . . . . . . 1827.2 The Case of T (N) = N2 +N + 1 . . . . . . . . . . . . . . . . . . . . 1887.3 The Primality of L . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1927.4 The Case of T (N) = N2 − 1 . . . . . . . . . . . . . . . . . . . . . . . 1977.5 Primality Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

8 Conclusion 2048.1 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2048.2 Improvements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2058.3 Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

Bibliography 207

A 215

vi

Chapter 1

The Problem

1.1 Introduction

Let P and Q be coprime integers and α, β be the zeros of the polynomial f(x) =

x2 − Px+Q where α 6= P . The Lucas functions Un and Vn are defined by:

Un = (αn − βn)/(α− β), Vn = αn + βn.

Since both Un and Vn are symmetric functions of the zeros of a polynomial with

integer coefficients they must be integers for all non-negative integral values of n.

Furthermore, they must both satisfy the simple linear recurrence:

Xn+1 = PXn −QXn−1.

Since U0 = 0, U1 = 1, V0 = 2, V1 = P , this recurrence can be used to compute Un

and Vn for any integral value of n.

From 1876 until about 1880, Edouard Lucas discovered many properties of these

functions. Indeed, it was during this period that he used these properties to develop

tests for the primality of large integers, including what is now called the Lucas-

Lehmer test for the primality of Mersenne numbers. (See section 5.4 of [Wil98].)

These tests were usually sufficiency tests, which could be used to prove whether a

number N of a certain special form is a prime. As Lucas well realized these test

were quite novel for their time, because instead of having to trial divide N by a large

1

2

number of integers, for example all the primes less than√N , it was only necessary

to compute some integer S and test whether N | S.

It is important to recognize, however, that Lucas found many other applications

of his functions. He was particularly struck by the similarity of his functions with

the sine and cosine functions; in fact, he noticed that if i is used to denote a zero of

x2 + 1, then

Un = (2Qn/2/√−∆) sin[(ni/2) log(α/β)] and (1.1)

Vn = 2Qn/2 cos[(ni/2) log(α/β)], (1.2)

where ∆ = (α−β)2 = P 2−4Q. As both sine and cosine are singly periodic functions

with period 2π, Lucas (see Section 26 of [Luc78]) regarded Un and Vn as simply

periodic numerical functions, where for any particular modulus m, the (numerical)

period in this case is the least positive integer p such that both

Un+p ≡ Un (mod m) and Vn+p ≡ Vn (mod m)

hold.

Throughout his several papers on Un and Vn, Lucas alluded to the problem of

extending or generalizing these functions and offered various suggestions by which

this might be done. However, in spite of these ideas, he seems never to have produced

any consistent theory that was analogous to his work on the Lucas functions. The

purpose of this thesis is to provide an extension of the Lucas functions which makes

use of the zeros of a cubic polynomial and to develop a theory which is very much

analogous to that of the Lucas functions. The functions that we will discuss were

almost certainly known to Lucas and the techniques that we will employ would have

3

been available to him; thus, it is conceivable that he might have developed some or

much of this theory himself. However, we must emphasize here that the evidence

that Lucas was thinking exactly along these lines is at best circumstantial.

1.2 Sources

In much of his published work on the Un and Vn functions, Lucas mentioned the

problem of extending or generalizing them. In what follows, we present the most im-

portant quotes, which are relevant to this theme. It should be pointed out that Lucas

frequently repeated himself, so we will give only one of any repeated statements.

The first of these comes from [Luc76].

The objective that we intend in this note is to show the identity of for-

mulas concerning certain numerical functions of the roots of an equation

of the second degree with rational coefficients with those which connect

to the circular functions, and to indicate, more generally, the identity

of formulas concerning the numerical functions of the roots of an alge-

braic equation of the fourth or any degree with those which connect to the

elliptic or abelian transcendentals.

In his memoir [Luc78], the most extensive and important work which Lucas

devoted to the Lucas functions, we have several interesting quotes.

This memoir has as its objective the study of symmetric functions of the

roots of an equation of the second degree, and its application to the theory

of prime numbers. We will first show the complete analogy of these sym-

metric functions with the circular and hyperbolic functions. We then show

4

the connection that exists between the symmetric functions and the theory

of determinants, combinations, continued fractions, divisibility, divisors

of quadratic forms, continued radicands, division of the circumference of

the circle, indeterminate analysis of the second degree, quadratic residues,

decomposition of large numbers into their prime factors, etc. This method

is the point of departure of a more complete study, of the properties of

the symmetric functions of the roots of an algebraic equation of any de-

gree with rational coefficients, in their relation to the theories of elliptic

and Abelian functions, of power residues, and indeterminate analysis of

higher degrees. (Section 1)

We will complete this section with the proof of formulas of extreme im-

portance, because these will serve later as a basis for the theory of the

numerical functions of double period, derived from the consideration of

the symmetric functions of the roots of third and fourth degree equations

with rational coefficients. (Section 9)

We see that the coefficients of the binomial raised to the power p are

integers and divisible by p, whenever p denotes a prime number, except

for the coefficients of the pth powers. On denoting by α, β, γ, . . . λ, any

n integers one has therefore

[α + β + γ + · · ·+ λ]p − [αp + βp + γp + · · ·+ λp] ≡ 0 (mod p),

and, for α = β = γ = · · · = λ = 1, one obtains

np − n ≡ 0 (mod p).

5

It is this congruence which contains Fermat’s theorem that one can gen-

eralize in the following manner, which is different from Euler’s approach.

If α, β, γ, . . . λ, denote the qth powers of the roots of an equation with

integral coefficients, and Sq their sum, the first part of the preceding con-

gruence represents the product by p of a symmetric function, integral and

with integral coefficients, of the roots, and, as a consequence of the coef-

ficients of the proposed equation. One has therefore

Spq ≡ Spq (mod p),

and by applying the Theorem of Fermat,

Spq ≡ Sq (mod p).

The study of the prime divisors of the numerical function Sn and of some

others which are analogous is very important; one has in particular, for

n = 1 and S1 = 0, as in the equation

x3 = x+ 1,

the congruence

Sp ≡ 0 (mod p);

and thence deduces conversely that if, in the case of S1 = 0, one has

Sn divisible by p for n = p and not previously, the number p is prime.

Indeed, suppose p is equal, for example, to the product of two primes g

and h. One has

Sgh ≡ Sh (mod g)

Sgh ≡ Sg (mod h);

6

as a consequence, if one finds that

Sgh ≡ 0 (mod gh),

one will also have

Sg ≡ 0 (mod h),

Sh ≡ 0 (mod g),

and, by the demonstrated theorem,

Sg ≡ Sh ≡ 0 (mod gh).

Thus Sgh would not be the first of the numbers Sn divisible by gh. One

can obtain, in this fashion, a great many theorems serving, like that of

Wilson, to verify prime numbers. We will leave aside, for the moment,

the curious and new developments that we have thus found, in order

to consider only those which are derived from simply periodic numerical

functions. (Section 21)

We have further indicated (Sections 9 and 21) a first generalization of the

principal idea of this memoir in the study of recurrence sequences which

arise from the symmetric functions of the roots of algebraic equations of

the third and fourth degree and, more generally, of the roots of equations

of any degree with rational coefficients. One finds, in particular, in the

study of the function

Un = ∆(an, bn, cn, . . . )/∆(a, b, c, . . . )

7

in which a, b, c designate the roots of the equation, and ∆(a, b, c, . . . ), the

alternating function of the roots, or the square root of the discriminant

of the equation, the generalization of the principal formulas contained in

the first part of this work. (Section 29)

In later writing concerning this memoir, Lucas [Luc80] remarked,

Since the publication of this work, the author has added to it twenty

other sections, as yet unpublished, which altogether form the arithmetical

theory of the symmetric functions of the roots of equations of the second

degree. The author hopes to find the time to write up in a similar manner

the theory of doubly periodic functions, in their connection to symmetric

functions of the roots of equations of the third and fourth degree, and with

elliptic functions.

Finally, in the year of his death, Lucas [Luc91a] wrote at much greater length

concerning the problem of generalizing his functions.

But we think that we should stress in particular research concerning

linear recurring sequences of various orders and its connection to the

theory of elliptic and abelian functions. In several papers published in

the Comptes rendues de l’association, from the meetings in Clermont,

Nancy, Paris and le Havre, in the Actes de l’Academie royale des Sci-

ences de Turin and of Saint Petersburg, in Nouvelle Correspondance

mathematique, in the Journal de Sylvester in Baltimore [American

Journal of Mathematics], etc., we have demonstrated the analogy and,

8

as it were, the identity of the circular and hyperbolic functions with the

numerical functions of the second order, whose characteristic polynomi-

als are of degree two. (See Chapters 17 and 18 in our book.) To every

trigonometric formula corresponds a formula for these functions, and

conversely.

We had hoped to find in this study, through the prime decomposition of

the expressions (an ± bn), a demonstration of Fermat’s last theorem con-

cerning the impossibility of solving in integers the indeterminate equation

xp + yp + zp = 0

in which it suffices to assume that p denotes a prime. Although Kum-

mer treated this equation masterfully some time ago, it has still not been

completely solved, since many of the exponents p cannot be dealt with by

his admirable analysis.

But if this research plan has not up to now provided the solution of this

celebrated problem, it does allow us to obtain a number of Wilsonian the-

orems, that is to say the necessary and sufficient conditions that a given

p of twenty or thirty digits must satisfy to be prime when one knows the

decomposition in prime factors of one of the numbers p ± 1. Further-

more, this method guides us to the notion of periodicity of the residues

for prime or composite moduli. Therefore, there is every reason to search

for formulas analogous to the addition and multiplication formulas for

the numerical functions which are derived from recurrences whose char-

acteristic polynomials are of degree three and four. These formulas find

9

their origin in the theory of elliptic functions, and we encounter some of

them in a beautiful memoir of Moutard.

One rediscovers these recurrence sequences by generalizing the theory of

linear substitutions, described by Serret in his Cours d’Algebre superieure

(4th edition, vol II, pp. 356-412) in a very particular form. If one con-

siders n linear homogeneous forms in n variables provided by the linear

substitutions in which the coefficients λ, µ, ν, . . . , are constants, the

forms xp+1, yp+1, zp+1, . . . , are expressed as functions of x, y, z, . . . ,

and the coefficients of the variables x, y, z, . . . , this produces a sequence

of linear recurrences having for their characteristic equation

U =

∣∣∣∣∣∣∣∣∣∣∣∣∣

λ1 − u µ1 ν1 . . .

λ2 µ2 − u ν2 . . .

λ3 µ3 ν3 − u . . .

......

.... . .

∣∣∣∣∣∣∣∣∣∣∣∣∣= 0

where u denotes the variable. The ratios of consecutive functions, or

of coefficients corresponding to two consecutive functions, have for their

limits under certain conditions of convergence, the root of largest modulus

of the equation U = 0. One can therefore generalize in an infinitude

of ways Bernoulli’s approximation technique for calculating the roots of

equations. This method is developed in the first volume of Legendre’s

Theorie des nombres, but only for a very particular case.

In Addition X Sur l’extraction des racines pour les moyennes (p. 506)

of our book, we have pointed out a new process for obtaining roots of any

10

index.

Furthermore, this process relates to the preceding theories and to lin-

ear substitutions. We think that, by developing these new methods, by

searching for the addition and multiplication formulas of the numerical

functions which originate from recurrence sequences of the third and of

the fourth degree, and by studying in a general way the laws of the residues

of these functions for prime moduli, according to their aspect, their char-

acter (cubic or biquadratic) for the discriminant of the equation U = 0,

that we would arrive at important new properties of prime numbers. And

perhaps the complete proof of Fermat’s last theorem is just a consequence

of the famous theorem of Jacobi concerning the impossibility of more than

two periods for holomorphic functions of a single complex variable.

This is pretty much all that has survived of Lucas’ writings on this problem. He

may have been contemplating doing more in later volumes of his book, Theorie des

nombres [Luc91b], only the first volume of which he completed (see Chapter 6 of

[Dec99]), but there is little evidence to suggest this. However, in the introduction to

this book, he wrote,

The theory of recurrent sequences is an inexhaustible mine which con-

tains all the properties of numbers; by calculating the successive terms of

such sequences, decomposing them into their prime factors and seeking

out by experimentation the laws of appearance and reproduction of the

prime numbers, one can advance in a systematic manner the study of the

11

properties of numbers and their application to all branches of mathemat-

ics.

After Lucas’ untimely death, there seemed to be little interest in the problem of

generalizing his functions. His old friend, C.-A. Laisant tried to kindle some interest

through a question in L’Intermediaire des mathematiciens [Lai96].

We know how much the famous theorem of Fermat concerning the im-

possibility of the identity xn + yn = zn, in integers, has so preoccupied

mathematicians. We can no longer ignore that the greater part of the

work of Ed. Lucas on the theory of numbers had for its object, direct

or indirect, the quest for a demonstration of this theorem. In particu-

lar, he published a very interesting memoir Sur la theorie des fonctions

numeriques simplement periodiques. These functions Un and Vn arise

from the equation of the second degree and present striking analogies with

the sine and cosine functions.

In seeking to generalize these ideas, Lucas later sent a Communication

to the Societe mathematique de France, which unfortunately was not

inserted, no Note having been put in by the author, on three numerical

functions, arising from the equation of the third degree, offering very great

analogies with the elliptic functions, and exhibiting the property of being

doubly periodic.

Shortly after, during a conversation that I had with him, Lucas said

to me: “if one could establish that my doubly periodic functions only

have two distinct periods, the theorem of Fermat would be proved.” And

12

during a meeting, assuming this hypothesis, he justified his statement by

a demonstration which was easy for me to follow, but of which I have

totally lost any memory, being far from suspecting his approaching death.

I recall only that the case of the exponent 2 was isolated, just as it should

be, in a very precise manner.

It is possible that Lucas had made similar Communications to other

colleagues, more favoured from the point of view of memory and more

attentive than I had been. In this case, I make an appeal to them through

the present question to consult their memories. It may also be possible

that the members of the Societe mathematique might be able to recover

the three numerical functions of which I have spoken, and concerning

which I have not found any indication in the papers left by the author.

This would assuredly be a very interesting gap to fill.

Later in 1913, Laisant [Bel24] updated his 1896 question through a response to

a letter written by D. E. Smith at the behest of E. T. Bell. Bell wanted to know

whether there was anything beyond what is asserted in the above-mentioned memoir

concerning the connection between recurring series and elliptic functions.

The mathematical papers of Lucas, after his death, were entrusted to a

commission of three members: M. Delannoy, Lemoine and myself. We

found in them the necessary elements for the publication of the last two

volumes (iii and iv) of the Recreations Mathematiques, and of the

volume Arithemique Amusant.

The remainder consisted of scattered notes which, in our estimation,

13

were not available for publication . . . . I found no trace of the subject

about which you particularly enquire, and I regret it keenly. I had studied

with great interest the memoir Sur les fonctions numeriques simplement

periodiques, and I often chatted over it with the author. These functions

U , V , derived from the equation of the second degree present curious

analogies with the sine and cosine. Lucas has also considered three func-

tions derived from the equation of the third degree, on which he once made

a communication to the Mathematical Society of France. I can find no

trace of this communication, and I have lost all memory of it. I recall

only the definition of one of these functions; it was an + bn + cn, a, b,

c, being the roots of the equation. From the point of view of periodicity

these functions exhibited the closest analogies with sn, cn, dn of elliptic

functions. They presented certain characters of double periodicity.

Lucas, in a conversation at his house, said to me: ‘if we could prove that

these functions admit only a single system of periods, Fermat’s Theorem

would be demonstrated.’ And, making this assumption, he developed this

proof for me in less than a quarter of an hour. Now I have completely

lost all recollection of it. That was some months before his death, which

I did not in the least anticipate. Since then I proposed a question on this

subject in this Intermediare. It has remained unanswered . . . . I am

more and more confirmed in my conviction that Lucas’ premature death

was an irreparable loss to the science of numbers. . . .

In response to a 1930 letter written by Duncan Harkin, Bell [Bel30] responded,

14

. . . you ask about a feasible generalization through elliptic functions of

the Functions of Lucas. If I knew how to do this, I [s]hould be far from

telling you, as I have tried my darndest to make some real progress on

it myself for the past twentyeight [sic] years. It is a tough nut of the

first order. Whoever cracks it will make a contribution to the theory of

numbers on a par with anything Fermat did. Go to it!

1.3 Commentary

On examining the material in the previous section, we note several properties of

Lucas’ investigation into his functions and those that he might have considered as

proper generalizations. We certainly see that he was interested in functions satisfying

linear recurring sequences; these functions should be symmetric functions of the zeros

of a defining polynomial with rational (in practice, usually integral) coefficients, and

there is more than one function to be considered. He seems to have been particularly

interested in defining polynomials of degree three or four. He indicated the need to

find addition and multiplication formulas involving these functions; this is certainly

what he did in order to prove the many properties of his own functions. His method

of approach was to use empirical methods to attempt to elucidate what the laws

of apparition and repetition for these functions would be, and from this material

he should be able, as he did in the case of Un and Vn, to derive primality testing

algorithms.

However, there was another aspect of this study: periodicity. As mentioned in

Section 1.1, Lucas was very impressed by the close analogy between his functions

15

and the circular functions—functions that are singly periodic—and derived a kind

of numerical periodicity for his functions. It is clear from what little he did write

on this matter that he considered an attempt at generalizing his functions should

begin by looking at doubly periodic functions, such as elliptic functions. It seems

that Lucas believed (probably by analogy) that the numerical functions that would

be derived through this analysis should exhibit the property of being doubly (nu-

merically) periodic. However, it is not clear what this property would have been.

Suppose we try the following definition of such a function.

Definition 1.1. Let H(x) be a function of an integer variable x such that H(x) is

also an integer. We will say that H(x) is doubly numerically periodic modulo m, if

there exists a pair of positive integers p1 and p2 such that p1 < p2, p1 does not divide

p2 and p1 is the least positive integer such that

H(n+ p1) ≡ H(n) (mod m) and H(n+ p2) ≡ H(n) (mod m)

for all sufficiently large values of n.

This seems to be the direct doubly periodic analog of a singly periodic numerical

function like Un or Vn, but no such function can exist. For suppose that for some m

such a function H(x) does exist. By the definition, we must have positive integers

p1 and p2 such that

H(n+ rp1 + sp2) ≡ H(n) (mod m)

for any fixed pair of integers r, s and all n sufficiently large. Suppose we specify

r, s to be integers such that rp1 + sp2 = d, where d = (p1, p2). Since p1 does not

16

divide p2, we must have d < p1. However, for all sufficiently large n we must have

H(n+ d) ≡ H(n) (mod m) with 0 < d < p1, a contradiction to the definition of p1.

Thus, there are no doubly period numerical functions that would be similar to

the Lucas functions. Nevertheless, Lucas’ intuition that elliptic functions would be

helpful was moving him in a very productive direction. Unfortunately, he did not

possess the mathematical knowledge, nor did such knowledge exist until the 20th

century, to take advantage of his rather vague ideas. This is explained in some detail

in Chapter 17 of [Wil98] and needs no further elaboration here. What is important

to note is that his belief that linear recurring sequences would play a role in this

approach led him to a dead end.

If, instead, we focus our attention on certain symmetric functions of the zeros α,

β, γ of a cubic polynomial which also satisfy linear recurrences, we would certainly

examine Sn = αn + βn + γn, which is what Lucas did for the particular polynomial

x3−x−1. This particular sequence S ′n is now called the Perrin sequence, and was the

focus of much attention by Adams and Shanks [AS82]. It seems first to have been

considered by Catalan in 1861 (see Chapter 8 of [Dec99]). Catalan, who denoted the

sequence by An, believed that An is or is not divisible by n according to whether

or not n is a prime. This could easily be converted into a primality test that would

execute in polynomial time in log n, but unfortunately, Catalan’s assertion is untrue.

For example, we see in [AS82] that 271441(= 5212) divides A271441; many other

examples of this phenomenon are also provided. We should remark here that some

of the work in [AS82] was later extended by Szekeres [Sze96]. Decaillot [Dec99] has

raised the interesting possibility that Lucas was aware of Catalan’s work before he

(Lucas) embarked on his work in primality testing and that it might have inspired him

17

in his investigations. However, Lucas, who knew Catalan and who is usually most

punctilious about assigning priority, nowhere mentions Catalan’s work. Also, his

result concerning this matter is more carefully stated than Catalan’s, even though the

proof is incomplete. Perhaps Lucas did not want to embarrass Catalan by mentioning

his less circumspect work.

Laisant raised the intriguing possibility that Lucas was considering three func-

tions that were symmetric functions of α, β, γ, one of which was Sn; what were

the other two functions? In his attempt to interpret Lucas’ writings, Bell [Bel24]

considered three functions, which he denoted as xn, yn, zn. These can be most easily

described by the equation

αn = xn + ynα + znα2,

with similar expressions involving β and γ. Clearly, these functions are symmetric

functions of α, β, γ. However, none of these functions is Sn; furthermore, these

functions were known to Lucas (see pp. 305–306 of [Luc91b]), who mentioned them

in a more general context without further comment. If these were the functions he

was thinking about, it seems peculiar that he would not have mentioned something

about them. Further properties of Bell’s xn, yn, zn were discussed by Ward [War31a]

and Mendelsohn [Men62].

It is possible that Lucas had intended to publish his findings concerning the

extension of his functions in one of the later volumes of Theorie des nombres. We

know that he was considering the publication of additional books in this series (see

the latter part of Chapter 6 of [Dec99]), and Harkin [Har57] has pointed out a short

table of contents for Volume II: Divisibility and Algebraic Irreducibility, Binomial

Congruences and Primitive roots. However, in response to a question raised by G.

18

de Rocquigny concerning the possible appearance of the second and third volumes

of Theorie des nombres, Delannoy, Laisant and Lemoine [DLL95] replied,

A careful examination of the papers left by Ed. Lucas has led us to this

conclusion, that contrary to our first hopes, it would be very difficult to

publish a continuation to the Theorie des nombres, of which only the

first volume has appeared. Nevertheless, the author’s notes, certain pas-

sages of his correspondence, and the reprinting of some of his little known

memoirs, would constitute an interesting volume for those interested in

the higher arithmetic. This is a project which has not been completely

abandoned, but whose realization will not be soon, whenever it does hap-

pen.

In spite of the lack of information concerning it, the problem of extending or

generalizing the Lucas functions has inspired a great deal of work. Some early

attempts at this are mentioned in Chapter XVII of the first volume of [Dic19]. In

the next section, we will briefly describe some of these and some of the more modern

investigations into this problem.

1.4 Previous Extensions of the Lucas Functions

One of the earliest attempts to extend the Lucas functions was done in 1880 by de

Longchamps [dL80]. If we put R = αβγ, where α, β, γ are the zeros of a cubic

19

polynomial f(x), de Longchamps considered Dn, En and Sn, where

RnDn = (αn + βn)(βn + γn)(γn + αn),

RnEn =(αn − βn)(βn − γn)(γn − αn)

(α− β)(β − γ)(γ − α),

Sn = αn + βn + γn,

to be the degree three recurring function analogs of the Lucas functions. Are these

the three functions that Laisant mentioned? They would certainly have been known

to Lucas because he was the session chair for the talk in which de Longchamps

presented his results. In fact he (de Longchamps) showed how to express Dn and

En in terms of the coefficients of f(x). However, Lucas would likely not have been

comfortable with the fact that the first two of these functions are not necessarily

integral. Also, as we have seen in Section 1.2, Lucas had certainly mentioned the

function ∆(αn, βn, γn)/∆(α, β, γ) that de Longchamps denoted by RnEn. This seems

to be all that de Longchamps wrote concerning this topic because the list of his papers

in [Laz07] does not contain any other paper devoted to this subject.

The next work done on this problem was that of Pierce [Pie16] in 1916. He

defined the two functions

Sm =∏

(1 + αmi ) and ∆m =∏

(1− αmi ),

where the product is taken over all the zeros of a given polynomial f(x) with integral

coefficients. Pierce obtained several number theoretic results concerning these func-

tions, particularly when the degree of f(x) is three. Later, Lehmer [Leh33] extended

some of Pierce’s results, showing, among other things, that each satisfies a linear

recurrence relation. Indeed, Lehmer [Leh71], [Leh68] made use of these functions in

20

a test, which makes use of the factors of N2 + N + 1 to demonstrate the primality

of N . Unfortunately, Pierce’s functions are difficult to compute, which means that

using them is not very practical. While Pierce’s work represents a kind of extension

of the Lucas functions, it is very unlikely that Lucas was thinking in this direction,

because nowhere in his work does he allude to anything like these functions. Also,

Pierce’s functions do not become Un and Vn when f(x) is of degree two.

In 1929, Carmichael advocated to study the functions which he denoted by Gn

and Hn. Although it appears that Carmichael was unaware of this, Lucas had

mentioned both in his published work. The function Gn occurs as Un on page 306

of [Luc91b] (also, in the cubic case Gn is the same as Bell’s zn) and Hn is the same

as ∆(an, bn, cn, . . . )/∆(a, b, c, . . . ). Carmichael stated that an investigation of the

properties of these two functions would lead to two generalizations of Lucas’ Un

function, but he did not follow up on this remark.

Lehmer [Leh30] extended the Lucas functions by replacing the parameter P by√R, where R is an integer coprime to Q; however, the resulting sequences are no

longer integers for all n. Lehmer’s functions were later generalized by Williams

[Wil76], but in spite of the successes of the theory of Lehmer’s extension and its

generalization, there is no reason to believe that this was the direction in which

Lucas was looking to extend his functions.

Carmichael [Car20], Engstrom [Eng31] and Ward [War31b], [War31c], [War33],

[War36], [War37], [War55] investigated the arithmetical theory of linear recurring

sequences, but they did not produce a set of functions which were analogous to

Lucas’ Un and Vn. One of the most important properties of Lucas’ function Un is

that it satisfies the condition of being a divisibility sequence; that is, the sequence

21

of integers {Un} (n > 0) is such that if m | n, then Um | Un. Lucas was very

aware of this property of Un and made heavy use of it in developing his theory.

Both Hall [Hal36] and Ward were interested in the problem of whether any function

satisfying a linear recurrence could also be a divisibility sequence. While they did

not succeed in answering this question completely (this was done later by Bezivin,

Petho and van der Poorten [BPvdP90]), they did show that this would be a very

unlikely property for a sequence satisfying a third order recurrence unless it was a

very uninteresting sequence, such as a special sequence satisfying Un+3 = RUn. In-

deed, one of the simplest, non-trivial, linear divisibility sequence after Lucas’ Un is

∆(αn, βn, γn)/∆(α, β, γ), where α, β, γ are the zeros of a cubic polynomial with inte-

gral coefficients. Ward, who was Bell’s PhD student, seems to have contracted Bell’s

enthusiasm for extending Lucas’ functions. In fact, he coined the term “lucasian”

for any function satisfying a linear recurrence which was also a divisibility sequence.

In [War38] he discussed two candidates for lucasian sequences, one of which we will

discuss in great detail in this work. Of all the individuals who worked on the problem

of extending the Lucas functions, he seems to have made the most progress. While

we have mentioned only a few of his publications here, there are many more that are

also of some relevance to this discussion and we urge the interested reader to consult

the list of his published papers in [Leh93].

Williams [Wil69], [Wil72a], [Wil77] generalized the Lucas functions, but while his

functions satisfy a linear recurrence, they are not symmetric functions of the zeros of

a polynomial f(x). Furthermore, they are not always integers unless the coefficients

of f(x) obey certain properties. Again, these functions do not seem to be those for

which Lucas was searching. Although in the case where f(x) is of degree three, it is

22

possible to use certain of these functions to extend (1.1) and (1.2), by making use of

the tresine and cotresine functions of Graves [Gra47].

1.5 Our Objective

While many researchers have looked directly or peripherally at the problem of ex-

tending Lucas’ functions, none of them seems to have produced the kind of results

that Lucas was seeking. In what follows, we will offer a new suggestion as to how

Lucas might have wanted to extend his functions. This is based on a very simple

variant of Longchamps’ original suggestion.

We begin with a cubic polynomial f(x) = x3 − Px2 + Qx − R, where P , Q, R

are integers and we put

δ = (α− β)(β − γ)(γ − α), ∆ = δ2 = P 2Q2 − 4Q3 − 4RP 3 + 18PQR− 27R2,

where α, β, γ are the zeros of f(x). We will assume that δ 6= 0. We next define Cn

and Wn by

δCn = (αn − βn)(βn − γn)(γn − αn)

= (αnβ2n + βnγ2n + γnα2n)− (α2nβn + β2nγn + γ2nαn)

Wn = (αnβ2n + βnγ2n + γnα2n) + (α2nβn + β2nγn + γ2nαn).

Note that Cn is the same as Lucas’ ∆(αn, βn, γn)/∆(α, β, γ)(= RnEn) and Wn =

Ln − 2Rn, where

Ln = RnDn = (αn + βn)(βn + γn)(γn + αn).

23

Both Cn and Wn are symmetric functions of α, β, γ and are therefore integers for

all non-negative values of n. It is these functions that we will use as our extensions

of the Lucas functions Un and Vn. Observe that {Cn} is a divisibility sequence.

In Chapter 2 we will list the most important properties of the Lucas functions Un

and Vn; most of these were known to Lucas, and can be found in his memoir [Luc78].

It would be reasonable to expect that he would want to extend these results, and

this seems to be the tenor of his remarks in Section 1.2 above. In the succeeding

chapters we will develop analogous results involving Cn and Wn. These will include,

among several other items, the addition formulas, the multiplication formulas, the

laws of apparition and repetition and some primality testing results. What is most

remarkable in this entire investigation is the need for only two functions, not three.

The main tools that we will employ would have been known to Lucas. For

example, he would have needed the fundamental theorem of symmetric polynomials,

but he indicated in several places (see, for example, Section 21 of [Luc78] above),

that he was well aware of this result. We will make a great deal of use of Waring’s

theorem, but this was described in great detail by Lucas in Chapter XV of [Luc91b].

We will also use the theory of finite fields, but this would have been known (at least

the amount that he would need) to Lucas through the second volume of Serret’s

Cours d’Algebre superieure [Ser79], with which Lucas was quite familiar (see p.

vii of [Luc91b]). To develop our law of repetition, we require a small amount of

algebraic number theory to prove Theorem 4.18. Lucas might have been aware of

some of this material because he claims in part CLIX of [Luc80] that he was working,

together with a M. Tastavin, on producing a French translation of the third edition of

Dirichlet–Dedekind’s Vorlesungen uber die Zahlentheorie. Unfortunately, this volume

24

never appeared, but the result that we need could easily have been deduced by Lucas,

even though the proof might not have been completely rigorous. In the Appendix, we

provide an alternate, more elementary proof of Theorem 4.18, which Lucas should

have been able to deduce. We also make use of derivatives to establish a certain

identity that will be useful in our investigation into the law of repetition, but Lucas

often did this himself. See, for example, Section XVII of [Luc78].

Chapter 2

Lucas Sequences

Given the polynomial x2 − Px + Q, where P , Q are coprime integers, the Lucas

functions Un and Vn are defined by

Un = Un(P,Q) = (αn − βn)/(α− β), (2.1)

Vn = Vn(P,Q) = αn + βn, (2.2)

where α and β are the zeros of the given polynomial. Further let ∆ = δ2 = (α−β)2 =

P 2 − 4Q.

2.1 Identities

Lucas sequences satisfy many well-known identities, several of which will be men-

tioned herein. For further information the reader is referred to standard works such

as [Wil98] and [Rib89].

For a fixed m both Un, Vn satisfy the following equality

Xn+2m = VmXn+m −QmXn, (2.3)

where U0 = 0, U1 = 1, V0 = 2 and V1 = P .

Substituting n−m for n in (2.3) gives both

Un+m = VmUn −QmUn−m and Vn+m = VmVn −QmVn−m. (2.4)

25

26

If the mth and nth terms are known, the (m + n)th term of a Lucas function

may be found using the addition formulas presented below:

2Um+n = VmUn + UmVn, (2.5)

2Vm+n = VmVn + ∆UmUn. (2.6)

Now, if the facts QnU−n = −Un and QnV−n = Vn are used in (2.5) and (2.6), the

following subtraction formulas can be derived:

2QmUn−m = UnVm − VnUm, (2.7)

2QmVn−m = VnVm −∆UnUm. (2.8)

Subtracting (2.5) and (2.7) yields

Un+m = VnUm +QmUn−m. (2.9)

Doing the same with (2.6) and (2.8) gives

Vm+n = ∆UnUm +QmVn−m. (2.10)

Furthermore, by writing Un, Vn, ∆ in terms of α, β, it is clear that

V 2n −∆U2

n = 4Qn (2.11)

A doubling formula for Un follows from (2.5) and a doubling formula for Vn follows

from (2.6) and (2.11) to yield

U2n = VnUn, (2.12)

V2n = V 2n − 2Qn = ∆U2

n + 2Qn. (2.13)

27

Of key importance to later sections, the following multiplication formulas for Un

and Vn can be obtained by use of the fact that Vn+δUn = 2αn. From this we see that

2m−1[Vmn + δUmn] = [Vn + δUn]m and then expanding using the binomial theorem,

we obtain

2m−1Umn =

b(m−1)/2c∑i=0

(m

2i+ 1

)∆iU2i+1

n V m−2i−1n , (2.14)

2m−1Vmn =

bm/2c∑i=0

(m

2i

)∆iU2i

n Vm−2in . (2.15)

2.2 Computation of Un, Vn

Often we are interested in calculating Un or Vn, for some particular value of n.

Although this can clearly be done via the formulas (2.1), (2.2) or equation (2.3),

both methods are too slow for practical purposes. A faster method is presented

here.

Let (b0b1 . . . bk)2 be the binary representation of m ∈ Z+ such that b0 = 1,

bi ∈ {0, 1} for (1 ≤ i ≤ k) and k = blog2mc. The following formulas for Lucas

sequences depend on identities (4.2.22) and (4.2.24) from [Wil98]:

U2n = 2Un+1Un − PU2n,

U2n+1 = U2n+1 −QU2

n,

U2n+2 = PU2n+1 − 2QUnUn+1.

Now, if P0 = {1, P} and

Pi+1 =

{2AB − PA2, B2 −QA2} if bi+1 = 0,

{B2 −QA2, PB2 − 2QAB} if bi+1 = 1,

28

where Pi = {A,B}, then Pk = {Um, Um+1}. Moreover, one may use Pk to compute

Vm, as

Vm = 2Um+1 − PUm.

Hence Um (Vm) can be computed in O(logm) multiplications and additions. Note

that this result can be employed to compute Um (Vm) (mod N) quickly by defining

Pi+1 =

{2AB − PA2, B2 −QA2} (mod N) if bi+1 = 0,

{B2 −QA2, PB2 − 2QAB} (mod N) if bi+1 = 1.

This is a more useful result as the growth of Um is exponential.

Now let

Wn ≡ Q−nV2n (mod N).

Then clearly

W1 ≡ P 2Q−1 − 2 (mod N)

and by (2.13)

W2n ≡ W 2n − 2 (mod N).

Further, by (2.4) replacing n by 2n+ 2 and m by 2n, we have

V4n+2 = V2nV2n+2 −Q2nV2;

so then

W2n+1 ≡ WnWn+1 −W1 (mod N).

In this case define P0 = {W1,W2} and

Pi+1 =

{A2 − 2, AB −W1} (mod N) if bi+1 = 0,

{AB −W1, B2 − 2} (mod N) if bi+1 = 1,

29

where Pi = {A,B}, then

Pk = {Wm,Wm+1}.

This method for finding {Wn,Wn+1} (mod N) is faster to compute than the

previous method for Un or Vn (mod N). Moreover, this method may also be used

to find a particular value of Un or Vn (mod N) as follows. First,

V2h ≡ QhWh (mod N),

and by (2.3) with m = 1 and n = 2h+ 1 we have

PV2h+1 ≡ Qh+1(Wh+1 +Wh) (mod N).

Also, by (2.10) with m = 1 and n = 2h+ 1,

∆U2h+1 ≡ Qh+1(Wh+1 −Wh) (mod N).

To complete this a formula for U2h (mod N) is needed in terms of Wh and Wh+1.

This is achieved by the use of (2.6) with m = 2h and n = 2 to see

2V2+2h = (P 2 − 2Q)V2h + ∆PU2h,

and hence

∆PU2h ≡ Qh(2QWh+1 − (P 2 − 2Q)Wh) (mod N).

2.3 Arithmetic Results

The identities from the previous section may be employed to construct arithmetic

results for Lucas sequences. The global arithmetic results presented here are stan-

dard.

30

Definition 2.1. If a and b are not both zero, then the greatest common divisor (gcd)

of a and b is defined to be the largest integer that divides both a and b, denoted by

(a, b).

Definition 2.2. If a and b are nonzero integers, then the least common multiple

(lcm) of a and b is defined to be the least positive integer l such that a | l and b | l,

denoted by [a, b].

To begin, by the use of equation (2.11) it may be shown that

(Un, Vn) | 2Qn. (2.16)

If (P,Q) = 1, then by setting m = 1 in equation (2.3) and by induction it is clear

that for n > 0, we have

(Un, Q) = (Vn, Q) = 1, (2.17)

and hence for any n ≥ 0

(Un, Vn) | 2. (2.18)

Furthermore, it is not difficult to show that {Un} is a divisibility sequence; i.e.

Um | Un, when m | n. (2.19)

Note that if n = ms, then

Un(P,Q) =αn − βn

α− β=αms − βms

α− β

=αm − βm

α− β· α

ms − βms

αm − βm= Um(P,Q) · Us(Vm, Qm).

31

Definition 2.3. Given m ∈ Z, let ω be the least positive integer, if it exists, such

that m | Uω. This value is called the rank of apparition of m, denoted by ω(m).

The next theorem is the first local theorem to be seen here. It does however have

a global result as a corollary.

Theorem 2.4. Let (Q,m) = 1 and ω = ω(m). If m | Un for some n > 0, then

ω | n.

Proof. Put

n = qω + r where 0 ≤ r < ω.

If r = 0, then we are done. Thus we assume r > 0. Also,

n = (q + 1)ω − (ω − r).

Since either q or q + 1 is even, without loss of generality, let

n = qω + s where 2 | q and |s| < ω.

Setting m = qω2

+ s and n = qω2

in (2.4) produces

Un = U( qω2

+s)+ qω2

= U qω2V qω

2+s −Q

qω2

+sU−s.

Now since m | Un and m | U qω2⇒ m | Q qω

2+sU−s. Note that s may be positive or

negative. If s > 0, then U−s = Q−sUs ⇒ m | Q qω2 Us. Hence m | U|s|. But |s| < ω,

so by the minimality of ω, it must be that s = 0⇒ ω | n.

Corollary 2.4.1. If m, n > 0 and d = (m,n), then

(Um, Un) = |Ud|.

32

Proof. Let G = (Um, Un), then Ud | G, since Ud | Um and Ud | Un. Let ω = ω(G)

be the rank of apparition of G. Then by Theorem 2.4 ω | m and ω | n⇒ ω | d⇒

G | Ud. Thus, G = |Ud|.

The following theorem is a global result of Carmichael, and may be found as a

corollary to Theorem 17 in [Car13].

Theorem 2.5. If m, n ≥ 1, then

(Umn/Un, Un) | m.

Proof. Let r = bm/2c, then it may be shown that

(Umn/Un, Un) | mQnr.

From the identity (4.2.41) of [Wil98]

U(2r+1)n = Un

r∑j=0

2r + 1

r − j

(r + j

r − j − 1

)Qn(r−j)∆jU2j

n

with m = 2r + 1, we get

U(2r+1)n/Un ≡ (2r + 1)Qnr (mod Un).

Thus, if 2 - m, then

(Umn/Un, Un) | mQnr.

Also, from the identity (4.2.43) of [Wil98] we can write

U2rn = Vn

r−1∑j=0

(r + j

r − j − 1

)Qn(r−j−1)∆jU2j+1

n .

So, if m = 2r, then

U2rn/Un ≡ rVnQn(r−1) (mod Un);

33

thus

(U2rn/Un, Un) = (rVnQn(r−1), Un).

Now

(rVnQn(r−1), Un) | rQn(r−1)(Vn, Un);

hence, it follows from (2.16) that

(Umn/Un, Un) | mQnr

when 2 | m. Lastly, since (P,Q) = 1, we have

(Umn/Un, Un) | m

from (2.17).

We are often interested in values of n for which a prime p divides Un. It will be

assumed that p - Q. Notice that if p | Q, then p - P and

Un ≡ P n−1 (mod p).

Thus, p | U0 and p - Un for n ≥ 1. The following theorem provides us with what is

called the law of repetition for a prime p.

Theorem 2.6. If p is a prime and for λ > 0, we have pλ 6= 2 and pλ || Un, then

pλ+1 || Upn. If pλ = 2, then pλ+1 | Upn.

Proof. Let pλ || Un for some λ ≥ 1. If p = 2, then by (2.11) 2 | Vn, and since

U2n = UnVn, we get 2λ+1 | U2n.

Now if λ > 1, and p = 2, then Q is odd, hence by (2.11) 2 || Vn; thus in this case

2λ+1 || U2n.

34

If p is an odd prime, then by (2.14) with m = p, one has

2p−1Upn ≡ pUnVp−1n (mod pλ+2).

Since p - Q, then p - Vn by (2.11); thus

pλ+1 || Upn.

Definition 2.7. Let ε(n) be the Jacobi symbol of (∆/n).

The following theorem is called the law of apparition for a prime p. Let ε = ε(p)

for the remainder of the chapter.

Theorem 2.8. If p is a prime such that p - 2Q, then p | Up−ε.

Proof. First, note that p |(pi

)for i 6= 0, p. So by (2.14) and (2.15), with n = 1 and

m = p, the following congruences exist

2p−1Up ≡ ∆p−12 (mod p), 2p−1Vp ≡ P p (mod p).

So by Fermat’s little theorem and Euler’s criterion for quadratic residuacity

Up ≡ ε (mod p), Vp ≡ P (mod p). (2.20)

Thus if ε = 0, then p | Up−ε. If ε 6= 0, then we can use (2.5), (2.6), (2.7)and (2.8) to

deduce

2Q1+ε2 Up−ε ≡ PUp − εVp (mod p), (2.21)

2Q1+ε2 Vp−ε ≡ PVp − ε∆Up (mod p). (2.22)

Thus by (2.20) p | Up−ε when p is odd.

35

In the sequel we will need the following result.

Vp−ε ≡ 2Q1−ε2 (mod p). (2.23)

This follows easily from (2.22).

We have similar arithmetic results for {Vn}; many of these were possibly not

known to Lucas, but might have appeared in the subsequent literature (see, for

example, [Mul01]). In any event, we make no claims of originality of these results.

Observe the following short lemma that will be called upon in the next two theorems.

Lemma 2.9. If 2 | P , then 2 | Vn for all n ≥ 0. If 2 - P and 2 | Q, then 2 | Vn

only for n = 0. If 2 - P and 2 - Q, then 2 | Vn ⇔ 3 | n.

Proof. Certainly, if 2 | P , then 2 | Vk for all k ≥ 0. If 2 - P , then since V1 = P and

Vk+1 ≡ PVk (mod Q), we see that if 2 | Q, then 2 - Vn for n > 0. If 2 - P and 2 - Q,

it follows by using induction on (2.3) that 2 | Vk if and only if 3 | k.

Note that from the above lemma we can easily see that if 2 | Vn, then 2 | Vtn for

all t ∈ N. It is known that {Un} is a divisibility sequence, but this is not necessarily

true for {Vn}; however, we have the following weaker results provided by the next

two theorems.

Theorem 2.10. If m | n and 2 - nm

, then Vm | Vn.

Proof. Since m | n and 2 - nm

, then n = km where k odd, i.e. k = 2r + 1 for some

r ∈ Z. From (2.6)

2Vn = 2Vkm = 2V(2r+1)m = 2V2rm+m = VmV2rm + ∆UmU2rm.

36

Since V 2m −∆U2

m = 4Qm, if 2 | Vm, then 2 | ∆Um. Also, Vm | U2m and U2m | U2rm

implies Vm | U2rm, hence, Vm | 2Vkm. If 2 | Vm, then by Lemma 2.9, 2Vm | 2Vkm ⇒

Vm | Vkm. On the other hand, if 2 - Vm, then Vm | Vkm.

Note that if r | Vn and n > 0, then since (Vn, Q) = 1, it must be that (r,Q) = 1.

Also, since V−n = Vn/Qn, we may write r | V−n. This simply means that r divides

the integral numerator of the fraction V−n.

Theorem 2.11. If m | n and 2 | nm

, then (Vm, Vn) | 2.

Proof. We first employ (2.13) and (2.17) to observe that

V2m = V 2m − 2Qm ⇒ (Vm, V2m) = (2Qm, Vm) = (2, Vm).

Hence (Vm, V2m) | 2. Now assume (Vm, V2km) | 2, this is certainly true for k = 1,

then since

V(2k+2)m = VmV(2k+1)m −QmV2km,

we find that (V(2k+2)m, Vm) = (QmV2km, Vm) = (V2km, Vm). Thus the result follows

by induction.

Corollary 2.11.1. If 2 | m, then (Vn, Vmn) = (2, Vn).

The rank of apparition has been introduced for {Un}, and we might expect to

have something similar for {Vn}. But the situation may exist where r - Vn for every

n ∈ Z, hence the following modified definition for the {Vn} case is needed.

Definition 2.12. Suppose r | Vn (n > 0). Denote by ρ(r) the least positive integer

ρ such that r | Vρ.

37

In order to say something about ρ(r) for r | Vn, the result below is needed first.

Lemma 2.13. If r | Vn and r | Vm, then r | V2km+n (k ≥ 1).

Proof. By (2.6) we have

2V2km+n = V2kmVn + ∆U2kmUn.

Now r | Vm ⇒ r | U2m by (2.12). Consequently, r | U2km for any integral k ≥ 1.

Thus, since r | Vn and r | U2km we have r|2V2km+n. If 2 - r, the desired result is

obtained. On the other hand, if 2 | r, then 2 | Vm, and by Lemma 2.9, 2 | V2km.

Also, 2 | Vn and 2 | ∆Un, as V 2n + ∆U2

n = 4Qn. Hence 2r | 2V2km+n ⇒ r | V2km+n.

The theorem below is a local arithmetic result for {Vn} and is very similar to

Theorem 2.4 as seen for {Un}, though the method of proof is different.

Theorem 2.14. If r | Vn (n > 0), then ρ(r) | n.

Proof. Let 2µ || (n, ρ). Then 2µ || n or 2µ || ρ. Suppose, 2µ || n, then 2µ || d,

where d = (2ρ, n). There exist x, y ∈ Z such that

d = 2ρx+ ny ⇒ d

2µ=

2ρx

2µ+ny

2µ.

Now, 2 | 2ρ2µ

and 2 - d2µ⇒ 2 - y so by Theorem 2.10 r | Vyn. Let 2k || 2x, then

Vd = V2k 2x

2kρ+yn, and since r | Vρ again, by Theorem 2.10 r | V 2x

2kρ because 2 - 2x/2k.

Thus, r | Vd by the previous lemma. Hence, d ≥ ρ. But d | 2ρ⇒ d2µ| 2ρ

2µand since

d2µ

is odd, we get d2µ| ρ

2µwhich means that d | ρ ⇒ d = ρ. Since d | n, we have

completed the proof for this case.

38

Next suppose that 2µ || ρ and 2µ+1 | n. Put d = (ρ, 2n)⇒ 2µ || d. There exist

x, y ∈ Z such that

d = ρx+ 2ny ⇒ d

2µ=ρx

2µ+

2ny

2µ.

Here, 2 - d2µ

and 2 | 2n2µ⇒ 2 - x⇒ r | Vxρ. Let 2k || 2y (y ≥ 1). So, Vd = V2k 2ny

2k+xρ,

and since r | V 2y

2kn we get r | Vd by the previous lemma. This implies d ≥ ρ. But

d | ρ ⇒ d = ρ. Also, as d | 2n we must have ρ | 2n, which means that ρ2µ| 2n

2µ.

Since 2 - ρ2µ

, then ρ2µ| n

2µ⇒ ρ | n.

A consequence of Corollary 2.11.1 is the following helpful result which will be

called upon in three of the next four theorems.

Lemma 2.15. If r > 2 and r | Vn, then 2 - nρ(r)

.

Proof. Suppose 2 | nρ(r)

, then n = mρ, where m is even. Thus, by Corollary 2.11.1

(Vρ, Vn) = (Vρ, Vmρ) = (2, Vρ) ≤ 2.

This is a contradiction since r | (Vρ, Vn) and r > 2.

The next two theorems cover what can be said about r1r2 | Vs for some s, when

we know that r1 | Vn and r2 | Vm. The results here really depend on how many

factors of 2 the quantities m and n have and hence there are two cases: the first

case, 2µ || n and 2µ || m, is addressed in Theorem 2.16 and the second case, 2µ || n

and 2ν || n (µ 6= ν), in Theorem 2.17.

Theorem 2.16. If r1 | Vm, r2 | Vn, (r1, r2) = 1, 2µ || m and 2µ || n, then

r1r2 | V[m,n].

39

Proof. Note, [m,n]m

and [m,n]n

are both odd. It follows from Theorem 2.10 that

r1 | Vm [m,n]m

and r2 | Vn [m,n]n

.

Since (r1, r2) = 1, r1r2 | V[m,n].

Theorem 2.17. If r1 | Vm, r2 | Vn, (r1, r2) = 1, r1, r2 > 2, 2µ || m and 2ν || n

(µ 6= ν), then r1r2 - Vk for every k ∈ Z.

Proof. Let ρ1 = ρ(r1) and ρ2 = ρ(r2). Since, r1, r2 > 2, by Theorem 2.14 and Lemma

2.15,

ρ1 | m, ρ2 | n and 2 -m

ρ1

, 2 -n

ρ2

.

Thus, 2µ || ρ1, 2ν || ρ2. If r1r2 | Vs, then 2 - sρ1

and 2 - sρ2⇒ 2µ || s and 2ν || s.

This is obviously a contradiction, so the wanted result is obtained.

It has been established that for m, n > 0, (Um, Un) = |U(m,n)|. A similar, but

more complicated result exists for (Vm, Vn) and again it is dependent on the number

of factors of 2 for m and n. The two cases are covered in the next two theorems.

Theorem 2.18. If 2µ || m and 2µ || n, then

(Vm, Vn) = |V(m,n)|.

Proof. Since 2µ || m and 2µ || n, then 2 - m(m,n)

, 2 - n(m,n)

⇒ V(m,n) | Vm, V(m,n) | Vn

by Theorem 2.10. Put d = (Vm, Vn), then V(m,n) | d.

Now note that ρ(d) | m, ρ(d) | n ⇒ ρ(d) | (m,n). By Lemma 2.15, for d > 2,

it must be that 2 - mρ(d)

, 2 - nρ(d)⇒ 2 - (m,n)

ρ(d). So then, by Theorem 2.10, d | V(m,n) ⇒

(Vm, Vn) = |V(m,n)|.

40

If d = 1, then we are done. If d = 2, then 2 | Vn and 2 | Vm so that, by Lemma

2.9, either 2 | P and 2 | V(m,n), or 2 - P and 3 | (m,n). Thus 2 | V(m,n).

Theorem 2.19. If 2µ || m and 2ν || n (µ 6= ν), then (Vm, Vn) | 2.

Proof. Let d = (Vm, Vn), then for d > 2, by Lemma 2.15 one must have 2 - mρ(d)

,

2 - nρ(d)⇒ 2λ || m, 2λ || n, when 2λ || ρ(d). This is clearly a contradiction, thus

(Vm, Vn) | 2, if µ 6= ν.

The short theorem below, which was known to Lucas, is of interest because it

provides some insight into the characteristics of an odd prime p, when p | Vn.

Theorem 2.20. If p is an odd prime and p | Vn, then p ≡ ±1 (mod 2ν+1) where

2ν || n.

Proof. If p | Vn, then p - Q by (2.17). Also, by (2.11), we see that p - ∆ and p - Un.

However, we do have p | U2n by (2.12). Thus, if ω is the rank of apparition of p, then

ω | 2n by Theorem 2.4 and ω - n. Also, ω | p ± 1 by Theorem 2.8. So if 2ν || n,

then 2ν+1 | ω ⇒ p ≡ ±1 (mod 2ν+1).

We have shown, in Theorem 2.8, that for a prime p where p - 2∆Q, we have

p | Up−ε. Thus, Up−ε = U p−ε2V p−ε

2and so p | U p−ε

2or p | V p−ε

2, but not both. The

question of which one is divisible by p is answered by the following theorem, called

Euler’s criterion for the Lucas functions. This result was not known to Lucas and

was first proved in a more general setting by Lehmer [Leh30].

41

Theorem 2.21. If p is a prime such that p - 2∆Q, then

p | U p−ε2⇔ (Q/p) = 1,

p | V p−ε2⇔ (Q/p) = −1.

Proof. Setting n = p− ε in (2.13) yields

Vp−ε = V 2p−ε2

− 2Qp−ε2 .

So then by (2.23),

V 2p−ε2

≡ 2Q1−ε2 + 2Q

p−ε2 ≡ 2Q

1−ε2 + 2Q

1−ε2 Q

p−12 ≡ 2Q

1−ε2 (1 + (Q/p)) (mod p).

Thus p | V p−ε2

if and only if (Q/p) = −1.

2.4 Primality Testing

Lucas’ main purpose for his investigation into the sequences now named for him was

to find new methods for the discovery of primes. This can be seen in the following

result, which Lucas called his fundamental theorem.

Theorem 2.22. Suppose N is an odd integer. Let T = T (N) = N + 1 or N − 1. If

N | UT but N - UT/d for all d such that d < T and d | T , then N is a prime.

It was Lehmer [Leh27], who realized that this theorem could be rewritten as

follows.

Theorem 2.23. Suppose N is an odd integer. Let T = T (N) = N + 1 or N − 1. If

N | UT but N - UT/q for each prime divisor q of T , then N is a prime.

42

We also have the following corollary.

Corollary 2.23.1. Suppose N is an odd integer and T = T (N) = N + 1 or N − 1.

If N | UT and N | UT/UT/q for each prime divisor q of T , then N is a prime.

Proof. By Theorem 2.5, we have

(UT/UT/q, UT/q) | q.

Thus if N | UT/UT/q, then N - UT/q. The result follows by the theorem.

The following theorem is called the Lucas-Lehmer theorem. Lucas used a result

similar to this one to implement a primality test for Mersenne numbers.

Theorem 2.24. If N = A2n − 1, n ≥ 3, 0 < A < 2n, 2 - A, and the Jacobi symbols

(∆/N) = (Q/N) = −1, then N is a prime if and only if

N | VN+12

(P,Q).

Proof. Suppose N | VN+12

(P,Q). Let p be some prime such that p | N , then p ≡ ±1

(mod 2n) by Theorem 2.20. So p = k2n±1 for some k ∈ Z. Assume N is composite,

then without loss of generality N = pq, where p = k2n + 1, q = l2n− 1 and l, k > 0.

Thus

A2n − 1 = N = pq = (k2n + 1)(l2n − 1) = (kl2n + l − k)2n − 1;

in particular

A = kl2n − k + l.

43

Now if l ≥ k, then (kl2n−k+ l) ≥ 2n ⇒ A ≥ 2n, which is a contradiction as A < 2n.

On the other hand, if l < k, then l + 1 ≤ k and since l2n − 1 > 0 we have

kl2n − k + l = k(l2n − 1) + l ≥ (l + 1)(l2n − 1) + 1

= (l2 + l)2n − l ≥ 2n+1 − 1 ≥ 2n.

Again, this is a contradiction as A < 2n. So N is a prime.

Now suppose N is a prime. Since (∆/N) = −1, then

UN+1 ≡ 0 (mod N)

by the law of apparition Theorem 2.8. Further, since (Q/N) = −1 we have

N | VN+12

(P,Q)

by Euler’s criterion in Theorem 2.21.

Corollary 2.24.1. Suppose A = 1 and 2 - n, n ≥ 3. Put Q = −2, P ≡ 2 (mod N).

Then N is a prime if and only if

N | VN+12

(2,−2).

Proof. Since ∆ ≡ 12 (mod N), we have (∆/N) = (Q/N) = −1.

Put S0 = 4, Sj+1 = S2j − 2. Then

N | VN+12

(2,−2)⇔ N |Sn−2,

as

V2j(2,−2) = 22j−1

Sj−1.

44

Thus, if N is a Mersenne number, we have that N is a prime if and only if N | Sn−2.

It is this corollary that provides an efficient test for Mersenne primes; for further

information see [Leh35]. In fact, the largest prime ever found by hand calculation1,

M127 = 2127 − 1, was found by Lucas in 1876 using a result similar to this corollary.

This was most remarkable as M127 is a 39 digit number. Strangely Lucas seems to

have lacked confidence in this result, despite the robustness of a positive outcome. It

is believed that Lucas only performed this test once as it has been estimated that he

spent between 170 and 300 hours performing the necessary calculations. The world’s

largest known primes are still of the Mersenne form and continue to be found by use

of the Lucas-Lehmer test via the GIMPS project (the great internet Mersenne prime

search) [gim]. There are only 46 Mersenne primes known to date, the largest being

M43112609 which is an impressive 12978189 decimal digit number. For large primes of

other forms we direct the reader to the website the prime pages maintained by Boris

Iskra [Isk].

We conclude this chapter by characterizing all the values of P and Q modulo a

prime p ≡ −1 (mod 4) for which (∆p

) = (Qp

) = −1. We use the notation α to denote

the conjugate of α ∈ Q(√

∆) and we use N(α) = αα to denote the norm of α and

Tr(α) = α + α, to denote the trace of α.

Theorem 2.25. Let p be a prime such that p ≡ −1 (mod 4). There exist P , Q such

that (∆p

) = (Qp

) = −1 if and only if Q ≡ N(λ), P ≡ Tr(λ) (mod p), where λ ∈ Z[i]

and (N(λ)p

) = −1.

Proof. Suppose Q ≡ N(λ), P ≡ Tr(λ) (mod p) and (N(λ)p

) = −1. Then (Qp

) =

1Lucas did not actually write out the calculations but made a game of it, for details see [WS94]

45

(N(λ)p

) = −1 and ∆ = P 2 − 4Q = Tr(λ)2 − 4 N(λ). If λ = a + bi, then Tr(λ) = 2a

and N(λ) = a2 + b2, hence Tr(λ)2 − 4 N(λ) = −4b2 and (∆p

) = (−4b2

p) = (−1

p) = −1.

Now, suppose (∆p

) = (Qp

) = −1. We have (P2−4Qp

) = −1, thus (4Q−P 2

p) = 1.

Hence, there exists some c (mod p) such that 4Q − P 2 ≡ c2 (mod p) and hence

Q ≡ (2−1P )2 + (2−1c)2 (mod p). Putting λ = a + bi, where a ≡ 2−1P , b ≡ 2−1c

(mod p), we get λ ∈ Z[i], P ≡ Tr(λ), Q ≡ N(λ) (mod p) and (N(λ)p

) = −1.

Chapter 3

A New Attempt to Generalize the Lucas

Sequences

3.1 De Longchamps’ Method

Perhaps the oldest cubic generalization of Lucas sequences was provided by Gastone

Gohierre de Longchamps. If we let α, β, γ be the zeros of X3 − PX2 + QX − R,

where P,Q,R are integers, then we can define the following sequences suggested by

de Longchamps (1880)[dL80],

RnDn = (αn + βn)(βn + γn)(γn + αn),

RnEn = (αn − βn)(βn − γn)(γn − αn)/[(α− β)(β − γ)(γ − α)],

Sn = αn + βn + γn,

Notice that if we let δ = (α − β)(β − γ)(γ − α), then δ2 = ∆ = Q2P 2 − 4Q3 −

4RP 3 + 18PQR− 27R2, where ∆ is the discriminant of the above mentioned cubic.

Also note that α + β + γ = P , αβ + βγ + γα = Q and αβγ = R.

For the sake of clarity let us denote RnDn = Ln, RnEn = Cn and Sn = An, so

this notation will match the other generalizations. De Longchamps’ work yielded a

few interesting results, including the multiplicative formula

C2n = LnCn.

46

47

He also developed the following identities

Ln = Rn(σn + τn + 2) and δCn = Rn(σn − τn),

where

σn =αn

βn+βn

γn+γn

αnand τn =

βn

αn+αn

γn+γn

βn.

However, it should be stated that neither σn nor τn are integer sequences.

If we let Sn = αnβ2n + βnγ2n + γnα2n, Tn = α2nβn + β2nγn + γ2nαn, then

δCn = Sn − Tn and Ln = Sn + Tn + 2Rn.

Also,

SnTn = RnA3n +B3

n − 6RnAnBn + 9R2n

= RnA3n +B3n + 3R2n (see Theorem 3.5),

where Bn is defined in the next section.

3.2 Another Cubic Generalization

In an attempt to develop a theory analogous to that of Lucas functions the following

method was proposed by Williams (1998) [Wil98]. Again, there are three sequences

defined in this generalization. As in the last method, let α, β, γ be the zeros of

X3 − PX2 +QX −R, where P,Q,R are integers. Now define

An = αn + βn + γn, (3.1)

Bn = αnβn + βnγn + γnαn, (3.2)

Cn =

(αn − βn

α− β

)(βn − γn

β − γ

)(γn − αn

γ − α

). (3.3)

48

Rather than a second order linear recurrence as in Theorem 2.3 for the Lucas

case, there is the following result for An and Bn.

Theorem 3.1. The sequences An and Bn respectively satisfy the following third order

recurrence formulas,

tn+3 = Ptn+2 −Qtn+1 +Rtn,

tn+3 = Qtn+2 −RPtn+1 +R2tn.

Proof. First let

tn = c1αn + c2β

n + c3γn,

where ci are constants. Then

tn+3 = c1αn+3 + c2β

n+3 + c3γn+3

= c1αnα3 + c2β

nβ3 + c3γnγ3.

Using the fact that α, β and γ are the roots of the polynomial X3−PX2 +QX−R,

we can observe

α3 = Pα2 −Qα +R

β3 = Pβ2 −Qβ +R

γ3 = Pγ2 −Qγ +R.

49

Substituting these equalities into our original equation for tn+3 gives

tn+3 = c1αn(Pα2 −Qα +R) + c2β

n(Pβ2 −Qβ +R)

+ c3γn(Pγ2 −Qγ +R)

= P (c1αn+2 + c2β

n+2 + c3γn+2)−Q(c1α

n+1 + c2βn+1 + c3γ

n+1)

+ R(c1αn + c2β

n + c3γn)

= Ptn+2 −Qtn+1 +Rtn.

The recurrence relation for Bn follows from substituting αβ for α, βγ for β and

γα for γ in the above argument.

The recurrence relation for Cn is not as simple as that for An and Bn and will be

covered in a later section.

Some easily verified identities for generalized Lucas functions follow in the theo-

rems below. The next result is a nice generalization of the facts Un = −QnU−n and

Vn = QnV−n.

Theorem 3.2.

An = RnB−n,

Bn = RnA−n and

Cn = −R2nC−n.

Proof.

RnB−n = αnβnγn(α−nβ−n + β−nγ−n + γ−nα−n

)= γn + αn + βn = An.

50

The proof for Bn and Cn follow by the same method, that is, writing both sides of

the equation in terms of α, β and γ.

If we write (2.11) as ∆U2n = V 2

n − 4Qn, then the following theorem is a useful

generalization for this cubic case.

Theorem 3.3.

∆C2n = A2

nB2n + 18AnBnR

n − 4B3n − 4A3

nRn − 27R2n,

27∆C2n = 4(A2

n − 3Bn)3 − (27Rn + 2A3n − 9AnBn)2.

There are also doubling formulas analogous to (2.12) and (2.13) and tripling

formulas.

Theorem 3.4.

A2n = A2n − 2Bn,

B2n = B2n − 2RnAn,

C2n = Cn(AnBn −Rn).

Theorem 3.5.

A3n = A3n − 3AnBn + 3Rn,

B3n = B3n − 3RnAnBn + 3R2n,

C3n = Cn(A2nB

2n −B3

n −RnA3n).

More general than the doubling or tripling formulas, there is the following theo-

rem that provides some addition formulas for An and Bn.

51

Theorem 3.6.

An+m = AnAm − (BnAm−n −RnAm−2n)

Bn+m = BnBm −Rn(AnBm−n −RnBm−2n).

Proof.

AnAm − (BnAm−n −RnAm−2n) = (αn + βn + γn)(αm + βm + γm)

− [(αnβn + βnγn + γnαn)(αm−n + βm−n + γm−n)

− αnβnγn(αm−2n + βm−2n + γm−2n)]

= [αn+m + βn+m + γn+m + αmβn + γnαm + αnβm + γnβm + αnγm + βnγm]

− [(αmβn + αm−nβnγn + γnαm + αnβm + βmγn + γnαnβm−n

+ γm−nαnβn + βnγm + γmαn)− (αm−nβnγn + γnαnβm−n + γm−nαnβn)]

= αn+m + βn+m + γn+m = An+m.

Similar methods are used to show Bn+m = BnBm −Rn(AnBm−n −RnBm−2n).

Corollary 3.6.1.

σn+m = σnσm − τnσm−n + σm−2n

τn+m = τnτm − σnτm−n + τm−2n.

Proof. These identities follow from the previous theorem by setting R = 1 and

replacing α by α/β, β by β/γ and γ by γ/α.

52

Note that historically Corollary 3.6.1 was discovered by de Longchamps in his

original paper [dL80].

The addition identities together with the doubling and tripling formulas may

then be used to derive the following results.

Theorem 3.7.

A5n = A5n − 5A3

nBn + 5A2nR

n + 5AnB2n − 5BnR

n

B5n = B5n − 5B3

nAnRn + 5B2

nR2n + 5BnA

2nR

2n − 5AnR3n.

Proof. Replace the identities A2n, B2n and A3n from Theorem 3.5 into the addition

formula for An+m from Theorem 3.6 while setting m = 4n. Similarly use identi-

ties B2n, A2n, and B3n from Theorem 3.5 into the addition formula for Bn+m from

Theorem 3.6 setting m = 4n.

Corollary 3.7.1.

σ5n = σ5n − 5σ3

nτn + 5σnτ2n + 5σ2

n − 5τn

τ5n = τ 5n − 5τ 3

nσn + 5τnσ2n + 5τ 2

n − 5σn.

Proof. These identities follow from the previous theorem by setting R = 1 and

replacing α by α/β, β by β/γ and γ by γ/α.

3.3 Our Generalization

Let α1, α2, . . . , αm be the roots of the degree m polynomial Xm − Pm−1Xm−1 +

Pm−2Xm−2 − · · · + (−1)mP0, where Pm−1, . . . , P0 are integers. Further if we let

53

δ =∏

1≤i<j≤m(αj − αi) then ∆ = δ2 is the discriminant of the above polynomial. It

will be assumed that ∆ 6= 0. Lastly, let

V =

1 αn1 α2n1 . . . α

(m−1)n1

1 αn2 α2n2 . . . α

(m−1)n2

1 αn3 α2n3 . . . α

(m−1)n3

......

.... . .

...

1 αnm α2nm . . . α

(m−1)nm

where V is a Vandermonde matrix. Then we can define generalized Lucas sequences

of degree m as follows:

δCn = detV

=∏

1≤i<j≤m

(αnj − αni ).

Or, using the Leibniz formula,

δCn =∑σ∈Sm

sgn(σ)αn(σ(1)−1)1 . . . αn(σ(m)−1)

m

and we define Wn by

Wn =∑σ∈Sm

αn(σ(1)−1)1 . . . αn(σ(m)−1)

m ,

where Sm denotes the set of permutations of {1, 2, . . . ,m}, and sgn(σ) denotes the

sign of the permutation σ.

It can be readily verified for the case m = 2 that this generalization is, in fact,

just the historic Lucas sequence, that is, Cn = Un and Wn = Vn. Note that this

54

generalization only relies on the use of two sequences as in the original case, not the

expected three as for the cubic case.

In an effort to achieve simplicity and clarity with the new generalization, we will

restrict ourselves to the case where m = 3. As usual, let α, β, γ be the same as

in the previous generalizations where P , Q and R are their elementary symmetric

functions. We can put

δCn =(αnβ2n + βnγ2n + γnα2n

)−(α2nβn + β2nγn + γ2nαn

)and

Wn =(αnβ2n + βnγ2n + γnα2n

)+(α2nβn + β2nγn + γ2nαn

).

Theorem 3.8. For a fixed m, the sequences Cn and Wn satisfy the recurrence for-

mula

Xn+6m = a1Xn+5m − a2Xn+4m + a3Xn+3m − a4Xn+2m + a5Xn+m − a6Xn,

where

a1 = Wm, a2 = (W 2m −∆C2

m)/4 +RmWm,

a3 = Rm(W2m + 2RmWm + 2R2m), a4 = R2ma2,

a5 = R4ma1, a6 = R6m.

The proof of the above theorem follows on noting that both Cn and Wn are

linear combinations of αmβ2m, βmγ2m, γmα2m, α2mβm, β2mγm, γ2mαm and these 6

quantities are the zeros of

x6 − a1x5 + a2x

4 − a3x3 + a4x

2 − a5x+ a6.

55

Returning to de Longchamps’ work, we can make the following observations.

Letting Sn = αnβ2n+βnγ2n+γnα2n, Tn = α2nβn+β2nγn+γ2nαn as before, we have

δCn = Sn − Tn, Wn = Sn + Tn,

Ln = Sn + Tn + 2Rn and Wn = Ln − 2Rn = AnBn − 3Rn.

It is also true that

Sn = Rnσn and Tn = Rnτn.

The relations above combine to yield the important formulas

Sn = Rnσn =Wn + δCn

2and Tn = Rnτn =

Wn − δCn2

. (3.4)

One can easily verify that

W 2n −∆C2

n

4∈ Z

by noting

W 2n −∆C2

n

4= SnTn = 3R2n +RnA3n +B3n (3.5)

= RnA3n +B3

n − 6RnAnBn + 9R2n ∈ Z. (3.6)

Theorem 3.9.

R2nC−n = −Cn and R2nW−n = Wn.

Note that in the above theorem R2n is the logical analogue to Qn in the identities

QnU−n = −Un and QnV−n = Vn

for the quadratic case.

56

3.4 Addition Formulas for Wn and Cn

As in the historic generalizations for Lucas sequences, there exist addition formulas

for Cn and Wn. These formulas build on de Longchamps’ work, and are analogues

of (2.5) and (2.6).

Theorem 3.10.

2W2n+m = WnWn+m + ∆CnCn+m −Rn(WnWm −∆CnCm − 2R2mWn−m)

2C2n+m = Cn+mWn + CnWn+m −Rn(CmWn − CnWm + 2R2mCn−m).

Proof. First, it is clear that

(Wn + δCn)(Wn+m + δCn+m) = WnWn+m + δCnWn+m + δCn+mWn + ∆CnCn+m.

Using the fact that Rnσn = Wn+δCn2

we have

(Wn + δCn)(Wn+m + δCn+m) = (2Rnσn)(2Rn+mσn+m)

= 4R2n+mσnσn+m.

Corollary 3.6.1 and the fact σ−n = τn yield

σnσn+m = σ2n+m + τnσm − τn−m.

Hence

(Wn + δCn)(Wn+m + δCn+m) = 4R2n+m(σ2n+m + τnσm − τn−m)

= 4R2n+m

(W2n+m + δC2n+m

2R2n+m+Wn − δCn

2Rn

Wm + δCm2Rm

− Wn−m − δCn−m2Rn−m

)= 2W2n+m + 2δC2n+m +Rn(WnWm − δCnWm + δCmWn −∆CnCm

− 2R2mWn−m + 2δR2mCn−m).

57

Thus we may conclude

WnWn+m + δCnWn+m + δCn+mWn + ∆CnCn+m

= 2W2n+m + 2δC2n+m + δRn(−CnWm + CmWn + 2R2mCn−m)

+ Rn(WnWm −∆CnCm − 2R2mWn−m).

We next use the identity Rnτn = Wn−δCn2

and manipulate (Wn−δCn)(Wn+m−δCn+m)

with the additive identity for τ in Corollary 3.6.1. By adding and subtracting the

resulting formula from that given above, we get

2W2n+m = WnWn+m + ∆CnCn+m −Rn(WnWm −∆CnCm − 2R2mWn−m)

and

2C2n+m = Cn+mWn + CnWn+m −Rn(CmWn − CnWm + 2R2mCn−m).

There are the following special cases of the previous theorem.

Corollary 3.10.1.

2W2n = ∆C2n +W 2

n − 4RnWn,

C2n = Cn(Wn + 2Rn) = CnLn,

4W3n = 3∆C2n(Wn + 2Rn) +W 2

n(Wn − 6Rn) + 24R3n,

4C3n = Cn(∆C2n + 3W 2

n).

The next corollary is only a slight modification of the previous theorem, but it

does put the identities in a nicer form by removing the subtractions in the subscripts.

58

Corollary 3.10.2.

2Wn+3m = ∆CmCn+2m +WmWn+2m −RmWmWn+m +Rm∆CmCn+m + 2R3mWn

2Cn+3m = WmCn+2m + CmWn+2m −RmWmCn+m +RmCmWn+m − 2R3mCn.

Proof. Use Theorem 3.10 and replace n by m and m by n+m.

Theorem 3.11.

4R2n−1PQ = W 2n −∆C2

n + 2(Wn+1Cn − Cn+1Wn)− 2R(Wn+1Cn−1

− Wn−1Cn+1) + 2R2(WnCn−1 −Wn−1Cn).

Proof. Replacing n with n+ r in the equations from Theorem 3.10 returns

2W2n+2r+m = Wn+rWn+r+m + ∆Cn+rCn+r+m

− Rn+r(Wn+rWm −∆Cn+rCm − 2R2mWn+r−m) (3.7)

2C2n+2r+m = Cn+r+mWn+r + Cn+rWn+r+m

− Rn+r(CmWn+r − Cn+rWm + 2R2mCn+r−m). (3.8)

Put m = −r in (3.7) and (3.8) to obtain

2W2n+r = Wn+rWn + ∆Cn+rCn

− Rn+r(Wn+rW−r −∆Cn+rC−r − 2R−2rWn+2r)

2C2n+r = CnWn+r + Cn+rWn

− Rn+r(C−rWn+r − Cn+rW−r + 2R−2rCn+2r).

59

Then use W−r = Wr/R2r, C−r = −Cr/R2r to get

2W2n+r = Wn+rWn + ∆Cn+rCn

− Rn−rWn+rWr −Rn−r∆Cn+rCr + 2Rn−rWn+2r (3.9)

2C2n+r = CnWn+r + Cn+rWn

− Rn−rCrWn+r +Rn−rCn+rWr − 2Rn−rCn+2r. (3.10)

Setting m = 0 in (3.7) and (3.8) we get

2W2n+2r = W 2n+r + ∆C2

n+r − (W0 − 2)Rn+rWn+r (3.11)

2C2n+2r = 2Wn+rCn+r + (W0 − 2)Rn+rCn+r. (3.12)

If we put m = n in (3.8) and m = n+ 2r in the second identity in Theorem 3.10 we

get

2C3n+2r = Wn+rC2n+r + Cn+rWn+2r −Rn+rWn+rCn +Rn+rCn+rWn − 2R3n+rCr

2C3n+2r = WnC2n+2r + CnW2n+2r −RnWnCn+2r +RnCnWn+2r + 2R3nC2r.

It follows by equating the right hand sides of the previous two equations and doubling

that

2Wn+rC2n+r +2Cn+rWn+2r − 2Rn+rWn+rCn + 2Rn+rCn+rWn − 4R3n+rCr

= 2WnC2n+2r + 2CnW2n+2r − 2RnWnCn+2r + 2RnCnWn+2r + 4R3nC2r.

Rearrange this to obtain

4R3n(C2r +RrCr)

= 2Wn+rC2n+r + 2Cn+rWn+2r − 2WnC2n+2r − 2CnW2n+2r

− 2Rn+r(Wn+rCn − Cn+rWn) + 2Rn(WnCn+2r − CnWn+2r). (3.13)

60

Now, using (3.9) and (3.10), notice that

2Wn+rC2n+r + 2Cn+rWn+2r

= Wn+r(Wn+rCn + Cn+rWn +Rn−rWn+rCr +Rn−rCn+rWr − 2Rn−rCn+2r)

+ Cn+r(Wn+rWn + ∆Cn+rCn −Rn−rWn+rWr −Rn−r∆Cn+rCr + 2Rn−rWn+2r)

= Cn(W 2n+r + ∆C2

n+r) + 2WnWn+rCn+r +Rn−r(W 2n+r −∆C2

n+r)Cr

+ 2Rn−r(Wn+2rCn+r − Cn+2rWr).

Similarly by (3.11) and (3.12)

2WnC2n+2r + 2CnW2n+2r

= Cn(W 2n+r + ∆C2

n+r) + 2WnWn+rCn+r +Rn+r(W0 − 2)(WnCn+r − CnWn+r).

Using the last two identities we see

2Wn+rC2n+r + 2Cn+rWn+2r − 2WnC2n+2r − 2CnW2n+2r

= Rn−r(W 2n+r −∆C2

n+r)Cr + 2Rn−r(Wn+2rCn+r − Cn+2rWr)

+ Rn+r(W0 − 2)(CnWn+r −WnCn+r).

Now use the above to modify (3.13) as follows

4R3n(C2r +RrCr)

= Rn−r(W 2n+r −∆C2

n+r)Cr + 2Rn−r(Wn+2rCn+r − Cn+2rWr)

+ Rn+r(W0 − 2)(CnWn+r −WnCn+r)− 2Rn+r(Wn+rCn − Cn+rWn)

+ 2Rn(WnCn+2r − CnWn+2r)

= Rn−r(W 2n+r −∆C2

n+r)Cr + 2Rn−r(Wn+2rCn+r − Cn+2rWr)

+ Rn+r(W0 − 4)(CnWn+r −WnCn+r) + 2Rn(WnCn+2r − CnWn+2r).

61

Dividing both sides of this equation by Rn−r gives

4R2n+r(C2r +RrCr)

= (W 2n+r −∆C2

n+r)Cr + 2(Wn+2rCn+r − Cn+2rWr)

+ R2r(W0 − 4)(CnWn+r −WnCn+r)− 2Rr(CnWn+2r −WnCn+2r). (3.14)

Putting r = 1 and replacing n by n− 1 in (3.14) yields

4R2n−1(C2 +RC1) = W 2n −∆C2

n + 2(Wn+1Cn − Cn+1Wn)− 2R(Wn+1Cn−1

− Wn−1Cn+1) +R2(W0 − 4)(WnCn−1 −Wn−1Cn).

Noting C2 = W1 + 2R, W1 = PQ − 3R, W0 = 6 and C1 = 1 ⇒ C2 + RC1 = PQ

completes the proof.

This formula is an extension the Lucas identity (2.11)

V 2n −∆U2

n = 4Q′n

where Vn = Vn(P ′, Q′) and Un = Un(P ′, Q′). This can be justified as follows. Since

V−n = Vn/Q′n and U−n = −Un/Q′n, we see that R2 corresponds to Q′. Using the

identity

2Q′mUn−m = VmUn − UmVn

we can see

−2Q′n = Vn+1Un − Un+1Vn when m = n+ 1 and n = n,

−2Q′n−1 = VnUn−1 − UnVn−1 when m = n and n = n− 1,

−2Q′n−1P ′ = Vn+1Un−1 − Un+1Vn−1 when m = n+ 1 and n = n− 1.

62

Replacing Q′ by R2 in the above returns

Vn+1Un − Un+1Vn = −2R2n,

VnUn−1 − UnVn−1 = −2R2n−2,

Vn+1Un−1 − Un+1Vn−1 = −2R2n−2P ′.

Also note that U2 + RU1 = P ′ + R. Using the above and replacing Wm by Vm and

Cm by Um into the identity in Theorem 3.11 we see

V 2n −∆U2

n = 4R2n−1(P ′ +R)− 2(−2R2n) + 2R(−2R2n−2P ′) + 2R2(−2R2n−2)

= 4R2n.

It is not surprising that Theorem 3.11 involves 6 objects: Wn−1, Wn, Wn+1, Cn−1,

Cn, Cn+1, as one may recall that both {Wn} and {Cn} satisfy a degree 6 recurrence.

By similar methods we can develop and justify another generalization of the same

Lucas identity V 2n −∆U2

n = 4Q′n in the following theorem.

Theorem 3.12.

4R2n−1(P 2Q2 − 2Q3 − 2RP 3 + 5PQR− 6R2) =

−(W 2n −∆C2

n)W1 + 2(Wn+1Wn −∆Cn+1Cn) + 2R(Wn+1Wn−1 −∆Cn−1Cn+1)

+2R2(WnWn−1 −∆Cn−1Cn).

Proof. If we put m = n in (3.7) and m = n+ 2r in the first identity of Theorem 3.10

we get

2W3n+r = Wn+rW2n+r + ∆Cn+rC2n+r

− Rn+rWn+rWn +Rn+r∆Cn+rCn + 2R3n+rWr

63

2W3n+r = WnW2n+2r + ∆CnC2n+2r

− RnWn+2rWn +Rn∆Cn+2rCn + 2R3nW2r.

Equate the right hand sides, double, then rearrange to obtain

4R3n(W2r −RrWr) = 2Wn+rW2n+r + 2∆Cn+rC2n+r − 2WnW2n+2r − 2∆CnC2n+2r

− 2Rn+r(Wn+rWn −∆Cn+rCn)

+ 2Rn(Wn+2rWn −∆Cn+2rCn). (3.15)

Similarly, use (3.9) and (3.10) to see

2Wn+rW2n+r + 2∆Cn+rC2n+r

= (W 2n+r + ∆C2

n+r)Wn + 2∆Wn+rCn+rCn −Rn−r(W 2n+r −∆C2

n+r)Wr

+ 2Rn−r(Wn+2rWn+r −∆Cn+rCn+2r).

By (3.11) and (3.12) we have

2WnW2n+2r + 2∆CnC2n+2r

= Wn(W 2n+r + ∆C2

n+r) + 2∆Wn+rCn+rCn

− (W0 − 2)Rn+r(WnWn+r −∆CnCn+r).

Using (3.15) and the above, we have

4R2n+r(W2r +RrWr) = −(W 2n+r −∆C2

n+r)Wr + 2(Wn+2rWn+r −∆Cn+2rCn+r)

+ 2R(Wn+2rWn −∆Cn+2rCn) +R2r(W0 − 4)

(Wn+rWn −∆Cn+rCn).

64

Using the above and replacing r by 1 and n by n− 1 we have

4R2n−1(W2 +RW1) = −(W 2n −∆C2

n)W1 + 2(Wn+1Wn −∆Cn+1Cn) + 2R(Wn+1Wn−1

− ∆Cn−1Cn+1) +R2(W0 − 4)(WnWn−1 −∆Cn−1Cn).

Using the identities W2 = 12∆+ 1

2W 2

1−2RW1 and W1 = PQ−3R to show W2+RW1 =

P 2Q2 − 2Q3 − 2RP 3 + 5PQR− 6R2 completes the proof.

The formula from the above theorem is another logical extension of the Lucas

identity

V 2n −∆U2

n = 4Q′n

where Vn = Vn(P ′, Q′) and Un = Un(P ′, Q′). Again, this can be justified as follows.

Since V−n = Vn/Q′n and U−n = −Un/Q′n we see that R2 corresponds to Q′. Using

the identity

2Q′mVn−m = VnVm −∆UnUm

we can see

2Q′nV1 = Vn+1Vn −∆Un+1Un when m = n and n = n+ 1,

2Q′n−1V2 = Vn+1Vn−1 −∆Un+1Un−1 when m = n− 1 and n = n+ 1,

2Q′n−1V1 = VnVn−1 −∆UnUn−1 when m = n− 1 and n = n.

Again, replace Q′ by R2 in the above to obtain

Vn+1Vn −∆Un+1Un = 2R2nV1,

Vn+1Vn−1 −∆Un+1Un−1 = 2R2n−2V2,

VnVn−1 −∆UnUn−1 = 2R2n−2V1.

65

It is easily verified that V2−RV1 = P ′2− 2R2−RP ′. The above facts and replacing

Wm by Vm and Cm by Um into the equation in Theorem 3.12 yield

P ′(V 2n −∆U2

n) = −4R2n−1(P ′2 − 2R2 −RP ′) + 2(2R2nP ′

+ 2R(2R2n−2(P ′2 − 2R2))− 2R2(2R2n−2P ′)

= 4R2nP ′.

Replacing R2 with Q′ and dividing both sides by P ′ completes the analogy, giving

V 2n −∆U2

n = 4Q′n.

In view of the importance that the quantity Wn − 6Rn will assume in later

chapters, we also point out that from Theorem 3.3 it is easy to deduce that

(Wn − 6Rn)2 + 3∆C2n = 4(A2

n − 3Bn)(B2n − 3RnAn).

3.5 Multiplication Formulas for Wn and Cn

Theorem 3.13.

16C5n/Cn = ∆2C4n + 20R2n∆C2

n + 20Rn∆C2nWn + 10∆C2

nW2n + 80R3nWn

−20R2nW 2n − 20RnW 3

n + 5W 4n + 80R4n,

16W5n = 5∆2C4n(Wn + 2Rn) + 10∆C2

n(W 3n − 2R2nWn + 4R3n) +Wn(W 4

n

−10RnW 3n + 20R2nW 2

n + 40R3nWn − 80R4n).

Proof. Replace σn and τn with the equations in (3.4) and place them in the first

66

identity from Corollary 3.7.1 to see

W5n + δC5n

2R5n=

(Wn + δCn

2Rn

)5

− 5

(Wn + δCn

2Rn

)3(Wn − δCn

2Rn

)+ 5

(Wn + δCn

2Rn

)(Wn − δCn

2Rn

)2

+ 5

(Wn + δCn

2Rn

)2

− 5

(Wn − δCn

2Rn

).

Multiply both sides by 32R5n to see

16(W5n + δC5n) = (Wn + δCn)5 − 10R(Wn + δCn)3(Wn − δCn) + 20R2n(Wn + δCn)

(Wn − δCn)2 + 40R3n(Wn + δCn)2 − 80R4n(Wn − δCn)

= W 5n + 5δW 4

nCn + 10∆W 3nC

2n + 10δ∆W 2

nC3n + 5∆2WnC

4n + δ∆2C5

n

−10R(W 4n −∆2C4

n) + 20R2n(W 3n −∆WnC

2n − δW 2

nCn + δ∆C3n)

+40R3n(W 3n + ∆C2

n + 2δWnCn)− 80R4n(Wn − δCn).

Equating the irrational parts and rearranging completes the proof. If δ ∈ Z, then

we may use W5n−δC5n

2R5n and the second identity from Corollary 3.7.1 to complete the

proof.

A more general multiplicative result is shown in the following theorem and this

result is our analogue to (2.14) and (2.15). It is at this point where our generalization

begins to outperform the others. This is because other generalizations are missing

the necessary multiplication formulas needed in order to develop arithmetic results.

Theorem 3.14. For any integers m ≥ 0 we have

Wmn =∑ (−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!Rn(λ0+λ3)Qλ2

n Vλ1−λ2(Pn, Qn) (3.16)

CmnCn

=∑ (−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!Rn(λ0+λ3)Qλ2

n Uλ1−λ2(Pn, Qn). (3.17)

67

Here the sum is extended over the values λi ∈ Z such that

λ0, λ1, λ2, λ3 ≥ 0, λ0 + λ1 + λ2 + λ3 = m, λ1 + 2λ2 + 3λ3 = m,

Uk is the Lucas function Uk(Pn, Qn) and Pn = Wn, Qn = (W 2n −∆C2

n)/4.

Proof. First note σ1 = α/β + β/γ + γ/α =∑ri, where the sum is over the three

quantities r1 = α/β, r2 = β/γ, and r3 = γ/α. Thus σ1 is the first elementary

function of degree three involving these three terms. Also τ1 = β/α + γ/β + α/γ =∑i 6=j rirj. Thus τ1 is the second elementary function of degree three. Finally note∑i 6=j 6=k rirjrk = r1r2r3 = 1. Hence we can use Waring’s theorem (see, for example,

[Mac15]) to see that

σn = (α/β)n + (β/γ)n + (γ/α)n =∑

λ1,λ2,λ3

(−1)n+kn(k − 1)!

λ1!λ2!λ3!σλ1

1 τλ21 ,

where λ1, λ2, λ3 ≥ 0, λ1 + λ2 + λ3 = k and λ1 + 2λ2 + 3λ3 = n.

Setting λ0 = n − k so (−1)n+k = (−1)n−k = (−1)λ0 we can write the previous

identity as

σn =∑

λ0,λ1,λ2,λ3

(−1)λ0n(n− λ0 − 1)!

λ1!λ2!λ3!σλ1

1 τλ21 ,

where λ0, λ1, λ2, λ3 ≥ 0, λ0 + λ1 + λ2 + λ3 = n and λ1 + 2λ2 + 3λ3 = n.

Similarly, we can use Waring’s theorem to derive

σmn =∑

λ0,λ1,λ2,λ3

(−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!σλ1n τ

λ2n , (3.18)

τmn =∑

λ0,λ1,λ2,λ3

(−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!τλ1n σλ2

n , (3.19)

where λ0, λ1, λ2, λ3 ≥ 0, λ0 + λ1 + λ2 + λ3 = m and λ1 + 2λ2 + 3λ3 = m. This is the

sum as stated in the theorem. Now, since Smn = Rmnσmn and Tmn = Rmnτmn, we

68

obtain

Wmn = Smn + Tmn =∑

λ0,λ1,λ2,λ3

(−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!Rmn(σλ1

n τλ2n + σλ2

n τλ1n ).

Or considering the term following the coefficient we obtain

Rmn(σλ1n τ

λ2n + σλ2

n τλ1n ) = R(m−λ1−λ2)n(Rnλ1σλ1

n Rnλ2τλ2

n +Rnλ2σλ2n R

nλ1τλ1n )

= R(λ0+λ3)n(Sλ1n T

λ2n + Sλ2

n Tλ1n ).

Now we will employ some well-known results for Lucas sequences; that is,

Sλn =

(Wn + δCn

2

)λ=Vλ + δnUλ

2,

T λn =

(Wn − δCn

2

)λ=Vλ − δnUλ

2,

where U = U(Pn, Qn), V = V (Pn, Qn), ∆n = ∆C2n, δn = δCn and Pn, Qn are as

stated in the theorem.

So

Sλ1n T

λ2n + Sλ2

n Tλ1n =

Vλ1 + δnUλ1

2

Vλ2 − δnUλ2

2+Vλ2 + δnUλ2

2

Vλ1 − δnUλ1

2

=Vλ1Vλ2 − ∆nUλ1Uλ2

2.

To complete the proof of the first identity, use the following identity known for Lucas

sequences:

2QmVn−m = VnVm −∆UnUm,

replacing n = λ1, m = λ2 and ∆ = ∆n. Hence

Sλ1n T

λ2n + Sλ2

n Tλ1n = Qλ2

n Vλ1−λ2 .

69

The second identity is proven similarly by expanding δCn = Sn−Tn via Waring’s

theorem as follows,

δCmn = Smn − Tmn =∑

λ0,λ1,λ2,λ3

(−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!Rmn(σλ1

n τλ2n − σλ2

n τλ1n ).

Now, using substitutions for the U , V as before,

Rmn(σλ1n τ

λ2n − σλ2

n τλ1n ) = R(m−λ1−λ2)n(Rnλ1σλ1

n Rnλ2τλ2

n −Rnλ2σλ2n R

nλ1τλ1n )

= R(λ0+λ3)n(Sλ1n T

λ2n − Sλ2

n Tλ1n )

=Vλ1 + δnUλ1

2

Vλ2 − δnUλ2

2− Vλ2 + δnUλ2

2

Vλ1 − δnUλ1

2

=δn2

(Uλ1Vλ2 − Uλ2Vλ1).

Again, we will use an identity for Lucas sequences to complete the proof,

Uλ1Vλ2 − Uλ2Vλ1 = 2Qλ2n Uλ1−λ2

and we replace δn by δCn.

The corollary below states some special cases.

Corollary 3.14.1.

C5n/Cn = 5R4n + 5R3nPn − 5R2nQn − 5RnPnQn + P 4n − 3P 2

nQn + Q2n,

C6n/Cn = 8R3n(P 2n − Qn)− 6RnQn(P 2

n − Qn) + P 5n − 4P 3

nQn + 3PnQ2n,

C7n/Cn = 7R6n − 7R5nPn − 21R4nQn + 7R3nPn(P 2n − Qn) + 14R2nQ2

n

+7RnPnQn(2Qn − P 2n) + P 6

n − 5P 4nQn + 6P 2

nQ2n − Q3

n.

It is of interest that we can use the multiplicative identity for Cmn to show we

may calculate Cm as a sum of Lucas sequences.

70

Corollary 3.14.2.

Cm = m∑

λ0,λ1,λ2,λ3

(−1)λ0(m− λ0 − 1)!

λ1!λ2!λ3!Rλ0+λ3Qλ2

1 Uλ1−λ2(P1, Q1),

where P1 = PQ− 3R and Q1 = (P 21 −∆)/4 = RP 3 +Q3 − 6PQR + 9R2.

It is the general multiplication formulas that allow us to proceed with this cubic

generalization. With them we are able to develop arithmetic properties for Cn and

Wn in Chapter 4. Once we have arithmetic properties some primality testing can be

done.

3.6 Calculating Generalized Lucas Sequences

By Theorem 3.8 we know that both the {Wn} and {Cn} sequences satisfy the recur-

rence formula

Znm+6n = a1Znm+5n − a2Znm+4n + a3Znm+3n − a4Znm+2n + a5Znm+n − a6Znm,

where

a1 = Wn, a2 = (W 2n −∆C2

n)/4 +RnWn,

a3 = Rn(W2n + 2RnWn + 2R2n), a4 = R2na2,

a5 = R4na1, a6 = R6n.

Also, we have in Corollary 3.10.1 the result that 2W2n = ∆C2n +W 2

n − 4RnWn. Now

put

Xk =Wk

2Rkand Dk =

∆k

4R2k=

∆C2k

4R2k,

71

then

a1

Rn= 2Xn,

a2

R2n=

(W 2n −∆C2

n)

4R2n+RnWn

R2n= X2

n + 2Xn − Dn,

a3

R3n=

Rn(W2n + 2RnWn + 2R2n)

R3n=W2n

R2n+

2Wn

Rn+ 2

=∆C2

n +W 2n − 4RnWn

2R2n+

2Wn

Rn+ 2 =

∆C2n +W 2

n

2R2n− 2Wn

Rn+

2Wn

Rn+ 2

= 2∆C2

n

4R2n+ 2

W 2n

4R2n+ 2 = 2(X2

n + Dn + 1),

a4

R4n=

R2na2

R4n=

a2

R2n= X2

n + 2Xn − Dn,a5

R5n=R4na1

R5n=

a1

Rn= 2Xn,

a6

R6n= 1.

Hence

X(m+6)n = 2XnX(m+5)n − (X2n − Dn + 2Xn)X(m+4)n + 2(X2

n + Dn + 1)X(m+3)n

− (X2n − Dn + 2Xn)X(m+2)n + 2XnX(m+1)n −Xmn,

where

X0 = 3, Xn =Wn

2Rn, X2n =

W2n

2R2n=

∆C2n +W 2

n − 4RnWn

4R2n= X2

n + Dn − 2Xn,

X3n =W3n

2R3n=

W 3n

8R3n+ 3

∆C2nWn

8R3n+ 3

Rn∆C2n

4R3n− 3

RnW 2n

4R3n+

6R3n

2R3n

=W 3n

8R3n+ 3

∆C2n

4R2n

Wn

2Rn+ 3

∆C2n

4R2n− 3

W 2n

4R2n+ 3

= X3n + 3DnXn + 3Dn − 3X2

n + 3.

Also,

X−mn =W−mn2R−mn

=Wmn

R2(mn)/2R−mn =

Wmn

2Rmn= Xmn.

It follows that Xmn = Fm(Xn, Dn), where

Fm+6 = 2XnFm+5 − (X2n − Dn + 2Xn)Fm+4 + 2(X2

n + Dn + 1)Fm+3

− (X2n − Dn + 2Xn)Fm+2 + 2XnFm+1 − Fm,

72

and

F0 = 3, F1 = Xn, F2 = X2n + Dn − 2Xn, F3 = X3

n + 3DnXn + 3Dn − 3X2n + 3,

Fm(Xn, Dn) = F−m(Xn, Dn).

Furthermore, if we put Ym,n = CmnCnRmn−n

, then Ym,n = Gm(Xn, Dn) where

Gm+6 = 2XnGm+5 − (X2n − Dn + 2Xn)Gm+4 + 2(X2

n + Dn + 1)Gm+3

− (X2n − Dn + 2Xn)Gm+2 + 2XnGm+1 −Gm,

and

G0 = 0, G1 = Y1,n =Cn

CnR1−1= 1,

G2 =C2n

CnRn=CnWn

CnRn+

2RnCnCnRn

= 2Xn + 2,

G3 =C3n

CnR2n=

3CnW2n

4CnR2n+

∆C3n

4CnR2n=

3W 2n

4R2n+

∆C2n

4R2n= 3X2

n + Dn.

Also,

Y−m,n =C−mn

CnR−mn−n=−CmnR2mn

/CnR−mn−n = −CmnR

mn+n

CnR2mn= − Cmn

CnRmn−n = −Ym,n.

So

Gm(Xn, Dn) = −G−m(Xn, Dn).

Note that

Dmn =∆C2

mn

4R2mn= ∆

C2n

C2n

C2mn

4R2mn=

∆C2n

4R2n

C2mn

C2n(Rmn−n)2

= DnG2m.

Thus we have found that if we put

Xn =Wn

2Rn, Dn =

∆C2n

4R2nand Ym,n =

CmnCnRmn−n ,

73

then

Xmn = Fm(Xn, Dn), Ym,n = Gm(Xn, Dn),

where Fm, Gm ∈ Z[x, y] satisfy

Zm+6 = 2xZm+5 − (x2 + 2x− y)Zm+4 + 2(x2 + y + 1)Zm+3

− (x2 + 2x− y)Zm+2 + 2xZm+1 − Zm, (3.20)

and

F0 = 3, F1 = x, F2 = x2 − 2x+ y, F3 = x3 + 3yx+ 3y − 3x2 + 3,

G0 = 0, G1 = 1, G2 = 2x+ 2, G3 = 3x2 + y,

F−m(x, y) = Fm(x, y) and G−m(x, y) = −Gm(x, y).

Consider the equation

z3 − (x+√y)z2 + (x−√y)z − 1 = 0. (3.21)

Let λ, µ, ν be the zeros of (3.21). Then clearly λ−1, µ−1, ν−1 are the zeros of

z3 − (x−√y)z2 + (x+√y)z − 1 = 0

and therefore λ, µ, ν, λ−1, µ−1, ν−1 are the zeros of

(z3 − (x+√y)z2 + (x−√y)z − 1)(z3 − (x−√y)z2 + (x+

√y)z − 1)

= z6 − 2xz5 + (x2 + 2x− y)z4 − (2x2 + 2y + 2)z3 + (x2 + 2x− y)z2 − 2xz + 1.

Note that λµν = 1. From this, the boundary conditions on Fm and Gm and the

fact that they satisfy the recurrence (3.20), we find that

Fm = Fm(x, y) =1

2(λm + µm + νm + λ−m + µ−m + ν−m),

74

Gm = Gm(x, y) =λm + µm + νm − λ−m − µ−m − ν−m

2√y

for y 6= 0.

Since λ, µ, ν satisfy

zn+3 = (x+√y)zn+2 − (x−√y)zn+1 + zn,

z−(n+3) = (x−√y)z−(n+2) − (x+√y)z−(n+1) + z−n,

we can see that

Fn+3 =1

2(λn+3 + µn+3 + νn+3 + λ−(n+3) + µ−(n+3) + ν−(n+3))

=1

2((x+

√y)λn+2 − (x−√y)λn+1 + λn + (x+

√y)µn+2

−(x−√y)µn+1 + µn + (x+√y)νn+2 − (x−√y)νn+1

+νn + (x−√y)λ−(n+2) − (x+√y)λ−(n+1) + λ−n

+(x−√y)µ−(n+2) − (x+√y)µ−(n+1) + µ−n + (x−√y)ν−(n+2)

−(x+√y)ν−(n+1) + ν−n)

=1

2(xλn+2 +

√yλn+2 − xλn+1 +

√yλn+1 + λn

+xµn+2 +√yµn+2 − xµn+1 +

√yµn+1 + µn

+xνn+2 +√yνn+2 − xνn+1 +

√yνn+1 + νn

+xλ−(n+2) −√yλ−(n+2) − xλ−(n+1) −√yλ−(n+1) + λ−n

+xµ−(n+2) −√yµ−(n+2) − xµ−(n+1) −√yµ−(n+1) + µ−n

+xν−(n+2) −√yν−(n+2) − xν−(n+1) −√yν−(n+1) + ν−n)

= xFn+2 − xFn+1 + yGn+2 + yGn+1 + Fn.

By the same method one can easily verify that

Gn+3 = xGn+2 − xGn+1 + Fn+2 + Fn+1 +Gn.

75

If we put Hn = λn + µn + νn then by Theorem 3.6 we have that

Hn+m = HnHm −H−nHm−n +Hm−2n. (3.22)

Putting m = n+ 1 in equation (3.22), we get

H2n+1 = HnHn+1 −H−nH1 +H−n+1. (3.23)

Now using the facts

Fn =1

2(Hn +H−n) and Gn =

1

2√y

(Hn −H−n),

yields

Hn = Fn +√yGn and H−n = Fn −

√yGn.

Substituting this into (3.23) we get

F2n+1 = FnFn+1 + yGnGn+1 − xFn + yGn + Fn−1,

G2n+1 = GnFn+1 +Gn+1Fn + xGn − Fn −Gn−1.

Or more generally, using (3.22) and replacing m by m+ n, we can derive

F2n+m = yGn(Gm+n +Gm) + Fn(Fm+n − Fm) + Fm−n,

G2n+m = Gn(Fm+n + Fm) + Fn(Gm+n −Gm) +Gm−n.

The cost of computing F2n+m, G2n+m from

{Fm+n, Gm+n, Fm, Gm, Fn, Gn, Fm−n, Gm−n}

is 5 multiplications.

76

Now since

Fn+2 = xFn+1 − xFn + yGn+1 + yGn + Fn−1,

Gn+2 = xGn+1 − xGn + Fn+1 + Fn +Gn−1,

we get

Fn−1 = Fn+2 − xFn+1 + xFn − yGn+1 − yGn,

Gn−1 = Gn+2 − xGn+1 + xGn − Fn+1 − Fn.

Hence

F2n+1 = FnFn+1 + yGnGn+1 − xFn+1 − yGn+1 + Fn+2

= Fn+1(Fn − x) + yGn+1(Gn − 1) + Fn+2, (3.24)

G2n+1 = GnFn+1 +Gn+1Fn −Gn+2 + xGn+1 + Fn+1

= Fn+1(Gn + 1) +Gn+1(Fn + x)−Gn+2. (3.25)

Also, if we replace n by n+ 1 in the above, then

F2n+3 = Fn+1(Fn+2 − x)− yGn+1(Gn+2 − 1) + Fn, (3.26)

= G2n+3 = Fn+1(Gn+2 + 1) +Gn+1(Fn+2 + x)−Gn. (3.27)

We can also set m = n in (3.22) to get

H2n = H2n − 3H−n +H−n = H2

n − 2H−n and H−2n = H2−n − 2Hn.

77

Using this, we can easily obtain

F2n = F 2n + yG2

n − 2Fn = Fn(Fn − 2) + yG2n, (3.28)

G2n = 2Gn(Fn + 1), (3.29)

F2n+2 = Fn+1(Fn+1 − 2) + yG2n+1, (3.30)

G2n+2 = 2Gn+1(Fn+1 + 1). (3.31)

Thus given the sextet

Sn = {Fn, Fn+1, Fn+2, Gn, Gn+1, Gn+2}

we can compute

S2n+1 = {F2n+1, F2n+2, F2n+3, G2n+1, G2n+2, G2n+3}

using (3.24), (3.25), (3.26), (3.27), (3.30), (3.31) with 12 multiplications. If one is

not careful it may appear as though we need to do 14 multiplications, but yGn+1

occurs 3 times and we need only calculate it once. We are also able to compute

S2n = {F2n, F2n+1, F2n+2, G2n, G2n+1, G2n+2}

using (3.24), (3.25), (3.28), (3.29), (3.30), (3.31) with 12 multiplications.

These observations can now be used to compute (by F ) Xmn, Ym,n (mod r) for a

given modulus r, given Xn, Dn in O(logm) modular multiplications. We begin with

S1 = {F1, F2, F3, G1, G2, G3} (mod r),

which can be computed using Xn, Dn only. We then compute

Sm = {Fm, Fm+1, Fm+2, Gm, Gm+1, Gm+2} (mod r)

78

as follows. Let (b0b1 . . . bk)2 = m be the binary representation of m such that b0 6= 0.

Set P0 = S1 and for i = 0 to i = k − 1

Pi+1 =

S2n (mod r) if bi+1 = 0

S2n+1 (mod r) if bi+1 = 1.

Then Pk = Sm. This gives us Xmn ≡ Fm (mod r), Ym,n ≡ Gm (mod r) and Dmn =

DnG2m (mod r).

Thus, if k = dlogme we need to perform 12k modular multiplications to com-

pute Sm (mod r). To compute tm ≡ am (mod r) requires on average 32k modular

multiplications. Thus, for a given m computing Sm is 8 times more expensive than

computing tm.

Chapter 4

Arithmetic Properties of {Cn} and {Wn}

4.1 Introductory Arithmetic Results

To continue our generalization we need to develop arithmetic results, both global

and local, that are logical analogues of the arithmetic results seen in Chapter 2 for

Lucas sequences.

Lemma 4.1. If (Q,R) = 1, then (Bn, R) = 1, for n > 0.

Proof. First note B0 = 3, B1 = Q and B2 = Q2 −RP . Also for n ≥ 0

Bn+3 = QBn+2 −RPBn+1 +R2Bn.

By induction Bn ≡ Qn (mod R) for n > 0. The result follows immediately.

We also can produce a somewhat similar result which involves An instead of Bn.

Theorem 4.2. If (Q,R) = 1, then (An, R,∆) | 4.

Proof. Let p be any odd prime such that p | (∆, R). From the formula for ∆, we see

that p | Q2P 2 − 4Q3, and since (Q,R) = 1, we must have that p | P 2 − 4Q. Now

A0 = 3, A1 = P , A2 = P 2 − 2Q and

Ak+2 ≡ PAk+1 −QAk (mod R).

Since Q ≡ P 2/4 (mod p), we get

An ≡ P n/2n−1 (mod p)

79

80

by induction on n. Since p - P , we have p - An and p - (An, R,∆).

Next, suppose that 2ν || (An, R,∆). When ν > 2, we see that 2 | P and since Q

is odd, we must have P/2 odd. Since

Ak+2 ≡ PAk+1 −QAk (mod 4)

and Q ≡ P/2 ≡ 1 (mod 2), 2 || A1 and 2 || A2, we find by induction on n that

2 || An. This is a contradiction to the fact ν > 2.

Corollary 4.2.1. If (Q,R) = 1, then (Wn, R,∆) | 4.

Proof. Since Wn = AnBn − 3Rn, we get (Wn, R) = (AnBn, R) = (An, R) by Lemma

4.1. Hence, (Wn, R,∆) | 4 by the previous theorem.

To prove the next lemma we first note that since

27∆ = 4(P 2 − 2Q)3 − (2P 3 − 9PQ+ 27R)2,

we must have ∆ ≡ 0, 1 (mod 4).

Lemma 4.3. If 2 - R and 2α || (Wn, Cn), then α ∈ {0, 1}, and if 2 | Wn, then

Qn = W 2n−∆C2

n

4is odd.

Proof. If 2 - Wn we are done. Suppose 2 | Wn. Since 2 - R we must have 2 - AnBn,

as AnBn = Wn + 3Rn. So both An and Bn are odd and then 2 | A2n − 3Bn.

By Theorem 3.3 we have

27∆C2n = 4(A2

n − 3Bn)3 − (2A3n − 9Wn)2.

Hence

27∆C2n ≡ −(2A3

n − 9Wn)2 (mod 8).

81

Now, since 2A3n − 9Wn ≡ 2−Wn (mod 4), we get

27∆C2n ≡ −(2−Wn)2 (mod 8).

If 2 || Wn, then 8 | ∆C2n and Qn is odd. If 4 | Wn, then 27∆C2

n ≡ −4 (mod 8) and

Qn is odd, thus, if 4 | ∆ then Cn ≡ 1 (mod 2). If ∆ ≡ 1 (mod 4), then Cn ≡ 2

(mod 4).

Lemma 4.4. If 2 | R, 2 - Q and 2α || (Wn, Cn), then α ∈ {0, 1}, and if 2 | Wn,

then Qn is odd.

Proof. If 2 - Wn we are done. If 2 | Wn, then 2 | AnBn. Since 2 - Q we know 2 - Bn

and 2 | An. We may then observe that A2n − 3Bn is odd and thus

27∆C2n ≡ 4− (2A3

n − 9Wn)2 (mod 8)

≡ 4− (−9Wn)2 (mod 8)

≡ 4−W 2n (mod 8).

If 2 || Wn, then 8 | ∆C2n and Qn is odd. If 4 | Wn, then 27∆C2

n ≡ 4 (mod 8) or

−∆C2n ≡ 4 (mod 8) which implies Cn is odd or 2 || Cn; in either case Qn is odd.

From the above results we have the following theorem.

Theorem 4.5. If (Q,R) = 1 and 2α || (Wn, Cn), then α ∈ {0, 1}. If 2 | Wn, then

Qn is odd.

The following result is a clear analogue of (2.18).

82

Lemma 4.6. If (Q,R) = 1, then

(Wn, Cn, R) | 2.

Proof. If (Wn, Cn, R) = 1, we are done. Let p be any prime such that p | (Wn, Cn, R).

Since p | Wn and p | Cn we must have p | W 2n − ∆C2

n. Observe by equation (3.5)

W 2n − ∆C2

n = 4SnTn = 4(3R2n + RnA3n + B3n) ⇒ p | 4B3n. Also, B3n = B3n −

3RnAnBn + 3R2n, so p | 4B3n. Since (Q,R) = 1, we have (Bn, R) = 1 by Lemma

4.1 but this implies p - Bn ⇒ p = 2. Indeed, by Lemma 4.4, we must have

(Wn, Cn, R) | 2.

Furthermore, it is not difficult to show that, like {Un}, {Cn} is a divisibility

sequence; i.e.

Cm | Cn, when m | n. (4.1)

Note that if n = ms, then

Cn(P,Q,R) =(αn − βn)(βn − γn)(γn − αn)

(α− β)(β − γ)(γ − α)=

(αms − βms)(βms − γms)(γms − αms)(α− β)(β − γ)(γ − α)

=(αm − βm)(βm − γm)(γm − αm)

(α− β)(β − γ)(γ − α)· (αms − βms)(βms − γms)(γms − αms)

(αm − βm)(βm − γm)(γm − αm)

= Cm(P,Q,R) · Cs(Am, Bm, Rm).

Definition 4.7. Given m ∈ Z, let r be the least positive integer, if it exists, such

that m | Cr. This value is called the rank of apparition of m for the sequence {Cn}

and will be denoted by r(m).

In Theorem 2.4 for the classic Lucas case, we had that if m | Uk, then r(m) | k.

However, this is not necessarily true for {Cn}. It may be that m | Ck, yet r(m) - k.

83

Definition 4.8. Let r1 be the least positive integer for which p | Cr1. For i =

1, 2, . . . , k define ri+1, if it exists, to be the least positive integer such that p | Cri+1,

ri+1 > ri and rj - ri+1 for any j ≤ i+ 1. We define r1, r2, . . . , rk to be the ranks of

apparition for {Cn}.

It will become clear that the number of ranks of apparition is finite.

For example, if we let P = 1, Q = 2, R = 3 and p = 7, then {Cn} has two ranks

of apparition for the prime 7. In fact, C3 ≡ 0 (mod p) and C7 ≡ 0 (mod p). Also,

if we let P = 3, Q = 9, R = 7 and p = 31, then {Cn} has three ranks of apparition.

Here, C6 ≡ 0 (mod p), C10 ≡ 0 (mod p) and C15 ≡ 0 (mod p).

Our sequence {Cn} also fails to satisfy the generalization of Corollary 2.4.1 where

if d = (m,n) then

(Um, Un) = |Ud|.

It can be that

(Cm, Cn) 6= |Cd|,

and d = (m,n). For example, if P = 3, Q = 9, R = 7, then (C6, C10) = 22 · 5 · 31

and C2 = 22 · 5.

We can, however, reach a relatively close analogue to Carmichael’s result seen in

Theorem 2.5. To do so we must first derive several preliminary arithmetic results in

the next lemmas and theorems.

Lemma 4.9. If

Im =∑ m(−1)λ0(m− λ0 − 1)!

λ1!λ2!λ3!(4.2)

is summed over all λ0, λ1, λ2, λ3 ∈ Z≥0 such that λ0+λ1+λ2+λ3 = m, λ1+2λ2+3λ3 =

m and λ1 6≡ λ2 (mod 2), then Im ≡ m (mod 2).

84

Proof. Put I ′m equal to the right side of (4.2), where we insist that λ1 ≡ λ2 (mod 2).

By Waring’s theorem

αm + βm + γm =∑ m(−1)λ0(m− λ0 − 1)!

λ1!λ2!λ3!P λ1Qλ2Rλ3 ,

where the sum is over all λi satisfying the constraints in the statement of the lemma

except that of λ1 6≡ λ2 (mod 2). This is true as P = α + β + γ, Q = αβ + βγ + γα

and R = αβγ.

Putting P = Q = R = 1 we get

αm + βm + γm = I ′m + Im

so I ′m + Im = 1m + im + (−i)m ≡ 1 (mod 2), where i2 = −1. Putting P = Q = −1,

R = 1 we have α = −1, β = −1,γ = 1, yielding I ′m − Im = 2(−1)m + 1. It then

follows that

Im =im + (−i)m

2− (−1)m ≡ m (mod 2).

Let Pn = Wn and Qn = (W 2n −∆C2

n)/4 for the remainder of this section. We will

now give a series of results that will be useful in the next chapter.

Theorem 4.10. If 2 | Pn and 2 - Qn, then CmnCn≡ m (mod 2).

Proof. First note that, by equation (2.7), we can derive

−Qλ2n Uλ1−λ2 = Qλ1

n Uλ2−λ1 (4.3)

and if k ≥ 0, Uk(Pn, Qn) ≡ Uk(2, 1) ≡ k (mod 2). It follows that

Qλ1n Uλ2−λ1 ≡ λ2 − λ1 (mod 2)

85

for λ2 ≥ λ1 and λ2 < λ1. Hence by the multiplication formula for Cmn, we get

CmnCn≡∑ m(−1)λ0(m− λ0 − 1)!

λ1!λ2!λ3!R(λ0+λ3)n (mod 2)

where the sum is as in (3.17) with the extra condition λ1 6≡ λ2 (mod 2).

If 2 | R, then each of the terms in the expression for Cmn/Cn modulo 2 is even

unless λ0 + λ3 = 0, λ0 + λ1 + λ2 + λ3 = m, λ1 + 2λ2 + 3λ3 = m and 2 - (λ1 + λ2).

These conditions imply λ0 = λ3 = 0⇒ λ2 = 0⇒ λ1 = m⇒ 2 - m. In this case

CmnCn

≡ m(−1)λ0(m− λ0 − 1)!

λ1!λ2!λ3!R(λ0+λ3)n

≡ 1 (mod 2).

If 2 - R, then by Lemma 4.9

CmnCn≡ Im ≡ m (mod 2).

Lemma 4.11. If CrnCn≡ 0 (mod k) for all n > 0, then Cmn

Cn≡ 0 (mod k) if r | m.

Proof. Let m = rs. Then it is easy to see that

CmnCn

=CrsnCn

=CrsnCrn

.CrnCn≡ 0 (mod k).

Theorem 4.12. If 2 - R, 2 | Pn, 2 | Qn, then

CmnCn

≡ m (mod 2) if 3 - m

CmnCn

≡ 0 (mod 2) if 3 | m.

86

Proof. Note that Qλ2n Uλ1−λ2 ≡ 0 (mod 2) unless λ2 = 0, λ1 = 1 or λ1 = 0, λ2 = 1.

We now consider these cases where Qλ2n Uλ1−λ2 6≡ 0 (mod 2).

It follows that if m ≡ 1 (mod 3) it must be that λ2 = 0, λ1 = 1, λ3 = m−13

and

λ0 = 2m−13

. So m− λ0 − 1 = m−13

. This implies

CnmCn≡ m

(m−1

3

)!(

m−13

)!≡ m (mod 2).

Similarly, if m ≡ −1 (mod 3), we can only have λ2 = 1, λ1 = 0, λ3 = m−23

and

λ0 = 2m−13

. Hence m− λ0 − 1 = m−23

, which gives

CmnCn≡ m (mod 2).

If 3 | m we can never have λ2 = 0, λ1 = 1, or λ1 = 0, λ2 = 1. Thus,

CnmCn≡ 0 (mod 2).

Theorem 4.13. If 2 - R, 2 - Pn and 2 - Qn, then

CmnCn≡ 1 (mod 2) if 3 - m

CmnCn≡ 0 (mod 2) if 3 | m.

Proof. First note that

Ut(Pn, Qn) ≡ Ut(1, 1) ≡

0 if 3 | t

1 if 3 - t(mod 2).

Hence

Qλ2n Uλ1−λ2 ≡

0 if 3 | λ1 − λ2

1 if 3 - λ1 − λ2

(mod 2).

87

Since λ1 + 2λ2 + 3λ3 = m, we see that if λ1 ≡ λ2 (mod 3), then 3 | m. It follows

that if 3 - m, we know λ1 6≡ λ2 (mod 3) and we get

CmnCn≡∑ (−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!= Im + I ′m ≡ 1 (mod 2).

We note that C3n

Cn= (∆C2

n + 3W 2n)/4 = P 2

n − Qn ≡ 0 (mod 2), thus in the case

where 3 | m, it follows from Lemma 4.11 that CmnCn≡ 0 (mod 2).

Theorem 4.14. If 2 - R, 2 - Pn and 2 | Qn, then

CmnCn≡ 1 (mod 2) if 7 - m

CmnCn≡ 0 (mod 2) if 7 | m.

Proof. First, we use the fact that

Ut(Pn, Qn) ≡ Ut(1, 0) (mod 2).

This implies

Ut(Pn, Qn) ≡ 1 (mod 2) if t > 0.

Now, Qλ2n Uλ1−λ2 ≡ 0 (mod 2) if λ1 ≥ λ2 > 0 or λ2 ≥ λ1 > 0. Thus, if λ1 = 0, λ2 ≥ 1

or λ2 = 0, λ1 ≥ 1, then

Qλ2n Uλ1−λ2 ≡ 1 (mod 2)

as QtnU−t = −Ut.

88

It follows that

CmnCn

≡∑

λ0 + λ1 + λ3 = m

λ1 + 3λ3 = m

(−1)λ0m(m− λ0 − 1)!

λ1!λ3!

+∑

λ0 + λ2 + λ3 = m

2λ2 + 3λ3 = m

(−1)λ0m(m− λ0 − 1)!

λ2!λ3!(mod 2).

Now let us restrict ourselves to the case where P = 1, Q = 0, R = 1 and α, β, γ

are the zeros of X3 − PX2 +QX −R. Then

Am = αm + βm + γm =∑

λ0 + λ1 + λ3 = m

λ1 + 3λ3 = m

(−1)λ0m(m− λ0 − 1)!

λ1!λ3!.

Also let P ′ = 0, Q′ = 1, R′ = 1 and α′, β′, γ′ be the zeros of X3−P ′X2 +Q′X −R′,

then

A′m = α′m + β′m + γ′m =∑

λ0 + λ2 + λ3 = m

2λ2 + 3λ3 = m

(−1)λ0m(m− λ0 − 1)!

λ2!λ3!.

Clearly, the above two identities follow from Waring’s theorem.

Now it is clear that A′0 = 3, A′1 = P ′ = 0, A′2 = P ′2 − 2Q′ = −2 and

A′n+3 = P ′A′n+2 −Q′A′n+1 +R′A′n ≡ A′n+1 + A′n (mod 2).

89

Thus,

A′m ≡ 1 (mod 2) if m ≡ 0, 3, 5, 6 (mod 7)

A′m ≡ 0 (mod 2) if m ≡ 1, 2, 4 (mod 7).

Similarly, one can show that

Am ≡ 1 (mod 2) if m ≡ 0, 1, 2, 4 (mod 7)

Am ≡ 0 (mod 2) if m ≡ 3, 5, 6 (mod 7).

Hence

Am + A′m ≡ 1 (mod 2) if 7 - m

Am + A′m ≡ 0 (mod 2) if 7 | m.

Since CmnCn≡ Am + A′m (mod 2), we are done.

We now assume that (Q,R) = 1, and we recall from Theorem 4.5 that if 2 | Pn,

then 2 - Qn. Putting n = 1, from these results, we now know that

Cm ≡ m (mod 2)

when 2 | P1, 2 - Q1. Put r = 2 in this case.

If 2 - R, 2 - P1, 2 - Q1, put r = 3 and note that

Cm ≡ 1 (mod 2) if 3 - m,

Cm ≡ 0 (mod 2) if 3 | m.

90

If 2 - R, 2 - P1, 2 | Q1, put r = 7 and note that

Cm ≡ 1 (mod 2) if 7 - m,

Cm ≡ 0 (mod 2) if 7 | m.

There remains the case of 2 - P1 and 2 | R. We have 2 - Q; thus, by Corollary

3.14.2, we get 2 - Q1 and

Cm ≡ Um(1, 1) (mod 2).

In this case, we put r = 3 and we have

Cm ≡ 1 (mod 2) if 3 - m,

Cm ≡ 0 (mod 2) if 3 | m.

Hence, we have proved the following theorem.

Theorem 4.15. If (Q,R) = 1, there always exists a minimal r > 1 such that 2 | Cr.

Furthermore, if 2 | Cn, then r | n.

Lemma 4.16. If m, n ≥ 1, then

CmnCn≡ 0 (mod (Pn, Qn)) if 3 | m

CmnCn≡ mR(m−1)n (mod (Pn, Qn)) if 3 - m.

Proof. Recall for Uk(Pn, Qn), U−k = −Uk/Qkn. By (4.3) we see that if λ1 ≥ λ2, then

Qλ2n Uλ1−λ2 ≡ 0 (mod Qn) if λ2 ≥ 1,

Qλ2n Uλ1−λ2 ≡ Uλ1 ≡ P λ1−1

n (mod Qn) if λ2 = 0 6= λ1,

Qλ2n Uλ1−λ2 ≡ 0 (mod Qn) if λ1 = λ2.

91

For λ2 ≥ λ1, we have the following results:

−Qλ1n Uλ2−λ1 ≡ 0 (mod Qn) if λ1 ≥ 1,

−Qλ1n Uλ2−λ1 ≡ −Uλ2 ≡ −P λ2−1

n (mod Qn) if λ1 = 0 6= λ2,

−Qλ1n Uλ2−λ1 ≡ 0 (mod Qn) if λ1 = λ2.

Thus Qλ2n Uλ1−λ2 ≡ 0 (mod Qn) unless λ1 ≥ 1, λ2 = 0 or λ1 = 0, λ2 ≥ 1. But since

(Pn, Qn) will divide Pn, we get Qλ2n Uλ1−λ2 ≡ 0 (mod (Pn, Qn)) unless λ1 = 1, λ2 = 0

or λ1 = 0, λ2 = 1.

If 3 | m, then 3 | λ1 + 2λ2 and neither λ1 = 1, λ2 = 0 nor λ1 = 0, λ2 = 1 can

occur. Thus, from Theorem 3.14 we can conclude that

CmnCn≡ 0 (mod (Pn, Qn)) if 3 | m.

If 3 - m, then either m ≡ 1 (mod 3) or m ≡ 2 (mod 3). If m ≡ 1 (mod 3),

then we must have λ1 = 1, λ2 = 0, in which case Qλ2n Uλ1−λ2 = 1, λ3 = m−1

3,

λ3 = m− λ0 − 1, and λ0 = 2m−13

. Applying the above to Theorem 3.14 yields

CmnCn≡ mR(m−1)n (mod (Pn, Qn)) if m ≡ 1 (mod 3).

Similarly, if m ≡ 2 (mod 3), then it must be that λ1 = 0, λ2 = 1. Then we can

see −Qλ1n Uλ2−λ1 = Qλ2

n Uλ1−λ2 = QnU−1 = −U1 = −1, λ3 = m−23

, λ0 = 2m−13

which is

odd. Again use Theorem 3.14 to get

CmnCn≡ mR(m−1)n (mod (Pn, Qn)) if m ≡ 2 (mod 3).

Which completes the proof.

Corollary 4.16.1. If 3 - m, then

(Cmn/Cn, Pn, Qn) | mRn(m−1).

92

Proof. This follows immediately from Lemma 4.16.

Theorem 4.17. If 2 - R or if 2 | R and 2 - Q, then

(Cmn/Cn,Wn, Cn) | mRn(m−1) when 3 - m.

Proof. We have by Corollary 4.16.1 that if 3 - m, then

(Cmn/Cn, Pn, Qn) | mRn(m−1) ⇒ (Cmn/Cn,Wn, (W2n −∆C2

n)/4) | mRn(m−1).

We divide our proof into 2 cases.

Case 1: 2 - Wn. In this case

(4Cmn/Cn, 4Wn,W2n −∆C2

n) | 4mRn(m−1) ⇒ (Cmn/Cn,Wn, Cn) | 4mRn(m−1).

But since Wn is odd (Cmn/Cn,Wn, Cn) | mRn(m−1).

Case 2: 2 | Wn. In this case we have 2 | Pn and Qn is odd, so by Theorem 4.10

CmnCn≡ m (mod 2).

Also, we know by Theorem 4.5 that if 2s || (Wn, Cn) then s ∈ {0, 1}.

Since (Cmn/Cn,Wn, Cn) | 4mRn(m−1) we see that (Cmn/Cn,Wn, Cn) | mRn(m−1)

when m is odd. Otherwise, if m is even and 2s || (Cmn/Cn,Wn, Cn), then s ∈ {0, 1}.

If s = 0, then (Cmn/Cn,Wn, Cn)|mRn(m−1). If s = 1, then 2 || (Cmn/Cn,Wn, Cn),

which implies (Cmn/Cn,Wn, Cn) | mRn(m−1).

We have been working towards the following corollary which is somewhat anal-

ogous to Carmichael’s result seen in Theorem 2.5. We will derive a closer analogue

in Chapter 5.

93

Corollary 4.17.1. If (Q,R) = 1, then

(Cmn/Cn,Wn, Cn) | m when 3 - m.

Proof. If (Wn, Cn, R) = 1, we are done. Let p be any prime such that p | (Wn, Cn, R).

By Lemma 4.6 we can only have p = 2. Since 2 | Wn we have Qn odd and then by

Theorem 4.10 we have

CmnCn≡ m (mod 2).

Now if 2 - m, then

(Cmn/Cn,Wn, Cn, R) = 1⇒ (Cmn/Cn,Wn, Cn) | m.

If 2 | m, then

(Cmn/Cn,Wn, Cn, Rn(m−1)) = 2⇒ (Cmn/2Cn,Wn/2, Cn/2, R

n(m−1)/2) = 1.

We have then

(Cmn/2Cn,Wn/2, Cn/2) | mRn(m−1)/2⇒ (Cmn/2Cn,Wn/2, Cn/2) | m

⇒ (Cmn/Cn,Wn, Cn) | 2m.

But 2 || (Cmn/Cn,Wn, Cn) and 2 | m ⇒ (Cmn/Cn,Wn, Cn) | m.

We have seen that many of Lucas’ results have analogues when we assume that

(Q,R) = 1. This is similar to Lucas’ condition that (P,Q) = 1, and we will assume

for the remainder of this work that (Q,R) = 1.

94

4.2 Preliminary Results for the Law of Repetition for {Cn}

We will now require some elementary results in algebraic number theory to develop

the proof of the following theorem. This result will be of some importance in estab-

lishing a law of repetition for {Cn}.

Theorem 4.18. If p is a prime and p - 6R∆, p | Cn and p | Wn−6Rn, then p3 | Cn

and p2 | Wn − 6Rn.

Proof. Let α, β, γ be the distinct (p - ∆) zeros of x3 − Px2 + Qx − R and put

L = Q(α). If we put K = Q(α, β), then K is the normal closure of L and is, of

course, Galois. Put δ = (α − β)(β − γ)(γ − α), λ1 = αn − βn, λ2 = βn − γn and

λ3 = γn − αn. Note λ1 + λ2 + λ3 = 0.

Since δCn = λ1λ2λ3, we note that if p is prime ideal divisor of (p) in K, then

p | (λ1λ2λ3). We also note that the discriminant of L must divide ∆. It follows

(see for example, Theorem 86 of [Hil98]), that since p - ∆, then p cannot divide the

discriminant of K. Thus, in K we must have

(p) =k∏i=1

pi,

where the prime ideals pi (i = 1, 2, . . . , k) are all distinct, that is (pi, pj) = O, the

maximal order of K, for i 6= j. Since p | (λ1λ2λ3), we must have p | (λ1) or p | (λ2)

or p | (λ3). Without loss of generality, suppose p | (λ1).

Since

Wn − 6Rn = 2βn(αn − γn)2 − (αn − βn)(βn − γn)(αn + γn)

and p | Wn − 6Rn, we have

2βn(αn − γn)2 ≡ (αn − βn)(βn − γn)(αn + γn) ≡ 0 (mod p).

95

Since (p, 2R) = 1 and R = αβγ, we must have p - β and hence αn ≡ γn

(mod p) ⇒ p | (λ3). Also since λ2 = −λ1 − λ3 and (λ1) ≡ (λ3) ≡ 0 (mod p) we

must have (λ2) ≡ 0 (mod p). Hence p3 | (λ1λ2λ3). Since ((p), (δ)) = O, we get

p3 | ((λ1λ2λ3)/δ).

Since p3i | ((λ1λ2λ3)/δ) for i = 1, 2, . . . , k and the pi for i = 1, 2, . . . , k are distinct

prime ideals, we must have

k∏i=1

p3i | ((λ1λ2λ3)/δ).

Thus (p3) | (Cn)⇒ p3 | Cn. Note that we also have p2 | Wn − 6Rn.

Suppose again that the prime p - 6∆R and that pµ || Cn, pν || Wn− 6Rn, where

µ, ν ≥ 1. Let p be any of the distinct ideals which lie over (p) in K. Since

pµ || (αn − βn)(βn − γn)(γn − αn),

we may assume without loss of generality that

pµ1 || αn − βn, pµ2 || βn − γn, pµ3 || γn − αn,

where µ1 + µ2 + µ3 = µ and µ1 ≥ µ2 ≥ µ3. Since

γn − αn = −(βn − γn)− (αn − βn),

we see that pµ2 | γn − αn and µ2 ≤ µ3. Hence µ2 = µ3. If µ1 > µ2, then since

Wn − 6Rn = 2βn(αn − γn)2 − (αn − βn)(βn − γn)(αn + γn)

and µ1 +µ2 > 2µ2, we must have p2µ2 || Wn−6Rn and 2µ2 < µ. Thus, ν = 2µ2 < µ.

If µ1 = µ2, then 3 | µ and µ1 = µ2 = µ3 = µ/3. This seems to suggest that the case

of ν > µ would occur less frequently than the case of ν ≤ µ.

96

Put Dn = (Wn − 6Rn, Cn). Since (Wn, Cn, R) | 2 by Lemma 4.6, we see that

if p 6= 2 and p | Dn, then p - R. Further results on Dn will be developed in the

following chapter. We also have the following theorem for the case where p = 2.

Theorem 4.19. If 2 - R∆ and 16 | Cn, then 8 | Wn − 6Rn.

Proof. We note that if 2 - R and 4 | Cn, then 2 || Wn by Theorem 4.5. Thus Wn −

6Rn ≡ 0 (mod 4). Now since 2 - ∆, we have (2) is the product of distinct prime ideals

in K. Let p be any one of these prime ideals. Since 16 | (αn−βn)(βn−γn)(γn−αn),

we must have p4 | (αn − βn)(βn − γn)(γn − αn). Without loss of generality, let

pµ1 || (αn−βn), pµ2 || (βn−γn) and pµ3 || (γn−αn), where µ1 ≥ µ2 ≥ µ3. We must

have µ1 + µ2 + µ3 = 4⇒ µ1 ≥ 2 and µ2 ≥ 1.

Since p2 | Wn − 6Rn, we get

p2 | 2βn(αn − γn)2 − (αn − βn)(βn − γn)(αn + γn).

Since p3 | (αn − βn)(βn − γn), we get p2 | 2βn(αn − γn)2 ⇒ p | (αn − γn)2 ⇒

p | αn − γn ⇒ p3 | Wn − 6Rn. It follows that 8 | Wn − 6Rn.

4.3 The Polynomial Km(x)

We now introduce the polynomials Hm(X, Y ) and Km(X). We will develop some

properties of these polynomials which will help us to produce the law of repetition

for {Cn}. Put

Hm(X, Y ) =∑ (−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!Xλ1−λ2Y 2λ2

and

Km(X) =∑ (−1)λ0m(m− λ0 − 1)!(λ1 − λ2)

λ1!λ2!λ3!Xλ1+λ2−1. (4.4)

97

As before, the sums are extended over the values λi ∈ Z such that

λ0, λ1, λ2, λ3 ≥ 0, λ0 + λ1 + λ2 + λ3 = m, λ1 + 2λ2 + 3λ3 = m.

Note that

Km(X) =∂Hm(X, Y )

∂X

∣∣∣Y=X

.

Consider

FX,Y (Z) = Z3 −XZ2 + (Y 2/X)Z − 1

and let α1(X, Y ), α2(X, Y ), α3(X, Y ) be the three (not necessarily distinct) zeros of

FX,Y (Z). By Waring’s theorem

Hm(X, Y ) = α1(X, Y )m + α2(X, Y )m + α3(X, Y )m.

Hence

∂Hm(X, Y )

∂X= m

(α1(X, Y )m−1∂α1(X, Y )

∂X+ α2(X, Y )m−1∂α2(X, Y )

∂X

+α3(X, Y )m−1∂α3(X, Y )

∂X

).

For convenience we will now write α1, α2 and α3, to denote α1(X, Y ), α2(X, Y ) and

α3(X, Y ), respectively. One can easily see then,

α1 + α2 + α3 = X, α1α2 + α2α3 + α3α1 = Y 2/X and α1α2α3 = 1.

Following Lagrange, we put

θ1(X, Y ) = α1 + ζα2 + ζ2α3,

θ2(X, Y ) = α1 + ζ2α2 + ζα3,

98

where ζ is a primitive cube root of unity (ζ2 + ζ + 1 = 0). Then

α1 =1

3(X + θ1 + θ2) ,

α2 =1

3

(X + ζ2θ1 + ζθ2

),

α3 =1

3

(X + ζθ1 + ζ2θ2

).

Further, one can easily show

θ1θ2 = X2 − 3Y 2/X and θ31 + θ3

2 = 2X3 − 9Y 2 + 27.

With these two identities we can show

θ31(X, Y ) =

2X3 − 9Y 2 + 27 +√

∆(X, Y )

2and

θ32(X, Y ) =

2X3 − 9Y 2 + 27−√

∆(X, Y )

2,

where

∆(X, Y ) = (2X3 − 9Y 2 + 27)2 − 4(X2 − 3Y 2/X)3

= −27(Y 4 − 4Y 6/X3 − 4X3 + 18Y 2 − 27).

Note that

∆(X, Y )∣∣∣Y=X

= −27(X4 − 8X3 + 18X2 − 27

)= −27 (X − 3)2 (X2 − 2X − 3

).

Put

D(X) = X2 − 2X − 3 = (X − 3)(X + 1).

Now, observe

∂∆(X, Y )

∂X= −27(−12X2 + 12Y 6/X4)⇒ ∂∆(X, Y )

∂X

∣∣∣Y=X

= 0.

99

It is easy to see that

∂θ1(X, Y )

∂X=

1

3θ−2

1

(6X2 +

1

2∆−1/2(X, Y )

∂∆(X, Y )

∂X

)/2.

Thus

∂θ1(X, Y )

∂X

∣∣∣Y=X

= θ−21 (X,X)X2.

Similarly

∂θ2(X, Y )

∂X

∣∣∣Y=X

= θ−22 (X,X)X2.

Since

θ1(X,X)θ2(X,X) = X2 − 3X,

we get

θ−11 (X,X) =

θ2(X,X)

X2 − 3Xand θ−1

2 (X,X) =θ1(X,X)

X2 − 3X.

So we then have

∂θ1(X, Y )

∂X

∣∣∣Y=X

=θ2

2(X,X)

(X − 3)2and

∂θ2(X, Y )

∂X

∣∣∣Y=X

=θ2

1(X,X)

(X − 3)2.

Now

θ1(X,X) = α1(X,X) + ζα2(X,X) + ζ2α3(X,X) and

FX,X(Z) = Z3 −XZ2 +XZ − 1.

Hence, we may put

α1(X,X) = 1, α2(X,X) =X − 1 +

√D(X)

2, α3(X,X) =

X − 1−√D(X)

2.

We then have

θ1(X,X) =3−X + (ζ − ζ2)

√D(X)

2,

θ2(X,X) =3−X − (ζ − ζ2)

√D(X)

2.

100

Upon squaring and using the fact (ζ − ζ2)2 = −3, we get

θ21(X,X) = (X − 3)

[−X − 3− (ζ − ζ2)

√D(X)

2

],

θ22(X,X) = (X − 3)

[−X − 3 + (ζ − ζ2)

√D(X)

2

].

Hence

∂θ1(X, Y )

∂X

∣∣∣Y=X

=−X − 3 + (ζ − ζ2)

√D(X)

2(X − 3),

∂θ2(X, Y )

∂X

∣∣∣Y=X

=−X − 3− (ζ − ζ2)

√D(X)

2(X − 3).

Now

∂α1(X, Y )

∂X

∣∣∣Y=X

=1

3

(1 +

∂θ1(X, Y )

∂X+∂θ2(X, Y )

∂X

) ∣∣∣Y=X

=1

3

(1 +−X − 3 + (ζ − ζ2)

√D(X)

2(X − 3)

+−X − 3− (ζ − ζ2)

√D(X)

2(X − 3)

)

=1

3

(1− X + 3

X − 3

)=−2

X − 3.

Also,

∂α2(X, Y )

∂X

∣∣∣Y=X

=1

3

(1 + ζ2∂θ1(X, Y )

∂X+ ζ

∂θ2(X, Y )

∂X

) ∣∣∣Y=X

=1

3

(1 +

X + 3 + (2− ζ − ζ2)√D(X)

2(X − 3)

)

=X − 1 +

√D(X)

2(X − 3)=α2(X,X)

X − 3,

∂α3(X, Y )

∂X

∣∣∣Y=X

=1

3

(1 +

X + 3− (2− ζ − ζ2)√D(X)

2(X − 3)

)=α3(X,X)

X − 3.

101

It follows that

Km(X) =∂Hm(X, Y )

∂X

∣∣∣Y=X

= mαm−11 (X,X)

(−2

X − 3

)+mαm−1

2 (X,X)

(α2(X,X)

X − 3

)+mαm−1

3 (X,X)

(α3(X,X)

X − 3

)=

m

X − 3[−2 + α2(X,X)m + α3(X,X)m] .

We then get the identity

Km(X) =m

X − 3

[(X − 1 +

√D(X)

2

)m

+

(X − 1−

√D(X)

2

)m

− 2

].

If m is odd, then using identity (4.2.44) from [Wil98] gives

Vm(X − 1, 1) = α2(X,X)m + α3(X,X)m (4.5)

= (X − 1)

(m−1)/2∑j=0

((m− 1)/2 + j

(m− 1)/2− j

)Dj(X). (4.6)

In this case

Km(X) = m

1 +

(m−1)/2∑j=1

((m− 1)/2 + j

(m− 1)/2− j

)(X − 1)(X + 1)j(X − 3)j−1

.If m is even, then identity (4.2.42) from [Wil98] yields

Vm(X − 1, 1) = αm2 (X,X) + αm3 (X,X) =

m/2∑j=0

m

m/2− j

(m/2 + j

m/2− j

)Dj(X) (4.7)

= 2+

m/2∑j=1

m

m/2− j

(m/2 + j

m/2− j

)Dj(X).(4.8)

In this case

Km(X) = m

m/2∑j=1

m

m/2− j

(m/2 + j − 1

m/2− j − 1

)(X + 1)j(X − 3)j−1.

102

Theorem 4.20. Let X ∈ Z and p be a prime such that p > 3. If p - X − 3, then

Kp(X) ≡ p (mod p2).

If p | X − 3, then

Kp(X) ≡ p3 (mod p4).

Proof. From the top of page 100 we have

(X − 3)Kp(X) = p(−2 + αp2(X,X) + αp3(X,X)).

Since α2α3 = 1 and α2 + α3 = X − 1, we have

αp2(X,X) + αp3(X,X) = Vp(X − 1, 1) ≡ X − 1 (mod p).

Thus pVp(X − 1, 1) ≡ p(X − 1) (mod p2) and so we have

(X − 3)Kp(X) ≡ p(X − 3) (mod p2)⇒ Kp(X) ≡ p (mod p2)

when p - X − 3.

Now suppose that p | X − 3. We know that

Kp(X) = p

1 +

(p−1)/2∑j=1

((p− 1)/2 + j

(p− 1)/2− j

)(X − 1)(X + 1)j(X − 3)j−1

.Hence, if p ≥ 7,

Kp(X) ≡ p

[1 +

((p+ 1)/2

2

)(X2 − 1) +

((p+ 3)/2

4

)(X − 3)(X + 1)2(X − 1)

+

((p+ 5)/2

6

)(X − 3)2(X + 1)3(X − 1)

](mod p4).

Next note that

6

((p+ 1)/2

2

)+ 32

((p+ 3)/2

4

)= 6

(p2 − 1)

8+

(p2 − 9)(p2 − 1)

12

= p2

(p2 − 1

8

)≡ 0 (mod p2).

103

Also, ((p+ 1)/2

2

)+ 32

((p+ 3)/2

4

)+ 128

((p+ 5)/2

6

)

=p2 − 1

8+

(p2 − 9)(p2 − 1)

12+

(p2 − 25)(p2 − 9)(p2 − 1)

360

=

(p2 − 1

8

)(1 +

2(p2 − 9)

3+

(p2 − 25)(p2 − 9)

45

)=

(p2 − 1

8

)(p4 − 4p2

45

)=p2(p2 − 1)(p2 − 4)

360≡ 0 (mod p2).

Since p 6= 3, by using

X2 − 1 = (X − 3)2 + 6(X − 3) + 8,

(X + 1)2(X − 1) = (X − 3)3 + 10(X − 3)2 + 32(X − 3) + 32,

(X + 1)3(X − 1) = (X − 3)4 + 14(X − 3)3 + 72(X − 3)2 + 160(X − 3) + 128,

we can rewrite Kp(X) (mod p4) as follows

Kp(X) ≡ p

[1 +

((p+ 1)/2

2

)((X − 3)2 + 6(X − 3) + 8

)+

((p+ 3)/2

4

)(X − 3)

((X − 3)3 + 10(X − 3)2 + 32(X − 3) + 32

)+

((p+ 5)/2

6

)(X − 3)2

((X − 3)4 + 14(X − 3)3 + 72(X − 3)2 + 160(X − 3) + 128

)]≡ p

[1 +

((p+ 1)/2

2

)8 + (X − 3)

(6

((p+ 1)/2

2

)+ 32

((p+ 3)/2

4

))+(X − 3)2

(((p+ 1)/2

2

)+ 32

((p+ 3)/2

4

)+ 128

((p+ 5)/2

6

))]≡ p

[1 + (p2 − 1) + (X − 3)

(p2(p2 − 1)

12

)+ (X − 3)2

(p2(p2 − 1)(p2 − 4)

360

)]≡ p3 (mod p4).

In the case of p = 5 we get

Kp(X) ≡ 5

[1 +

(3

2

)(X2 − 1) +

(4

4

)(X − 3)(X + 1)2(X − 1)

](mod p4).

104

So

Kp(X) ≡ 5[1 + 3

((X − 3)2 + 6(X − 3) + 8

)+(X − 3)

((X − 3)3 + 10(X − 3)2 + 32(X − 3) + 32

)]≡ 5

[25 + 50(X − 3) + 35(X − 3)2

]≡ 125 (mod p4).

4.4 The Law of Repetition for {Cn}

Note that

P 2n − 4Qn = ∆n = ∆C2

n and 4Qn ≡ W 2n (mod C2

n).

If m > 0, we can easily derive from the multiplicative properties of Lucas sequences

mentioned in Chapter 2 that

2m−1Um(P,Q) ≡ mPm−1 (mod ∆) and 2m−1Vm(P,Q) ≡ Pm (mod ∆).

Thus,

2m−1Um(Pn, Qn) ≡ mPm−1n (mod ∆C2

n) and

2m−1Vm(Pn, Qn) ≡ Pmn (mod ∆C2

n).

We may use the identity

2Qλ2n Uλ1−λ2(Pn, Qn) = Uλ1(Pn, Qn)Vλ2(Pn, Qn)− Vλ1(Pn, Qn)Uλ2(Pn, Qn)

105

as follows:

2λ1+λ2−1Qλ2n Uλ1−λ2(Pn, Qn) = 2λ1−1Uλ1(Pn, Qn)2λ2−1Vλ2(Pn, Qn)

−2λ1−1Vλ1(Pn, Qn)2λ2−1Uλ2(Pn, Qn)

≡ λ1Pλ1−1n P λ2

n − P λ1n λ2P

λ2−1n (mod ∆C2

n)

≡ (λ1 − λ2)P λ1+λ2−1n (mod ∆C2

n).

Similarly, we have

2λ1+λ2−1Qλ2n Vλ1−λ2(Pn, Qn) ≡ P λ1+λ2

n (mod ∆C2n).

If 2 - ∆Cn, then

Qλ2n Uλ1−λ2(Pn, Qn) ≡ (λ1 − λ2)(Pn/2)λ1+λ2−1 (mod ∆C2

n),

Qλ2n Vλ1−λ2(Pn, Qn) ≡ 2(Pn/2)λ1+λ2 (mod ∆C2

n).

When 2 | ∆Cn, we have 2 | Pn and Qn ≡ (Pn/2)2 (mod ∆C2n/4). In this case

we can show by induction that

Um(Pn, Qn) ≡ m(Pn/2)m−1 (mod ∆C2n/4),

Vm(Pn, Qn) ≡ 2(Pn/2)m (mod ∆C2n/4)

and therefore

Qλ2n Uλ1−λ2(Pn, Qn) ≡ (λ1 − λ2)(Pn/2)λ1+λ2−1 (mod ∆C2

n/4), (4.9)

Qλ2n Vλ1−λ2(Pn, Qn) ≡ 2(Pn/2)λ1+λ2 (mod ∆C2

n/4). (4.10)

It follows by (3.17) that

CmnCn

≡∑

λ0,λ1,λ2,λ3

((−1)λ0m(m− λ0 − 1)!Rn(m−λ1−λ2)

λ1!λ2!λ3!

)(

(λ1 − λ2)P λ1+λ2−1n

2λ1+λ2−1

)(mod Fn),

106

where

Fn =

∆C2n if 2 - Cn

∆C2n/4 if 2 | Cn.

(4.11)

This symbol Fn and the symbol Gn introduced near the beginning of Chapter 5

should not be confused with the Fn and Gn defined in Section 3.6. So we can use

the above and equation (4.4) to see

CmnCn≡ Rn(m−1)Km(Wn/2R

n) (mod Fn).

Thus, by our earlier results, if m is odd,

CmnCn

≡ mRn(m−1) +m

(m−1)/2∑j=1

((m− 1)/2 + j

(m− 1)/2− j

)2−2jRn(m−2j−1)(Wn − 2Rn)

(Wn + 2Rn)j(Wn − 6Rn)j−1 (mod Fn)

and if m is even,

CmnCn

≡ m

m/2∑j=1

m

m/2− j

(m/2 + j − 1

m/2− j − 1

)2−2j+1Rn(m−2j)(Wn + 2Rn)j

(Wn − 6Rn)j−1 (mod Fn).

If p 6= 2, p | Cn and p | R, then since p | p(p−λ0−1)!λ1!λ2!λ3!

and λ0 + λ3 6= 0 whenever

λ1 6= p, we get by equation (3.17) that

CpnCn≡ Up(Pn, Qn) (mod p2).

Then we may use (2.14) with m = p and n = 1 to see

Up(Pn, Qn) ≡ p(Wn/2)p−1 ≡ p (mod p2).

Thus if pλ || Cn, then pλ+µ || Cpµn.

107

Now suppose that p - 6R and p | Cn. It is easy to show by use of (3.16) that

Wpn ≡ Vp(Pn, Qn) ≡ Pn ≡ Wn (mod p).

Clearly, then, Wpn − 6Rpn ≡ Wn − 6Rn (mod p). If p - Wn − 6Rn, then p2 | Fn and

by Theorem 4.20

CpnCn≡ Rn(p−1)Kp(Wn/2R

n) ≡ Rn(p−1)p ≡ p (mod p2).

In this case if pλ || Cn, then pλ+µ || Cpµn.

On the other hand, if p | Wn − 6Rn and p - ∆, then by Theorem 4.18 we have

p3 | Cn ⇒ p3 | Fn. Also, if p | Wn − 6Rn and p | ∆, then p3 | Fn. We then have

by (4.9)

Qλ2n Uλ1−λ2(Pn, Qn) ≡ (λ1 − λ2)(Pn/2)λ1+λ2−1 (mod p3).

Further, since p | p(p−λ0−1)!λ1!λ2!λ3!

when λ1 6= p, we can say

CpnCn

= Up(Pn, Qn) +∑λ1 6=p

(−1)λ0p(p− λ0 − 1)!

λ1!λ2!λ3!Rn(p−λ1−λ2)Qλ2

n Uλ1−λ2(Pn, Qn)

≡ Up(Pn, Qn) +∑λ1 6=p

(−1)λ0p(p− λ0 − 1)!

λ1!λ2!λ3!Rn(p−λ1−λ2)(λ1 − λ2)(Pn/2)λ1+λ2−1

≡ Up(Pn, Qn) +Rn(p−1)[Kp(Wn/2R

n)− p(Pn/2Rn)p−1]

(mod p4).

Now,

2p−1Up(Pn, Qn) ≡ pP p−1n + ∆n

(p

3

)P p−3n (mod p4).

Since p3 | Cn we have p4 | ∆n, yielding

Up(Pn, Qn) ≡ p(Pn/2)p−1 (mod p4),

108

and so

CpnCn

≡ p(Pn/2)p−1 +Rn(p−1)Kp(Wn/2Rn)− p(Pn/2)p−1 (mod p4)

≡ Rn(p−1)Kp(Wn/2Rn) (mod p4)

≡ (Rn)p−1p3 ≡ p3 (mod p4).

Thus, in this case, if pλ || Cn, then pλ+3µ || Cpµn.

We can now state the law of repetition for {Cn}. Let pλ || Cn (λ ≥ 1).

• If p = 2, 3 then pλ+µ | Cpµn.

• If p 6= 2 and p | R, then pλ+µ || Cpµn.

• If p - R and p - Wn − 6Rn, then pλ+µ || Cpµn.

• If p - R and p | Wn − 6Rn, then pλ+3µ || Cpµn.

We next provide a closer examination of the case of p = 3: If 3 | Cn, 3 - R and

3 - Wn − 6Rn, then 3 - Wn. By Corollary 3.10.1 we have

4W3n = 3∆C2n(Wn + 2Rn) +W 2

n(Wn − 6Rn) + 24R3n,

4C3n = Cn(∆C2n + 3W 2

n).

We then can see

4C3n

Cn≡ ∆C2

n + 3W 2n (mod 9)⇒ 4

C3n

Cn≡ 3W 2

n (mod 9).

Hence 3 || C3n

Cnand 3 - W3n. Thus, if 3λ || Cn, then 3λ+µ || C3µn.

If 3 | Cn, 3 - R, 3 | Wn − 6Rn, then 3 | Wn ⇒ 3 | AnBn − 3Rn ⇒ 3 | AnBn.

Since

∆C2n = A2

nB2n + 18AnBnR

n − 4B3n − 4A3

nRn − 27Rn,

109

we see that if 3 | An, then 3 | Bn and if 3 | Bn, then 3 | An. So we have 9 | AnBn

and this implies 3 || Wn. Further 27 || 3W 2n and 3 | W3n. Thus, if 3λ || Cn and

λ ≥ 2, then

4C3n

Cn≡ 3W 2

n (mod 81).

It follows that 33 || C3n/Cn ⇒ 3λ+3µ || C3µn. If 3λ = 3, then 3λ+3µ | C3µn.

We next provide a similar examination of the p = 2 case.

Case 1. 2 | R.

If 4 | Cn, then since (W 2n −∆C2

n)/4 ∈ Z we have 2 | Wn; furthermore, 2 | AnBn as

AnBn = Wn + 3Rn. Also

27∆C2n = 4(A2

n − 3B2n)3 − (27Rn + 2A3

n − 9AnBn)2

= 4(A2n − 3B2

n)3 − (2A3n − 9Wn)2.

We have in this case that A2n−3B2

n is odd since 2 - Q⇒ 2 - Bn and therefore 2 | An.

We can use this to see that

27∆C2n ≡ 4−W 2

n (mod 8).

Thus 2 || Wn ⇒ 2 || Wn + 2Rn. But this means 2 || C2n/Cn since C2n = Cn(Wn +

2Rn). We may conclude 2µ+λ || C2µn, if 2λ || Cn and λ ≥ 2. If 2 | R and λ = 1, we

can only show that 2µ+λ | C2µn.

Case 2. 2 - R.

If 4 | Cn, then 2 | Wn ⇒ 2 | AnBn − 3Rn ⇒ 2 - AnBn ⇒ 2 | A2n − 3Bn. So

27∆C2n ≡ −(2A3

n − 9Wn)2 (mod 8).

110

Now

2A3n − 9Wn ≡ 2−Wn (mod 4)

⇒ (2A3n − 9Wn)2 ≡ (2−Wn)2 (mod 8)

⇒ 27∆C2n ≡ −(2−Wn)2 (mod 8).

If 4 | Cn, then 8 | (2 −Wn)2 ⇒ 4 | 2 −Wn ⇒ 2 || Wn. Hence 4 | Wn + 2Rn ⇒

4 | C2n/Cn ⇒ 2λ+2µ | C2µn if 2λ || Cn and λ ≥ 2. If 2 || Cn, then 2 | C2n/Cn and

4 | C2n. Hence 22µ | C2µn when µ ≥ 1.

Since (Wn, Cn, R) | 2, we can now revise the law of repetition for {Cn} in the

theorem below.

Theorem 4.21. Let pλ || Cn and (λ ≥ 1)

• If p = 2 we have two cases:

– If 2 | R, then

2λ+µ || C2µn for λ > 1,

2λ+µ | C2µn for λ = 1.

– If 2 - R, then

2λ+2µ | C2µn for λ > 1,

22µ | C2µn for λ = 1.

• If p 6= 2, then

pλ+µ || Cpµn when p - Wn − 6Rn.

111

• If p 6= 2 and pλ 6= 3, then

pλ+3µ || Cpµn when p | Wn − 6Rn.

• If pλ = 3, then

3λ+3µ | C3µn when p | Wn.

If p = 2 and 2 - ∆R, we have some additional special cases. When λ ≥ 4, we can

only have 8 || Wn−6Rn or 16 | Wn−6Rn by Theorem 4.19. If 16 | Wn−6Rn, then

since Wn + 2Rn = Wn− 6Rn + 8Rn, we see that 23 || Wn + 2Rn ⇒ 23 || C2n/Cn, as

C2n = Cn(Wn + 2Rn). Also, 16 | W2n − 6R2n. Thus 2λ+3µ || C2µn by induction.

On the other hand, if 8 || Wn − 6Rn, then since

2(W2n − 6R2n) = ∆C2n + (Wn − 6Rn)(Wn + 2Rn),

we observe that 32 | W2n − 6R2n. Further, since W2n − 6R2n + 8R2n = W2n + 2R2n

and 32 | W2n − 6R2n, we have 8 || W2n + 2R2n. Now, both (Wn − 6Rn)/8 and Rn

are odd. Thus

2 | Wn − 6Rn

8+Rn ⇒ 8 · 2 | 8

(Wn − 6Rn

8+Rn

)⇒ 16 | Wn + 2Rn.

Finally, if 2ν || Wn + 2Rn, we get 2ν || C2n/Cn ⇒ 2ν+λ || C2n, then

23(µ−1)+ν+λ || C2µn

by our earlier observation and induction on µ.

In the case of the law of repetition for the Lucas functions Un, we know that

pλ+µ || Unmpµ , if p - m and pλ || Un. Unfortunately, this result does not generalize

112

to Cn. For example, if p - Wn − 6Rn and p - 2R, it is possible for pλ || Cn and

pλ+1 | Cmn, where p - m. We note that(Wn

2Rn− 3

)CmnCn≡ m (−2 + Vm(Wn/2R

n − 1, 1)) (mod p).

If Vm(Wn/2Rn − 1, 1) ≡ 2 (mod p) and p - m, then p | Cmn/Cn. We also have the

equality of the two Legendre symbols((Wn/2R

n − 1)2 − 4

p

)=

((Wn − 6Rn)(Wn + 2Rn)

p

).

So if ((Wn − 6Rn)(Wn + 2Rn)

p

)= 1,

then

Vp−1(Wn/2Rn − 1, 1) ≡ 2 (mod p)⇒ p | C(p−1)n/Cn and p - p− 1.

Also, if ((Wn − 6Rn)(Wn + 2Rn)

p

)= −1,

then

Vp+1(Wn/2Rn − 1, 1) ≡ 2 (mod p)⇒ p | C(p+1)n/Cn and p - p+ 1.

Lastly, if ((Wn − 6Rn)(Wn + 2Rn)

p

)= 0,

then p | (Wn/2Rn−1)2−4⇒ Wn/2R

n−1 ≡ ±2 (mod p). In this case, V2(Wn/2Rn−

1, 1) ≡ (Wn/2Rn−1)2−2 ≡ (Wn/2R

n−1)2−4+2 ≡ 2 (mod p). So then p | C2n/Cn

and 2 - p.

113

4.5 The Law of Apparition for {Cn}

If a prime p divides R, it is easy to see that

Cn ≡ Qn−1Un(P,Q) (mod p),

in which case the theory reduces to that of the Lucas function Un(P,Q). We will

therefore assume p - R in what follows.

We recall that

27∆ = 4(P 2 − 3Q)3 − (27R + 2P 3 − 9QP )2.

When p | ∆ and p 6= 2, the splitting field of f(x) = x3 − Px2 + Qx − R ∈ Fp[x] is

Fp, and we have two possible cases.

Case one occurs when p | P 2−3Q. Here f(x) ≡ (x−a)3 (mod p) where a ≡ P/3

(mod p) (if p = 3, then 3 | P ). In this case we can put α = β = γ = a in Fp. Now

in Fp,

αn − βn

α− β= αn−1 + βαn−2 + β2αn−3 + · · ·+ βn−1

= nan−1,

it follows that

Cn ≡ n3a3(n−1) (mod p) and Wn ≡ 6a3n (mod p).

We may then conclude that p | Cn ⇔ p | n. Also, if p | Cn, then p | Wn − 6Rn.

Case two occurs when p - P 2 − 3Q. In this case f(x) ≡ (x− a)2(x− b) (mod p),

where

a ≡ PQ− 9R

2(P 2 − 3Q)(mod p) and b ≡ P 3 − 4PQ+ 9R

P 2 − 3Q(mod p). (4.12)

114

Hence we can put α = β = a 6= 0 and γ = b 6= 0 in Fp. Put P ′ ≡ P − a (mod p)

and Q′ ≡ a2 − Pa + Q (mod p). One can see that since a2b ≡ R (mod p), we get

ab ≡ R/a ≡ a2 − Pa + Q (mod p). Also, 2a + b ≡ P (mod p) ⇒ a + b ≡ P − a

(mod p). We use these results to obtain

Cn =

(αn − βn

α− β

)(βn − γn

β − γ

)(γn − αn

γ − α

)= nan−1

(an − bn

a− b

)2

in Fp. Thus,

Cn ≡ nan−1U2n(P ′, Q′) (mod p).

It is also true that (∆′

p) = 1, as

∆′ = P ′2 − 4Q′ ≡ (a− b)2 ≡ (27R + 2P 3 − 9PQ)2

4(P 2 − 3Q)2≡ P 2 − 3Q (mod p).

Thus p | Cn ⇔ p | n or p | Un(P ′, Q′) since p - a. If p | a or p | Q′, then p | R,

which is a contradiction. Since the rank of apparition of p in Un(P ′, Q′) is a divisor

r of p− 1, we can say p | Cn ⇔ either p | n or r | n. Since (r, p) = 1 we have two

ranks of apparition in this case. We also note that since

Wn − 6Rn ≡ 2an∆′U2n(P ′, Q′) (mod p)

we see that p | Wn − 6Rn if and only if n is a multiple of r.

We have already shown that r(2) always exists and is unique. The case for p = 3

can be handled explicitly by calculation. The results are given in Table 4.1, where

we assume 3 - R and P , Q, R are given modulo 3.

115

Table 4.1: Ranks of apparition for p = 3P Q R ∆ (mod 3) Cm ≡ 0 (mod 3) iff Corresponding Wm

2 2 2 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)2 2 1 0 m ≡ 0 (mod 2) or W2 ≡ 0 (mod 3) and

m ≡ 0 (mod 3) W3 ≡ 1 (mod 3)2 1 2 2 m ≡ 0 (mod 2) W2 ≡ 1 (mod 3)2 1 1 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)2 0 2 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)2 0 1 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)1 2 2 0 m ≡ 0 (mod 2) or W2 ≡ 0 (mod 3) and

m ≡ 0 (mod 3) W3 ≡ 2 (mod 3)1 2 1 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)1 1 2 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)1 1 1 2 m ≡ 0 (mod 2) W2 ≡ 1 (mod 3)1 0 2 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)1 0 1 2 m ≡ 0 (mod 4) W4 ≡ 1 (mod 3)0 2 2 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)0 2 1 1 m ≡ 0 (mod 13) W13 ≡ 0 (mod 3)0 1 2 2 m ≡ 0 (mod 4) W4 ≡ 1 (mod 3)0 1 1 2 m ≡ 0 (mod 4) W4 ≡ 1 (mod 3)0 0 2 0 m ≡ 0 (mod 3) W3 ≡ 0 (mod 3)0 0 1 0 m ≡ 0 (mod 3) W3 ≡ 0 (mod 3)

From this we see that there must always exist at least one rank of apparition for

3 in {Cn} as long as (Q,R) = 1. Also, r(3) ≤ 13 = 32 + 3 + 1. Note also that if 3 - ∆

and 3 | Cn, then r(3) | n.

4.6 Solutions of the Cubic

We now deal with those primes p such that p - 6∆R. The law of apparition for {Cn}

is more difficult than that for {Un}. This is largely due to the fact that {Cn} can have

multiple ranks of apparition, as has been seen. Just how many ranks of apparition

{Cn} actually has modulo a prime p, is dependent on the splitting behaviour of f(x)

116

modulo p. Following Adams and Shanks [AS82] we will characterize the primes that

do not divide 6∆R as follows.

Let f(x) = x3−Px2 +Qx−R and p - 6R∆. There are three possibilities for the

splitting field K of f(x) ∈ Fp[x] :

1. if K = Fp, we say that p is an S prime.

2. if K = Fp2 , we say that p is a Q prime.

3. if K = Fp3 , we say that p is an I prime.

Suppose G is the galois group of the polynomial f(x). There are four possibilities

for G: G1 = {1}, G2 a group of order 2, G3 a group of order 3 or G6 the dihedral

group of order 6. In G1 there is only one conjugacy class, in G2 there are two

conjugacy classes, in G3 there are two conjugacy classes and in G6 there are three

conjugacy classes. Let π(x) denote the number of primes up to x and for a fixed f(x)

denote by πS(x), πQ(x), πI(x) the number of S, Q, I primes up to x, respectively.

By the Chebotarev density theorem [SHWL96] we know that if G ' G1, then all the

primes are S primes. If G ' G2, then

limx→∞

πS(x)

π(x)=

1

2, and lim

x→∞

πQ(x)

π(x)=

1

2.

If G ' G3, then

limx→∞

πS(x)

π(x)=

1

3, and lim

x→∞

πI(x)

π(x)=

2

3.

Finally, if G ' G6, then

limx→∞

πS(x)

π(x)=

1

6, lim

x→∞

πQ(x)

π(x)=

1

2and lim

x→∞

πI(x)

π(x)=

1

3.

117

Determining which of these types of prime p is important, since its type dictates

where Cn = 0 in K. This is also an old problem and several references to how it

can be solved are mentioned in Chapter VIII of the first volume of [Dic19] (see also

[WZ74] and [Mul04]). As the results concerning this problem are widely scattered, we

will present a self-contained version here. Remember that δ = (α−β)(β−γ)(γ−α),

∆ = (α− β)2(β − γ)2(γ − α)2, and −27∆ = (2P 3 − 9QR + 27R)2 − 4(P 2 − 3Q)3.

If p is a Q prime, we may assume that α ∈ Fp, β, γ ∈ Fp2\Fp. Hence αp = α,

βp = γ and γp = β. Then, in K,

δp = (α− β)p(β − γ)p(γ − α)p

= (αp − βp)(βp − γp)(γp − αp)

= (α− γ)(γ − β)(β − α) = −δ.

So δp−1 = −1⇔ ∆(p−1)/2 = −1⇔ (∆p

) = −1.

In the other cases, we get either αp = α, βp = β, γp = γ or αp = β, βp = γ,

γp = α. In either case δp = δ ⇒ (∆p

) = 1. Thus p is a Q prime if and only if

(∆p

) = −1.

Now assume (∆p

) = 1, and let A = 2P 3 − 9QR + 27R and B = P 2 − 3Q. Then

−27∆ = A2 − 4B3. Let K be the splitting field of f(x) ∈ Fp[x]. Then K = Fp or

Fp3 . In either case K is a subfield of L = Fp6 . Now let F∗p6 = 〈λ〉 for some primitive

element λ of F∗p6 . Put ζ = λ(p6−1)/3, then ζ 6= 1 and ζ3 = 1⇒ (ζ − 1)(ζ2 + ζ + 1) =

0 ⇒ ζ2 + ζ + 1 = 0. This implies that αβ(ζ2 + ζ + 1) = 0 ⇒ ζ2αβ + ζαβ = −αβ.

Put

L1 = α + ζβ + ζ2γ ∈ L

L2 = α + ζ2β + ζγ ∈ L.

118

Then, L1L2 = B ∈ Fp and L31 + L3

2 = A ∈ Fp. Thus L31, L3

2 are the zeros of

x2 − Ax+B3 ∈ Fp[x].

Now since, (∆p

) = 1 and δ2 = ∆ we have δ ∈ Fp. If we put ρ = 3(ζ − ζ2)δ, then

ρ2 = −27∆ and ρ ∈ Fp2 . So ρ2 = −27∆ = A2 − 4B3 = (L31 + L3

2)2 − 4L31L

32. Notice

(L31−L3

2)2 = (L31)2− 2L3

1L32 + (L3

2)2 = (L31)2 + 2L3

1L32 + (L3

2)2− 4L31L

32 = A2− 4B3 =

−27∆⇒ L31 − L3

2 = ±ρ. This implies

2L31 = A± ρ

2L32 = A∓ ρ.

Now since L31, L3

2 are the zeros of x2 − Ax+B3 we have

Vn(A,B3) = (L31)n + (L3

2)n and Un(A,B3) =(L3

1)n − (L32)n

L31 − L3

2

.

From this we see that

2(L31)n = Vn(A,B3)± ρUn(A,B3) (4.13)

and

2(L32)n = Vn(A,B3)∓ ρUn(A,B3). (4.14)

Suppose p ≡ 1 (mod 3). If K = Fp (p is an S prime) then

Lp1 = αp + ζpβp + ζ2pγp = α + ζ3nζβ + ζ6nζ2γ = L1.

Similarly Lp2 = L2. This gives us L3( p−1

3)

1 = 1 and L3( p−1

3)

2 = 1.

If B 6= 0, then L1 6= 0 and L2 6= 0. If B = 0, then L1 = 0 or L2 = 0, but not both

since −27∆ = (L31 − L3

2)2. Without loss of generality assume L1 = 0, then L32 = A.

119

In this case p is an S prime if and only if Ap−13 ≡ 1 (mod p). Thus, if (∆

p) = 1,

p | B, then p is an S prime if and only if Ap−13 ≡ 1 (mod p).

If p - B, then L3( p−1

3)

1 = L3( p−1

3)

2 . Use the above and equations (4.13) and (4.14)

to see that U p−13

(A,B3) = 0.

Now suppose U p−13

(A,B3) ≡ 0 (mod p) ⇒ L3( p−1

3)

1 = L3( p−1

3)

2 in L by (4.13) and

(4.14). Since p - B, we have (L1/L2)p−1 = 1 ⇒ (L1/L2)p = (L1/L2), and therefore

L1/L2 ∈ Fp. Since L1L2 ∈ Fp we get L21, L2

2 ∈ Fp. Also (L31 − L3

2)2 ≡ −27∆

(mod p) and (−27∆p

) = (−3p

)(32

p)(∆

p) = 1 gives us L3

1 − L32 ∈ Fp and L3

1 − L32 6= 0. So

L31−L3

2 = (L1−L2)(L21 +L2

2 +L1L2) and L21 +L2

2 +L1L1 ∈ Fp ⇒ L1−L2 ∈ Fp. But

L21 − L2

2 ∈ Fp ⇒ L1 + L2 ∈ Fp ⇒ L1, L2 ∈ Fp. Since ζp = ζ, we have ζ ∈ Fp, and it

then follows that α, β, γ ∈ Fp ⇒ p is a S prime.

Suppose p ≡ −1 (mod 3) and K = Fp. Then B 6= 0. For if B = 0 we get

−27∆ = A2 − 4B3 = A2 and (−27∆p

) = 1. This is a contradiction since p ≡ −1

(mod 3)⇒ (−27∆p

) = −1. Now,

Lp1 = (α + ζβ + ζ2γ)p = αp + ζpβp + ζ2pγp.

Thus if p is an S prime, then Lp1 = L2 and Lp2 = L1 ⇒ Lp2

1 = L1 and Lp2

2 =

L2 ⇒ (L3(p−1)1 )(p+1)/3 = 1, (L

3(p−1)2 )(p+1)/3 = 1⇒ (L3

2/L31)(p+1)/3 = 1⇒ (L3

2)(p+1)/3 =

(L31)(p+1)/3 ⇒ U p+1

3(A,B3) = 0.

Now if U p+13

(A,B3) ≡ 0 (mod p), then L3 p+1

32 = L

3 p+13

1 in L⇒ (L2/L1)p+1 = 1⇒

Lp+12 = Lp+1

1 ⇒ (L2/L1)p = (L1/L2). So (L2/L1)p2

= (L2/L1). Hence L2/L1 ∈ Fp2

and L1/L2 ∈ Fp2 . Since ζp2

= ζ, we can employ our previous reasoning to establish

that αp2

= α, βp2

= β, γp2

= γ. Since (∆p

) = 1, we must have that α, β, γ are in

Fp or Fp3 . If p is an I prime, then αp = β, βp = γ, γp = α ⇒ αp2

= γ, βp2

= α,

120

γp2

= β. But then αp2

= γ = α, and p | ∆, which is a contradiction. Hence p is not

an I prime ⇒ p is an S prime.

It follows that if p ≡ ε (mod 3) (ε ∈ {−1, 1}), then if (∆p

) = 1 and p - B, we have

that p is an S prime if and only if U p−ε3

(A,B3) ≡ 0 (mod p).

Thus, we have proved the following theorem.

Theorem 4.22. Suppose p is a prime and p - 6∆R. If (∆p

) = −1, then p is a Q

prime. If (∆p

) = 1, p ≡ ε (mod 3), A = 2P 3 − 9QR + 27R, B = P 3 − 3Q and

p | U p−ε3

(A,B3), then p is an S prime; otherwise, p is an I prime.

We will now develop the law of apparition for a prime p in {Cn}. We first prove

the following simple lemma.

Lemma 4.23. Let p be a prime such that p - 2R∆, K be the splitting field for

f(x) ∈ Fp[x] and α, β, γ be the zeros of f(x) in K. If αn = βn and αm = βm, then

αs = βs, where n ≡ s (mod m).

Proof. We have n = qm + s and αqm+s = βqm+s. Since αqm = βqm and R 6= 0, we

get αs = βs.

We next determine the number of ranks of apparition of a Q prime.

Theorem 4.24. Let p be a Q prime and α, β, γ be the zeros of f(x) in Fp2, where

β 6∈ Fp. Then p | Cm if and only if βm = βpm.

Proof. Since p is a Q prime, f(x) has the 3 zeros, namely, α in Fp and β, γ in Fp2

such that αp = α, βp = γ, γp = β.

121

Assume p | Cm. It follows that αm = βm, βm = γm or γm = αm. If αm = βm,

then βm ∈ Fp ⇒ βpm = βm. If βm = γm, then βpm = γpm = βm. If γm = αm, then

βpm = γm = αm ⇒ βpm ∈ Fp ⇒ βp2m = βpm ⇒ βm = βpm.

On the other hand if βm = βpm, then βm = γm ⇒ p | Cm.

Corollary 4.24.1. If p is a Q prime, then p | Cp+1.

Proof. This follows directly from the theorem by noting,

βp+1 = ββp = βp2

βp = βp2+p = βp(p+1).

Corollary 4.24.2. Let p be a Q prime, then p can only have one rank, r, of appari-

tion in {Cn} and r | p+ 1.

Proof. Suppose r is the minimal rank of apparition of p and p | Cm. We must

have βm = βpm by Theorem 4.24. Put m = qr + s such that 0 < s < r. We

have βqr+s = βpqr+ps. Now βr = βrp ⇒ βqr = βqrp 6= 0. So βs = βps ⇒ p | Cs

contradicting the definition of r. It follows that s = 0 and r | m. Thus, there can

only be one rank of apparition r for p and r | p+ 1.

Corollary 4.24.3. If p is a Q prime and r is its rank of apparition in {Cn}, then

if p | Cn, we must have r | n.

Proof. The proof follows at once from Corollary 4.24.2; if there existed an n with

r - n then there would be another rank of apparition.

122

Note that if p is a Q prime then

Wp+1 ≡ 2α4βγ + 2β3γ3 + 2R3 (mod p).

This will not be useful to us here; however, we can see that

Wp2−1 ≡ 6 (mod p). (4.15)

Theorem 4.25. If p is an I prime, then p | Cp2+p+1.

Proof. Since p is an I prime,

αp = β, βp = γ and γp = α.

So

αp2+p+1 = βp

2+p+1 = γp2+p+1 = R.

Hence the result follows.

Under the same conditions we can see

Wp2+p+1 ≡ 6R3 (mod p).

Corollary 4.25.1. Let p be an I prime, then p can only have one rank, r, of appari-

tion in {Cn} and r | p2 + p+ 1.

Proof. Suppose r is the minimal rank of apparition of p for {Cn}. So, without loss

of generality αr = βr ⇒ αpr = βpr ⇒ βr = γr ⇒ αr = βr = γr. If p | Cm, then

αm = βm, βm = γm or γm = αm. Thus, if m = qr + s and 0 ≤ s < r, then p | Cs

by Lemma 4.23. By the definition of r, we must have s = 0 and r | m. Thus, there

can only be one rank of apparition of p in {Cn}. Furthermore, since p | Cp2+p+1, we

get r | p2 + p+ 1.

123

Corollary 4.25.2. If p is an I prime and r is its rank of apparition in {Cn}, then

if p | Cn, we must have r | n.

Thus, the situation with Q and I primes parallels that concerning primes that

divide Un. That is, we know that if a prime p divides Un, then the rank of apparition

ω = ω(p) of p in {Un} must divide n. However, the situation with S primes can be

different from this as we see below.

Theorem 4.26. Let p - 6∆R and p be an S prime, then p | Cp−1.

Proof. Since α, β, γ ∈ Fp ⇒ αp−1 = βp−1 = γp−1 = 1. Hence, p | Cp−1.

Once more we note that

Wp−1 ≡ 6 (mod p),

under these circumstances.

Corollary 4.26.1. Let p - 6∆R and p be an S prime, then p may have at most 3

ranks of apparition in {Cn} and each rank of apparition divides p− 1.

Proof. Let r1 be any rank of apparition of p in {Cn}. If p | Cr1 , then αr1 = βr1 or

βr1 = γr1 or γr1 = αr1 . Without loss of generality assume αr1 = βr1 . Since p is an S

prime, we know that αp−1 = βp−1. Suppose r1 is the least positive integer such that

αr1 = βr1 . Let p−1 = qr1+s such that 0 < s < r1. Then αqr1+s = βqr1+s ⇒ αs = βs.

This is a contradiction, thus r1 | p− 1.

Now suppose r1 is the minimal rank of apparition. Further, suppose p | Cr2 and

r1 - r2. We know αr2 6= βr2 , for if αr2 = βr2 , then r2 = r1q + s where 0 < s < r1 and

by Lemma 4.23 αs = βs, which is a contradiction. Thus without loss of generality

124

βr2 = γr2 and let us assume r2 is the least positive integer such that this is true.

We know by the above reasoning that r2 | p − 1. Continue in this fashion, letting

r3 | Cr3 and r1 - r3, r2 - r3. Again, by Lemma 4.23 γr3 = αr3 . Also, if we assume r3

to be the least positive integer such that γr3 = αr3 , then r3 | p − 1. Now if we try

to define r4 such that r4 | Cr4 and r1 - r4, r2 - r4, r3 - r4, then αr4 = βr4 , βr4 = γr4

or γr4 = αr4 . None of which is possible by Lemma 4.23. Thus there can be at most

3 ranks of apparition in this case.

Corollary 4.26.2. If p is an S prime and p | Cn, then at least one of the ranks of

apparition of p in {Cn} must divide n.

Proof. Without loss of generality we may suppose that αn = βn in K. We have

already seen that if r1 is the least positive integer for which αr1 = βr1 , then r1 is a

rank of apparition of p in {Cn}. Furthermore, by Lemma 4.23 we must have αs = βs,

where n = qr1 + s (0 ≤ s < r1). If s > 0, we get a contradiction to the definition of

r1; thus, s = 0 and r1 | n.

In the next theorem, we show that the case of 3 ranks of apparition can occur

infinitely often.

Theorem 4.27. Let p be a prime such that p = 2κk1k2 + 1, where k1, k2 are odd,

(k1, k2) = 1, and k1, k2 > 1. There exists a set of values for P , Q, R such that {Cn}

has 3 ranks of apparition.

Proof. We select any primitive root g of p and any integer r. In Fp, put

α = gr, β = gr+k1 , γ = gr+k1+k2 .

125

From these we can easily produce corresponding P , Q, R (mod p). Note that if we

put

r1 = 2κk2, r2 = 2κk1, r3 =2κ−1k1k2

(2κ−1k1k2, (k1 + k2)/2),

then

αr1 = βr1 , βr2 = γr2 , γr3 = αr3 .

Thus

Cr1 , Cr2 , Cr3 ≡ 0 (mod p).

Furthermore, none of r2, r2, r3 divides any of the others. Thus, there must be three

ranks of apparition for {Cn}.

We remark that there exists an infinitude of distinct primes satisfying the con-

ditions of the theorem. For if we put k2 = 2xk1 + 1 for a fixed odd k1 > 1, then by

Dirichlet’s theorem there must exist an infinitude of values of x for any fixed κ ≥ 1

such that

2κk1k2 + 1 = 2κ(2xk1 + 1)k1 + 1 = 2κ+1k21x+ 2κk1 + 1

is a prime.

Theorem 4.28. Let p be an S prime and p ≡ 1 (mod 3). Suppose that Rp−13 6≡ 1

(mod p). Then there is one and only one rank of apparition, r, of p such that r | p−13

.

Proof. Since p ≡ 1 (mod 3) we can let ζ2 +ζ+1 = 0 in Fp. Since Rp−13 6≡ 1 (mod p),

we know that

αp−13 = ζ i, β

p−13 = ζj, γ

p−13 = ζk,

where 3 - i+ j + k. Since i, j, k cannot all be the same or all different modulo 3, we

must have exactly two equal modulo 3. Without loss of generality suppose i = j 6= k.

126

Then αp−13 = β

p−13 ⇒ p | C p−1

3⇒ ∃ a rank of apparition r of p such that r | p−1

3.

Thus if αn = βn, then r | n. Suppose r1 is another rank of apparition of p such that

r1 6= r. If αr1 = βr1 , then r | r1 ⇒ r = r1, which is a contradiction.

Thus we must have βr1 = γr1 or αr1 = γr1 . If r1 | p−13

, then βp−13 = γ

p−13 or

αp−13 = γ

p−13 , neither of which is possible, as k is distinct modulo 3 from i and j.

Chapter 5

Arithmetic Properties of {Dn}

5.1 Preliminary Results for the Law of Repetition for {Dn}

While, as we have seen, Cn is analogous to the Lucas function Un in many respects,

there are a number of significant differences between the arithmetic behaviour of

Cn and Un. This is particularly the case in the law of repetition and the law of

apparition, where it is possible to have more than one rank of apparition for {Cn}.

In the law of repetition for a prime p such that p | Cn, it is important to know

whether or not p divides the quantity Wn − 6Rn. We also noted in Section 4.6 that

we often have the case of a prime p dividing both Cn and Wn− 6Rn. In view of this,

we define

Dn = gcd(Cn,Wn − 6Rn).

This is not as peculiar as it might seem at first. For if we look at the formula for

Wn − 6Rn in terms of α, β, γ, we see that the corresponding formula involving α, β

of the Lucas functions would be

α2n + β2n − 2αnβn = V2n − 2Qn.

This is because if we consider Wn − 6Rn to be a polynomial in αn, βn, γn, then it

is of degree three and the αnβnγn term is subtracted as many times as there are

terms in the expression for Wn. Hence, the degree two counter part to this would be

127

128

α2n + β2n − 2αnβn. However,

V2n − 2Qn = V 2n − 4Qn = ∆U2

n and (V2n − 2Qn, Un) = Un.

Notice that by Theorem 4.5 and Lemma 4.6 we have

(Dn, R) | 2. (5.1)

As we shall see below, it turns out that Dn has arithmetic properties which are much

more analogous to those of Un than does Cn. In order to derive the law of repetition

for {Dn} we will first develop some results for the sequence {Lm} in the same way

we derived a law of repetition for {Cn} by first using the sequence {Km}.

Let

Lm(X) =∑ (−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!Xλ1+λ2 ,

where the sum is extended over the values λi ∈ Z such that

λ0, λ1, λ2, λ3 ≥ 0, λ0 + λ1 + λ2 + λ3 = m, λ1 + 2λ2 + 3λ3 = m.

By Waring’s theorem,

Lm(X) = αm1 + αm2 + αm3 ,

where α1, α2, α3 are the zeros of Z3 −XZ2 +XZ − 1 such that

α1 = 1, α2 =X − 1 +

√D(X)

2, α3 =

X − 1−√D(X)

2,

and D(X) = (X − 3)(X + 1). We can then write

Lm(X) = 1 + αm2 + αm3 = 1 + Vm(X − 1, 1).

So if 2 - m, then by (4.6) we have

Vm(X − 1, 1) = V1

(m−1)/2∑j=0

((m− 1)/2 + j

(m− 1)/2− j

)Dj(X).

129

On the other hand, if 2 | m, then by (4.7) we have

Vm(X − 1, 1) =

m/2∑j=0

m

m/2− j

(m/2 + j − 1

m/2− j − 1

)Dj(X).

Now, by using results similar to those in Section 4.4 and noting Pn = Wn, we have

Wmn ≡ 2∑

λ0,λ1,λ2,λ3

(−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!

(Pn2

)λ1+λ2

Rn(m−λ1−λ2) (mod Fn)

where

Fn =

∆C2n if 2 - Cn

∆C2n/4 if 2 | Cn.

Let 2γ || Dn. Then Dn/2γ | Fn, 2 - Dn/2

γ and (Dn/2γ, R) = 1. Put Gn =

Dn/2γ. Then

Wmn ≡ 2Rmn∑ (−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!

(Wn

2Rn

)λ1+λ2

(mod Fn) (5.2)

≡ 2RmnLm(Wn/2Rn) (mod Fn). (5.3)

If m is odd,

Lm(Wn/2Rn) = 1 + (Wn/2R

n − 1)

(m−1)/2∑j=0

((m− 1)/2 + j

(m− 1)/2− j

)(Wn/2R

n + 1)j

(Wn/2Rn − 3)j.

Since Wn/2Rn − 3 ≡ 0 (mod Gn),

Lm(Wn/2Rn) ≡ 1 +Wn/2R

n − 1 ≡ 3 (mod Gn).

If m is even,

Lm(Wn/2Rn) = 1 +

m/2∑j=0

m

m/2− j

(m/2 + j − 1

m/2− j − 1

)(Wn/2R

n + 1)j

(Wn/2Rn − 3)j

≡ 3 (mod Gn).

130

Thus, Wmn ≡ 6Rmn (mod Gn)⇒ Gn | Wmn − 6Rmn.

It follows that if γ = 0, then Dn | Dnm.

If γ = 1, then since 2 | (Wn, Cn), we have 2 | Cmn, and since Qmn is an integer,

2 | Wmn; thus, Dn | Dmn.

If γ > 1, then 4 | Cn and 4 | Wn−6Rn. Recall that if 2α || (Wn, Cn), then α = 0

or 1 by Theorem 4.5. In this case α = 1 and 2 || Wn. It follows from 4 | Wn − 6Rn

that R must be odd. Thus, since 2γ ≥ γ + 2, we have 2γ | Fn and

Wmn ≡ 2Rmn∑ (−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!

(Wn

2Rn

)λ1+λ2

(mod 2γ)

≡ 6Rmn (mod 2γ),

so that 2γ | Wmn − 6Rmn ⇒ Dn | Wmn − 6Rmn ⇒ Dn | Dmn. Thus, if n | m, we

get Dn | Dm. Therefore, like {Un} and {Cn}, {Dn} is a divisibility sequence.

5.2 The Law of Repetition for {Dn}

The goal in this section is to develop a law of repetition for {Dn}.

Suppose pµ || Dn, µ ≥ 1 and p - 6∆. There are three cases to consider.

Case 1. pµ || Wn − 6Rn and pν || Cn such that ν > µ ≥ 1.

Note that µ ≥ 2 by Theorem 4.18, and hence 3µ ≥ 2µ + 2. We then have p2ν | Fn.

Thus, by (5.3)

Wpn ≡ 2RpnLp(Wn/2Rn) (mod p2ν) and

Lp(Wn/2Rn) ≡ 1 + (Wn/2R

n − 1)

[1 +

(p2 − 1

8

)(Wn/2R

n + 1)(Wn/2Rn − 3)+

(p2 − 1)(p2 − 9)

16 · 24(Wn/2R

n + 1)2(Wn/2Rn − 3)2

](mod p3µ).

131

Now, since

Wn/2Rn + 1 ≡ 4 (mod pµ) and p2µ | (Wn/2R

n − 3)2,

we have

(Wn/2Rn + 1)2(Wn/2R

n − 3)2 ≡ 16(Wn/2Rn − 3)2 (mod p3µ).

Thus

Lp(Wn/2Rn) ≡ 1 + (Wn/2R

n − 1)

[1 +

(p2 − 1

8

)(Wn/2R

n + 1)(Wn/2Rn − 3)+

(p2 − 1)(p2 − 9)

24(Wn/2R

n − 3)2

]≡ Wn/2R

n +

(p2 − 1

8

)((Wn/2R

n)2 − 1)(Wn/2Rn − 3) +

(p2 − 1)(p2 − 9)

24(Wn/2R

n − 1)(Wn/2Rn − 3)2 (mod p3µ).

Now

Wn/2Rn − 1 ≡ 2 (mod pµ) and p2µ | (Wn/2R

n − 3)2

so

(Wn/2Rn − 1)(Wn/2R

n − 3)2 ≡ 2(Wn/2Rn − 3)2 (mod p3µ).

Thus,

Lp(Wn/2Rn) ≡ Wn/2R

n +

(p2 − 1

8

)((Wn/2R

n)2 − 1)(Wn/2Rn − 3) +

(p2 − 1)(p2 − 9)

12(Wn/2R

n − 3)2 (mod p3µ).

Also, we have

(Wn/2Rn − 3)2 ≡ 0 (mod p2µ)⇒ (Wn/2R

n)2 − 6Wn/2Rn + 9 ≡ 0 (mod p2µ)

132

⇒ (Wn/2Rn)2 − 1 ≡ 6Wn/2R

n − 10 (mod p2µ).

Note, then, that

((Wn/2Rn)2 − 1)(Wn/2R

n − 3) ≡ (6Wn/2Rn − 10)(Wn/2R

n − 3) (mod p3µ).

This yields

Lp(Wn/2Rn) ≡ Wn/2R

n +

(p2 − 1

8

)(6Wn/2R

n − 10)(Wn/2Rn − 3) +

(p2 − 1)(p2 − 9)

12(Wn/2R

n − 3)2

≡[(Wn/2R

n − 3) +

(p2 − 1

8

)(6Wn/2R

n − 10)(Wn/2Rn − 3)+

(p2 − 1)(p2 − 9)

12(Wn/2R

n − 3)2

]+ 3 (mod p3µ).

This can be rewritten as

Lp(Wn/2Rn)− 3 ≡ (Wn/2R

n − 3)

[1 +

(p2 − 1)

4(3Wn/2R

n − 9 + 4)+

(p2 − 1)(p2 − 9)

12(Wn/2R

n − 3)

]≡ (Wn/2R

n − 3)

[1 + 3

(p2 − 1)

4(Wn/2R

n − 3) + 4(p2 − 1)

4+

(p2 − 1)

4

(p2 − 9)

3(Wn/2R

n − 3)

]≡ p2(Wn/2R

n−3) + (Wn/2Rn−3)2

[3

(p2 − 1)

4+

(p2 − 1)

4

(p2 − 9)

3

]≡ p2(Wn/2R

n − 3) +(p2 − 1)

4(Wn/2R

n − 3)2

[3 +

(p2 − 9)

3

]≡ p2(Wn/2R

n − 3) + p2 (p2 − 1)

12(Wn/2R

n − 3)2 (mod p3µ)

≡ p2(Wn/2Rn − 3) (mod p2µ+2).

Observe that

Wpn − 6Rpn ≡ 2RpnLp(Wn/2Rn)− 6Rpn (mod p2ν)

133

and that ν > µ⇒ ν ≥ µ+ 1⇒ 2ν ≥ 2µ+ 2. So

Wpn − 6Rpn ≡ 2Rpn

[p2(Wn/2R

n − 3) + p2 (p2 − 1)

12(Wn/2R

n − 3)2

](mod p2µ+2)

≡ 2Rpn[p2(Wn/2R

n − 3)]

(mod p2µ+2)

≡ (Rn)p−1p2(Wn − 6Rn) (mod p2µ+2)

≡ (1 + kp)p2(Wn − 6Rn) (mod p2µ+2),

for some k ∈ Z. Since pµ || Wn−6Rn and 2µ+2 > µ+2 we have pµ+2 || Wpn−6Rpn.

We also know that pµ+3 | Cpn; hence, pµ+2 || Dpn.

Case 2. pµ || Wn − 6Rn and pµ || Cn.

We then have p2µ | Fn which gives us

Wpn ≡ 2RpnLp(Wn/2Rn) (mod p2µ) and

Lp(Wn/2Rn) ≡ 1 + (Wn/2R

n − 1)

[1 +

((p+ 1)/2

2

)(Wn/2R

n + 1)(Wn/2Rn − 3)

]≡ Wn/2R

n +(p2 − 1)

8((Wn/2R

n)2 − 1)(Wn/2Rn − 3) (mod p2µ).

Now, note that since

Wn/2Rn + 1 ≡ 4 (mod pµ) and Wn/2R

n − 1 ≡ 2 (mod pµ),

we have

((Wn/2Rn)2 − 1) ≡ (Wn/2R

n − 1)(Wn/2Rn + 1) ≡ 8 (mod pµ).

Together with the fact that pµ | Wn/2Rn − 3, we see that

((Wn/2Rn)2 − 1)(Wn/2R

n − 3) ≡ 8(Wn/2Rn − 3) (mod p2µ).

134

So

Lp(Wn/2Rn) ≡ Wn/2R

n + (p2 − 1)(Wn/2Rn − 3) (mod p2µ).

Thus

Wpn − 6Rpn ≡ 2Rpn[Wn/2R

n + (p2 − 1)(Wn/2Rn − 3)

]− 6Rpn

≡ Rn(p−1)[Wn + (p2 − 1)(Wn − 6Rn)− 6Rn

]≡ p2(Wn − 6Rn) (mod p2µ).

Since by Theorem 4.18, µ ≥ 3, we get 2µ ≥ µ+ 3 and we can write

Wpn − 6Rpn ≡ p2(Wn − 6Rn) (mod pµ+3).

Using the fact pµ || Wn − 6Rn, we get pµ+2 || Wpn − 6Rpn. Again we know that

pµ+3 || Cpn ⇒ pµ+2 || Dpn.

Case 3. pν || Cn and pµ || Wn − 6Rn such that ν < µ.

In this case pν+3 || Cpn and

Wpn − 6Rpn ≡ p2(Wn − 6Rn) (mod p2ν)

≡ p2(Wn − 6Rn) ≡ 0 (mod pν+3).

Thus pν+3 | Wpn − 6Rpn ⇒ pν+3 || Dpn.

If p > 3, p | ∆, pµ || Wn−6Rn and pν || Cn such that ν ≥ µ, we have p2ν+1 | Fn.

Since

Qλ2n Vλ1−λ2(Pn, Qn) ≡ 2(Pn/2)λ1+λ2 (mod 22ν+1)

and p | p(p−λ0−1)!λ1!λ2!λ3!

except for λ1 = p we find, by using the reasoning in Section 4.4,

that

Wpn ≡ 2Rpn [Lp(Wn/2Rn)− (Wn/2R

n)p] + Vp(Pn, Qn) (mod p2ν+2).

135

We may use the fact that

Vp(Pn, Qn) ≡ Pn(Pn/2)p−1 + P p−2n ∆n

(p

2

)(mod p2ν+2)

≡ Pn(Pn/2)p−1 (mod p2ν+2)

to get

Wpn ≡ 2RpnLp(Wn/2Rn)− 2Rpn(Wn/2R

n)p +Wn(Wn/2)p−1

≡ 2RpnLp(Wn/2Rn) (mod p2ν+2).

Since

Wpn − 6Rpn ≡ p2R(p−1)n(Wn − 6Rn) (mod p2µ+2),

we get pµ+2 || Dpn when ν ≥ µ(≥ 1).

If ν < µ, then pν+3 | Wpn − 6Rpn and pν+3 || Dpn.

If p = 3, 3µ || Dn, 3µ || Wn − 6Rn and 3ν || Cn where ν ≥ µ, then we use

4(W3n − 6R3n) = 3∆C2n(Wn + 2Rn) +W 2

n(Wn − 6Rn).

We have 32ν+1 | 3∆C2n and

4(W3n − 6R3n) ≡ W 2n(Wn − 6Rn) (mod 32ν+1).

Suppose µ > 1. Since ν ≥ µ ≥ 2⇒ 2ν + 1 ≥ µ+ 3 we have

4(W3n − 6R3n) ≡ W 2n(Wn − 6Rn) (mod 3µ+3).

Since (Dn, R) | 2, and 9 | Wn − 6Rn, we must have 3 || Wn. Thus for µ > 1 we

have 3µ+2 || D3n. If µ = 1, all we can say is that 3µ+2 | D3n.

136

Now if p = 3, µ > 1, 3µ || Cn and 3ν || Wn− 6Rn such that ν > µ, we have that

3µ+3 || C3n and

4(W3n − 6R3n) ≡ W 2n(Wn − 6Rn) (mod 32µ+1)

≡ W 2n(Wn − 6Rn) (mod 3µ+3)

≡ 0 (mod 3µ+3),

hence 3µ+3 || D3n. If µ = 1, we can only say that 3µ+3 | D3n.

For the case where p = 2 and 2 - R we have the following. If 2 | Dn, then

2 | Wn and 2 | Cn, hence 2 | C2n/Cn as C2n = Cn(Wn + 2Rn). Also, since Qn =

(W 2n −∆C2

n)/4 is odd in this case, we get W 2n −∆C2

n ≡ 4 (mod 8) and

W 2n + ∆C2

n ≡ 4 + 2∆C2n ≡ 4 (mod 8).

Now

2W2n = ∆C2n +W 2

n − 4RnWn ≡ ∆C2n +W 2

n ≡ 4 (mod 8).

This means

W2n ≡ 2 (mod 4)⇒ 2 || Wn ⇒ 4 | W2n − 6R2n.

Now suppose further that 2m | Dn and m ≥ 2. So then

22m | (Wn − 6Rn)2 ⇒ W 2n ≡ 12RnWn − 36R2n (mod 22m).

This, together with the identity 2W2n = ∆C2n +W 2

n − 4RnWn yields,

2W2n ≡ −36R2n + 12RnWn − 4RnWn ≡ 8RnWn − 36R2n (mod 22m).

137

So then

2(W2n − 6R2n) ≡ 8RnWn − 36R2n − 12R2n (mod 22m)

≡ 8RnWn − 48R2n (mod 22m)

≡ 8Rn(Wn − 6Rn) (mod 22m).

Hence,

W2n − 6R2n ≡ 4Rn(Wn − 6Rn) (mod 22m−1),

and since m ≥ 2, we get 2m− 1 ≥ m+ 1, so

W2n − 6R2n ≡ 4Rn(Wn − 6Rn) (mod 2m+1).

Thus we can see 2m+1 | W2n − 6R2n. Thus, if 2m | Dn, then 2m+1 | D2n.

When m ≥ 4 we can say more. Here we have 16 | Wn−6Rn, so 2 - R, since other-

wise 4 | (Wn, Cn) and this is a contradiction because (Wn, Cn) | 2 when (Q,R) = 1

by Theorem 4.5. Now

Wn + 2Rn = Wn − 6Rn + 8Rn ⇒ 8 || Wn + 2Rn ⇒ 8 || C2n/Cn.

Also, if 2ν || Cn, then

2W2n ≡ W 2n − 4RnWn (mod 22ν),

or

2(W2n − 6R2n) ≡ (Wn − 6Rn)2 + 8Rn(Wn − 6Rn) (mod 22ν).

Now let 2µ || Wn − 6Rn.

Case 1. If ν ≥ µ, we get

2(W2n − 6R2n) ≡ 8Rn(Wn − 6Rn) (mod 22µ).

138

Hence

W2n − 6R2n ≡ 4Rn(Wn − 6Rn) (mod 22µ−1).

Now we use the fact µ ≥ 4 to see that

W2n − 6R2n ≡ 4Rn(Wn − 6Rn) (mod 2µ+3).

Hence 2µ+2 || W2n − 6R2n and 2ν+3 || C2n ⇒ 2m+2 || D2n ⇒ 4 || D2n/Dn when

m = µ. Thus by induction we get 22k || D2kn/Dn.

Case 2. If ν < µ, then we have 2ν+3 | W2n−6R2n and 2ν+3 || C2n ⇒ 2ν+3 || D2n ⇒

23 || D2n/Dn. However, since we do not know that 2ν+4 | W2n − 6R2n, the best we

can say in general is that, if 2γ || D2kn/Dn, then γ ≤ 3k.

Theorem 5.1. If p is a prime, pλ || Dn (λ ≥ 1) and p - m, then pλ || Dmn.

Proof. Suppose first that p 6= 2. If 2 - m, then since p | Dn, we get

CmnCn

≡ mRn(m−1) +m

(m2 − 1

8

)Rn(m−3)

4(Wn − 2Rn)(Wn + 2Rn) (mod p)

≡ mRn(m−1) +m

(m2 − 1

8

)Rn(m−3)

432R2n (mod p)

≡ m3Rn(m−1) (mod p).

If 2 | m, then

CmnCn

≡ m

(m

m/2− 1

(m/2

2

)Rn(m−2)

2(Wn + 2Rn)

)(mod p)

≡ m

((m

m/2− 1

)((m/2)(m/2− 1)

2

)Rn(m−2)

28Rn

)(mod p)

≡ m3Rn(m−1) (mod p).

Suppose pλ || Dn. There two cases: either pλ || Cn or pλ || Wn − 6Rn. If pλ || Cn

and p - m, then pλ+1 - Cmn, and we have pλ || Dmn.

139

On the other hand, if pλ || Wn − 6Rn, then pλ | Cn and p2λ | Fn. If 2 - m,

Lm(Wn/2Rn) ≡ 1 + (Wn/2R

n − 1) +

(m2 − 1

8

)(Wn/2R

n − 1)(Wn/2Rn + 1)

(Wn/2Rn − 3) (mod p2λ)

≡ Wn/2Rn +

(m2 − 1

8

)8(Wn/2R

n − 3) (mod p2λ)

≡ Wn/2rn +m2(Wn/2R

n − 3)− (Wn/2Rn − 3) (mod p2λ)

≡ m2(Wn/2Rn − 3) + 3 (mod pλ+1).

If 2 | m we get

Lm(Wn/2Rn) ≡ 1 +

m

m/2+

m

m/2− 1

(m/2

2

)(Wn/2R

n + 1)(Wn/2Rn − 3) (mod p2λ)

≡ 3 +

(m

m/2− 1

)(m/2(m/2− 1)

2

)4(Wn/2R

n − 3) (mod p2λ)

≡ m2(Wn/2Rn − 3) + 3 (mod pλ+1).

We can then see

Wmn − 6Rmn ≡ 2RmnLm(Wn/2Rn)− 6Rmn (mod pλ+1)

≡ 2Rmn (Lm(Wn/2Rn)− 3) (mod pλ+1)

≡ 2Rmn(m2(Wn/2R

n − 3))

(mod pλ+1).

But then pλ+1 | Wmn−6Rmn ⇒ p | m. Hence if p - m and pλ || Dn, then pλ || Dmn.

Now for the case of p = 2. Since 2 | Dn, we have 2 | Cn and 2 | Wn. This means

that 2 | Pn and 2 - Qn by Theorem 4.5. By Theorem 4.10, we have

Cmn/Cn ≡ m (mod 2).

Thus, if 2 - m, then 2 - Cmn/Cn. If 2λ || Cn, then 2λ+1 - Cmn and so 2λ+1 - Dmn

140

when 2 - m. If 2λ+1 | Cn, then 2λ || Wn− 6Rn. In this case 22λ | Fn. If 2 - m, then

Wmn ≡ 2RmnLm(Wn/2Rn) (mod 22λ)

≡ 2Rmn(m2(Wn/2Rn − 3) + 3) (mod 2λ+1).

Thus

Wmn − 6Rmn ≡ 2m2Rmn(Wn/2Rn − 3) (mod 2λ+1)

≡ m2Rm(n−1)(Wn − 6Rn) (mod 2λ+1).

Since 2λ || Wn − 6Rn, we get 2λ || Wmn − 6Rmn when m is odd. Hence 2λ+1 - Dmn

when 2 - m.

Thus, we have shown that if p is any prime and pλ || Dn, then pλ || Dmn when

p - m.

Our Law of Repetition for {Dn} is stated in the following theorem.

Theorem 5.2. If pλ || Dn (p 6= 2, pλ 6= 3), then

pλ+2 || Dpn when pλ || Wn − 6Rn and

pλ+3 || Dpn otherwise .

Also, pλ+2 | Dpn when pλ = 3 and pλ+1 | Dpn when p = 2. Furthermore, pλ+1 - Dmn

if p - m.

Notice that if p 6= 2, 3 and pλ || Wn − 6Rn, pλ || Dn, then pλ+2 || Wpn − 6Rpn

and therefore

pλ+2µ || Dpµn.

141

However, if pλ || Dn and pλ+1 | Wn − 6Rn, it is not necessarily the case that

pλ+4 || Wpn − 6Rpn.

The best we are able to show is that pλ+3 | Wpn−6Rpn. If pλ+3 || Wpn−6Rpn, then

we return to the previous condition and by induction we get

(pλ+1+2µ =)pλ+3+2(µ−1) || Dpµn.

Of course, this latter situation would never occur if the case of

pλ || Dn and pλ+1 | Wn − 6Rn

could not happen. We have given some reason in Chapter 4 to believe that this

might be an infrequent occurrence, but unfortunately it does happen. For example,

if P = 257, Q = 2004 and R = 5389, then 73 || C6 and 74 | W6 − 6R6.

Thus, we cannot provide as complete a law of repetition for {Dn} as we were

able to do for {Cn}. However, if pλ || Cn, pλ+κ || Wn − 6Rn and κ < λ − 2, it can

be shown that

pλ+3µ || Cpµn and pλ+κ+2µ || Wpµn − 6Rpµn.

Hence, we get

pλ+3µ || Dpµn for µ ≤ κ.

Note that if µ = κ, then λ+ κ+ 2µ = λ+ 3µ and we return to the previous case. It

follows, then, that

pλ+κ+2µ || Dpµn

when µ > κ. Unfortunately if κ ≥ λ − 2, it seems to be difficult to formulate a

comprehensive law of repetition.

142

5.3 The Law of Apparition for {Dn}

Definition 5.3. Let p be a prime and ω(p) be the least positive integer n, if it exists,

such that p | Dn. We call this the rank of apparition of p in {Dn}.

The next theorems build towards a result very comparable to Theorem 2.4. What

is remarkable is that this is a result that did not hold for {Cn}. Hence, with the help

of {Dn} we are able to establish a more convincing analogue.

Lemma 5.4. Suppose p is a prime, p - 2R∆ and K is the splitting field of f(x) in

Fp[x]. If α, β, γ are the zeros of f(x) in K, then p | Dn if and only if αn = βn = γn

in K.

Proof. (⇒) If p | Dn, then p | Cn and we may assume with no loss of generality

that αn = βn. Since

Wn − 6Rn = 2βn(αn − γn)2 − (αn − βn)(βn − γn)(αn + γn),

it follows that

2βn(αn − γn)2 ≡ (αn − βn)(βn − γn)(αn + γn) (mod p),

as p | Wn − 6Rn and hence αn = γn.

(⇐)On the other hand, if αn = βn = γn, then it is clear that p | Dn.

We are now able to present an important result concerning ω(p).

Theorem 5.5. Suppose p is a prime such that p - 2R∆ and suppose further that

ω = ω(p) exists for {Dn}. If p | Dn, then ω | n.

143

Proof. Since p | Dn and p | Dω we have αn = βn = γn and αω = βω = γω in K by

Lemma 5.4. If ω - n, then n = ωq + r, where 0 < r < ω. By Lemma 4.23, we get

αr = βr = γr and p | Dr by Lemma 5.4. Since this contradicts the definition of ω,

we must have r = 0 and ω | n.

We next note that if p | R and p 6= 2, then ω(n) does not exist. If 2 | R, then

2 - Q. By our results in Theorem 4.15, there exists a rank of apparition r for {Cn}

of 2. Since 2 | Cn ⇒ 2 | Wn, we see that ω = ω(2) = r and 2 | Dn if and only if

r | n.

Theorem 5.6. If p - R, then ω(p) must exist. Further, if p | Dn, then ω(p) | n.

Proof. We have already seen that this holds if p = 2. We now turn our attention to

the case of p = 3. If 3 | Cn, then by Corollary 3.10.1 we have

8W4n ≡ 4W 22n − 16R2nW2n (mod 3)

≡ 2W2n(2W2n − 8R2n) (mod 3)

≡ (W 2n − 4RnWn)((W 2

n − 4RnWn − 8R2n) (mod 3)

≡ Wn(Wn −Rn)((W 2n + 2RnWn +R2n) (mod 3)

≡ Wn(Wn −Rn)(Wn +Rn)2 (mod 3)

≡ 0 (mod 3),

since 3 - R, then 3 must divide one of Wn, Wn − Rn or Wn + Rn. Thus, since r(3)

exists and is unique when 3 - ∆, we see by Table 4.1 that ω(3) exists. By Theorem

5.5, we also see that ω(3) | n if p | Dn. If p - 6∆R, then p is either an S prime,

I prime or Q prime. If p is an I prime, then αp = β, βp = γ, γp = α in Fp3 , the

144

splitting field of f(x). Hence

αp2+p+1 = βp

2+p+1 = γp2+p+1 = R

and p | Dp2+p+1.

If p is an S prime, then αp−1 = βp−1 = γp−1 = 1 in the splitting field Fp of f(x).

Hence, p | Dp−1 by Lemma 5.4.

If p is a Q prime, then αp2−1 = 1, βp

2−1 = (βp+1)p−1 = (γβ)p−1 = (R/α)p−1 = 1,

γp2−1 = 1 and p | Dp2−1. Thus, if p - 6∆R, then ω(p) exists, and if p | Dn, then

ω(p) | n by Theorem 5.5.

When p | ∆ and p 6= 2, we have seen that there are two cases:

Case 1. p | P 2 − 3Q.

In this case, we have seen in Section 4.3 that there is a unique r = r(p) = p, and

p | Cn if and only if r | n. Also, since, in this case, p | Wn− 6Rn whenever p | Cn,

we have ω = r and ω | n if p | Dn.

Case 2. p - P 2 − 3Q.

In this case, we know that by our results in Section 4.3 that p | Dn if and only if r | n.

Here r is the least positive integer such that p | ar−br, where a, b are as in equation

(4.12). Thus ω(p) = r and r | p− 1. Note that in both cases p | Dn ⇒ ω(p) | n.

Thus if p - R, we have shown that ω(p) always exists and ω(p) | n whenever

p | Dn.

Corollary 5.6.1. If p is a prime and ω(p) exists, then ω(p) ≤ p2 + p+ 1.

If p is an I prime, we have a very simple result connecting divisibility of Cn and

Dn by p.

145

Theorem 5.7. If p is an I prime, then p | Cn ⇔ p | Dn.

Proof. Clearly, if p | Dn, then p | Cn. Let α, β, γ be the zeros of f(x) in K = Fp3 .

In K we must have αn = βn ⇒ αpn = βpn ⇒ βn = γn ⇒ αn = βn = γn. It follows

that Wn = 6αnβnγn = 6Rn in K. Thus, p | Wn − 6Rn ⇒ p | Dn.

Since for any Q prime p we know that p | Cp+1, it is of some interest to determine

under what conditions p | Dp+1. We require a simple lemma.

Lemma 5.8. Let α, β, γ be the zeros of f(x) in Fp2 where αp = α, βp = γ, γp = β.

If p | Q3 −RP 3, then α3 = R in Fp2.

Proof. Suppose p | Q3 − RP 3. If p | P , then p | Q and f(x) ≡ x3 − R (mod p);

hence, α3 = R in Fp ⊆ Fp2 . If p - P , then

f(x) ≡ x3 − Px2 +Qx− (Q/P )3 (mod p)

and

P 3f(x) ≡ (Px−Q)(P 2x2 + (PQ− P 3)x+Q2) (mod p).

Because β, γ 6∈ Fp it follows that Pα−Q = 0 in Fp; hence,

α3 = Q3/P 3 = R in Fp2 .

We are now able to show that for a fixed triple P , Q, R, there can only be a finite

number of Q primes such that p | Dp+1.

Theorem 5.9. If p is a Q prime, then p | Dp+1 if and only if p | Q3 −RP 3.

146

Proof. Since p is a Q prime by Corollary 4.24.1, we have p | Cp+1. In Fp2 , we have

Wp+1 = 2R2 + 2(βγ)3 + 2Rα3;

hence, since α 6= 0 (R 6= 0), we have

Wp+1 − 6Rp+1 = 2(βγ)3 + 2Rα3 − 4R2

= 2((R/α)3 +Rα3 − 2R2)

=2R

α3(α3 −R)2.

It follows that p | Wp+1−6Rp+1 if and only if α3 = R in Fp2 . By Lemma 5.8 we know

that if p | Q3 − RP 3, then α3 = R. If α3 = R, then since α3 − Pα2 +Qα− R = 0,

we get Pα2 −Qα = 0 and hence Pα−Q = 0. If P = 0 in Fp2 , then p | Q3 − RP 3.

If P 6= 0, then α = Q/P and R = (Q/P )3 ⇒ p | Q3 −RP 3.

Suppose p - R and pα | Dn. Let ω = ω(p) and let ω(pα) denote the least positive

integer k such that pα | Dk. If pα | Dω, put ν = 0; otherwise define ν ∈ Z≥0 by

pα | Dpνω(p), pα - Dpν−1ω(p).

By our previous results concerning the law of repetition for {Dn} such a ν must

exist.

Theorem 5.10. If p - R, pα | Dn, then ω(pα) = pνω(p) and ω(pα) | n.

Proof. Since p | Dn, we must have n = mω(p) for some m ∈ N. Suppose pγ || m

and put m = m′pγ (p - m′). Since pα - Dpν−1ω(p) we must have pα - Dm′pν−1ω(p) ⇒

γ > ν − 1. But pα | Dm′pνω(p) ⇒ γ = ν.

Furthermore, since pα | Dpνω(p) we must have ω(pα) = pνω(p) and ω(pα) | n.

147

Theorem 5.11. Suppose m | Dn. Denote by ω(m) the least positive integer such

that m | Dω(m). Let

m =k∏i=1

pαii ,

then ω(m) = lcm[ω(pαii ); i = 1, 2, . . . k].

Proof. Clearly ω(pαii ) | ω(m) for i = 1, 2, . . . k. Since Dn is a divisibility sequence

the result follows.

We may now prove the following theorem, which very much resembles Corollary

2.4.1. Again, as seen in Chapter 4, this is another result that did not hold for {Cn}.

Theorem 5.12.

(Dn, Dm) = D(m,n).

Proof. Since Dn is a divisibility sequence we have D(m,n) | Dn, D(m,n) | Dm ⇒

D(m,n) | (Dn, Dm). Let pα || (Dn, Dm), then ω(pα) exists and ω(pα) | n, ω(pα) | m

hence ω(pα) | (m,n)⇒ pα | D(m,n). Thus (Dn, Dm) = D(m,n).

In Chapter 4 we were able to develop a result somewhat akin to Carmichael’s

result in Theorem 2.5. Surprisingly, if we look at {Dn} rather than {Cn}, we can in

fact do better. We have that

CmnCn≡ m3Rn(m−1) (mod (Fn,Wn − 6Rn)),

where Fn is as in (4.11). If 2ν || Cn and ν > 1, then Cn | Fn and

CmnCn≡ m3Rn(m−1) (mod Dn). (5.4)

148

If 2 || Cn, then CmnCn≡ m (mod 2) (Theorem 4.10) and Cn/2 | Fn so

CmnCn≡ m3Rn(m−1) (mod (Cn/2,Wn − 6Rn)).

Now (Dn, R) | 2. If (Dn, R) = 1, then (CmnCn

, Dn) | m3.

If (Dn, R) = 2, then by Lemma 4.4, 2 || Dn. We then have (CmnCn

, Dn/2) | m3.

Since (Dn/2, R) = 1 and (CmnCn

, 2) | m we have (CmnCn

, Dn) | m3.

We next examine (DmnDn

, Dn). Let pα || (DmnDn

, Dn). We will show that pα | m3

when p is a prime and α ≥ 1. This means of course that

(Dmn

Dn

, Dn) | m3.

We first observe that if p - m, then p - Dmn/Dn, which is a contradiction to Theorem

5.1, so p | m. If α < 4, then pα | m3. If α ≥ 4, then by the law of repetition for

Dn, we know that pλ || Dmn with λ ≤ 3µ + ν where pν || Dn (ν ≥ 4) and pµ | m.

Thus if pγ || Dmn/Dn, then γ = λ − ν ≤ 3µ ⇒ pγ | m3. We now have the desired

analogue of Theorem 2.5.

Theorem 5.13. If m, n ≥ 1, then

(Dmn/Dn, Dn) | m3.

The next two theorems will be needed in Chapter 6 to produce an analogue to

Euler’s criterion for the Lucas functions.

Theorem 5.14. Let p be an I prime and p ≡ 1 (mod 3), then

C p2+p+13

≡ 0 (mod p) and W p2+p+13

≡ 6Rp2+p+1

3 (mod p)

if and only if Rp−13 ≡ 1 (mod p).

149

Proof. We have α, β, γ as the zeros of f(x) in Fp3 where β = αp, γ = βp. In Fp3 we

get R = αp2+p+1 and

Rp−13 = (αp

2+p+1)p−13 = α

p3−13 .

Hence

αp2+p+1

3 Rp−13 = α

p2+p+13 α

p3−13 = α

p2+p+13

+ p3−13 = α

p3+p2+p3 = αp(

p2+p+13

).

This yields

βp2+p+1

3 = αp2+p+1

3 Rp−13 .

Similarly, one may show

γp2+p+1

3 = βp2+p+1

3 Rp−13 or γ

p2+p+13 = α

p2+p+13 R

2(p−1)3 .

Now if Rp−13 ≡ 1 (mod p), then p | C p2+p+1

3

. If Rp−13 6≡ 1 (mod p), then p - C p2+p+1

3

.

Then Rp−1 = 1 and

αp2+p+1

3 − βp2+p+1

3 = αp2+p+1

3 [1−Rp−13 ],

βp2+p+1

3 − γp2+p+1

3 = αp2+p+1

3 [Rp−13 −R

2(p−1)3 ],

γp2+p+1

3 − αp2+p+1

3 = αp2+p+1

3 [R2(p−1)

3 − 1]

imply

∆C2p2+p+1

3

= −27R2 6= 0.

150

Now note

W p2+p+13

= α2(p2+p+1)

3 βp2+p+1

3 + β2(p2+p+1)

3 αp2+p+1

3 + β2(p2+p+1)

3 γp2+p+1

3 +

γ2(p2+p+1)

3 βp2+p+1

3 + γ2(p2+p+1)

3 αp2+p+1

3 + α2(p2+p+1)

3 γp2+p+1

3

= α2(p2+p+1)

3 αp2+p+1

3 Rp−13 + α

2(p2+p+1)3 α

p2+p+13 R

2(p−1)3 +

α2(p2+p+1)

3 R2(p−1)

3 αp2+p+1

3 R2(p−1)

3 + α2(p2+p+1)

3 Rp−13 α

p2+p+13 R

p−13 +

α2(p2+p+1)

3 Rp−13 α

p2+p+13 + α

2(p2+p+1)3 α

p2+p+13 R

2(p−1)3

= 3R[Rp−13 +R

2(p−1)3 ].

If Rp−13 ≡ 1 (mod p), then WP2+p+1

3

− 6R ≡ 0 (mod p). Also, since 3 | p + 2, we

have R(p+2)(p−1)

3 ≡ 1 (mod p) ⇒ Rp2+p+1

3 ≡ R (mod p). Thus if Rp−13 ≡ 1 (mod p),

we get p | W p2+p+13

− 6Rp2+p+1

3 . If Rp−13 6≡ 1 (mod p), then R

2(p−1)3 +R

p−13 ≡ −1 6≡ 2

(mod p) and p - W p2+p+13

− 6Rp2+p+1

3 as W p2+p+13

≡ −3Rp2+p+1

3 (mod p).

Theorem 5.15. Let p (≡ 1 (mod 3)) be a Q prime, then p | D p2−13

if and only if

Rp−13 ≡ 1 (mod p).

Proof. We have α, β, γ as the zeros of f(x) in Fp2 such that

αp = α, βp = γ and γp = β

and

αp−1 = 1, αp+1 = α2, βp+1 = βγ and γp+1 = βγ.

Now βγ = R/α and

βp2−1

3 = γp2−1

3 = (βγ)p−13 =

(R

α

) p−13

=R

p−13

αp−13

· α2(p−1)

3

α2(p−1)

3

= Rp−13 α

2(p−1)3

= Rp−13 (α2)

p−13 = R

p−13 (αp+1)

p−13 = R

p−13 α

p2−13 .

151

We know that p | C p2−13

and

W p2−13

≡ 2αp2−1(1 +R

p−13 +R

2(p−1)3 ) ≡ 2(1 +R

p−13 +R

2(p−1)3 ) (mod p).

Hence

W p2−13

− 6Rp2−1

3 ≡ W p2−13

− 6R2(p−1)

3 ≡ 2(

1−R2(p−1)

3

)2

(mod p).

Thus p | W p2−13

− 6Rp2−1

3 ⇔ Rp−13 ≡ 1 (mod p).

A companion result to Theorem 5.15 for p ≡ −1 (mod 3) is given below.

Theorem 5.16. If p is a Q prime and p ≡ −1 (mod 3), then

p | D p2−13

if and only if p | C p+13.

Proof. If p | C p+13

, then

αp+13 = β

p+13 , β

p+13 = γ

p+13 or α

p+13 = γ

p+13 .

In the first case, since αp+13 ∈ Fp, we get β

p2−13 = 1 = α

p2−13 . In the second case,

βp2−1

3 = (βp−1)p+13 = (γ/β)

p+13 = 1 = (β/γ)

p+13 = (γp−1)

p+13 = γ

p2−13 .

In the third case, since αp+13 ∈ Fp, we get γ

p2−13 = 1 = α

p2−13 . Since (αβγ)

p2−13 = 1,

we get αp2−1

3 = βp2−1

3 = γp2−1

3 in all of the three cases. It follows that p | C p2−13

and

p | W p2−13

− 6Rp2−1

3 .

Conversely, if p | D p2−13

, then

1 = αp2−1

3 = βp2−1

3 , βp2−1

3 = γp2−1

3 or 1 = αp2−1

3 = γp2−1

3 .

152

In the first and the last of these cases we get

(βp−1)p+13 = (γ/β)

p+13 = 1 or (γp−1)

p+13 = (β/γ)

p+13 = 1.

In either case p | C p+13

. In the remaining case we get

(γ/β)p+13 = (β/γ)

p+13 or β

2(p+1)3 = γ

2(p+1)3 .

If βp+13 = γ

p+13 , we have p | C p+1

3. If β

p+13 = −γ p+1

3 , then

βp+1 = −γp+1 ⇒ βγ = −γβ,

which is impossible.

There remains the problem of dealing with S primes. If p is an S prime and p ≡ 1

(mod 3) the determination of when p | D p−13

is provided in the following result.

Theorem 5.17. Let p be an S prime and p ≡ 1 (mod 3). Then p | D p−13

if and

only if p | C p−13

and Rp−13 ≡ 1 (mod p).

Proof. Since p is an S prime, we have zeros α, β, γ of f(x) in Fp such that

αp−13 = ζ i, β

p−13 = ζj, γ

p−13 = ζk,

where ζ2 + ζ + 1 = 0. If p | D p−13

, then p | C p−13

and two of i, j, k are the same

modulo 3. Suppose without loss of generality that i ≡ j (mod 3). If k 6≡ i (mod 3),

then

W p−13≡ 0 6≡ 6R

p−13 (mod p),

which is impossible. Thus, we must have i ≡ j ≡ k (mod 3) and Rp−13 ≡ 1 (mod p).

If Rp−13 ≡ 1 (mod p) and p | C p−1

3, then 3 | i + j + k and two of i, j, k are

the same modulo 3. Hence, all three of them must be the same modulo 3 and

p | W p−13− 6R

p−13 .

153

In the following corollary we will use the sequence An as defined in (3.1).

Corollary 5.17.1. If p is an S prime, p ≡ 1 (mod 3) and Rp−13 ≡ 1 (mod p), then

p | D p−13

if and only if p - A p−13

.

Proof. Since Rp−13 ≡ 1 (mod p), we know that i, j, k in the proof of the theorem are

either all the same or all distinct modulo 3. If they are all the same, then p - A p−13

and p | C p−13

, hence p | D p−13

by the above theorem. If they are all distinct, then

p | A p−13

and p - C p−13

, so p - D p−13

.

Chapter 6

Arithmetic Properties of {En}

6.1 Preliminary Results for {En}

While working on the sequences {Wn} and {Cn}, several results developed concern-

ing the sequence {En}, where En = gcd(Wn, Cn). This sequence has a number of

properties analogous to those of the Lucas sequence {Vn}. In the next several sections

we will develop these properties. We begin with a result analogous to (Un, Vn) | 2

for Lucas functions.

Theorem 6.1. If (Q,R) = 1, then (Dn, En) | 6.

Proof. Suppose p is any prime such that p | Dn and p | En. Since p | Wn−6Rn, we

must have p | 6Rn. Since (Dn, R) = (En, R) and (Dn, R) | 2 by (5.1), we can only

have p = 2, 3. If 32 | (Dn, En), then 3 | R, which is impossible. If 22 | (Dn, En), then

22 | En, which we have seen by Theorem 4.5 is also impossible. Hence (Dn, En) | 6.

It is readily apparent that equation (2.12) implies Vn | U2n. Similarly we have

En | D3n, (6.1)

which can be seen as follows. We can rework the identity

4W3n = 3∆C2n(Wn + 2Rn) +W 2

n(Wn − 6Rn) + 24R3n

154

155

to see that

W3n − 6R3n = (Wn − 6Rn)

(W 2n −∆C2

n

4

)+ ∆WnC

2n. (6.2)

Recall again from Theorem 4.5 that if (Q,R) = 1, then 2α || (Wn, Cn)⇒ α = 0 or 1

and if α = 1, then Qn = W 2n−∆C2

n

4is odd.

We are now ready to show En | D3n. Clearly we have Cn | C3n. If 2 - En,

then En | Qn ⇒ En | W3n − 6R3n by equation (6.2). If 2 | En, then En/2 is

odd and En/2 | Qn. Since 2 | Wn, then 2 | Wn − 6Rn ⇒ En | (Wn − 6Rn)Qn ⇒

En | W3n − 6R3n by equation (6.2). Since En | Cn and Cn | C3n, we get En | C3n

and En | W3n − 6R3n ⇒ En | D3n.

We next derive some useful results concerning the primes which can divide En.

Theorem 6.2. If (Q,R) = 1 and p > 3 is a prime dividing En, then p ≡ 1 (mod 3).

Proof. First note p | Wn ⇒ AnBn ≡ 3Rn (mod p). Also remember p - R as

(Wn, Cn, R) | 2. Since p | (Wn, Cn) and W 2n−∆C2

n

4∈ Z we have p | W

2n−∆C2

n

4. Replac-

ing W 2n by (AnBn− 3Rn)2 and ∆C2

n by A2nB

2n + 18AnBnR

n− 4B3n− 4A3

nRn− 27R2n

yields

p | 3B3n + A4

nBn − 3A2nB

2n ⇒ p | Bn(A4

n − 3A2nBn + 3B2

n).

Notice p - Bn; for p | Bn ⇒ p | R and this is not possible by Lemma 4.1. Thus,

p | A4n − 3A2

nBn + 3B2n ⇒ p | 4A4

n − 12A2nBn + 12B2

n ⇒ p | (2A2n − 3Bn)2 + 3B2

n.

But this implies (2A2n − 3Bn)2 ≡ −3B2

n (mod p)⇒ (−3B2n

p) = 1⇒ (−3

p) = 1.

156

Thus, if p is a prime such that p > 3, p ≡ −1 (mod 3) and p | D3n, we know

that p - En. However, as shown in the next theorem we can say that if p | D3n and

p | Cn, then p | Dn or p | En.

Theorem 6.3. Let p be a prime such that p > 3. If p | D3n and p | Cn, then p | Dn

or p | En.

Proof. From Corollary 3.10.1, we see that

4(W3n − 6R3n) = 3∆C2nWn + 6∆C2

nRn +W 3

n − 6W 2nR

n.

Thus, if p | Cn and p | D3n, then p | W 2n(Wn − 6Rn). If p - En, then p - Wn. It

follows, then, that p | Wn − 6Rn and, therefore, p | Dn.

Corollary 6.3.1. Let p be a prime such that p > 3 and p ≡ −1 (mod 3). If p | D3n,

then

p | Dn ⇔ p | Cn.

Proof. Since p ≡ −1 (mod 3), we cannot have p | En by Theorem 6.2. Thus, if

p | Cn, then p | Dn by Theorem 6.3.

Theorem 6.4. Let p be a prime such that p > 3 and p ≡ −1 (mod 3). If p | D3n,

then

p - Dn ⇔ p | C3n/Cn.

Proof. First we will show (⇐). Assume that p | C3n/Cn and p | Dn. Since p | Dn,

we have p | Cn. By Corollary 3.10.1, we have

4C3n/Cn = ∆C2n + 3W 2

n ;

157

thus, since p | Cn and p | C3n/Cn, we have p | Wn. Hence p | En and p ≡ −1

(mod 3), which is a contradiction to Theorem 6.2.

Now suppose that p | D3n and p - Dn. By Corollary 6.3.1, we cannot have p | Cn;

hence, if p | C3n, then p | C3n/Cn.

We also have a result which tells us when p | D3n and p - Cn.

Theorem 6.5. Suppose p is a prime such that p - 6∆. Then p | D3n and p - Cn if

and only if

Wn ≡ −3Rn, ∆C2n ≡ −27R2n (mod p).

Proof. Suppose Wn ≡ −3Rn, ∆C2n ≡ −27R2n (mod p). Since

4C3n/Cn = ∆C2n + 3W 2

n ≡ 0 (mod p),

we get p | C3n/Cn. If p | R, then p | (Cn,Wn), which is impossible because

(Wn, Cn, R) | 2

by Lemma 4.6 and p 6= 2. Thus, p - 3R and hence p - Cn. Also, since

4(W3n − 6R3n) = 3∆C2nWn + 6∆C2

nRn +W 2

n(Wn − 6Rn),

we find that p | W3n − 6R3n and hence p | D3n.

Now suppose that p | D3n and p - Cn. Since p | C3n, we get

p | Cn(∆C2n + 3W 2

n)

and it follows that

∆C2n ≡ −3W 2

n (mod p).

158

Since p - Cn and p - ∆, we cannot have p | Wn. Also, p | W3n − 6R3n implies

p | 3∆C2nWn + 6∆C2

nRn +W 2

n(Wn − 6Rn)

and therefore p | 8W 2n(Wn + 3Rn). Thus, we must have

Wn ≡ −3Rn, and ∆C2n ≡ −27R2n (mod p).

We next eliminate the possibility that an I prime could divide En.

Theorem 6.6. If p is an I prime, then p - En.

Proof. Consider the zeros, α, β, γ, of f(x) in Fp3 . We have αp = β, βp = γ, γp = α.

If p | Cn, then without loss of generality αn = βn in Fp3 . So then αpn = βpn ⇒ βn =

γn ⇒ αn = βn = γn. Hence Wn = 6α3n = 6Rn in Fp3 . It follows that Wn ≡ 6Rn

(mod p). Since (Wn, Cn, R) | 2 by Lemma 4.6, we must have p - En.

Theorem 6.2 can now be generalized.

Theorem 6.7. If p is a prime, p > 3 and p | En, then p ≡ 1 (mod 3ν+1), where

3ν || n.

Proof. Since (En, R) | 2, we must have p - R. Suppose p | En. Then we know that

p cannot be an I prime by Theorem 6.6 and p ≡ 1 (mod 3) by Theorem 6.2. We

also have p | D3n. If p | Dn, then p | Wn and p | Wn− 6Rn ⇒ p | 6Rn, which is a

contradiction. Hence p - Dn. Let ω be the rank of apparition of p in {Dn}; we have

ω(p) | 3n, ω(p) - n, as p | D3n and p - Dn. So if 3ν || n, then 3ν+1 | ω(p). Also,

159

since p is not an I prime and p - 6R, we have ω(p) | p or ω(p) | p2 − 1, by results

seen in Theorem 5.6 for the S and Q prime cases. Hence,

ω(p) | (p− 1)(p+ 1)⇒ 3ν+1 | (p− 1)(p+ 1).

Since 3 | ω(p), we know ω(p) - p. Also, since 3 | p− 1, we must have

3ν+1 | p− 1⇒ p ≡ 1 (mod 3ν+1).

Lucas showed (Theorem 2.20) that if p is an odd prime such that p | Vn, then p ≡

±1 (mod 2ν+1), where 2ν || n. We next produce an analogue of this result. Recall

that Vn = U2n/Un. We will consider those primes p 6= 2, 3 such that p | D3n/Dn.

Theorem 6.8. If p is a prime, p > 3 and p | D3n/Dn, then p ≡ ±1 (mod 3ν+1),

where 3ν || n.

Proof. Since p | D3n, we see that if p | R, then p = 2 by (5.1), which is not possible.

Thus, p - R. Also, since (D3n/Dn, Dn) | 27 by Theorem 5.13, we cannot have p | Dn.

It follows by the same reasoning used in the proof of Theorem 6.7, that

3ν+1 | (p− 1)(p+ 1).

Hence p ≡ ±1 (mod 3ν+1).

We next produce some conditions on those primes p such that p - En and p is not

an I prime.

Theorem 6.9. If p (≡ 1 (mod 3)) is a Q prime or an S prime, then p - En if 3 - p−13

and Rp−13 ≡ 1 (mod p).

160

Proof. Let K be the splitting field of f(x) ∈ Fp[x] and let α(∈ Fp), β, γ, be the

zeros of f(x) in K. If p | Cn, then either αn = βn or βn = γn or αn = γn in K.

Without loss of generality assume βn = γn; then Wn = 2α2nβn + 2αnβ2n + 2β3n =

2βn(β2n + αnβn + α2n) in K. Suppose p | En. Since p - R we have βn 6= 0; thus,

since p | Wn, there exists ζ ∈ K such that ζ2 + ζ + 1 = 0 and βn = ζαn. We can

then see that

Rn = αnβnγn = ζ2α3n ⇒ Rn( p−13

) = ζ2( p−13

).

Thus, if 3 - p−13

, we cannot have Rp−13 = 1 in Fp. Hence, if R

p−13 ≡ 1 (mod p) and

3 - p−13

, then p - En.

Notice also that if 3 | n and 3 - p−13

, then p - En. Further, if 3 - n, 3 | p−13

and

Rp−13 6≡ 1 (mod p), then p - En.

We will next derive a law of repetition for {En}.

Theorem 6.10. If r | En and (r, 2) = 1, then

Wmn ≡

mWnR

(m−1)n (mod r2) if m ≡ 1 (mod 3)

−mWnR(m−1)n (mod r2) if m ≡ −1 (mod 3)

6Rmn (mod r2) if m ≡ 0 (mod 3)

and

CmnCn≡

mR(m−1)n (mod r) if m ≡ 1,−1 (mod 3)

0 (mod r) if m ≡ 0 (mod 3).

Proof. Our assumptions directly imply r | Wn and r2 | W2n−∆C2

n

4. As before, let Pn =

Wn and Qn = W 2n−∆C2

n

4. Then it can be shown by induction that rk−1 | Uk(Pn, Qn)

and rk | Vk(Pn, Qn).

161

Now use equations (2.7) and (2.8) as written below

2Qλ2n Uλ1−λ2 = Uλ1Vλ2 − Vλ1Uλ2 and 2Qλ2

n Vλ1−λ2 = Vλ1Vλ2 − ∆nUλ1Uλ2

to see that

rλ1+λ2−1 | Qλ2n Uλ1−λ2(Pn, Qn) and rλ1+λ2 | Qλ2

n Vλ1−λ2(Pn, Qn).

We now consider Wmn and Cmn/Cn under the three conditions for m (mod 3).

If λ1 + λ2 ≥ 2, then certainly

r | Qλ2n Uλ1−λ2(Pn, Qn) and r2 | Qλ2

n Vλ1−λ2(Pn, Qn).

The remaining cases are:

1. if λ1 = 1, λ2 = 0, then m ≡ 1 (mod 3),

2. if λ1 = 0, λ2 = 1, then m ≡ −1 (mod 3),

3. if λ1 = 0, λ2 = 0, then m ≡ 0 (mod 3).

Case 1. m ≡ 1 (mod 3). By use of Theorem 3.14 we have

Wmn ≡ (−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!Qλ2n Vλ1−λ2(Pn, Qn) (mod r2).

This can be further simplified using the facts λ1 = 1, λ2 = 0 and λ0 +λ1 +λ2 +λ3 =

m⇒ λ0 +λ3 = m−1⇒ m−1−λ0 = λ3. Further, we can see λ1 +2λ2 +3λ3 = m⇒

m = 3λ3 − 1 ⇒ m ≡ λ3 − 1 (mod 2). Since λ0 = m − 1 − λ3 ⇒ λ0 ≡ m − 1 − λ3

(mod 2) and m ≡ λ3 − 1 (mod 2)⇒ 2 | λ0. So

Wmn ≡ mWnR(m−1)n (mod r2).

162

Similarly,

CmnCn≡ mR(m−1)n (mod r).

Case 2. m ≡ −1 (mod 3).

Now since λ1 = 0, λ2 = 1 and λ0+λ1+λ2+λ3 = m⇒ λ0+λ3 = m−1⇒ m−1−λ0 =

λ3. Also λ0 = m − 1 − λ3 ⇒ λ0 ≡ m− 1 − λ3 (mod 2) but λ1 + 2λ2 + 3λ3 = m ⇒

2 + 3λ3 = m⇒ λ3 ≡ m (mod 2)⇒ 2 - λ0. So

Wmn ≡ (−1)mR(m−1)nQnV−1(Pn, Qn) (mod r2)

≡ −mR(m−1)nV1(Pn, Qn) (mod r2)

≡ −mR(m−1)nWn (mod r2).

Similarly,

CmnCn≡ mR(m−1)n (mod r).

Case 3. m ≡ 0 (mod 3).

Here λ1 = λ2 = 0⇒ λ0 +λ3 = m and 3λ3 = m. So λ3 ≡ m (mod 2) and λ0 ≡ m−λ3

(mod 2) yields λ0 ≡ 0 (mod 2). Hence

Wmn ≡ (−1)λ0m(m− λ0 − 1)!

λ1!λ2!λ3!R(λ0+λ3)nQλ2

n Vλ1−λ2(Pn, Qn) (mod r2)

≡ (−1)0 3λ3(λ3 − 1)!

λ3!R(m)nQ0

nV0(Pn, Qn) (mod r2)

≡ 6Rmn (mod r2).

Similarly, since U0 = 0

CmnCn≡ 0 (mod r).

163

Corollary 6.10.1. If p is any prime such that p > 3 and p | En, then p - Emn when

3 | m.

Corollary 6.10.2. If p is any odd prime and pµ || En (µ ≥ 1), then pµ || Emn when

p - m and 3 - m.

The case of p = 3 is contained in the next two corollaries.

Corollary 6.10.3. If 3µ || En, then 3 || Emn when 3 | m.

Corollary 6.10.4. If 3µ || En, then 3µ || Emn when 3 - m.

Of course, when p = 2, we know that 2µ || En ⇒ µ = 0 or 1. From this,

Corollary 3.10.1 and Theorem 4.5, we see that if 2 || En, then 2 || E2n. Also, since

4 | W 2n −∆C2

n, we see that 2 | En ⇔ 2 | Cn.

Theorem 6.11. If p is a prime such that p > 3 and pµ || En, then pµ+1 || Epn for

µ ≥ 1.

Proof. We note that p | p(p−λ0−1)!λ1!λ2!λ3!

unless λ1 = p, λ0 = λ2 = λ3 = 0.

If λ1 = p, then

Qλ2n Uλ1−λ2(Pn, Qn) = Up(Pn, Qn) and Qλ2

n Vλ1−λ2(Pn, Qn) = Vp(Pn, Qn).

Also, by the previous theorem, (pµ)p | Vp(Pn, Qn) and (pµ)p−1 | Up(Pn, Qn). Now

since p > 3 we know that 2µ+ 1 < µ(p− 1) and thus

Up(Pn, Qn) ≡ Vp(Pn, Qn) ≡ 0 (mod p2µ−1).

Furthermore, p2µ | Qλ2n Vλ1−λ2 for λ1 + λ2 ≥ 2⇒ p2µ+1 | p(p−λ0−1)!

λ1!λ2!λ3!Qλ2n Vλ1−λ2 .

164

Similarly, pµ | Qλ2n Uλ1−λ2(Pn, Qn) for λ1 + λ2 ≥ 2⇒ pµ+1 | p(p−λ0−1)!

λ1!λ2!λ3!Qλ2n Uλ1−λ2 .

Thus we need only concern ourselves with the case λ1 + λ2 = 1, which yields

Wpn ≡ ±pR(p−1)nWn (mod p2µ+1) andCpnCn≡ pR(p−1)n (mod pµ+1).

We may then conclude that if pµ || En, then pµ+1 || Epn.

Also, notice that if p > 3, p - En and p - ∆, then p - Epn. For if p - Cn, then

∆n = ∆C2n ⇒ p - ∆n ⇒ Up(Pn, Qn) 6≡ 0 (mod p). But Cpn

Cn≡ Up(Pn, Qn) (mod p),

so p - Cpn. Also, since if p | Cn, then p | Wn ⇔ p | Wpn, we see that p - Epn when

p - Wn.

6.2 A Law of Apparition for {En}

It will be seen here that {En} behaves in much the same way as {Vn}. By employing

{En}, we will be able to extend more of the results for {Vn} from Chapter 2 that

were, until now, missing. We must first deal with the case of p = 2. Since 2 | En ⇔

2 | Cn, from our results in Section 4.1, there always exists some minimal ρ such

that 2 | Eρ and 2 | En if and only if ρ | n. We next consider the case of a general

modulus. We note by the first two identities in Corollary 3.10.1 that Cn | C2n and

W2n ≡ (∆C2n +W 2

n)/2 (mod Wn). Since En is either odd or 2 || En, it is easy to see

that En | E2n.

Lemma 6.12. If r | Em, r | En and (r, 2) = 1, then r | E3n+m.

165

Proof. We interchange m and n in Corollary 3.10.2 to obtain

2W3n+m = WnW2n+m + ∆CnC2n+m −RnWnWn+m +Rn∆CnCn+m + 2R3nWm

2C3n+m = WnC2n+m + CnW2n+m −RnWnCn+m +RnCnWn+m − 2R3nCm.

The proof follows immediately.

We will say r | En when n < 0 if r | E|n|.

Lemma 6.13. If r | Em and r | En, then r | E3n+m.

Proof. Note that if 2 | r, then 2 - r2

as 4 - (Wn, Cn) as seen in Theorem 4.5. By

the previous lemma we have r2| E3n+m. Now since W 2

n−∆C2n

4∈ Z we have 2 | En ⇔

2 | Cn. So we need only show 2 | Cm and 2 | Cn implies 2 | C3n+m. There exists

some minimal ρ such that 2 | Cρ. Moreover, if 2 | Cn, then ρ | n. Thus n = k1ρ

and m = k2ρ. Hence 2 | C(3k1+k2)ρ ⇒ 2 | C3n+m, which concludes the proof.

Theorem 6.14. If r | Em and r | En, then r | E3kn+m for k ≥ 1.

Proof. Proceed by induction using the previous lemma. Clearly it is true for k = 1.

Assume this is true for r | E3kn+m. We also have r | En. So by the previous lemma,

if we replace m by 3kn+m, we have r | E3n+3kn+m ⇒ r | E3(k+1)n+m.

Corollary 6.14.1. If 3 - m, then En | Emn.

Proof. Since 3 - m, we must have m = 3k + 1 or m = 3k + 2. Since En | E2n, we

have En | E3kn+n and En | E3kn+2n. Hence En | Emn, when 3 - m.

We are now able to provide an analogue to Theorem 2.14.

166

Theorem 6.15. Suppose r | En, (n > 0), then there must be a least positive ρ = ρ(r)

such that r | Eρ. Further, ρ | n.

Proof. Clearly, since r | En, such a value for ρ must exist. Let 3µ || (n, ρ).

Case 1: 3µ || n

In this case put d1 = (3ρ, n)⇒ 3µ || d1. Since

d1 = 3ρx+ ny where x, y ∈ Z

we get

d1

3µ=

3ρx

3µ+ny

3µ.

Thus 3 - y as 3 - d13µ

and 3 | 3ρ3µ

.

Let 3k || x. Since r | E xρ

3kand r | Eyn by Corollary 6.10.2 we see from Theorem

6.14 that

r | E3ρx+yn ⇒ r | Ed1 .

We know d1 ≥ ρ by minimality of ρ. But d1 | 3ρ ⇒ d13µ| 3ρ

3µ. Since 3 - d1

3µwe have

d1 | ρ⇒ ρ = d1. Since d1 | n, we have ρ | n.

Case 2: 3µ || ρ

In this case put d2 = (ρ, 3n)⇒ 3µ || d2. Hence

d2 = 3nz + ρw where z, w ∈ Z.

Thus

d2

3µ=

3z

3µ+ρw

3µ⇒ 3 - w.

Reasoning as before with 3k || z, we get

r | E3nz+ρw ⇒ r | Ed2 ⇒ d2 ≥ ρ.

167

Since d2 | ρ⇒ d2 = ρ. Also ρ | 3n⇒ ρ3µ| n

3µ⇒ ρ | n.

6.3 Further Observations on the Law of Apparition for {En}

We next examine the problem of determining some m such that p | Em when p is

a prime and (p, 6R∆) = 1. Although it may not seem obvious, the next theorem is

part of our extension of Euler’s criterion, this will become more apparent later in

the chapter as several theorems concerning {En} and {Dn} lend themselves to the

generalization of this result. We note that if p is a Q prime, then Wp2−1 ≡ 6 (mod p)

by (4.15). Hence p - Ep2−1. However, it is possible that p | E p2−13

. We also recall

from Theorem 6.2 that if p | En for any n, then p ≡ 1 (mod 3).

Theorem 6.16. If p (≡ 1 (mod 3)) is a Q prime, then

p | E p2−13

⇔ Rp−13 6≡ 1 (mod p).

Proof. We have α, β, γ as the zeros of f(x) in Fp2 such that

αp = α, βp = γ and γp = β.

Hence,

αp−1 = 1, αp+1 = α2, βp+1 = βγ and γp+1 = βγ.

Now βγ = R/α and

βp2−1

3 = γp2−1

3 = (βγ)p−13 =

(R

α

) p−13

=R

p−13

αp−13

· α2(p−1)

3

α2(p−1)

3

= Rp−13 α

2(p−1)3

= Rp−13 (α2)

p−13 = R

p−13 (αp+1)

p−13 = R

p−13 α

p2−13 .

168

It then follows that p | C p2−13

and

W p2−13

≡ 2αp2−1(1 +R

p−13 +R

2(p−1)3 ) ≡ 2(1 +R

p−13 +R

2(p−1)3 ) (mod p).

We can then see

Rp−13 6≡ 1 (mod p)⇔ 1 +R

p−13 +R2 p−1

3 ≡ 0 (mod p)⇔ p | W p2−13

.

We next consider the possibility that p | Ep+1 when p is a Q prime. In this

case p 6= 3 and p ≡ 1 (mod 3) by Theorem 6.2. We already know that p | Cp+1

by Corollary 4.24.1. Let α, β, γ be the zeros of f(x) in Fp2 , then αp = α, βp = γ,

γp = β. Then we have αp+1 = α2, βp+1 = γβ, γp+1 = γβ. So

Wp+1 = (αp+1β2p+2 + βp+1γ2p+2 + γp+1α2p+2) + (α2p+2βp+1 + β2p+2γp+1 + γ2p+2αp+1)

= (α2β2γ2 + β3γ3 + α4βγ) + (α4βγ + β3γ3 + β2γ2α2)

= 2(α4βγ + α2β2γ2 + β3γ3).

Hence, we can see that

Wp+1 ≡ 0 (mod p)⇔ α4βγ + (αβγ)2 + (βγ)3 = 0⇔ α2 = ζβγ or α2 = ζ2βγ

where ζ2 + ζ + 1 = 0. Note p ≡ 1 (mod 3)⇒ ζ ∈ Fp.

Now consider

M1 = (α2 − ζβγ)(β2 − ζαγ)(γ2 − ζαβ)

M2 = (α2 − ζ2βγ)(β2 − ζ2αγ)(γ2 − ζ2αβ).

Clearly then M1M2 = 0, when Wp+1 = 0, as Wp+1 = 0 if and only if α2 = ζβγ

or α2 = ζ2βγ. Also if M1M2 = 0, then Wp+1 = 0. This is obvious for the cases

169

α2−ζβγ = 0 and α2−ζ2βγ = 0, so suppose β2−ζαβ = 0⇒ β2(p+1) = ζp+1(αβ)p+1 ⇒

(βγ)2 = ζp+1α2βγ ⇒ βγ = ζp+1α2 ⇒ α2 = ζ−(p+1)βγ ⇒ α2 = ζβγ. Similarly for

any of the other 3 factors of M1M2.

Now

M1 = (α2 − ζβγ)(β2 − ζαγ)(γ2 − ζαβ)

= α2β2γ2 − ζα3γ3 − ζβ3γ3 − ζα3β3 + ζ2αβγ4 + ζ2α4βγ + ζ2αβ4γ − ζ3ζ2α2β2γ2

= ζ2(αβγ)(α3 + β3 + γ3)− ζ((αβ)3 + (βγ)3 + (γα)3)

= ζ2RA3 − ζB3.

Also M2 = ζRA3 − ζ2B3. So

M1M2 = (ζ2RA3 − ζB3)(ζRA3 − ζ2B3)

= ζ3R2A23 − ζ2RA3B3 − ζ4RA3B3 + ζ3B2

3

= R2A23 − ζ2RA3B3 − ζRA3B3 +B2

3

= (RA3)2 +RA3B3 +B23 .

Hence, p | Wp+1 ⇔ p | (RA3)2 + RA3B3 + B23 . Thus, if p is a Q prime, then

p | Ep+1 ⇔ p | M1M2.

Remember A3 = P 3−3PQ+3R and B3 = Q3−3PQR+3R2. Since only a finite

number of primes can divide M1M2, there can only be a finite number of Q primes

p such that p | Wp+1, which means that there can only be a finite number of primes

p such that p | Wr, where r is the rank of apparition of p for {Cn}. For if p | Wr,

then p | Ep+1 because r | p+ 1 and 3 - p+ 1, by Corollary 6.10.2.

We now produce a result similar to Theorem 6.16 for S primes. Let α, β, γ be

the zeros of f(x) in K, where p ≡ 1 (mod 3) is an S prime. Suppose that there exists

170

ρ = ρ(p) such that p | Eρ, ρ > 1 and ρ is minimal. Then ρ = kr1 where r1 is some

rank of apparition of p in {Cn}. Suppose further that r1 | 3n and

(α/β)n = ζ i, (β/γ)n = ζj, where ζ2 + ζ + 1 = 0 in K.

This is certainly the case when n = (p−1)/3. Note that αn = ζ iβn and βn = ζjγn ⇒

αn = ζ i+jγn. Also since (ζ − 1)(ζ2 + ζ + 1) = 0⇒ ζ3 = 1, we have α3n = β3n = γ3n.

Thus W3n = 6R3n (mod p).

We distinguish 3 cases:

Case 1: 3 - ij(i+ j)

Since αn = ζ iβn, βn = ζjγn, αn = ζ i+jγn and 3 - ij(i+ j) we see that αn, βn, γn

are pairwise distinct. Thus p - Cn ⇒ r1 - n ⇒ 3 - 3nr1

. Since p | Cρ and p | Wρ, we

must have p | W 3nr1ρ by Corollary 6.10.2 and hence p | W3nk. Since W3nk ≡ 6R3nk

(mod p), this is impossible.

Case 2: 3 divides only one of i, j, (i+ j)

In this case p | Cn, since without loss of generality αn = βn, so 3 | i and 3 - j.

Using αn = βn = ζjγn,

Wn = (α2nβn + β2nγn + γ2nαn) + (αnβ2n + βnγ2n + γnα2n)

= (ζ2jγ2nζjγn + ζ2jγ2nγn + γ2nζjγn) + (ζjγnζ2jγ2n + ζjγnγ2n + γnζ2jγ2n)

= 2γ3n(1 + ζj + ζ2j)

= 0 since 3 - j ⇒ 1 + ζj + ζ2j = 0.

So p | En. Thus ρ exists and ρ | n.

Case 3: 3 divides 2 of i, j, (i+ j)

171

Since 3 divides two of i, j, (i+ j), it divides all of them. Hence αn = βn = γn ⇒

p | Cn and Wn = 6Rn (mod p).

Subcase 1: r1 - n.

Since r1 | 3n, we know that 3 - 3nr1⇒ p | W3nk, which is impossible.

Subcase 2: r1 | n

If 3 - n, then 3 - nr1

and by Corollary 6.10.2 we have p | W nr1ρ ⇒ p | Wkn. This

is a contradiction, as Wkn ≡ 6Rkn (mod p). Thus, if ρ exists, then ρ = kr1, where

r1 | n and 3 | n and (α/β)n/3 = ζ i, (β/γ)n/3 = ζj. By repeating the above argument

we see that if ρ exists, then ρ | n/3.

The fact that whenever ρ exists, ρ | n implies in the case that n = (p − 1)/3,

that (α/β)(p−1)/3 = ζ i and (β/γ)(p−1)/3 = ζj and therefore ρ | p−13

. Also, Rp−13 =

(αβγ)p−13 = ζ i+2j, and 3 | ij(i2 − j2). So if 3 - ij(i + j), then 3 | (i − j). If 3

divides only one of i, j, i + j, then 3 - i − j. But if 3 divides all of i, j, i + j, then

3 | i− j. Thus ρ exists and ρ | p−13

if Rp−13 6≡ 1 (mod p). When R

p−13 ≡ 1 (mod p)

and 3 - p−13

, ρ does not exist. If Rp−13 ≡ 1 (mod p) and 3 | p−1

3, then, if ρ exists,

ρ | p−19

. We have proved the following theorem.

Theorem 6.17. If p is an S prime and p ≡ 1 (mod 3), then

p | E p−13⇔ R

p−13 6≡ 1 (mod p).

Also, if Rp−13 ≡ 1 (mod p) and p 6≡ 1 (mod 9), then ρ(p) cannot exist. If R

p−13 ≡ 1

(mod p) and p ≡ 1 (mod 9), then ρ | p−19

if ρ exists.

We also have the following result.

172

Theorem 6.18. If p is an S prime and p ≡ 1 (mod 3), then

p | E p2−13

⇔ p | E p−13.

Proof. Clearly, since 3 - p+1, p | E p2−13

when p | E p−13

. Next suppose that p | E p2−13

.

If α, β, γ are defined as above, we must have

αp2−1

3 = βp2−1

3 ⇒ (αp+1)p−13 = (βp+1)

p−13 ⇒ (α2)

p−13 = (β2)

p−13 .

So, we then have

(αp−13 )2 = (β

p−13 )2 ⇒ α

p−13 = ±β

p−13 .

If αp−13 = β

p−13 , then p | C p−1

3. If α

p−13 = −β p−1

3 , then

αp−1 = −βp−1 ⇒ 1 = −1,

which is impossible. Also, since p | W p2−13

, we get

2αp2−1

3 (α2(p2−1)

3 + γp2−1

3 αp2−1

3 + γ2(p2−1)

3 ) ≡ 0 (mod p).

Since, p - 2R, this means that

αp2−1

3 = ζγp2−1

3

for some ζ such that ζ2 + ζ + 1 = 0. It follows that

α2(p−1)

3 = ζγ2(p−1)

3 .

Since αp−1 = γp−1 = 1, we find by squaring that

αp−13 = ζ2γ

p−13 .

Since p | C p−13

, we get p | E p−13

.

173

By similar techniques we can also establish the next result.

Theorem 6.19. If p is an S prime and p ≡ 1 (mod 3), then

p | D p2−13

⇔ p | D p−13.

Let νp(x) denote that value of ν such that pν || x, where x ∈ Z>0 and p is a

prime.

Theorem 6.20. Let p be an S prime and suppose that ρ = ρ(p) exists. In this case

ρ = kr1, where r1 is a rank of apparition of p in {Cn}. If r2 is any other rank of

apparition of p in {Cn}, then ν3(r2) > ν3(r1).

Proof. Since p | Eρ, we must have (α/β)ρ = ζ i, (β/γ)ρ = ζj in Fp, where ζ2 +ζ+1 =

0. Here 3 divides only one of i, j, i + j. Without loss of generality suppose 3 | i,

then βρ = ζjγρ, αρ = ζjγρ and 3 - j.

If r2 is another rank of apparition of p, then it is either the order of β/γ or α/γ

in Fp. So r2 | 3ρ and r2 - ρ. Thus if k = ν3(ρ) then ν3(r2) = k + 1. Since r1 | ρ, we

must have ν3(r1) ≤ ν3(ρ) = k < ν3(r2).

To proceed any further we will need the following results for Lucas sequences.

Because they seem to be difficult to locate in the literature, proofs are provided here.

Theorem 6.21. Let Vn, Un be the Lucas functions Vn(P,Q), Un(P,Q) and let p be

a prime such that p > 3 and (Qp

) = 1. Let ω be the rank of apparition of p in {Un}.

Let S2 ≡ Q (mod p). Then there exists a minimal λ such that

Vλ(P,Q) ≡ −Sλ (mod p)

if and only if 3 | ω. Furthermore, λ = ω/3 or 2ω/3.

174

Proof. (⇒) Suppose λ exists. Then V 2λ ≡ Qλ (mod p) and using U3n = Un(V 2

n −Qn)

we see p | U3λ. Also p - Uλ, for if p | Uλ, then V 2λ ≡ 4Qλ (mod p), as V 2

n −∆U2n =

4Qn, which is not possible. Thus ω | 3λ and ω - λ, so 3 | ω.

Put µ = ω/3. Then µ | λ. Again, use U3µ = Uµ(V 2µ − Qµ), knowing U3µ ≡ 0

(mod p) and Uµ 6≡ 0 (mod p), so V 2µ ≡ Qµ (mod p). Hence

Vµ(P,Q) ≡ ±Sµ (mod p).

If Vµ ≡ −Sµ (mod p), then λ = µ by minimality of λ.

If Vµ ≡ Sµ (mod p), then V2µ ≡ −S2µ (mod p), as

V2µ = V 2µ − 2Qµ

≡ Qµ − 2Qµ

≡ −S2µ (mod p).

Thus λ = 2µ by the minimality of λ.

(⇐) If 3 | ω, we must have

Vµ ≡ ±Sµ (mod p)

where µ = ω/3 as before. Thus Vµ ≡ −Sµ (mod p) or Vµ ≡ Sµ (mod p). Thus,

there exists a minimal λ such that Vλ(P,Q) ≡ −Sλ (mod p) holds, and we have

already seen that λ = µ or λ = 2µ.

One may also notice that if (∆p

) = 1, where ∆ = P 2 − 4Q and p ≡ 1 (mod 3),

then p | Up−1 ⇒ ω | p − 1. Let 3µ || p − 1. If 3 | ω, then ω - p−13µ

. Also if ω - p−13µ

,

then 3 | ω. Thus λ exists if and only if p - U p−13µ

.

175

Theorem 6.22. If ω is the rank of apparition of p in {Un} and 6 | ω, then Vω3≡ Q

ω3

(mod p).

Proof. Put ω = 6k. We have U6k ≡ 0 (mod p) and using U2n = UnVn we get

U2(3k) = U3kV3k ≡ 0 (mod p). So V3k ≡ 0 (mod p) as U3k 6≡ 0 (mod p). Moreover,

Vω2≡ 0 (mod p) because p - Uω

2. Now, since V2n = V 2

n − 2Qn if we set n = ω2⇒

Vω = V 2ω2− 2Q

ω2 . So Vω ≡ −2Q

ω2 (mod p).

We also know that V3n = V 3n−3QnVn; by setting n = ω

3, we get Vω = V 3

ω3−3Q

ω3 Vω

3.

So

−2Qω2 ≡ V 3

ω3− 3Q

ω3 Vω

3(mod p)

−2Qω2

Qω2

≡V 3ω3

Qω2

−3Q

ω3 Vω

3

Qω2

(mod p)

−2 ≡(Vω

3

Qω6

)3

− 3

(Vω

3

Qω6

)(mod p).

On putting T =Vω

3

Qω6

, we get T 3−3T+2 ≡ 0 (mod p) or (T−1)2(T+2) ≡ 0 (mod p).

If T + 2 ≡ 0 ⇒Vω

3

Qω6≡ −2 ⇒ Vω

3≡ −2Q

ω6 ⇒ V 2

ω3≡ 4Q2ω

6 (mod p). Using

V 2n −∆U2

n = 4Qn with n = ω3⇒ V 2

ω3−∆U2

ω3

= 4Qω3 ⇒ Uω

3≡ 0 (mod p), which is a

contradiction. So T ≡ 1 (mod p)⇒ Vω3≡ Q

ω6 (mod p).

We next establish some results concerning the divisibility of Wmn by a prime p

such that p | Cn.

Theorem 6.23. Suppose p is a prime such that p - 6R∆ and p | Cn. Then p | Wmn

if and only if Vm(Wn

2−Rn, R2n) ≡ −Rmn (mod p).

Proof. Since p | Cn, in K we must have βn = γn ⇒ Wn = 2(α2nβn + β2nαn + β3n)

176

and Rn = αnβ2n. We may then notice

Wn − 2Rn = 2(α2nβn + β3n)

and β3n(α2nβn) = α2nβ4n = R2n. Further, since βmn = γmn, we have

Wmn − 2Rmn = 2((α2nβn)m + (β3n)m).

Note that Vm(Wn

2− Rn, R2n) = α′m + β′m, where α′ = α2nβn, β′ = β3n. It follows

that

Wmn − 2Rmn = 2Vm(Wn

2−Rn, R2n).

Thus p | Wmn if and only if Vm(Wn

2−Rn, R2n) ≡ −Rmn (mod p).

Corollary 6.23.1. Suppose p is a prime such that p - 6R∆, Wn ≡ 6Rn (mod p)

and Cn ≡ 0 (mod p), then Wmn ≡ 6Rmn (mod p).

Proof. We know that

Wmn ≡ 2Rmn + 2Vm(Wn − 2Rn

2, R2n) (mod p).

Now Wn−2Rn

2≡ 6Rn−2Rn

2= 2Rn (mod p) and Vm(2S, S2) = 2Sm. By letting S =

2Rmn we get

Wmn ≡ 2Rmn + 2(2Rmn) = 6Rmn (mod p).

Corollary 6.23.2. Suppose p is a prime such that p - 6R∆, Wn ≡ −2Rn (mod p)

and Cn ≡ 0 (mod p), then

Wmn ≡

−2Rmn (mod p) if 2 | m

6Rmn (mod p) if 2 - m.

177

Proof. Here Wn−2Rn

2≡ −2Rn (mod p). Also Vm(−2S, S2) = 2(−1)mSm. If S = Rmn

and 2 - m, then

Wmn ≡ 2Rmn + 2(−2Rmn) ≡ −2Rmn (mod p).

Leaving S the same, but with 2 | m, we have

Wmn ≡ 2Rmn + 2(2Rmn) ≡ 6Rmn (mod p).

Theorem 6.24. Let p be an S prime or a Q prime. If p | Cn and

∆′ =

(Wn

2−Rn

)2

− 4R2n,

then (∆′

p) = 1 or 0.

Proof. If p is an S prime, we have βn = γn in K (= Fp) since p | Cn. So ∆′ =

(Wn

2−Rn)2−4R2n = (α2nβn+β3n)2−4(αnβ2n)2 = α4nβ2n+2α2nβ4n+β6n−4α2nβ4n =

α4nβ2n − 2α2nβ4n + β6n = (α2nβn − β3n)2. Thus (∆′

p) = 1, 0.

If p is a Q prime, we must have β, γ ∈ Fp2 , α ∈ Fp.

If p | Cn and we have αn = βn ⇒ αn = βn = γn ⇒ Wn = 6Rn ⇒ ∆′ =

(2Rn)2 − 4R2n = 0.

If p | Cn and we have βn = γn, then ∆′ = (α2nβn−β3n)2. Now (α2nβn−β3n)p =

α2pnβpn − β3pn = α2nγn − γ3n = α2nβn − β3n. Thus α2nβn − β3n ∈ Fp and (∆′

p) = 1.

Theorem 6.25. If p is a prime such that p | Cn, p - 6R∆ and (∆′

p) = 0, there is no

value of m such that p | Wmn.

178

Proof. If (∆′

p) = 0, then p | ∆′. Also, p | ∆′ ⇒ p | (Wn

2− Rn)2 − 4R2n ⇒ (Wn

2−

Rn)2 ≡ 4R2n (mod p) ⇒ Wn

2− Rn ≡ ±2Rn (mod p) ⇒ Wn ≡ 6Rn or Wn ≡ −2Rn.

Thus, by Corollaries 6.23.1 and 6.23.2 we have p - Wmn for all m.

If p is a Q prime, we know that p | Cp+1. Suppose p | Eρ, ρ > 1, ρ minimal,

then we must have p ≡ 1 (mod 3) by Theorem 6.2 and r | ρ ⇒ ρ = kr, where r is

the rank of apparition of p in {Cn}. Also r | p + 1 and if k > 1, by the minimality

of ρ we must have p - Wr.

Put

P ′ ≡ Wr/2−Rr and Q′ ≡ R2r (mod p).

Then p | Wρ ⇔ Vk(P′, Q′) ≡ −Rkr (mod p) by Theorem 6.21. Let ω′ be the rank

of apparition of p in Un(P ′, Q′). Now Vk(P′, Q′) ≡ −Rkr ⇒ V 2

k (P ′, Q′) ≡ Q′k

(mod p) ⇒ p | U3k(P′, Q′), p - Uk(P ′, Q′). The least possible value for k is ω′/3

and ω′|p−12

, as (Q′

p) = 1. But if Vω′/3(P ′, Q′) ≡ Rkr (mod p), where k = ω′/3, then

V2ω′/3(P ′, Q′) ≡ −R2kr (mod p). Hence k = ω′/3 or 2ω′/3. In either event, k | p−13

.

Theorem 6.26. If p is a Q prime and ρ exists, then ρ = kr, where r | p + 1 and

k | p−13

.

As mentioned earlier, if ρ(p) exists, then ρ = kr, where r is some rank of ap-

parition of p in {Cn}. If p is an S prime, then kr | p−13⇒ k | p−1

3r. If we put

P ′ ≡ Wr

2− Rr, Q′ ≡ R2r (mod p), then k exists if and only if (∆′

p) = 1 and 3 | ω′,

where ω′ is the rank of apparition of p in Un(P ′, Q′). Furthermore, k = ω′/3 or

2ω′/3. In particular by Theorem 6.22 we know that k = 2ω′/3 if 2 | ω′.

179

The theorem below provides an analogue to both Theorems 2.16 and 2.17. Rather

than consider highest powers of 2 as in the Lucas case, we consider powers of 3.

Theorem 6.27. Suppose (r1, r2) = 1 and r1 | Em, r2 | En. If 3µ || m and 3µ || n

(µ ≥ 0), then

r1r2 | E[m,n].

If 3µ || m and 3ν || n and µ 6= ν, then r1r2 - Es for any s ∈ Z.

Proof. When 3µ || m and 3µ || n, we have 3µ || [m,n]. It follows that 3 - [n,m]n

and

3 - [n,m]m

. Thus r1 | E[m,n] and r2 | E[m,n] by Corollary 6.14.1. Since (r1, r2) = 1 we

get r1r2 | E[m,n].

Now suppose that 3µ || m and 3ν || n and µ 6= ν. Put ρ1 = ρ(r1) and ρ2 = ρ(r2).

We know that ρ1 | m and ρ2 | n. Also, 3 - mρ1

, 3 - nρ2⇒ 3µ || ρ1, 3ν || ρ2.

Suppose for some s ∈ Z we get r1r2 | Es. Then ρ1 | s, ρ2 | s and 3 - sρ1

,

3 - sρ2

. But then 3µ || s and 3ν || s and hence µ = ν which is contrary to our

assumptions.

Put ρ = ρ(3), 3µ || Eρ. If ρ - n then 3 - En. If ρ | n and 3 | nρ, then 3 || En by

Corollary 6.10.3. If ρ | n and 3 - nρ, then 3µ || En by Corollary 6.10.4.

The next theorems parallel Theorems 2.18 and 2.19.

Theorem 6.28. Let 3µ || m, 3ν || n. If µ = ν, then

(Em, En) = E(m,n).

Proof. Let (Wm, Cm,Wn, Cn) = 2λ3κd such that (d, 6) = 1. Since 3µ || m and

3µ || n, we have 3 - m(m,n)

, 3 - n(m,n)

. Hence E(m,n) | Em and E(m,n) | En; conse-

quently, E(m,n) | 2λ3κd.

180

We know that λ = 0 or 1 by Theorem 4.5. If λ = 0, then ρ(2) - m or ρ(2) - n⇒

ρ(2) - (m,n) ⇒ 2 - E(m,n). If λ = 1, then ρ(2) | m, ρ(2) | n ⇒ ρ(2) | (m,n) ⇒

2 | E(m,n). Thus E(m,n) = 2λ3κ′d′ where (d′, 6) = 1, and κ′ ≤ κ, d′ | d.

If κ = 0, then ρ(3) does not exist ⇒ 3 - E(m,n), or ρ(3) exists and ρ(3) - m or

ρ(3) - n⇒ ρ(3) - (m,n)⇒ κ′ = 0.

If κ = 1, then ρ = ρ(3) exists and ρ | m, ρ | n. Without loss of generality

assume 3 || En. Let 3δ || Eρ. If δ = 1, then 3 || 2λ3κ′d′ ⇒ κ′ = κ. If δ > 1, then

3 | mρ

, 3 | nρ⇒ 3 || 2λ3κ

′d′ ⇒ κ′ = κ.

If κ > 1, then ρ(3κ) | m, ρ(3κ) | n and 3 - mρ(3κ)

, 3 - nρ(3κ)

. Now ρ(3κ) | (m,n)

and 3 - (m,n)ρ(3κ)

⇒ 3κ | 3κ′ ⇒ κ′ ≥ κ > 1. Since κ′ ≤ κ, we get κ = κ′.

Now since ρ(d) | m and ρ(d) | n, we must have ρ(d) | (m,n). Also, since 3 - mρ(d)

and 3 - nρ(d)

, we get 3 - (m,n)ρ(d)

. Hence, d | E(m,n) and so d | d′. Thus d = d′.

Theorem 6.29. If 3µ || m, 3ν || n and ν 6= µ, then

(Em, En) | 6.

Proof. If we are given p | (Em, En) and (p, 6) = 1, then ρ(p) | m and ρ(p) | n. Since

3 - mρ(p)

and 3 - nρ(p)

, we get 3µ || ρ(p), 3ν || ρ(p), which is impossible. If 3λ | (Em, En)

and λ > 1, then ρ(3λ) | m, ρ(3λ) | n. Also, 3 - mρ(3λ)

, 3 - nρ(3λ)

⇒ 3µ || ρ(3λ) and

3ν || ρ(3λ). Again, this is impossible. Further, by Theorem 4.5, if 2λ || (Em, En),

then λ ∈ {0, 1}.

Thus, we also have a result comparable to Theorem 2.19.

We restate Euler’s criterion for Un, Vn as follows:

181

Theorem 6.30. If p - 2∆Q, then

p | UT (p)2

⇔ Qp−12 ≡ 1 (mod p)

p | VT (p)2

⇔ Qp−12 ≡ −1 (mod p),

where T (p) = p− 1 if p splits in Q(α) and T (p) = p+ 1 otherwise.

By using Theorems 6.2, 6.16, 6.18, 5.14, 5.15, and 5.17, we can prove our analogue

of this result.

Theorem 6.31. Suppose p is a prime such that p - ∆R and p ≡ 1 (mod 3). Let

T (p) = p2 − 1 if p splits in Q(α) and T (p) = p2 + p + 1 otherwise. If p is a Q or I

prime, then

p | DT (p)3

⇔ Rp−13 ≡ 1 (mod p).

If p is an S prime, then

p | DT (p)3

⇔ Rp−13 ≡ 1 (mod p) and p | C p−1

3.

Also, if T (p) = p2 − 1, then

p | ET (p)3

⇔ Rp−13 6≡ 1 (mod p).

If T (p) = p2 + p+ 1, then

p - ET (p)3

.

Chapter 7

Primality Testing

7.1 An Analogue of Lucas’ Fundamental Theorem

As we have seen in Chapter 1, one of Lucas’ main purposes in attempting to extend

his functions was to find new primality tests. In this chapter we will explore how

the Wn and Cn functions can be used for producing such tests. We will first develop

some analogues of Theorem 2.22 of Chapter 2. We begin with a simple lemma.

Lemma 7.1. If k ≥ 2 and ri ≥ 5 for i = 1, 2, . . . , k, then(k∏i=1

r2i

)− 1 > 2

k∏i=1

(r2i + ri + 1

2

).

Proof. We note that

1 >1

54+ 2

(7

10

)2

;

hence

1 >1

52k+ 2

(7

10

)kfor k ≥ 2.

Now

1

52k≥

k∏i=1

1

r2i

and7

5= 1 +

2

5> 1 +

1

ri+

1

r2i

182

183

imply that

1 >k∏i=1

1

r2i

+ 2k∏i=1

(1 + 1

ri+ 1

r2i

2

)

⇒

(k∏i=1

r2i

)>

(k∏i=1

r2i

)(k∏i=1

1

r2i

)+

(k∏i=1

r2i

)(2

k∏i=1

(1 + 1

ri+ 1

r2i

2

))

⇒

(k∏i=1

r2i

)− 1 > 2

k∏i=1

(r2i + ri + 1

2

).

Theorem 7.2. Let N be an integer such that (N, 6) = 1. If N | DN2−1, N - DN2−1q

for all primes q such that q | N2 − 1 and (DN2−1q′, N) = 1 for some prime divisor q′

of N2 − 1, then N is a prime.

Proof. Clearly ω(N) exists and ω(N) | N2−1. Also, if ω(N) 6= N2−1, then N2−1 =

kω(N) where k > 1. If q is any divisor of k, then ω(N) | N2−1q⇒ N | DN2−1

q

, which

is a contradiction. Hence ω(N) = N2 − 1. Let

N =k∏i=1

pαii ,

where the primes pi are all distinct and exceed 4. Also, since N | DN2−1, then by

equation (5.1) and the fact 2 - N , we must have (N,R) = 1. We know by Theorem

5.11 that

ω(N) = lcm[ω(pαii ); i = 1, 2, . . . , k].

Since (pi, ω(N)) = 1 and by Theorem 5.10 ω(pαii ) = pνi ω(pi), we must have

ω(N) | lcm[ω(pi); i = 1, 2, . . . , k].

184

Let p be any prime divisor of N . We have p | DN2−1 and p - DN2−1q′

. Hence

ω(p) | N2 − 1 and ω(p) - N2−1q′⇒ q′ | ω(p). Hence

lcm[ω(pi); i = 1, 2, . . . , k] | q′k∏i=1

ω(pi)

q′.

Now by Corollary 5.6.1, for k ≥ 2 we have,

q′k∏i=1

ω(pi)

q′≤ q′

k∏i=1

p2i + pi + 1

q′≤ 2

k∏i=1

p2i + pi + 1

2

we get (k∏i=1

p2i

)− 1 = N2 − 1 ≤ 2

k∏i=1

p2i + pi + 1

2,

which is impossible by the previous lemma.

If k = 1, then N = pα and by Theorem 5.10 ω(N) = ω(pα) = pνω(p)⇒ N2−1 =

ω(p), since (p,N2−1) = 1. If α ≥ 2, then p4−1 ≤ p2+p+1, which is a contradiction.

Thus N = p, a prime.

By similar methods used to prove the previous theorem we also have the following

result.

Theorem 7.3. Let N be an integer such that (N, 6) = 1. If N | DN2+N+1 and

N - DN2+N+1q

for each prime divisor q of N2 +N + 1 and (DN2+N+1q′

, N) = 1 for some

prime divisor q′ of N2 +N + 1, then N is a prime.

We have proved our analogue of Theorem 2.22.

Theorem 7.4. Let N be an integer such that (N, 6) = 1 and let T = T (N) = N2−1

or N2+N+1. If N | DT and N - DTq

for each prime divisor q of T and (D Tq′, N) = 1

for some prime divisor q′ of T , then N is a prime.

185

The difficulty in providing this as a complete analogue to Lucas’ result is the need

to involve the prime q′, which is not needed in Theorem 2.22. This is because 2 | N±

1 and 2 | pi± 1, and any proof of Theorem 2.22 makes use of these observations. In

what follows next, we will modify Theorems 7.2 and 7.3 to eliminate the need for q′

in certain cases.

Suppose p is a prime, p - 6∆ and 3 | T (p). By Theorem 6.5, we know that if

m = T (p)/3, then p - Cm if and only if

Wm ≡ −3Rm, ∆C2m ≡ −27R2m (mod p).

We also have a result for an arbitrary modulus.

Lemma 7.5. Let (N, 6) = 1. If ∆C2n ≡ −27R2n (mod N) and Wn ≡ −3Rn

(mod N), then N | D3n and N - Cn.

Proof. We have ∆C2n + 3W 2

n ≡ 0 (mod N). Since 4C3n = Cn(∆C2n + 3W 2

n) we get

N | C3n. Also, 4W3n = 3∆C2n(Wn + 2Rn) +W 2

n(Wn − 6Rn) + 24R3n implies

4(W3n − 6R3n) = 3∆C2n(Wn + 2Rn) +W 2

n(Wn − 6Rn)

≡ −9W 2n(Wn + 2Rn) +W 2

n(Wn − 6Rn) (mod N)

≡ −9W 3n − 18RnW 2

n +W 3n − 6RnW 2

n (mod N)

≡ −8W 3n − 24RnW 2

n (mod N)

≡ −8W 2n(Wn + 3Rn) ≡ 0 (mod N).

Thus N | W3n − 6R3n ⇒ N | D3n. Now since (Wn, Cn, R) | 2 by Lemma 4.6 and

(N, 6) = 1 we have (N,R) = 1⇒ N - Cn.

We can use the last result to prove the following theorem.

186

Theorem 7.6. If N is odd, 3 | T (N), ∆C2T (N)

3

≡ −27R2T (N)

3 (mod N), WT (N)3

≡

−3RT (N)

3 (mod N), and N - CT (N)q

for each prime divisor q of T (N)3

, then N is a

prime.

Proof. By the previous lemma, we know that N | DT (N), N - DT (N)q

for all prime

divisors of T (N). By our earlier reasoning we have ω(N) = T (N). Also, since

(T (N), N) = 1,

ω(N) = lcm[ω(pi); i = 1, 2, . . . , k] if N =k∏i=1

pαii .

Let p be any prime divisor of N . If p | R, then by the conditions of the theorem

p | WT (N)3

and p | ∆CT (N)3

. Since p - CT (N)3

by Lemma 7.5, we must have p | ∆.

However, by Corollary 4.2.1 we can only have p = 2, which is not possible because

N is odd. Thus, (N,R) = 1. Also, if p | N and p | ∆, then p | R and p | WT (N)3

,

which is also impossible. It follows that (N, 6∆R) = 1. Now since p | CT (N) and

p - CT (N)3

, we get

p | ∆C2T (N)

3

+ 3W 2T (N)

3

and we know that p - ∆CT (N)3

WT (N)3

. Thus (−3∆p

) = 1. If p is an I prime, then

ω(p) | p2 + p + 1, (∆p

) = 1 and (−3p

) = 1, which means that p ≡ 1 (mod 3) and

3 | p2 +p+1. If p is a Q prime, then ω(p) | p2−1 and 3 | p2−1. If p is an S prime,

then ω(p) | p− 1 and p− 1 < (p2 − 1)/3 < (p2 + p+ 1)/3.

Thus,

lcm[ω(pi); i = 1, 2, . . . , k] ≤ 3k∏i+1

p2i + pi + 1

3.

That N is a prime now follows from our previous reasoning.

187

Notice that 3 | T (N) when T (N) = N2 − 1 and 3 | T (N) when T (N) = N2 +

N + 1 and N ≡ 1 (mod 3).

A more general result than Theorem 7.6 and one that is more in line with Lucas’

precept that the primality of N be established by showing that N divides certain

integers is provided in Theorem 7.8 below. In order to demonstrate this result we

need a simple lemma.

Lemma 7.7. Suppose N is odd and let m be any positive integer such that (m,N) =

1. If N | Cmn/Cn, then (N,Dn) = 1.

Proof. Suppose p is any prime divisor of Dn and N . Since p | Cn and p | Wn−6Rn,

we see by our results in Section 4.4, in particular equation (5.4), we have that

Cmn/Cn ≡ m3Rn(m−1) (mod p).

It follows that since p - m and p - R ((Dn, R) | 2), we must have p - Cmn/Cn,

contradicting N | Cmn/Cn.

We are now able to produce an analogue of Corollary 2.23.1.

Theorem 7.8. Let N be an integer such that (N, 6) = 1. If N | DT (N) and

N | CT (N)/CT (N)q

for each prime divisor q of T (N), then N is a prime.

Proof. Since (T (N), N) = 1, we have (q,N) = 1. By Lemma 7.7 we know that if p is

any prime divisor of N , then (N,DT (N)q

) = 1. Thus, N - DT (N)q

for all prime divisors

q of T (N) and (N,DT (N)

q′) = 1 for some (any) prime divisor q′ of T (N). The result

follows by Theorem 7.4.

188

We note here that Lucas himself ([Luc78], pp. 310-311) made use of the divisi-

bility of U3n/Un to produce a primality test for N = A3n−1. Also, the computation

of CT/CTq

can be done by using the methods of Section 3.6.

Unfortunately, results like Theorems 7.4 and 7.8 are of limited utility in primality

testing because we need to know the complete factorization of T (N), and this is often

not available to us. In the next section, we will consider some special cases when

T (N) = N2 +N + 1.

7.2 The Case of T (N) = N 2 +N + 1

We will deal here with the case of N2 +N + 1 = tL, where L is a prime.

Theorem 7.9. Let N2 + N + 1 = tL, where L is a prime. If t < N −√N + 1,

(N,Dt) = 1 and N | DN2+N+1, then N is a prime.

Proof. If N is composite, there must be a prime p such that p | N and p ≤√N .

Also, since ω(p) | N2 + N + 1, we must have ω(p) | tL. Certainly ω(p) 6= 1 and

since p - Dt, ω(p) - t. It follows that since ω(p) | tL and ω(p) - t, we must have

L | ω(p). Now, ω(p) ≤ p2 + p+ 1 by Corollary 5.6.1; hence

p2 + p+ 1 ≥ L =N2 +N + 1

t

or

t(p2 + p+ 1) ≥ N2 +N + 1 = (N +√N + 1)(N −

√N + 1).

Since p ≤√N and N+

√N+1 ≥ p2 +p+1, we get t ≥ N−

√N+1, a contradiction.

189

Corollary 7.9.1. If N is odd, N ≡ 1 (mod 3), L = (N2 + N + 1)/3 is a prime,

(N,D3) = 1, and N | DN2+N+1, then N is a prime.

Proof. If N is composite, then N ≥ 52 and 3 < 52 − 5 + 1.

We also have the following result.

Theorem 7.10. Let N2 + N + 1 = tL, where L is a prime. If L > t2 + t + 1,

(N,Dt) = 1 and N | DN2+N+1, then N is a prime.

Proof. From our proof of Theorem 7.9 we know that if p is any prime divisor of N ,

then L | ω(p). Thus, N2 + N + 1 | tω(p). Hence, if N is composite, then N = pr

(r > 1) and

p2r2 + pr + 1 ≤ tω(p) ≤ t(p2 + p+ 1);

hence, r < t. Now let q be any prime divisor of r. Then since q | N we have L | ω(q)

and

t2 + t+ 1 < L ≤ q2 + q + 1.

It follows that r ≥ q > t, a contradiction.

Now, suppose S is a fixed positive integer and N = tS + u, where t = u2 + u+ 1

and u ∈ Z. Then N2 +N + 1 = tL, where

L = tS2 + (2u+ 1)S + 1.

For such numbers, we have the following result.

Theorem 7.11. If N = tS + u, where S ≥ 2 and N > 4, we have t < N −√N + 1.

190

Proof. Since

N −√N + 1 =

1

4(2√N − 1)2 +

3

4,

we see that N −√N + 1 is an increasing function of N for N > 0. Thus, since

N ≥ 2t+ u > 0, we get

N −√N + 1 ≥ 2t+ u−

√2t+ u+ 1.

If u = 0, −1, −2, then 1 ≤ t ≤ 3 and N −√N + 1 > 4 − 2 + 1 > t. If u 6= 0, −1,

−2, then (u+ 1)2 ≥ 4 and

(u+ 1)4 ≥ 4(u+ 1)2 > 2u2 + 3u+ 2 = 2t+ u.

Thus,

(u+ 1)2 >√

2t+ u and hence 2t+ u−√

2t+ u > t.

From Theorems 7.9 and 7.11 we see that if tS2 + (2u+ 1)S + 1 is a prime, then

we can use the test of Theorem 7.9 to prove that tS + u is a prime. Of course, if

N = tS+u is a prime, it might not be an I prime and therefore T (N) 6= N2 +N +1;

consequently, this test would not be successful. Thus, we need to find values for P , Q,

R such that ifN = tS+u is a prime, thenN is an I prime for f(x) = x3−Px2+Qx−R.

It is well-known (see [Wil72b], [Leh58]) that if N is a prime and

Np−13 6≡ 1 (mod p),

where p is a prime (≡ 1 (mod 3)), 4p = r2 + 27s2 with r ≡ 1 (mod 3) and N - spr,

then the cubic congruence

x3 − 3px− pr ≡ 0 (mod N)

191

is irreducible; that is, N is an I prime for

P ≡ 0, Q ≡ −3p, R ≡ pr (mod N).

Notice that since (pr,N) = 1, there always exists some x such that (pr+xN,−3p) =

1; hence, the fact that (−3p, pr) = p 6= 1 does not have any affect on the validity of

our results. We have proved the following theorem.

Theorem 7.12. Let L = tS2 + (2u + 1)S + 1 be a prime and put N = tS + u.

Suppose (N, 6) = 1, p is a prime such that p ≡ 1 (mod 3), Np−13 6≡ 1 (mod p) and

4p = r2 + 27s2 with r ≡ 1 (mod 3) and (N, prs) = 1. If we put

P ≡ 0, Q ≡ −3p, R ≡ pr (mod N),

then N is a prime if and only if N | DN2+N+1 and (Dt, N) = 1.

Corollary 7.12.1. Let 2 - S, L = 3S2 − 3S + 1 be a prime and suppose that p is a

prime such that p ≡ 1 (mod 3) and

Np−13 6≡ 1 (mod p),

where N = 3S − 2. If we define r and s as above and (N, prs) = 1, then N is a

prime if and only if N | DN2+N+1, where

P ≡ 0, Q ≡ −3p, R ≡ pr (mod N).

Proof. This follows easily from the theorem by putting u = −2 and noting that

∆ ≡ 27(4p3 − p2r2) (mod N),

W1 = PQ− 3R ≡ −3pr (mod N),

4C3 = ∆ + 3W 21 ≡ −4(27p3) (mod N).

Hence, (C3, N) = 1.

192

Notice, that we need only perform O(logN(M(log2N))) operations to establish

the primality of N , once we know that L is a prime. This is much faster than several

other tests because it is not necessarily all that easy to find enough factors of N ± 1

to use the techniques of Brillhart, Lehmer and Selfridge [BLS75], which generalized

those of Lucas, to establish the primality of N .

7.3 The Primality of L

If we put

L = tS2 + (2u+ 1)S + 1, t = u2 + u+ 1 and N = tS + u,

the results of the last section allow us to establish the primality of N , when we have

already proved L is a prime. This is not a vacuous result because we certainly expect

by the Bateman–Horn conjecture [BH62] that there exists an infinitude of values of

S such that for a fixed u, L and N will both be prime. Also, for a fixed value of

S, we would expect that there exists an infinitude of values of u such that both L

and N will be prime. There remains, however, the difficulty of proving that L is a

prime. We notice, however, that S | L − 1. Suppose S = FG, where we know the

complete factorization of F . It is then possible, by using the methods of [BLS75] to

prove that L is either prime or that all the prime factors of L must have the form

kF + 1.

Theorem 7.13. If L = tS2+(2u+1)S+1 (t = u2+u+1), S = FG and all the prime

factors of L have the form kF + 1, then L is a prime when F > tG2 + |2u+ 1|G+ 2.

193

Proof. Suppose L is composite. We must have

L = (k1F + 1)(k2F + 1),

where k1, k2 ≥ 1. Hence,

tS2 + (2u+ 1)S = k1k2F2 + (k1 + k2)F,

tSG+ (2u+ 1)G = k1k2F + k1 + k2

and

(tG2 − k1k2)F = k1 + k2 − (2u+ 1)G.

If tG2 − k1k2 = 0, then

(k1 − k2)2 = (k1 + k2)2 − 4k1k2 = (2u+ 1)2G2 − 4tG2 = −3G2,

which is impossible; hence |tG2 − k1k2| ≥ 1. Also, since k1, k2 ≥ 1, we get (k1 −

1)(k2 − 1) ≥ 0 and hence k1k2 ≥ k1 + k2 − 1. Now

|tG2 − k1k2|F ≤ k1 + k2 + |2u+ 1|G ≤ k1k2 + 1 + |2u+ 1|G

= k1k2 − tG2 + tG2 + |2u+ 1|G+ 1;

hence,

F ≤ tG2 + |2u+ 1|G+ 2,

which is impossible.

Suppose we now consider the following simple example, where we put F = 2n,

G = 1. We get

Ln = (u2 + u+ 1)22n + (2u+ 1)2n + 1, Nn = (u2 + u+ 1)2n + u.

194

In this case if 2n > u2 + 3|u| + 4, we can easily establish (when it is the case) that

Ln is a prime. We can next use our earlier results to prove that Nn is a prime, when

that is the case. In the table below we provide all instances for various values of u

(−500 ≤ u ≤ 500) and n ≤ 1000 such that both Ln and Nn are prime.

195

Table 7.1: Values of u such that Ln and Nn are both prime (n ≤ 1000)n u n u

1 3, -3, -5, 13, 25, 31, -33, 37, -39, 55, -57, 41 -27, 111, -141-71, 79, -87 -159, 181, -183, 219, -221, -243, 43 -173

255, -255, 279, -281, 289, -291, 307, 325, 333, 46 -449-353, -369, 375, -395, -423, -435, -495 51 273

2 -11, 31, 55, 115, -191, -221, 271, 361 53 -53, 267, -2673 -3, 7, 19, 25, -33, 39, -51, -65, 79, 105, 117, 177, 55 25, 87, -327, 475,

-231, 259, -401, 483, 499 56 307, -4194 -5,31, 223, 277, -323, 367, 415 57 217, 361, 4835 3,-17, 19, 39, -39, -45, -65, 73, -95, -101, 58 -221

129, -137, -153, 165, 207, -233, 295, -297, 61 475-323, 339, -389, 417, 463, 481 65 -195

6 -65, 145, 259, -311 69 -21, 4157 9, -15, -95, 109, -243, 297, -297, 457, 459, -477 70 -58 -179, -209, -263, -395 72 3619 9,-65, 91, -227, 397, 471 73 44710 -5, 349 74 16911 13, 15, 25, 87, -111, 159, 199, 285, 309, -381 75 49912 -119, 205, 271 77 3913 25,-33, -285, 325, 349, -449 79 -22714 199,-281, -359,439 83 -25115 9, 25, 39, 105, -105, -107, 235, 313, 397, 415, -471 84 -3516 -89,277, -389, -395, -407 95 -6517 61, 73, -135, -141, 255, 321, 481 96 -23318 -101 121 2719 -123, -221, -255, -311, 487 123 -46720 31,-185,-269 137 -29721 -17, 81,-149, -413, 445 140 19322 -215, 319 143 40323 127, 129, 265, -275, -323, -335, -401, -437 207 12325 -267, -309, 499 211 -14127 39,-161 231 16328 181 360 -46730 235 399 11732 187, -209 407 23735 -381, 463 417 -25738 361 533 -40740 -365 819 289

196

If we put F = qn, where q is a prime and G = 1, we can once again easily establish

the primality of

Ln = (u2 + u+ 1)q2n + (2u+ 1)qn + 1

when Ln is a prime. However, if we specify q and u, it seems a very rare event to

have both Ln and Nn = (u2 + u+ 1)qn + u prime simultaneously. We illustrate this

rarity in the next table, we provide those few values of n for which Ln and Nn are

both prime for all n < 1000.

Table 7.2: Prime pairs for specific choices of u and qu q Ln Nn n < 1000

2 3 7 · 32n + 5 · 3n + 1 7 · 3n + 2 1-2 3 3 · 32n − 3 · 3n + 1 3 · 3n − 2 1, 4, 5-2 5 3 · 52n − 3 · 5n + 1 3 · 5n − 2 1, 2-2 7 3 · 72n − 3 · 7n + 1 3 · 7n − 2 1-2 11 3 · 112n − 3 · 11n + 1 3 · 11n − 2 1, 73 2 13 · 22n + 7 · 2n + 1 13 · 2n + 3 1, 5-3 2 7 · 22n − 5 · 2n + 1 7 · 2n − 3 1, 3-4 3 13 · 32n − 7 · 3n + 1 13 · 3n − 4 26 5 43 · 52n + 13 · 5n + 1 43 · 5n + 6 156 11 43 · 112n + 13 · 11n + 1 43 · 11n + 6 18 3 73 · 32n + 17 · 3n + 1 73 · 3n + 8 112 5 157 · 52n + 25 · 5n + 1 157 · 5n + 12 114 3 211 · 32n + 29 · 3n + 1 211 · 3n + 14 1, 1715 2 241 · 22n + 31 · 2n + 1 241 · 2n + 15 11-15 2 211 · 22n − 29 · 2n + 1 211 · 2n − 15 7-18 11 307 · 112n − 35 · 11n + 1 307 · 11n − 18 11-21 2 421 · 22n − 41 · 2n + 1 421 · 2n − 21 6927 2 757 · 22n + 55 · 2n + 1 757 · 2n + 27 121-28 3 757 · 32n − 55 · 3n + 1 757 · 3n − 28 3, 9

197

In the particular case of u = −2, q = 3, row 2 of Table 7.2, we get

Ln = 32n+1 − 3n+1 + 1 and Nn = 3n+1 − 2.

For n > 3, we need only find some b such that

bLn−1 ≡ 1 (mod Ln) and (bLn−1

3 − 1, Ln) = 1, (7.1)

to establish that Ln is a prime. Note that 3n+1 || Ln − 1. Suppose p is some prime

such that p | Ln. By (7.1) we have

p | bLn−1 − 1 and p - bLn−1

3 − 1.

If ω is the order of b modulo p, then

ω | Ln − 1 and ω -Ln − 1

3.

So 3n+1 | ω and ω | p − 1; thus p ≡ 1 (mod 3n+1). Hence p = k3n+1 + 1 for some

k ∈ N. We then have p ≥ 2 · 3n+1 + 1 and we can conclude that Ln is a prime

since p >√Ln. Having done this we can use Corollary 7.12.1 to establish that Nn

is a prime. This sort of testing of pairs of numbers for primality might have pleased

Lucas.

7.4 The Case of T (N) = N 2 − 1

It is certainly possible to test numbers of the form Aqn±1 for primality by using the

Wn and Cn functions; however, we will confine our attention here to the case where

N = A3n − 1, as this is the analogous form to A2n − 1 mentioned in Chapter 2. We

can produce a theorem similar to Theorem 2.24, except for the necessity condition.

198

Theorem 7.14. Let N = A3n − 1, where 2 | A, A < 3n, n ≥ 2 and (N,R) = 1. If

N | CN+1/CN+13,

then N is prime.

Proof. Let p be any prime divisor of N . Since p | CN+1, we must have some rank of

apparition r(p) in {Cn} such that r(p) | N + 1. Also, since

4CN+1/CN+13

= ∆C2N+1

3

+ 3W 2N+1

3

,

we see that if p | CN+13

, then p | WN+13

. Hence p | EN+13

and by Theorem 6.7,

we know that p ≡ 1 (mod 3n). If p - CN+13

, then r(p) - N+13

. It follows that

3n | r(p). If p | ∆, then r(p) | p or p− 1 and the first case is impossible, as p | N

and r(p) | N + 1 ⇒ r(p) - p. If p is an I prime, then r(p) | p2 + p + 1, but this is

impossible because 9 - p2 + p+ 1. If p is an S prime or a Q prime, then r(p) | p2− 1

and p ≡ ±1 (mod 3n). Thus, any prime divisor of N must be at least as large as

2 · 3n − 1. Since (2 · 3n − 1)2 > N , N must be a prime.

Our next objective will be to produce conditions that are both necessary and

sufficient for N = A3n − 1 to be prime. We first need to produce a result analogous

to Theorem 2.25. We begin with the following theorem.

Theorem 7.15. Let p be an odd prime such that p ≡ −1 (mod 3). Then there exist

P , Q, R such that p is a Q prime if and only if

P ≡ a+ Tr(λ), Q ≡ aTr(λ) + N(λ), R ≡ aN(λ) (mod p),

where a ∈ Z, λ = r1 + r2ρ ∈ Z[ρ], ρ2 + ρ+ 1 = 0 and p - ar2 N(λ).

199

Proof. Suppose P , Q, R satisfy the conditions of the theorem. Then clearly

∆ ≡ (a− λ)2(a− λ)2(λ− λ)2

= N(a− λ)2r22(ρ− ρ2)2

= −3m2 (mod p),

where m ∈ Z. Since p ≡ −1 (mod 3) and p - ar2 N(λ), we cannot have p | m. Thus(∆

p

)=

(−3m2

p

)=

(−3

p

)= −1.

Since p - 6R∆, p is a Q prime for f(x) = x3 − Px2 +Qx−R.

Next, suppose that p is a Q prime for f(x) = x3−Px2 +Qx−R. Then Fp2 is the

splitting field of f(x) in Fp[x]. Let α, β, γ be the zeros of f(x) in Fp2 , where αp = α,

βp = γ 6= β = γp. Since F∗p2 = 〈θ〉 for some suitable θ ∈ Fp2 , we put ζ = θp2−1

3 and

note that ζ2 + ζ + 1 = 0. Now since p ≡ −1 (mod 3),(β − γζ − ζ2

)p=γ − βζ2 − ζ

=β − γζ − ζ2

.

Hence β−γζ−ζ2 ∈ Fp. If we put

a ≡ α, b ≡ β − γζ − ζ2

6≡ 0, c ≡ β + γ = P − α ≡ P − a (mod p),

then

β =b+ c

2+ bζ, γ =

b+ c

2+ bζ2.

Putting r1 ≡ (b+ c)/2 (mod p), r2 ≡ b (mod p) we see that

P ≡ a+ Tr(λ), Q ≡ aTr(λ) + N(λ), R ≡ aN(λ) (mod p),

for λ = r1 + r2ρ, p - r2. Since p - R, we must also have p - aN(λ).

200

We can now present our analogue of Theorem 2.25.

Theorem 7.16. Let p be an odd prime such that p ≡ −1 (mod 3). If P , Q, R

satisfy the conditions of Theorem 7.15 and(λp

)36= 1, then

p | Cp+1/C p+13.

Proof. By Theorem 7.15, we know that p is a Q prime and therefore p | Cp+1. Let

α, β, γ be the zeros of f(x) in Fp2 , where αp = α, βp = γ 6= β = γp. Since λp2−1

3 6≡ 1

(mod p), we may assume with no loss of generality that βp2−1

3 6= 1 in Fp2 . Since

βp2−1

3 = (βp−1)p+13 =

(γ

β

) p+13

,

we have βp+13 6= γ

p+13 . Also, since

βpp+13 = γ

p+13 6= β

p+13 ,

we cannot have αp+13 = β

p+13 because β

p+13 6∈ Fp. Similarly α

p+13 6= γ

p+13 . It follows

that C p+136= 0 in Fp2 or p - C p+1

3. Hence p | Cp+1/C p+1

3.

By combining Theorems 7.14 and 7.16 we get the following necessary and suffi-

cient condition for N = A3n − 1 (2 | A, A < 3n) to be prime.

Theorem 7.17. Let N = A3n − 1, where 2 | A and 3 < A < 3n. Futhermore, let

q ≡ 1 (mod 3) be a prime such that q - N and

Nq−13 6≡ 1 (mod q).

Let λ = r1 + r2ρ (ρ2 +ρ+ 1 = 0) be a primary prime divisor of q in Z[ρ] and suppose

that N - r2. Let

P ≡ a+ Tr(λ), Q ≡ aTr(λ) + q, R ≡ aq (mod N),

201

where (a,N) = 1. Then N is a prime if and only if

N | Cp+1/C p+13.

Proof. Since (N,R) = 1, we know by Theorem 7.14 that N is a prime if

N | Cp+1/C p+13.

Next, suppose that N is a prime. We know that

Nq−13 6≡ 1 (mod q),

and since λ is a primary prime divisor of q, we have by the cubic reciprocity law(N

λ

)3

6= 1⇒(λ

N

)3

6= 1.

Thus, N | Cp+1/C p+13

by Theorem 7.16.

7.5 Primality Test

We may now use Theorem 7.17 to produce a primality test, somewhat similar to the

Lucas and Lehmer test for numbers of the form A2n − 1, for numbers of the form

A3n − 1. Of course, this test is not as practical as that of [Wil72b], but it would

have been of some interest to Lucas that Cn could be used to produce such a test.

Let m ∈ Z+ such that (m, 2R) = 1. Compute

S0 ≡Wn

2Rn(mod m) and R0 ≡

∆C2n

4R2n(mod m)

and define

Si ≡W3in

2Rn3i(mod m) and Ri ≡

∆C23in

4R2n3i(mod m).

202

Notice then

Si+1 =1

2R−n3i+1

W3i+1n =1

2R−n3i+1

W3(3in)

=1

2R−n3i+1

[3

∆C23in

4(W3in + 2R3in) +

W 33in

4− 6R3inW3in

4+ 6R3i+1n

]

=1

2R−n3i+1

[3R2(3in)Ri(2R

3inSi + 2R3in) +(2R3inSi)

3

4− 6R3in(2R3inSi)

2

4+ 6R3i+1n

]=

1

2R−n3i+1

[3RiR

3i+1n2(Si + 1) + 2R3i+1nS3i − 6R3i+1nS2

i + 6R3i+1n]

= 3Ri(Si + 1) + S3i − 3S2

i + 3.

Similarly,

Ri+1 =∆C2

3i+1n

4R2(3i+1n).

Using Corollary 3.10.1 we have

C3i+1n =1

4C3in(∆C2

3in + 3W 23in).

Use this and the fact that

W3in ≡ 2R3inSi and ∆C23in ≡ 4R2(3in)Ri (mod m)

to obtain

C3i+1n =1

4C3in(4R2(3in)Ri + 3(2R3inSi)

2).

This yields

C23i+1n =

1

4C2

3in

1

4(4R2(3in)Ri + 3(2R3inSi)

2)2 =C2

3in

4

1

4(4R2(3in)Ri + 12R2(3in)S2

i )2

=C2

3in

4

1

4(4R2(3in)(Ri + 3S2

i ))2 =

C23in

44R4(3in)(Ri + 3S2

i )2.

We manipulate this as follows:

∆C23i+1n

4=

∆C23inR

4(3in)

4(Ri + 3S2

i )2.

203

So then we get

∆C23i+1n

4R2(3i+1)n=

∆C23inR

4(3in)

4R2(3i+1)n(Ri + 3S2

i )2.

Thus

Ri+1 =∆C2

3inR4(3in)

4R6(3i)n(Ri + 3S2

i )2 =

∆C23in

4R2(3i)n(Ri + 3S2

i )2

= Ri(Ri + 3S2i )

2.

We can now employ these observations to produce our primality test. If we satisfy

the conditions of Theorem 7.17 where N = 3nA− 1 and set

S0 ≡WA

2RA(mod N) and R0 ≡

∆C2A

4R2A(mod N),

then

Si ≡W3iA

2R3iA(mod N) and Ri ≡

∆C23iA

4R2(3iA)(mod N).

We can produce the sequence {Si} and {Ri} (mod N) as follows:

Si+1 ≡ 3Ri(Si+1)+S3i −3S2

i +3 (mod N) and Ri+1 ≡ Ri(Ri+3S2i ) (mod N).

Then N is a prime if and only if Rn−1 ≡ −3S2n−1 (mod N).

Chapter 8

Conclusion

8.1 Main Result

The purpose of this thesis was to develop a cubic extension of the Lucas functions

that Lucas himself might have discovered. What has emerged from this work is

a theory of functions that displays a number of pleasing similarities with Lucas’

original work. The main tools in Lucas’ investigation of his functions were the

multiplication formulas (2.14) and (2.15). The multiplication formulas, proved in

Section 3.5, allowed us to obtain arithmetic results that closely resemble those for

the Lucas case. Key results like the laws of repetition and apparition, and Euler’s

criterion, as described in Sections 4.4, 4.5, 5.2, 5.3, 6.2, and 6.3, have analogues

in our extension. Most remarkably, the extension relies on the use of only two

functions1, despite the fact that you would expect three for the cubic case. Further,

when restricted to the quadratic case, our generalization in Section 3.3 satisfyingly

reduces to that of Lucas sequences.

With all that in mind, it is difficult to point to a single ‘main’ result. However,

knowing that Lucas’ own goal in generalizing his sequences was to find and implement

new primality tests, Theorem 7.17 and the primality test of Section 7.5 based on it

stand out. The test makes use of {Cn}, a sequence known to Lucas that surely

would have been a part of any generalization he would have done, to test numbers

1It might be argued that we are really considering four functions here because of Dn and En,but these latter functions are simply a convenient way of representing certain divisors of Cn.

204

205

of the form A3n − 1. Certainly, even more important than just the primality test

is Theorem 7.4, a result that is our analogue of Theorem 2.22, which Lucas refered

to as his fundamental theorem. It should be stated, however, that today there exist

many sophisticated methods for primality proving (see, for example, Chapter 4 of

[CP01]). The primality conditions proved here are of mere historical interest and are

perhaps what Lucas had in mind.

8.2 Improvements

In terms of what might be done to refine or improve the results of the thesis, it

would by satisfying to have more elementary proofs of Theorems 4.18 and 4.19. As

it stands, both proofs use facts from algebraic number theory, and these are the only

two instances where such powerful tools are required. However, some key results from

Serret’s Cours d’algebre superieure Vol. II [Ser79] allow for the desired elementary

alternate proof of Theorem 4.18. For the details, the interested reader is directed to

Appendix A. In fact, Lucas was familiar with Serret’s work, so if Lucas had proved

this result, it is imaginable that he would have used the methods seen in Appendix

A.

It would also be interesting to develop further the law of repetition for {Dn} so

that it more closely matches the Lucas case, where if p is a prime and for λ > 0, we

have pλ 6= 2 and pλ || Um, then pλ+µ || Umnpµ when p - n and µ ≥ 0. This is how

the law of repetition is presented for {Un} in Chapter 4 of [Wil98].

206

8.3 Future Work

Beyond the thesis itself, work could be done to try to use {Cn} and {Wn} to im-

plement an RSA-type cryptosystem. Peter Smith’s LUC [Smi93]2, a well-known

example of such a cryptosystem using the Lucas functions, provides the motivation

for this. It might also be possible to use {Cn} and {Wn} to perform a Diffie-Hellman-

like key exchange.

There is also, of course, the possibility of exploring the idea of a quartic extension

as mentioned by Lucas. For such an extension, Cn and Wn would be as described

early in Section 3.3. Despite being an interesting problem, we would expect it to be

difficult to work with the quartic case due to the number of terms involved in the

recurrences.

2See [BBL95] for some useful comments concerning LUC.

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Appendix A

In this appendix we provide an elementary proof of Theorem 4.18. This proof could

easily have been discovered by Lucas through his admitted knowledge of Chapter III

of [Ser79]. We first recapitulate some of the preliminary results in Serret’s work.

Definition A.1. Let p be a prime and let f(x), g(x) ∈ Z[x]. If there exist h(x),

k(x) ∈ Z[x] such that

h(x)g(x) = f(x) + pk(x),

we say that f(x) is divisible by g(x) with respect to modulus p. We write this as

h(x)g(x) ≡ f(x) (mod p).

Definition A.2. If f(x) ∈ Z[x], f(x) is monic and f(x) is not divisible by any

g(x) ∈ Z[x] with respect to modulus p, we say that f(x) is irreducible with respect

to p.

Theorem A.3. If g(x), h(x) ∈ Z[x], g(x), h(x) have no common divisor (of degree

≥ 1) with respect to the prime modulus p, then there exist Y (x), Z(x) ∈ Z[x] such

that

Y (x)h(x)− Z(x)g(x) ≡ 1 (mod p).

The next result does not appear explicitly in [Ser79], but it would have been easy

for Lucas to derive because its proof is entirely analogous to that of the elementary

number theory result which states that if a, b, c ∈ Z, (a, b) = 1, a | c and b | c, then

ab | c. This is proved by Lucas on p. 340 in [Luc91b]. The proof of Theorem A.4

would follow in exactly the same manner by using Theorem A.3.

215

216

Theorem A.4. If f(x), g(x), h(x) ∈ Z[x], g(x) and h(x) have no common divisor

(of degree ≥ 1) with respect to the prime modulus p, and g(x), h(x) are both divisors

of f(x) modulo p, then g(x)h(x) is a divisor of f(x) with respect to the modulus p.

The next result (Theorem I in Section 346 of [Ser79]) is most important for our

subsequent work. We give it in a somewhat different form from Serret’s results, but

Serret certainly establishes it in his proof of Theorem I.

Theorem A.5. Let f(x) ∈ Z[x] be irreducible with respect to a prime modulus p and

of degree ν. Then f(x) is a divisor of xpν−1 − 1 with respect to the modulus p.

We now suppose that f(x) = x3 − Px2 + Qx − R and p is a prime such that

p - 6∆R. We know that when p is an S prime we have

f(x) ≡ f1(x)f2(x)f3(x) (mod p),

where f1(x), f2(x), f3(x) ∈ Z[x] are each monic of degree 1 and, because p - ∆, have

no common divisor with respect to p. If p is a Q prime, then

f(x) ≡ f1(x)f2(x) (mod p),

where f1(x), f2(x) ∈ Z[x]; f1(x), f2(x) are monic and irreducible with respect to the

modulus p, deg f1(x) = 1, deg f2(x) = 2. Finally, if p is an I prime, then f(x) is

irreducible with respect to the modulus p.

It is now easy to prove, by making use of Theorems A.4 and A.5, that f(x) is a

divisor of xpµ−1−1 with respect to the modulus p, where µ = 1 when p is an S prime,

µ = 2 when p is a Q prime, and µ = 3 when p is an I prime. Putting m = pµ − 1,

we have

xm − 1 = f(x)g(x) + ph(x)

217

for some g(x), h(x) ∈ Z[x]. Putting x = α, β, γ, where α, β, γ ∈ C are the zeros of

f(x), we get

αm − 1 = ph(α),

βm − 1 = ph(β),

γm − 1 = ph(γ),

from which it follows that

αm − βm = p(h(α)− h(β)),

βm − γm = p(h(β)− h(γ)),

γm − αm = p(h(γ)− h(α)).

Hence

∆C2m = (αm − βm)2(βm − γm)2(γm − αm)2

= p6S(α, β, γ)

where,

S(x, y, z) = (h(x)− h(y))2(h(y)− h(z))2(h(z)− h(x))2.

Since S(x, y, z) is a symmetric polynomial in Z[x, y, z], we must have S(α, β, γ) ∈ Z

by the fundamental theorem of symmetric polynomials, and therefore p3 | Cm. Also

Wm − 6Rm = αm(βm − γm)2 + βm(γm − αm)2 + γm(αm − βm)2

= p2T (α, β, γ),

where

T (x, y, z) = (h(y)− h(z))2 + (h(z)− h(x))2 + (h(x)− h(y))2

+p[h(x)(h(y)− h(z))2 + h(y)(h(z)− h(x))2 + h(z)(h(x)− h(y))2].

218

Since T (x, y, z) is also a symmetric polynomial in Z[x, y, z], we must have T (α, β, γ) ∈

Z and p2 | Wm − 6Rm.

We are now able to present another proof of Theorem 4.18.

Theorem A.6. If p - 6∆R and p | Cn, p | Wn− 6Rn, then p3 | Cn and p2 | Wn−

6Rn.

Proof. Let ω = ω(p). Clearly ω exists and ω divides both n and m by Theorem 5.5.

From the proof of Theorem 5.1, we know that

Cm/Cω ≡ (m/ω)3Rm−ω (mod p).

Thus, since p - m, we get p - Cm/Cω ⇒ p3 | Cω ⇒ p3 | Cn. Also from the proof of

Theorem 5.1, we have

Wm − 6Rm ≡ (m/ω)2Rm(Wω − 6Rω) (mod p2);

thus, we get p2 | Wω − 6Rω. Furthermore,

Wn − 6Rn ≡ (n/ω)2Rn(Wω − 6Rω) (mod p2)

means that p2 | Wn − 6Rn.

Although this proof requires Theorems 5.1 and 5.5, these results did not require

the result of Theorem 4.18 in their respective proofs.