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10/25/2015

FIBONACCI AND LUCAS NUMBERS

2015

ALFIYA A


By

ALFIYA A

(MSc Mathematics)

Preface

The Fibonacci sequence and the Lucas sequence are the two shining

stars in the vast array of integer sequences. This book is intended for

college undergraduate and graduate students often opt to study

Fibonacci and Lucas numbers because they find them challenging and

exciting. The book contains details about Fibonacci and Lucas

identities, its divisibility properties and also its generalization.

Suggestion for the further improvement of this book will

be highly appreciated.

ACKNOWLEDGEMENT

I would like to express my gratitude to many people

who gave me their comments and suggestions which influenced

the preparation of the text. I want to thank teachers and friends

for their effective cooperation and great care for preparing this

book.

CONTENTS

CHAPTER TITLE PAGE NO.

CHAPTER – 1 FIBONACCI AND LUCAS NUMBERS 1

CHAPTER – 2 FIBONACCI AND LUCAS IDENTITIES 7

CHAPTER – 3 DIVISIBILITY PROPERTIES 25

CHAPTER – 4 GENERALIZATIONS 37

REFERENCES 44

CHAPTER – 1


In the realm of mathematics, many concepts have applications

in multiple mathematical fields. Without these important concepts, every field of

mathematics would be seemingly disjointed and unrelated to topics from other

mathematical fields. One of these concepts was discovered by a man named

Leonardo Pisano during the early 13th century. This particular concept is known

today as the Fibonacci sequence. Since its official introduction to the world by

Leonardo Pisano, the Fibonacci sequence has become one of the most fascinating

concepts in the entire realm of mathematics through its remarkable characteristics;

its useful applications to various mathematical fields such as number theory,

discrete mathematics, and geometry; and its clear demonstration of the aesthetic

nature of God.

The Fibonacci series was derived from the solution to a problem

about rabbits. The problem is:

Suppose there are two new born rabbits, one male and the other female. Find

the number of rabbits produced in a year if

• Each pair takes one month to become mature:

• Each pair produces a mixed pair every month, from the second month:

• All rabbits are immortal

Suppose, that the original pair of rabbits was born on January 1. They take a

month to become mature, so there is still only one pair on February 1. On March 1,

they are two months old and produce a new mixed pair, so total is two pair. So

continuing like this, there will be 3 pairs in April, 5 pairs in May and so on.

It is then seen that at the end of the 𝑛th month, all of the pairs during the

𝑛 − 2 th month has given birth to a new pair of rabbits. So then, the total number

of pairs at the end of the 𝑛th month is the number from the 𝑛 − 1 th month plus a

new pair for each pair of rabbits at the end of the 𝑛 − 2 th month. In some

simpler terms, the number of pairs of rabbits at the end of 𝑛 months is the sum of

the numbers at the end of the two previous months.

Without using rabbits, it is said that in the Fibonacci sequence of numbers,

each number is the sum of the previous two numbers, where the first two numbers

of the sequence are 0 and 1. The numbers generated by the Fibonacci sequence are

one of the most famous examples of a recurrence relation. A recurrence relation is

a relation which uses previous values in the relation and is usually defined by some

fixed numbers in the relation, known as the initial conditions. Writing the

Fibonacci sequence as a function, it is shown that 𝐹0 = 0, 𝐹1=1, 𝐹𝑛+2 = 𝐹𝑛+1 +

𝐹𝑛 . The recurrence is 𝐹𝑛+2 = 𝐹𝑛+1 + 𝐹𝑛 and the initial conditions are 𝐹0 = 0,

𝐹1=1.

Then the Fibonacci numbers are 0,1, 1, 2, 3, 5, 8,13,...

This yield the following recursive definition of the 𝑛th Fibonacci number 𝐹𝑛 .

𝐹1 = 1

𝐹2 = 1

⋮

𝐹𝑛 = 𝐹𝑛−1 + 𝐹𝑛−2, 𝑛 ≥ 3

Using the formula given above where 𝐹𝑛−1 and 𝐹𝑛−2 must be calculated

before adding them together to find 𝐹𝑛 is the most obvious method for calculating

a Fibonacci number. This method requires calculating all of the Fibonacci numbers

with indexes less than 𝑛 before the value of 𝐹𝑛 can be calculated. There is a fairly

simple formula named after the French mathematician Jacques Phillipe Marie

Binet to find the value of a given Fibonacci number without performing all of the

tedious calculations. If 𝐹𝑛 is the 𝑛th Fibonacci number , then

𝐹𝑛 =1

5

1 + 5

2

𝑛

− 1 − 5

2

𝑛

. The “span” or “degree” of the recursion is the difference between the

highest and lowest subscripts in the recursion ( 𝑛 + 1 − 𝑛 − 1 = 2 in the

Fibonacci recursion). Lucas discovered that if the degree is d, there are always d

and at most d “independent” sequences satisfying the same recursion (with

different starting values).

In the case where the degree is 2, like the Fibonacci numbers, the recursion can

be written as:

𝑈(𝑛 + 1) = 𝐴 ∗ 𝑈(𝑛) + 𝐵 ∗ 𝑈(𝑛 − 1)

There are two independent sequences (that is, one is not a multiple of the other)

satisfying this. We can always pick one such that U(1) = 1 and U(2) = 1, and also a

second one such that U(1) = 1 and U(2) = 3 . The second sequence which is

independent of the Fibonacci sequence and starts 1, 3,... is now called the Lucas

sequence after Edouard Lucas. Why choose 2, 1, ... for the start? It is probably

because of the Binet formulas for the Fibonacci and Lucas numbers:

𝐹𝑛 = 𝛼𝑛 − 𝛽𝑛 𝛼 − 𝛽

𝐿𝑛 = 𝛼𝑛 + 𝛽𝑛

where 𝑛 ≥ 0, 𝛼 = (1 + 5)/2 , and 𝛽 = (1 − 5)/2. Then 𝛼 and 𝛽 are the two

solutions to 𝑥2 − 𝑥 − 1 = 0. The fact that the coefficients of 𝛼𝑛 and 𝛽𝑛 are equal

in magnitude and opposite in sign in the first formula, and equal in magnitude and

sign in the second, is probably why this is the simplest and easiest second,

independent sequence to take. The first few terms of the Lucas sequence look like

this: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, ...

Lucas numbers 𝐿𝑛 are defined recursively as follows

𝐿1 = 1

𝐿2 = 3

⋮

𝐿𝑛 = 𝐿𝑛−1 + 𝐿𝑛−2, 𝑛 ≥ 3

CHAPTER - 2

FIBONACCI AND LUCAS IDENTITIES

Both Fibonacci and Lucas numbers satisfy numerous identities that have been

discovered over the centuries. In this chapter we explore several of these

fundamental identities.

In doing so, we notice the following interesting pattern:

𝐹1 = 1 = 2 − 1 = 𝐹3 − 1

𝐹1 + 𝐹2 = 2 = 3 − 1 = 𝐹4 − 1

𝐹1 + 𝐹2 + 𝐹3 = 4 = 5 − 1 = 𝐹5 − 1

𝐹1 + 𝐹2 + 𝐹3 + 𝐹4 = 7 = 8 − 1 = 𝐹6 − 1

𝐹1 + 𝐹2 + 𝐹3 + 𝐹4 + 𝐹5 = 12 = 13 − 1 = 𝐹7 − 1

Following this pattern, we conjecture that

𝐹𝑖

𝑛

1

= 𝐹𝑛+2 − 1

We shall establish the validity of this formula in two ways, but we first state it as a

theorem.

Theorem 2.1 (Lucas, 1876)

𝐹𝑖

𝑛

1

= 𝐹𝑛+2 − 1 (2.1)

Proof:

Using the Fibonacci recurrence relation, we have:

𝐹1 = 𝐹3 − 𝐹2

𝐹2 = 𝐹4 − 𝐹3

𝐹3 = 𝐹5 − 𝐹4

⋮

𝐹𝑛−1 = 𝐹𝑛+1 − 𝐹𝑛

𝐹𝑛 = 𝐹𝑛+2 − 𝐹𝑛+1

Adding these equations, we get:

𝐹𝑖 = 𝐹𝑛+2 −

𝑛

1

𝐹2 = 𝐹𝑛+2 − 1

AN ALTERNATE METHOD

An alternate method of proving Identity (2.1) is to apply the Principle of

Mathematical Induction. Since 𝐹1 = 𝐹3 −1, the formula works for 𝑛 = 1.

Now assume it is true for an arbitrary positive integer 𝑘 ≥ 1 :

𝐹𝑖 = 𝐹𝑘+2 − 1

𝑘

1

Then

𝐹𝑖

𝑘+1

1

= 𝐹𝑖 +

𝑘

1

𝐹𝑘+1

= (𝐹𝑘+2 − 1) + 𝐹𝑘+1,by the inductive hypothesis

= (𝐹𝑘+1 + 𝐹𝑘+2) − 1

= 𝐹𝑘+3 − 1

Thus, by Principle of Mathematical Induction, the formula is true for every

positive integer 𝑛.

For example,

𝐹𝑖 =

10

1

𝐹12 − 1 = 144 − 1 = 143


𝐹2𝑖−1

𝑛

1

= 𝐹2𝑛 (2.2)

Proof:

Using the Fibonacci recurrence relation, we have

𝐹1 = 𝐹2 − 𝐹0

𝐹3 = 𝐹4 − 𝐹2

𝐹5 = 𝐹6 − 𝐹4

⫶

𝐹2𝑛−3 = 𝐹2𝑛−2 − 𝐹2𝑛−4

𝐹2𝑛−1 = 𝐹2𝑛 − 𝐹2𝑛−2

Adding these equations, we get

𝐹2𝑖−1 = 𝐹2𝑛 − 𝐹0 =

𝑛

1

𝐹2𝑛 , since 𝐹0 = 0

For example,

𝐹2𝑖−1 = 𝐹16 =

8

1

987

Corollary 2.1 (Lucas, 1876)

𝐹2𝑖 =

𝑛

1

𝐹2𝑛+1 − 1 (2.3)

Proof:

𝐹2𝑖 =

𝑛

1

𝐹𝑖 −

2𝑛

1

𝐹2𝑖−1

𝑛

1

= (𝐹2𝑛+2 −1) − 𝐹2𝑛 , by theorems 2.1 and 2.2

= (𝐹2𝑛+2 − 𝐹2𝑛) − 1

= 𝐹2𝑛+1 − 1 , by the Fibonacci recurrence relation

Theorem 2.3 (Cassini’s Formula)

𝐹𝑛−1𝐹𝑛+1 − 𝐹𝑛2 = (−1)𝑛 , where 𝑛 ≥ 1 (2.4)

Proof:

Since 𝐹0𝐹2 − 𝐹12 = 0 · 1 − 1 = −1 = (−1)1, the given statement is clearly

true when 𝑛 = 1.

Now we assume it is true for an arbitrary positive integer k: :

𝐹𝑘−1𝐹𝑘+1 − 𝐹𝑘2 = (−1)𝑘 . Then,

𝐹𝑘𝐹𝑘+2 − 𝐹𝑘+12 = (𝐹𝑘+1 − 𝐹𝑘−1)(𝐹𝑘 + 𝐹𝑘+1) − 𝐹𝑘+1

2

= 𝐹𝑘𝐹𝑘+1 + 𝐹𝑘+12 − 𝐹𝑘𝐹𝑘−1 − 𝐹𝑘−1𝐹𝑘+1 − 𝐹𝑘+1

2

= 𝐹𝑘𝐹𝑘+1 − 𝐹𝑘𝐹𝑘−1 −𝐹𝑘2 − (−1)𝑘 by the inductive hypothesis

= 𝐹𝑘𝐹𝑘+1 − 𝐹𝑘 𝐹𝑘−1 + 𝐹𝑘 + (−1)𝑘+1

= 𝐹𝑘𝐹𝑘+1 − 𝐹𝑘𝐹𝑘+1 + (−1)𝑘+1

= (−1)𝑘+1

Thus the formula works for 𝑛 = 𝑘 + 1. So, by the Principle of Mathematical

Induction, the statement is true for every integer 𝑛 ≥ 1.

Corollary 2.2

Any two consecutive Fibonacci numbers are relatively prime; that is,

(𝐹𝑛+1,𝐹𝑛 ) = 1 for every 𝑛.

Proof:

Let p be a prime factor of both 𝐹𝑛 and 𝐹𝑛+1. Then, by Cassini's formula,

p|±1 , which is a contradiction.

Thus (𝐹𝑛+1,𝐹𝑛 ) = 1.


𝐹𝑖2 = 𝐹𝑛𝐹𝑛+1

𝑛

1

Proof:

When 𝑛 = 1,

L. H. S = 𝐹𝑖2

1

1

= 𝐹12 = 1 = 1 · 1 = 𝐹1 · 𝐹2 = R. H. S

So the result is true when 𝑛 = 1.

Assume it is true for an arbitrary positive integer 𝑘 :

𝐹𝑖2

𝑘

1

= 𝐹𝑘 𝐹𝑘+1

Then,

𝐹𝑖2 =

𝑘+1

1

𝐹𝑖2 +

𝑘

1

𝐹𝑘+12

= 𝐹𝑘𝐹𝑘+1 + 𝐹𝑘+12 , by the inductive hypothesis

= 𝐹𝑘+1 (𝐹𝑘 + 𝐹𝑘+1)

= 𝐹𝑘+1𝐹𝑘+2 , by the Fibonacci recurrence relation.

So the statement is true when 𝑛 = 𝑘 + 1.Thus it is true for every positive integer 𝑛.

For example,

𝐹𝑖2 = 𝐹15

15

1

𝐹16 = 610 · 987 = 602070

Interestingly enough, identities 2.1 through 2.5 have analogous results for

Lucas number also.

Result 2.1

𝐿𝑖 =

𝑛

1

𝐿𝑛+2 − 3 (2.6)

Proof:

Using the Lucas recurrence relation, we have,

𝐿1 = 𝐿3 − 𝐿2

𝐿2 = 𝐿4 − 𝐿3

𝐿3 = 𝐿5 − 𝐿4

⫶

𝐿𝑛−1 = 𝐿𝑛+1 − 𝐿𝑛

𝐿𝑛 = 𝐿𝑛+2 − 𝐿𝑛+1

Adding these equations, we get ,

𝐿𝑖 =

𝑛

1

𝐿𝑛+2 − 𝐿2

= 𝐿𝑛+2 − 3

Result 2.2

𝐿2𝑖−1 =

𝑛

1

𝐿2𝑛 − 2 (2.7)

Proof:

Using the Lucas recurrence relations, we have,

𝐿1 = 𝐿2 − 𝐿0

𝐿3 = 𝐿4 − 𝐿2

𝐿5 = 𝐿6 − 𝐿4

⫶

𝐿2𝑛−3 = 𝐿2𝑛−2 − 𝐿2𝑛−4

𝐿2𝑛−1 = 𝐿2𝑛 − 𝐿2𝑛−2

Adding these equations, we get,

𝐿2𝑖−1 =

𝑛

1

𝐿2𝑛 − 𝐿0

= 𝐿2𝑛 − 2

For example,

𝐿2𝑖−1 = 15125 = 15127 − 2 = 𝐿20 − 2

10

1

Result 2.3

𝐿2𝑖 =

𝑛

1

𝐿2𝑛+1 − 1 (2.8)

Proof:

𝐿2𝑖 =

𝑛

1

𝐿𝑖

2𝑛

1

− 𝐿2𝑖−1

𝑛

1

= (𝐿2𝑛+2 − 3) − (𝐿2𝑛 − 2)

= 𝐿2𝑛+2 − 𝐿2𝑛 − 1

= 𝐿2𝑛+1 − 1 , by the Lucas recurrence relation.

Result 2.4

𝐿𝑛−1𝐿𝑛+1 − 𝐿𝑛2 = 5(−1)𝑛−1 (2.9)

Proof:

Since 𝐿𝑜𝐿2 − 𝐿12 = 2 ˑ 3 − 12

= 6 – 1

= 5

= 5(−1)1−1

The given statement is clearly true when 𝑛 = 1. Now we assume it is true for an

arbitrary positive integer 𝑘, 𝐿𝑘−1𝐿𝑘+1 − 𝐿𝑘2 = 5(−1)𝑘−1

Then,

, 𝐿𝑘𝐿𝑘+2 − 𝐿𝑘+12 = 𝐿𝑘+1 − 𝐿𝑘−1 𝐿𝑘 + 𝐿𝑘+1 − 𝐿𝑘+1

2

= 𝐿𝑘𝐿𝑘+1 + 𝐿𝑘+12 − 𝐿𝑘𝐿𝑘−1 − 𝐿𝑘−1𝐿𝑘+1 − 𝐿𝑘+1

2

= 𝐿𝑘𝐿𝑘+1 − 𝐿𝑘𝐿𝑘−1 − 𝐿𝑘2 − 5 −1 𝑘−1

= 𝐿𝑘𝐿𝑘+1 − 𝐿𝑘 𝐿𝑘−1 + 𝐿𝑘 + 5(−1)𝑘

= 𝐿𝑘𝐿𝑘+1 − 𝐿𝑘𝐿𝑘+1 + 5(−1)𝑘

= 5(−1)𝑘

Thus the formula works for 𝑛 = 𝑘 + 1.

So, by the Principle of Mathematical Induction, the statement is true for every

integer 𝑛 ≥ 1.

Result 2.5

𝐿𝑖2 =

𝑛

1

𝐿𝑛𝐿𝑛+1 − 2 (2.10)

Proof:

When 𝑛 = 1,

L. H. S = 𝐿𝑖2

1

1

= 𝐿12

= 1 = 1 · 3 − 2

= 𝐹1𝐹2 − 2

= R. H. S

So the result is true when 𝑛 = 1.

Assume it is true for an arbitrary positive integer 𝑘, 𝐿𝑖2 =

𝑘

1

𝐿𝑘𝐿𝑘+1 . Then,

𝐿𝑖2 =

𝑘+1

1

𝐿𝑖2 +

𝑘

1

𝐿𝑘+12

= 𝐿𝑘𝐿𝑘+1 − 2 + 𝐿𝑘+12 , by the inductive hypothesis

= 𝐿𝑘+1 𝐿𝑘 + 𝐿𝑘+1 − 2

= 𝐿𝑘+1𝐿𝑘+2 − 2 , by the Lucas recurrence relation

So the statement is true when 𝑛 = 𝑘 + 1.

Thus it is true for every positive integer 𝑛.

For example,

𝐿𝑖2 =

𝑛

1

24475 = 24477 − 2 = 123 · 199 − 2 = 𝐿10𝐿11 − 2

Theorem 2.5

𝐹𝑛+𝑚 = 𝐹𝑛−1𝐹𝑚 + 𝐹𝑛𝐹𝑚+1

Proof:

We shall prove the theorem by induction on 𝑚.

For 𝑚 = 1, we get 𝐹𝑛+1 = 𝐹𝑛−1𝐹1 + 𝐹𝑛𝐹1+1 = 𝐹𝑛−1 + 𝐹𝑛 which is true.

Suppose that it is true for 𝑚 = 𝑘 and 𝑚 = 𝑘 + 1,we shall prove it is also true for

𝑚 = 𝑘 + 2.

Let 𝐹𝑛+𝑘 = 𝐹𝑛−1𝐹𝑘 + 𝐹𝑛𝐹𝑘+1 and 𝐹𝑛+(𝑘+1) = 𝐹𝑛−1𝐹𝑘+1 + 𝐹𝑛𝐹𝑘+2.

Adding these two equations, we get

𝐹𝑛+(𝑘+2) = 𝐹𝑛−1𝐹𝑘+2 + 𝐹𝑛𝐹𝑘+3

Hence, 𝐹𝑛+𝑚 = 𝐹𝑛−1𝐹𝑚 + 𝐹𝑛𝐹𝑚+1.

Note:

To derive new identities, we now present an explicit formula for 𝐹𝑛 .

Let 𝛼 and 𝛽 be the roots of the quadratic equation 𝑥2 − 𝑥 − 1 = 0,

so, 𝛼 = (1 + 5)/2 and 𝛽 = (1 − 5)/2

Then 𝛼 + 𝛽 = 1 and 𝛼𝛽 = −1

Besides, 𝛼2 = 𝛼 (1 − 𝛽) = 𝛼 − 𝛼𝛽 = 𝛼 + 1

𝛼3 = 𝛼(𝛼 + 1) = 𝛼2 + 𝛼 = 2𝛼 + 1

and 𝛼4 = 𝛼 2𝛼 + 1 = 2𝛼2 + 𝛼 = 2(𝛼 + 1) + 𝛼 = 3𝛼 + 2

Thus we have;

𝛼 = 1𝛼 + 0

𝛼2 = 1𝛼 + 1

𝛼3 = 2𝛼 + 1

𝛼4 = 3𝛼 + 2

Notice an interesting pattern emerging. The constant term and the coefficient of

𝛼 on the RHS appear to be adjacent Fibonacci numbers.

Accordingly, we have the following result.

Lemma 2.1

𝛼𝑛 = 𝛼𝐹𝑛 + 𝐹𝑛−1 , where 𝑛 ≥ 0.

Corollary 2.3

𝛽𝑛 = 𝛽𝐹𝑛 + 𝐹𝑛−1 , where 𝑛 ≥ 0.

Note:

Let 𝑢𝑛 = 𝛼𝑛 − 𝛽𝑛 / 5 , where 𝑛 ≥ 1. Then,

𝑢1 = 𝛼−𝛽

5 =

5

5 = 1 and

𝑢2 = 𝛼2−𝛽2

5 =

𝛼+𝛽 (𝛼−𝛽)

5 = 1

Suppose 𝑛 ≥ 3. Then,

𝑢𝑛−1 + 𝑢𝑛−2 = 𝛼𝑛−1− 𝛽𝑛−1

5 +

𝛼𝑛−2− 𝛽𝑛−2

5

= 𝛼𝑛−2 𝛼+1 −𝛽𝑛−2 𝛽+1

5

= 𝛼𝑛−2𝛼2− 𝛽𝑛−2𝛽2

5

= 𝛼𝑛− 𝛽𝑛

5

= 𝑢𝑛

Thus 𝑢𝑛 satisfies the Fibonacci recurrence relation and the two initial conditions.

This gives us an explicit formula for 𝐹𝑛 : 𝐹𝑛 = 𝑢𝑛

Theorem 2.6

Let 𝛼 be the positive root of the quadratic equation 𝑥2 − 𝑥 − 1 = 0 and 𝛽 its

negative root. Then

𝐹𝑛 = 𝛼𝑛−𝛽𝑛

𝛼−𝛽 , where 𝑛 ≥ 1

This explicit formula for 𝐹𝑛 is called Binet’s formula, after the French

mathematician Jacques-Phillipe-Marie Binet, who discovered it in 1843.

Corollary 2.4 (Lucas, 1876)

𝐹𝑛+12 + 𝐹𝑛

2 = 𝐹2𝑛+1 (2.11)

𝐹𝑛+12 − 𝐹𝑛−1

2 = 𝐹2𝑛 (2.12)

For example,

𝐹92 + 𝐹8

2 = 342 + 212 = 1156 + 441 = 1597 = 𝐹17

𝐹112 − 𝐹9

2 = 892 − 342 = 7921−1156 = 6765 = 𝐹20

Corresponding to Binet’s formula for 𝐹𝑛 , there is one for 𝐿𝑛 also.

Theorem 2.7

Let 𝑛 ≥ 1. Then 𝐿𝑛 = 𝛼𝑛 + 𝛽𝑛

Corollary 2.5

𝐹2𝑛 = 𝐹𝑛𝐿𝑛 (2.13)

𝐹𝑛−1 + 𝐹𝑛+1 = 𝐿𝑛 (2.14)

𝐹𝑛+2 − 𝐹𝑛−2 = 𝐿𝑛 (2.15)

𝐿𝑛−1 + 𝐿𝑛+1 = 5𝐹𝑛 (2.16)

For example,

𝐹22 = 17711 = 89 · 199 = 𝐹11𝐿11

𝐹15 + 𝐹17 = 610 + 1597 = 2207 = 𝐿16

𝐹15 − 𝐹11 = 610 − 89 = 521 = 𝐿13

𝐿12 + 𝐿14 = 322 + 843 = 1165 = 5 · 233 = 5 · 𝐹13

Identity (2.13) implies that when 𝑛 ≥ 3, every Fibonacci number

𝐹2𝑛 with an even subscript has nontrivial factors. According to identity (2.14), the

sum of any two Fibonacci numbers that are two units away is a Lucas number.

Likewise, by identity (2.15), the difference of any two Fibonacci numbers that lie

four units away is also a Lucas number.

Identity (2.13) has an interesting by-product.

Let 2𝑛 = 2𝑚 , where 𝑚 ≥ 1. Then,

𝐹2𝑚 = 𝐿2𝑚−1 𝐹2𝑚−1

= 𝐿2𝑚−1 (𝐿2𝑚−2 𝐹2𝑚−2 )

= 𝐿2𝑚−1 𝐿2𝑚−2 𝐹2𝑚−2

= 𝐿2𝑚−1 𝐿2𝑚−2 (𝐿2𝑚−3 𝐹2𝑚−3 )

= 𝐿2𝑚−1 𝐿2𝑚−2 𝐿2𝑚−3 𝐹2𝑚−3

Continuing like this we get,

𝐹2𝑚 = 𝐿2𝑚−1 𝐿2𝑚−2 ………𝐿8 𝐿4 𝐿2 𝐿1

For example,

𝐹16 = 𝐿8𝐿4𝐿2𝐿1 = 47 · 7 · 3 · 1 = 987

Theorem 2.8

A positive integer 𝑛 is a Fibonacci number if and only if 5𝑛2 ± 4 is a perfect

square.

Proof:

We have,

(−1)𝑟 + 𝐹𝑟2 = 𝐹𝑟+1𝐹𝑟−1 [Cassini’s Formula]

and 𝐿𝑟 = 𝐹𝑟+1 + 𝐹𝑟−1

∴ 𝐿𝑟2 − 4 −1 𝑟 + 𝐹𝑟

2 = 𝐹𝑟+1 + 𝐹𝑟−1 2 − 4𝐹𝑟+1𝐹𝑟−1

= (𝐹𝑟+1 − 𝐹𝑟−1)2

= 𝐹𝑟2

𝐿𝑟2 = 5𝐹𝑟

2 +4(−1)𝑟

Thus if 𝑛 is a Fibonacci number, then 5𝑛2 ± 4 is a perfect square.

Conversely, let 5𝑛2 ± 4 be a perfect square 𝑚2 . Then,

𝑚2 − 5𝑛2 = ±4

𝑚+𝑛 5

2.𝑚−𝑛 5

2 = ±1

Since 𝑚 and 𝑛 have the same parity (both odd or both even), both

(𝑚 + 𝑛 5)/2 and (𝑚− 𝑛 5)/2 are integers in the extension field 𝑄( 5) =

{ 𝑥 + 𝑦 5|𝑥,𝑦 ∈ 𝑄}, where 𝑄 denotes the set of rational numbers. Since their

product is ±1, they must be units in the field. But the only integral units in 𝑄( 5)

are of the form ±𝛼±𝑖 . Then,

𝑚+𝑛 5

2 = 𝛼𝑖 =

1

2 [(𝛼𝑖 + 𝛽𝑖 ) + (𝛼𝑖 − 𝛽𝑖) ]

= 𝐿𝑖+𝐹𝑖 5

2

Thus 𝑛 = 𝐹𝑖 , a Fibonacci number.

NUMBER OF DIGITS IN 𝑭𝒏 AND 𝑳𝒏

Binet’s formula can be successfully employed to predetermine the number of

digits in 𝐹𝑛 and 𝐿𝑛 . We can show this by writing 𝐹𝑛 as 𝐹𝑛 = 𝛼𝑛

5 [1 −

𝛽

𝛼 𝑛

]

Since |𝛽| < |𝛼| , 𝛽 𝛼 𝑛 → 0 as 𝑛 → ∞. Therefore, when 𝑛 is sufficiently large,

𝐹𝑛 ≈ 𝛼𝑛

5

log 𝐹𝑛 ≈ 𝑛 log 𝛼 − (log 5)/2

Number of digits in 𝐹𝑛 = 1 + characteristic of log 𝐹𝑛 = log𝐹𝑛

= 𝑛 log𝛼 − (log 5 )/2

= 𝑛 log( 1 + 5 ) − log 2 − (log 5 )/2

For example,

The number of digits in 𝐹25 is given by,

25 log 1 + 5 − log 2 − log 5 /2 = 4.875206004 = 5

Note that F25 = 75025 , i.e, it exactly have 5 digits which establishes the

result.

Since 𝐿𝑛 = 𝛼𝑛 + 𝛽𝑛 , it follows that when 𝑛 is sufficiently large, 𝐿𝑛 ≈ αn, so that

log 𝐿𝑛 ≈ 𝑛 log 𝛼 . Thus the number of digits in 𝐿𝑛 is given by

log𝐿𝑛 = 𝑛 log𝛼 = 𝑛 [log 1 + 5 − log 2]

For example,

The number of digits in 𝐿28 is given by,

28 log 1 + 5 − log 2 = 5.851653927 = 6

Note that 𝐿28 = 710647, i.e, it exactly have 6 digits which establishes the result.

Theorem 2.9

A positive integer 𝑛 is a Lucas number if and only if 5𝑛2 ± 20 is a

perfect square.

Proof:

Let 𝑛 = 𝐿2𝑚+1. Then

5𝑛2 + 20 = 5(𝛼2𝑚+1 + 𝛽2𝑚+1)2 + 20

= 5[𝛼4𝑚+2 + 𝛽4𝑚+2 + 2 𝛼𝛽)2𝑚+1 + 20

= 5[𝛼4𝑚+2 + 𝛽4𝑚+2 − 2(𝛼𝛽)2𝑚+1]

= 5(𝛼2𝑚+1 − 𝛽2𝑚+1)2

= 5( 5𝐹2𝑚+1)2 = 25𝐹2𝑚+1 2

On the otherhand, let 𝑛 = 𝐿2𝑚 . Then

5𝑛2 − 20 = 5(𝛼2𝑚 + 𝛽2𝑚)2 − 20 = 5[𝛼4𝑚 + 𝛽4𝑚 + 2 𝛼𝛽)2𝑚 − 20

= 5[𝛼4𝑚 + 𝛽4𝑚 − 2 𝛼𝛽)2𝑚 = 25𝐹2𝑚 2

Thus, if 𝑛 is a Lucas number, then 5𝑛2 ± 20 is a perfect square.

Because the proof of the converse is a bit complicated, we omit it.

For example,

Let 𝑛 = 1364 = 𝐿15 . Then 5𝑛2 + 20 = 5 · 13642 + 20 = 9302500 = 30502, a

perfect square.

Let 𝑛 = 322 = 𝐿12 . Then 5𝑛2 − 20 = 5 · 3222 − 20 = 518400 = 7202, a

perfect square.

With Binet's formulas at hand, we can extend the definitions of 𝐹𝑛 and 𝐿𝑛 to

negative subscripts also. If we apply the Fibonacci recurrence relation to the

negative side, we get

. . . 𝐹−4 𝐹−3 𝐹−2 𝐹−1 𝐹0 𝐹1 𝐹2 𝐹3 𝐹4 . . . . . . −3 2 − 1 1 0 1 1 2 3 . . .

So, it appears that 𝐹−𝑛 = (−1)𝑛+1𝐹𝑛 ,𝑛 ≥ 1. To prove this, assume Binet's formula holds for negative exponents

𝐹−𝑛 = 𝛼−𝑛− 𝛽−𝑛

5 =

(−𝛽)𝑛− (−𝛼)𝑛

5 since 𝛼𝛽 = −1

= (−1)𝑛 (𝛽𝑛− 𝛼𝑛 )

5 =

(−1)𝑛+1 (𝛼𝑛− 𝛽𝑛 )

5

= (−1)𝑛+1𝐹𝑛 (2.18)

Likewise,

𝐿−𝑛 = (−1)𝑛 𝐿𝑛 (2.19)

Thus, 𝐹−𝑛 = 𝐹𝑛 if and only if 𝑛 is odd, and 𝐿−𝑛 = 𝐿𝑛 if and only if 𝑛 is even.

A formula for 𝛼−𝑛 can now be derived easily.

Since 𝛼𝑛 = 𝛼 𝐹𝑛 + 𝐹𝑛−1

(Lemma 2.1), it follows that

𝛼−𝑛 = 𝛼 𝐹−𝑛 + 𝐹−𝑛−1

= 𝛼 (−1)𝑛+1 𝐹𝑛 + (−1)𝑛+2 𝐹𝑛+1

= (−1)𝑛+1 (𝛼 𝐹𝑛 − 𝐹𝑛+1)

= 𝛼𝐹𝑛 − 𝐹𝑛+1 if 𝑛 is odd 𝐹𝑛+1 − 𝛼 𝐹𝑛 otherwise

(2.20)

For example,

α−12 = F11 − αF12 = 89 − 144α

Formula 2.20 can also be established by Principle of Mathematical Induction or by

showing that 𝛼𝑛 𝛼𝐹𝑛 − 𝐹𝑛+1 = −1 𝑛+1, using Binet's formula.

Likewise,

𝛽−𝑛 = 𝛽𝐹𝑛 − 𝐹𝑛+1 if 𝑛 is odd𝐹𝑛+1 − 𝛽𝐹𝑛 otherwise

(2.21)

Notice two intriguing patterns that emerge from 𝛼−𝑛 :

α−1 = 1 · α − 1

α−2 = 2 − 1 · 𝛼

α−3 = 2 · α − 3

α−4 = 5 − 3 · α

α−5 = 5 · α − 8

⋮

They are indicated by the two crisscrossing arrows: The absolute values of the

coefficients of 𝛼 are consecutive Fibonacci numbers, and so are the absolute values

of the various constants.

The summation Formulas (2.1) through (2.3) are a special case of the generalized

summation formula, given in the next theorem. In addition, the theorem yields an

array of fascinating formulas as by-products. Its proof is a consequence of Binet's

formulas and the geometric summation formula

𝑟𝑖𝑛−1

𝑖=0

=𝑟𝑛 − 1

𝑟 − 1 where 𝑟 ≠ 1

Theorem 2.10 (Koshy, 1998)

Let 𝑘 ≥ 1 and 𝑗 any integer. Then

𝐹𝑘𝑖+𝑗

𝑛

𝑖=0

=

𝐹𝑛𝑘+𝑘+𝑗 − −1 𝑘 𝐹𝑛𝑘+𝑗 − 𝐹𝑗 − −1 𝑗 𝐹𝑘−𝑗

𝐿𝑘 − −1 𝑘 − 1 if 𝑗 < 𝑘

𝐹𝑛𝑘+𝑘+𝑗 − −1 𝑘 𝐹𝑛𝑘+𝑗 − 𝐹𝑗 + −1 𝑘 𝐹𝑗−𝑘

𝐿𝑘 − −1 𝑘 − 1 otherwise

(2.22)

Proof:

𝐹𝑘𝑖+𝑗 = 𝛼𝑘𝑖+𝑗 − 𝛽𝑘𝑖+𝑗

5

𝑛

𝑖=0

𝑛

𝑖=0

= 1

5 𝛼𝑗 𝛼𝑘𝑖 − 𝛽𝑗 𝛽𝑘𝑖

=1

5 𝛼𝑗 .

𝛼𝑛𝑘+𝑘 − 1

𝛼𝑘 − 1− 𝛽𝑗 .

𝛽𝑛𝑘+𝑘 − 1

𝛽𝑘 − 1

= 𝛼𝑛𝑘 +𝑘+𝑗 − 𝛼 𝑗 𝛽𝑘− 1 − 𝛽𝑛𝑘 +𝑘+𝑗 − 𝛽 𝑗 𝛼𝑘 − 1

5 𝛼𝛽 𝑘 − 𝛼𝑘+𝛽𝑘 + 1

= −𝐹𝑛𝑘 +𝑘+𝑗 + (−1)𝑘 𝐹𝑛𝑘 +𝑗 + 𝐹𝑗+ (𝛼𝑘𝛽 𝑗 − 𝛼 𝑗𝛽𝑘)/ 5

(−1)𝑘 − 𝐿𝑘+ 1

But 𝛼𝑘𝛽𝑗 − 𝛼𝑗𝛽𝑘 = 𝛼𝛽 𝑗 𝛼𝑘−𝑗 − 𝛽𝑘−𝑗 if 𝑗 < 𝑘

𝛼𝛽 𝑘 𝛽𝑗−𝑘 − 𝛼𝑗−𝑘 otherwise

= −1 𝑗 5𝐹𝑘−𝑗 if 𝑗 < 𝑘

−1 𝑘+1 5𝐹𝑗−𝑘 otherwise

∴ 𝐹𝑘𝑖+𝑗 =

𝐹𝑛𝑘+𝑘+𝑗 − −1 𝑘𝐹𝑛𝑘+𝑗 − 𝐹𝑗 − −1 𝑗𝐹𝑘−𝑗

𝐿𝑘 − −1 𝑘 − 1 if 𝑗 < 𝑘

𝐹𝑛𝑘+𝑘+𝑗 − −1 𝑘𝐹𝑛𝑘+𝑗 − 𝐹𝑗 − −1 𝑗𝐹𝑗−𝑘

𝐿𝑘 − −1 𝑘 − 1 otherwise

𝑛

𝑖=0

Letting 𝑗 = 0 in this formula yields the following result.

Corollary 2.6 (Koshy, 1998)

𝐹𝑘𝑖 = 𝐹𝑛𝑘+𝑘 − (−1)𝑘 𝐹𝑛𝑘 − 𝐹𝑘

𝐿𝑘 − (−1)𝑘 − 1

𝑛

𝑖=1

(2.23)

CHAPTER - 3

DIVISIBILITY PROPERTIES

Theorem 3.1

𝐹𝑚 |𝐹𝑚𝑛

Proof:

We can prove this theorem by applying the Principle of Mathematical

Induction.

The given statement is clearly true when 𝑛 = 1.

Now assume it is true for all integers 1 through 𝑘, where 𝑘 ≥ 1.

i.e. 𝐹𝑚 |𝐹𝑚𝑖 for every 𝑖, where 1 ≤ 𝑖 ≤ 𝑘.

To show that 𝐹𝑚 |𝐹𝑚 𝑘+1 , we use the identity: 𝐹𝑟+𝑠 = 𝐹𝑟−1𝐹𝑠 + 𝐹𝑟𝐹𝑠+1

∴ 𝐹𝑚 (𝑘+1) = 𝐹𝑚𝑘 +𝑚 = 𝐹𝑚𝑘−1𝐹𝑚 + 𝐹𝑚𝑘 𝐹𝑚+1

Since 𝐹𝑚 |𝐹𝑚𝑘 , by the induction hypothesis, it follows that 𝐹𝑚 |𝐹𝑚 𝑘+1 .

Thus by the strong version of the Principle of Mathematical Induction,

the result is true for all integers 𝑛 ≥ 1.

For example,

𝐹7 = 13, and 𝐹28 = 317811. Since 7|28, it follows by the theorem that 13|317811.

Particularly,

𝐹3 ,𝐹6 ,𝐹9 ……… are all divisible by 𝐹3



Theorem 3.2

If 𝐹𝑚 |𝐹𝑛 , then 𝑚|𝑛.

Proof:

𝐹𝑚 ,𝐹𝑛 = 𝐹𝑞𝑛+𝑟 , 𝐹𝑛

= 𝐹𝑞𝑛−1𝐹𝑟 + 𝐹𝑞𝑛𝐹𝑟+1𝐹𝑛 , 𝐹𝑛 by theorem (2.5)

= 𝐹𝑞𝑛−1𝐹𝑟 , 𝐹𝑛

= 𝐹𝑟 ,𝐹𝑛 by Lemma 3.1

= 𝐹𝑛 , 𝐹𝑟

Theorem 3.3

𝐹𝑚 ,𝐹𝑛 = 𝐹 𝑚 ,𝑛

Proof:

Suppose 𝑚 ≥ 𝑛. Applying the Euclidean algorithm with 𝑚 as the dividend

and 𝑛 as the divisor, we get the following sequence of equations:

𝑚 = 𝑞0𝑛 + 𝑟1 0 ≤ 𝑟1 < 𝑛

𝑛 = 𝑞1𝑟1 + 𝑟2 0 ≤ 𝑟2 < 𝑟1

𝑟1 = 𝑞2𝑟2 + 𝑟3 0 ≤ 𝑟3 < 𝑟2

⫶

𝑟𝑛−2 = 𝑞𝑛−1𝑟𝑛−1 + 𝑟𝑛 0 ≤ 𝑟𝑛 < 𝑟𝑛−1

𝑟𝑛−1 = 𝑞𝑛𝑟𝑛 + 0

By Lemma 3.2, 𝐹𝑚 ,𝐹𝑛 = 𝐹𝑛 ,𝐹𝑟1 = 𝐹𝑟1

,𝐹𝑟2 = ⋯ ··· = 𝐹𝑟𝑛−1

,𝐹𝑟𝑛 .

But 𝑟𝑛 |𝑟𝑛−1, so 𝐹𝑟𝑛 |𝐹𝑟𝑛−1, by Theorem 3.1. Therefore, (𝐹𝑟𝑛−1

,𝐹𝑟𝑛 ) = 𝐹𝑟𝑛 . Thus

(𝐹𝑚 ,𝐹𝑛) = 𝐹𝑟𝑛 .

But, by Euclidean algorithm, 𝑟𝑛 = (𝑚,𝑛) ; therefore, (𝐹𝑚 ,𝐹𝑛) = 𝐹(𝑚 ,𝑛).

For example,

F16 , F24 = 𝐹 16,24 = 𝐹8 = 21. That is, 987, 46368 = 21

Corollary 3.3

If 𝑚 and 𝑛 are relatively prime, then so are 𝐹𝑚 and 𝐹𝑛 .

For example,

19, 27 = 1, so F19, F27 = 4181, 196418 = 1

Corollary 3.4

If 𝐹𝑚 |𝐹𝑛 , then 𝑚|𝑛.

Proof:

Suppose 𝐹𝑚 |𝐹𝑛 . Then 𝐹𝑚 ,𝐹𝑛 = 𝐹𝑚 = 𝐹(𝑚 ,𝑛), by Theorem 3.3;

∴ 𝑚 = 𝑚,𝑛 . Thus 𝑚ǀ𝑛.

Corollary 3.5

There are infinitely many primes.

Proof:

Suppose there is only a finite number of primes, 𝑝1 ,𝑝2 ,…,and 𝑝𝑘 .

Then consider the Fibonacci numbers 𝐹𝑝1,𝐹𝑝2

,…,and 𝐹𝑝𝑘 . Clearly, they are pair

wise relatively prime.

Since there are only 𝑘 primes, each of these Fibonacci numbers has exactly

one prime factor, that is, each is a prime.

This is a contradiction, since 𝐹19 = 4181 = 37 · 113.

Thus our assumption that there are only finitely many primes is false.

In other words, thus there are infinitely many primes.

A quick look at Lucas numbers shows that every third Lucas number is even,

that is, 2|𝐿3𝑛 . This is, in fact, always true.

Theorem 3.4

𝐿𝑚 |𝐹𝑛 if and only if 2𝑚|𝑛, where 𝑚 ≥ 2.

For example,

12|24, so 𝐿6|𝐹24 ; 18|46368

Theorem 3.5

𝐿𝑚 |𝐿𝑛 if and only if 𝑛 = 2𝑘 − 1 𝑚, where 𝑚 ≥ 2 and 𝑘 ≥ 1.

For example,

Let 𝑚 = 3 , and 𝑛 = 5 · 3 = 15. We have 𝐿3 = 4 and 𝐿15 = 1364. Clearly,

𝐿3|𝐿15 .

Theorem 3.6 (Freeman, 1967)

1. Let 𝑎: 𝑏 = 𝐹𝑛 :𝐹𝑛+1. Then 𝑎 + 𝑏 𝐹𝑛−1 = 𝑎, 𝑏 + −1 𝑛 𝑎, 𝑏 ,where

𝑛 ≥ 2.

2. Let (𝑐,𝑑) = 1 such that 𝑎: 𝑏 = 𝑐:𝑑. Let (𝑎 + 𝑏)𝐹𝑛−1 = [𝑎, 𝑏] +

−1 𝑛 𝑎, 𝑏 , where 𝑛 ≥ 3. Then the number of solutions of the ratio 𝑐:𝑑

equals one-half the number of positive factors of 𝐹𝑛𝐹𝑛−2, one of them being

𝐹𝑛𝐹𝑛+1.

Proof:

1. Let 𝑎: 𝑏 = 𝐹𝑛 :𝐹𝑛+1.

Then, since (𝐹𝑛 ,𝐹𝑛+1) = 1, 𝑎 = 𝐹𝑛𝑘 , 𝑏 = 𝐹𝑛+1𝑘, 𝑎, 𝑏 = 𝑘, 𝑎, 𝑏 =

𝐹𝑛𝐹𝑛+1𝑘 for some positive integer 𝑘.

∴ 𝑎 + 𝑏 𝐹𝑛−1 = 𝐹𝑛−1 𝐹𝑛 + 𝐹𝑛+1 𝑘 = 𝐹𝑛−1𝐹𝑛+2𝑘

= 𝐹𝑛+1 − 𝐹𝑛 𝐹𝑛+2𝑘 = 𝐹𝑛+1 𝐹𝑛 + 𝐹𝑛+1 𝑘 − 𝐹𝑛𝐹𝑛+2𝑘

= 𝐹𝑛𝐹𝑛+1𝑘 + 𝐹𝑛+12 − 𝐹𝑛𝐹𝑛+2 𝑘

= 𝑎, 𝑏 + −1 𝑛 𝑎, 𝑏 by Cassini’s

rule

2. Let 𝑎: 𝑏 = 𝑐:𝑑, where (𝑐 ,𝑑) = 1 .

Then 𝑎 = 𝑐𝑘, 𝑏 = 𝑑𝑘, (𝑎, 𝑏) = 𝑘 and [𝑎, 𝑏] = 𝑐𝑑𝑘 for some positive integer 𝑘.

Since (𝑎 + 𝑏) 𝐹𝑛−1 = [𝑎, 𝑏] + −1 𝑛 (𝑎, 𝑏), we have

𝑐 + 𝑑 𝐹𝑛−1 = 𝑐𝑑 + −1 𝑛

This yields

𝑐 =𝑑𝐹𝑛−1 − −1 𝑛

𝑑 − 𝐹𝑛−1

= 𝐹𝑛−1 +𝐹𝑛−1

2 − −1 𝑛

𝑑 − 𝐹𝑛−1

= 𝐹𝑛−1 +𝐹𝑛𝐹𝑛−2

𝑑 − 𝐹𝑛−1 (3.1)

If 0 < 𝑑 < 𝐹𝑛−1, then 𝑐 < 0; so 𝑑 > 𝐹𝑛−1.

Since 𝑐 is an integer, 𝑑 − 𝐹𝑛−1|𝐹𝑛𝐹𝑛−2.

Thus Eq. (3.1) yields a value of 𝑐 for every positive factor of 𝐹𝑛𝐹𝑛−2.

But, if 𝑐 = 𝐴, 𝑑 = 𝐵 is a solution of the ratio 𝑐:𝑑, then so is 𝑐 = 𝐵,𝑑 = 𝐴.

Thus the number of distinct values of the ratio 𝑐:𝑑 equals the number of

p

positive factors of 𝐹𝑛𝐹𝑛−2.

In particular, let 𝑑 = 𝐹𝑛+1. Then

𝑐 = 𝐹𝑛−1 +𝐹𝑛𝐹𝑛−2

𝐹𝑛+1 − 𝐹𝑛−1

= 𝐹𝑛−1 +𝐹𝑛𝐹𝑛−2

𝐹𝑛= 𝐹𝑛

Thus 𝑐:𝑑 = 𝐹𝑛 :𝐹𝑛+1 is also a value of the ratio.

Example 3.2

1. Let 𝑎: 𝑏 = 𝐹11 :𝐹12 = 89: 144 , so 𝑛 = 11. Let 𝑎 = 445,𝑏 = 720, so

𝑎: 𝑏 = 89: 144

𝑎, 𝑏 + −1 𝑛 𝑎, 𝑏 = 64080 − 5 = 64075 = 445 + 720 · 55 = 𝑎 + 𝑏 𝐹10

2. Since (𝑎 + 𝑏)𝐹10 = [𝑎, 𝑏] + −1 11 (𝑎, 𝑏), it follows that

𝑐 = 𝐹10 + 𝐹11𝐹9

𝑑−𝐹10

= 55 +89 · 34

𝑑 − 55= 55 +

3026

𝑑 − 55

Since 3026 = 2 · 17 · 89, 3026 has eight positive factors: 1, 2, 17, 34,89, 178, 1513 and 3026. So 𝑑 has eight possible values: 56, 57, 72,89, 144, 233, 1568 and 3081. Consequently, the various values of 𝑐:𝑑 are

3081: 56, 1568: 57, 233: 72, 144: 89, 89: 144, 72: 233, 57: 1568 and 56: 3081. Since one-half of them are duplicates, the four distinct values of

𝑐:𝑑 are 89: 144, 72: 233, 57: 1568 and 56: 3081, keeping the numerator to

be smaller. Notice that one of the ratios is 89: 144 = 𝐹11 :𝐹12, as expected.

Lemma 3.3

𝐹2𝑛−1 = 𝐹𝑛+1𝐿𝑛+2 − 𝐿𝑛𝐿𝑛+1 , 𝑛 ≥ 2.

Theorem 3.7

1. Let 𝑎: 𝑏 = 𝐿𝑛 : 𝐿𝑛+1. Then 𝑎 + 𝑏 𝐹𝑛+1 = 𝑎, 𝑏 + 𝑎, 𝑏 𝐹2𝑛−1, 𝑛 ≥ 2.

2. Let 𝑎: 𝑏 = 𝐹𝑛−2:𝐹𝑛−1. Then 𝑎 + 𝑏 𝐹𝑛+1 = 𝑎, 𝑏 + 𝑎, 𝑏 𝐹2𝑛−1, 𝑛 ≥ 3.

3. Let (𝑐,𝑑) = 1 such that 𝑎: 𝑏 = 𝑐:𝑑. If 𝑎 + 𝑏 𝐹𝑛+1 = 𝑎, 𝑏 + 𝑎, 𝑏 𝐹2𝑛−1

where 𝑛 ≥ 2, then the ratios 𝑐:𝑑 are determined by the positive factors of

𝐹𝑛+12 − 𝐹2𝑛−1, one of them being 𝐿𝑛 : 𝐿𝑛+1. For 𝑛 ≥ 3, 𝐹𝑛−2:𝐹𝑛−1 is also a

solution.

Proof:

1. Let 𝑎: 𝑏 = 𝐿𝑛 : 𝐿𝑛+1. Since 𝐿𝑛 , 𝐿𝑛+1 = 1, 𝑎 = 𝑘𝐿𝑛 , 𝑏 = 𝑘𝐿𝑛+1, 𝑎, 𝑏 =

𝑘, and 𝑎, 𝑏 = 𝐿𝑛𝐿𝑛+1𝑘 for some positive integer 𝑘. Then

(𝑎 + 𝑏)𝐹𝑛+1 = (𝐿𝑛 + 𝐿𝑛+1)𝑘𝐹𝑛+1 = 𝐹𝑛+1𝐿𝑛+2𝑘

= (𝐹2𝑛−1 + 𝐿𝑛𝐿𝑛+1)𝑘

= [𝑎, 𝑏] + (𝑎, 𝑏)𝐹2𝑛−1

as desired.

2. Suppose 𝑎 ∶ 𝑏 = 𝐹𝑛−2:𝐹𝑛−1. Then = 𝑘𝐹𝑛−2, 𝑏 = 𝑘𝐹𝑛−1, (𝑎 , 𝑏) = 𝑘 , and

[𝑎, 𝑏] = 𝐹𝑛−1𝐹𝑛−2𝑘 for some positive integer 𝑘. Then

(𝑎, 𝑏)𝐹𝑛+1 = (𝐹𝑛−2 + 𝐹𝑛−1)𝑘𝐹𝑛+1 = 𝐹𝑛𝐹𝑛+1𝑘

= 𝐹2𝑛−1 + 𝐹𝑛−1𝐹𝑛−2 𝑘

Since 𝐹2𝑛−1 = 𝐹𝑛𝐹𝑛+1 − 𝐹𝑛−2𝐹𝑛−1

= [ 𝑎, 𝑏 ] + ( 𝑎 , 𝑏 ) 𝐹2𝑛−1

again as desired.

3. Let 𝑎 ∶ 𝑏 = 𝑐 ∶ 𝑑, where 𝑐 ,𝑑 = 1. As before, 𝑎 = 𝑐𝑘 , 𝑏 = 𝑑𝑘, 𝑎 , 𝑏 = 𝑘, and [ 𝑎 , 𝑏 ] = 𝑐𝑑𝑘 for some positive integer 𝑘.

Since 𝑎 + 𝑏 𝐹𝑛+1 = 𝑎, 𝑏 + 𝑎 , 𝑏 𝐹2𝑛−1

𝑐 + 𝑑 𝐹𝑛+1 = 𝑐𝑑 + 𝐹2𝑛−1

𝑐 =𝑑𝐹𝑛+1 − 𝐹2𝑛−1

𝑑 − 𝐹𝑛+1 (3.2)

= 𝐹𝑛+1 + 𝐹𝑛+1

2 −𝐹2𝑛−1

𝑑−𝐹𝑛−1

Since 𝑐 and 𝑑 are positive integers, it follows that the ratio 𝑐:𝑑 is

determined by the positive factors of 𝐹𝑛+12 − 𝐹2𝑛−1. d

In particular, let 𝑑 = 𝐹𝑛+1. Then by Lemma 3.3,

𝑐 = 𝐹𝑛+1 + 𝐹𝑛+1

2 −𝐹𝑛+1𝐿𝑛+2+𝐿𝑛𝐿𝑛+1

𝐿𝑛+1−𝐹𝑛+1

= 𝐹𝑛+1 𝐿𝑛+1−𝐿𝑛+2 +𝐿𝑛𝐿𝑛+1

𝐿𝑛+1− 𝐹𝑛+1

= 𝐿𝑛𝐿𝑛+1−𝐿𝑛𝐹𝑛+1

𝐿𝑛+1−𝐹𝑛+1 = 𝐿𝑛

Thus 𝐿𝑛 : 𝐿𝑛+1 is a solution of the ratio 𝑐:𝑑. ( By symmetry, 𝐿𝑛+1: 𝐿𝑛 is also

a solution).

Unlike Theorem 3.7, not all solutions are obtained by considering the case

𝑑 > 𝐹𝑛+1.. For instance, let 𝑑 = 𝐹𝑛−1.Then by Eq.(3.2).

𝑐 =𝐹𝑛−1𝐹𝑛+1 − 𝐹2𝑛−1

𝐹𝑛−1 − 𝐹𝑛+1

= 𝐹𝑛−1𝐹𝑛+1− 𝐹𝑛𝐹𝑛+1− 𝐹𝑛−2𝐹𝑛−1

−𝐹𝑛

=−𝐹𝑛+1 𝐹𝑛−1− 𝐹𝑛 + 𝐹𝑛−2𝐹𝑛−1

𝐹𝑛

=−𝐹𝑛−2𝐹𝑛+1+ 𝐹𝑛−2𝐹𝑛−1

𝐹𝑛

= = 𝐹𝑛−2(𝐹𝑛−1−𝐹𝑛+1)

𝐹𝑛 = 𝐹𝑛−2

Thus 𝐹𝑛−2:𝐹𝑛−1 is also a solution of the ratio.

Example 3.3.

Let 𝑛 = 11. We have 𝐹𝑛+1 = 𝐹12 = 144 and 𝐹2𝑛−1 = 𝐹21 = 10946.

1. Let 𝑎: 𝑏 = 𝐿𝑛 : 𝐿𝑛+1 = 𝐿11 : 𝐿12 = 199: 322. Let 𝑎 = 597 and 𝑏 = 966.

Then

𝑎, 𝑏 + 𝑎, 𝑏 𝐹2𝑛−1 = 597, 966 + 597, 966 · 10946

= 192234 + 3 · 10946 = 225072

= 597 + 966 · 144

= (𝑎 + 𝑏)𝐹𝑛+1

2. Let 𝑎: 𝑏 = 𝐹𝑛−2:𝐹𝑛−1 = 𝐹9:𝐹10 = 34: 55. Let 𝑎 = 272 and 𝑏 = 440.

Then 𝑎, 𝑏 + 𝑎, 𝑏 𝐹2𝑛−1 = 272 ,440 + 8 · 10946

= 102528 = 272 + 440 · 144

= 𝑎 + 𝑏 𝐹𝑛+1

3. Let 𝑎: 𝑏 = 208: 240 = 13: 15, where 𝑐:𝑑 = 13: 15 and 15, 17 = 1.

Then 𝑐 = 𝐹𝑛+1 + 𝐹𝑛+1

2 −𝐹2𝑛−1

𝑑−𝐹𝑛+1

= 144 + 1442−10,946

𝑑−144

= 144 + 9790

𝑑−144

Since 9790 = 2 · 5 · 11 · 89, 9790 has 16 positive factors: 1, 2, 5, 10, 11,

22, 55, 89, 110, 178, 445, 890, 979, 1958, 4895 and 9790.The

corresponding ratios are 145: 9934, 146: 5039, 149: 2102, 154: 1123,

155: 1034, 166: 589, 199: 322, 233: 254, 254: 233, 322: 199, 589: 166,

1034: 155, 1123: 154, 2102: 149, 5039: 146 and 9934: 145. These yields 8

distinct ratios 𝑐:𝑑 with (𝑐, 𝑑) = 1,namely, 145:9934, 146:5039, 149:2102,

154:1123, 155:1034, 166:589, 199:322 and 233:254. Notice that 34: 55 is

also a solution. Among these ratios we find 𝐿11 : 𝐿12 = 199: 322 and

𝐹9:𝐹10 = 34: 55 as expected.

AN ALTERNATE FIBONACCI SEQUENCE

In 1971, Underwood Dudley and Bessie Tucker of DePauw University in Indiana

investigated a slightly altered Fibonacci sequence, defined by 𝐺𝑛 = 𝐹𝑛 + −1 𝑛 ,

where 𝑛 ≥ 1. They made an interesting observation ,as table 2.1 shows : The

1st,3

rd,5

th,…entries (see the circled numbers) in the (𝐺𝑛 ,𝐺𝑛+1)-row are the 2

nd , 4

th,

6th

,…. Fibonacci numbers; and the 2nd

, 4th

, 6th

,….. entries are the 3rd

, 5th , 7

th,…..

Lucas numbers.

TABLE 3.1.

To establish these two results, we need the following theorem.

Theorem 3.8 (Dudley and Tucker, 1971)

(1) 𝐹4𝑛 + 1 = 𝐹2𝑛−1𝐿2𝑛+1 (2) 𝐹4𝑛 − 1 = 𝐹2𝑛+1𝐿2𝑛−1

(3) 𝐹4𝑛+1 + 1 = 𝐹2𝑛+1𝐿2𝑛 (4) 𝐹4𝑛+1 − 1 = 𝐹2𝑛𝐿2𝑛+1

(5) 𝐹4𝑛+2 + 1 = 𝐹2𝑛+2𝐿2𝑛 (6) 𝐹4𝑛+2 − 1 = 𝐹2𝑛𝐿2𝑛+2

(7) 𝐹4𝑛+3 + 1 = 𝐹2𝑛+1𝐿2𝑛+2 (8) 𝐹4𝑛+3 − 1 = 𝐹2𝑛+2𝐿2𝑛+1

Proof:

The proof requires the following identities:

𝐹𝑚+𝑛 + 𝐹𝑚−𝑛 = 𝐹𝑛𝐿𝑚 if 𝑛 is odd𝐹𝑚𝐿𝑛 otherwise

𝐹𝑚+𝑛 − 𝐹𝑚−𝑛 = 𝐹𝑚𝐿𝑛 if 𝑛 is odd 𝐹𝑛𝐿𝑚 otherwise

Then

𝐹4𝑛 + 1 = 𝐹4𝑛 + 𝐹2 = 𝐹 2𝑛+1 + 2𝑛−1 + 𝐹 2𝑛+1 − 2𝑛−1

= 𝐹2𝑛−1𝐿2𝑛+1

and

𝐹4𝑛+1 + 1 = 𝐹4𝑛+1 + 𝐹1 = 𝐹 2𝑛+1 +2𝑛 + 𝐹 2𝑛+1 −2𝑛

= 𝐹2𝑛+1𝐿2𝑛

The other formulas can be established similarly.

The following corollary, observed in 1971 by Hoggatt, follows easily from this

theorem.

Corollary 3.6

(1) 𝐹4𝑛+1 + 1, 𝐹4𝑛+2 + 1 = 𝐿2𝑛 (2) 𝐹4𝑛+1 + 1,𝐹4𝑛+3 + 1 = 𝐹2𝑛+1

(3) 𝐹4𝑛+1 − 1, 𝐹4𝑛+2 − 1 = 𝐹2𝑛 (4) 𝐹4𝑛+1 − 1,𝐹4𝑛+3 − 1 = 𝐿2𝑛+1

(5) 𝐹4𝑛−1 − 1, 𝐹4𝑛+1 + 1 = 𝐹2𝑛 (6) 𝐹4𝑛−1 + 1,𝐹4𝑛+1 + 1 = 𝐿2𝑛

(7) 𝐹4𝑛+3 + 1, 𝐹4𝑛 − 1 = 𝐹2𝑛+1 (8) 𝐹4𝑛+3 + 1,𝐹4𝑛+2 − 1 = 𝐹2𝑛

(9) 𝐹4𝑛+4 − 1,𝐹4𝑛+3 − 1 = 𝐿2𝑛+1

Although it is not yet known whether or not the Fibonacci sequence contains

infinitely many primes, this theorem establishes their finiteness in the sequences

𝐹𝑛 + 1 and 𝐹𝑛 − 1 , as the next corollary shows.

Corollary 3.7

𝐹𝑛 + 1 is composite if 𝑛 ≥ 4, and 𝐹𝑛 − 1 is composite if 𝑛 ≥ 7.

Proof:

When 𝑛 = 1, 𝐹4𝑛+ 1 = 4 is composite. When 𝑛 ≥ 2, it follows from Theorem 2.9

that 𝐹4𝑛+1 + 1, 𝐹4𝑛+2 + 1, and 𝐹4𝑛+3 + 1 have nontrivial factors. Thus 𝐹𝑛 + 1 is

composite if 𝑛 ≥ 4. Likewise, 𝐹𝑛 − 1 is composite if 𝑛 ≥ 7.

Notice that 𝐹𝑛 + 1 is a prime if 𝑛 < 4 and 𝐹𝑛 − 1 is a prime if 𝑛 < 7.

Corollary 3.8

𝐺4𝑛 ,𝐺4𝑛+1 = 𝐿2𝑛+1 , 𝐺4𝑛+1,𝐺4𝑛+3 = 𝐿2𝑛+1, and 𝐺4𝑛+2,𝐺4𝑛+3 = 𝐹2𝑛+2

where 𝑛 ≥ 1

Proof:

By theorem 3.9,

𝐺4𝑛 ,𝐺4𝑛+1 = 𝐹4𝑛+1 + 1,𝐹4𝑛+1 − 1

= 𝐹2𝑛−1𝐿2𝑛+1 ,𝐹2𝑛𝐿2𝑛+1

= 𝐿2𝑛+1 𝐹2𝑛−1,𝐹2𝑛

= 𝐿2𝑛+1

Similarly, we can prove the other two parts.

Corollary 3.9

Let 𝐻𝑛 = 𝐹𝑛 − −1 𝑛 . Then 𝐻4𝑛 ,𝐻4𝑛+1 = 𝐹2𝑛+1 , 𝐻4𝑛+1,𝐻4𝑛+3 = 𝐹2𝑛+1 and

𝐻4𝑛+2,𝐻4𝑛+3 = 𝐿2𝑛+2, where 𝑛 ≥ 1.

CHAPTER – 4

GENERALIZATION

We can study properties common to Fibonacci and Lucas numbers by

investigating a number sequence that satisfies the Fibonacci recurrence relation,

but with arbitrary initial conditions.

GENERALIZED FIBONACCI NUMBERS

Consider the sequence 𝐺𝑛 , 𝐺1 = a, 𝐺𝑛 = b, and 𝐺𝑛 = 𝐺𝑛−1 + 𝐺𝑛−2, 𝑛 ≥ 3. The

ensuing sequence

𝑎, 𝑏, 𝑎 + 𝑏,𝑎 + 2𝑏, 2𝑎 + 3𝑏, 3𝑎 + 5𝑏, . ..

is called the generalized Fibonacci sequence (GFS).

Take a close look at the coefficients of 𝑎 and 𝑏 in the various terms of this

sequence. They follow an interesting pattern: The coefficients of 𝑎 and 𝑏 are

Fibonacci numbers. In fact, we can pinpoint these two Fibonacci coefficients, as

the following theorem shows.

Theorem 4.1

Let 𝐺𝑛 denote the nth term of the GFS. Then 𝐺𝑛 = 𝑎𝐹𝑛−2 + 𝑏𝐹𝑛−1, 𝑛 ≥ 3.

Proof:

Since 𝐺3 = 𝑎 + 𝑏 = 𝑎𝐹1 + 𝑏𝐹2 , the statement is true when 𝑛 = 3.

Let k be an arbitrary integer ≥ 3. Assume the given statement is true for all

integers 𝑖, where 3 ≤ 𝑖 ≤ 𝑘: 𝐺𝑖 = 𝑎𝐹𝑖−2 + 𝑏𝐹𝑖−1. Then:

𝐺𝑘+1 = 𝐺𝑘 + 𝐺𝑘−1

= 𝑎𝐹𝑘−2 + 𝑏𝐹𝑘−1 + 𝑎𝐹𝑘−3 + 𝑏𝐹𝑘−2

= 𝑎 𝐹𝑘−2 + 𝐹𝑘−3 + 𝑏 𝐹𝑘−1 + 𝑏𝐹𝑘−2

= 𝑎𝐹𝑘−1 + 𝑏𝐹𝑘

Thus, by the principle of mathematical induction the formula holds for every

integer 𝑛 ≥ 3.

Notice that this theorem is in fact true for all 𝑛 ≥ 1.

Theorem 4.2

𝐺𝑘+𝑖

𝑛

𝑖=1

= 𝐺𝑛+𝑘+2 − 𝐺𝑘+2

Proof:

By Theorem 4.1,

𝐺𝑘+𝑖

𝑛

𝑖=1

= 𝑎 𝐹𝑘+𝑖−2

𝑛

𝑖=1

+ 𝑏 𝐹𝑘+𝑖−1

𝑛

𝑖=1

= 𝑎 𝐹𝑛+𝑘 − 𝐹𝑘 + 𝑏 𝐹𝑛+𝑘+1 − 𝐹𝑘+1

= 𝑎𝐹𝑛+𝑘 + 𝑏𝐹𝑛+𝑘+1 − 𝑎𝐹𝑘 + 𝑏𝐹𝑘+1

= 𝐺𝑛+𝑘+2 − 𝐺𝑘+2

Theorem 4.3 (Koshy, 1998)

𝐺𝑖𝐺𝑖+1

𝑛

𝑖=1

= 𝑎2 𝐹𝑛−22 − 𝑣 + 𝑏2 𝐹𝑛−1

2 − 𝑣 + 1

+ 𝑎𝑏 𝐿2𝑛−1 + 5𝐹𝑛−1𝐹𝑛 + 𝑣 + 1 /5

where 𝑣 = 1 if 𝑛 is odd 0 otherwise

Theorem 4.4 (Binet's formula).

Let 𝑐 = 𝑎 + 𝑎 − 𝑏 𝛽 and 𝑑 = 𝑎 + (𝑎 − 𝑏)𝛼. Then

𝐺𝑛 =𝑐𝛼𝑛 − 𝑑𝛽𝑛

𝛼 − 𝛽

Proof:

By Theorem 4.1,

𝐺𝑛 = 𝑎𝐹𝑛−2 + 𝑏𝐹𝑛−1

5𝐺𝑛 = 𝑎 𝛼𝑛−2 − 𝛽𝑛−2 + 𝑏 𝛼𝑛−1 − 𝛽𝑛−1

= 𝛼𝑛 𝑎

𝛼2+

𝑏

𝛼 − 𝛽𝑛

𝑎

𝛽2+

𝑏

𝛽

= 𝛼𝑛 𝑎𝛽2 − 𝑏𝛽 − 𝛽𝑛 𝑎𝛼2 − 𝑏𝛼

= 𝛼𝑛 𝑎 + 𝑎 − 𝑏 𝛽 − 𝛽𝑛 𝑎 + 𝑎 − 𝑏 𝛼

∴ 𝐺𝑛 =𝑐𝛼𝑛 − 𝑑𝛽𝑛

𝛼 − 𝛽

as desired.

Notice that

𝑐𝑑 = [𝑎 + (𝑎 − 𝑏)𝛽][𝑎 + (𝑎 − 𝑏)𝛼]

= 𝑎2 + (𝑎 − 𝑏)2𝛼𝛽 + 𝑎(𝑎 − 𝑏)(𝛼 + 𝛽)

= 𝑎2 − (𝑎 − 𝑏)2 + 𝑎(𝑎 − 𝑏)

= 𝑎2 + 𝑎𝑏 − 𝑏2

This constant occurs in many of the formulas for generalized Fibonacci numbers. It

is called the characteristic of the GFS. We denote it by the Greek letter μ (mu):

𝜇 = 𝑎2 + 𝑎𝑏 − 𝑏2

The characteristic of the Fibonacci sequence is 1, and that of the Lucas sequence is

−5.

Theorem 4.5

𝐺𝑛+1𝐺𝑛−1 − 𝐺𝑛2 = 𝜇 −1 𝑛

Proof:

5 𝐺𝑛+1𝐺𝑛−1 − 𝐺𝑛2 = 𝑐𝛼𝑛+1 − 𝑑𝛽𝑛+1 𝑐𝛼𝑛−1 − 𝑑𝛽𝑛−1 − 𝑐𝛼𝑛 − 𝑑𝛽𝑛 2

= −𝑐𝑑 𝛼𝑛+1𝛽𝑛−1 + 𝛼𝑛−1𝛽𝑛+1 + 2𝑐𝑑 𝛼𝛽 𝑛

= −𝜇 𝛼𝛽 𝑛−1 𝛼2 + 𝛽2 + 2𝜇 𝛼𝛽 𝑛

= 5𝜇 −1 𝑛

Therefore, 𝐺𝑛+1𝐺𝑛−1 − 𝐺𝑛2 = 𝜇 −1 𝑛 .

In particular, 𝐿𝑛+1𝐿𝑛−1 − 𝐿𝑛2 = 5 −1 𝑛−1.

Theorem 4.6

Let 𝐴𝐵𝐶 be a triangle with AC = 𝐺𝑘𝐺𝑘+3, BC=2𝐺𝑘+1𝐺𝑘+2 and AB = 𝐺2𝑘+3. Then

∆ABC is a right triangle with hypotenuse AB.

In Chapter 2, we found that the sum of any 10 consecutive Fibonacci

numbers is 11 times the seventh number in the sequence. Also, notice that the first

10 terms of the generalized Fibonacci sequence are , 𝑏, 𝑎 + 𝑏,𝑎 + 2𝑏, 2𝑎 +

3𝑏, 3𝑎 + 5𝑏, 5𝑎 + 8𝑏, 8𝑎 + 13𝑏, 13𝑎 + 21𝑏, and 21𝑎 + 34𝑏. Their sum is

55𝑎 + 88𝑏, which is clearly 11 times the seventh term 5𝑎 + 8𝑏. Interestingly

enough, 11 = 𝐿5. Thus

Gi

10

1

= 𝐿5 . G7

where 𝐿5 = (55,89 – 1) = (𝐹10 ,𝐹11 − 1)

When 𝑛 = 10, this sum is divisible by 𝐿5, as we just observed. Consequently,

let us look for a way to factor this sum. Since 𝑎 and 𝑏 are arbitrary, we look for the

common factors of 𝐹𝑛 and 𝐹𝑛+1 − 1 . [Although (𝐹𝑛 ,𝐹𝑛+1) = 1,𝐹𝑛 and 𝐹𝑛+1 − 1

need not be relatively prime.]

Table 4.1 shows a few specific values of 𝐹𝑛 ,𝐹𝑛+1 − 1 , and their factorizations; we

have omitted the cases where (𝐹𝑛 ,𝐹𝑛+1 − 1) = 1.

TABLE 4.1

It is apparent from the table that when n is of the form 4k + 2, (𝐹𝑛 ,𝐹𝑛+1 − 1) is

a Lucas number and the various quotients are consecutive Fibonacci numbers; and

when n is of the form 4k, (𝐹𝑛 ,𝐹𝑛+1 − 1) is a Fibonacci number and the various

quotients are consecutive Lucas numbers.

Next we proceed to confirm these two observations, for which we need the

following facts from Theorem 2.9:

𝐹4𝑛+1 − 1 = 𝐿2𝑛+1𝐹2𝑛 and 𝐹4𝑛+2 − 1 = 𝐿2𝑛𝐹2𝑛+2

Case 1. Let n be of the form 4k + 2.Then

𝐺𝑖

4𝑘+2

1

= 𝑎𝐹4𝑘+2 + 𝑏(𝐹4𝑘+3 − 1)

= 𝑎𝐿2𝑘+1𝐹2𝑘+1 + 𝑏𝐿2𝑘+1𝐹2𝑘+2

= 𝐿2𝑘+1(𝑎𝐹2𝑘+2 + 𝑏𝐹2𝑘+2)

= 𝐿2𝑘+1𝐺2𝑘+3

Thus 𝐺𝑖

4𝑘+2

1

can be obtained by multiplying 𝐺2𝑘+3 with 𝐿2𝑘+1

In particular, 𝐺𝑖

10

1

= 𝐿5 .𝐺7 = 11 . 𝐺7

as observed earlier. This is an interesting case, since multiplication by 11 is

remarkably easy. Likewise,

we can compute 𝐺𝑖

30

1

by multiplying 𝐺17 with 𝐿15 = 1364

Case 2. Let n be of the form 4𝑘. Then

𝐺𝑖

4𝑘

1

= 𝑎𝐹4𝑘 + 𝑏 𝐹4𝑘+1 − 1

= 𝑎𝐿2𝑘𝐹2𝑘 + 𝑏𝐿2𝑘+1𝐹2𝑘

= 𝐹2𝑘 𝑎𝐿2𝑘 + 𝑏𝐿2𝑘+1

= 𝐹2𝑘 𝑎 𝐹2𝑘−1 + 𝐹2𝑘+1 + 𝑏 𝐹2𝑘 + 𝐹2𝑘+2

= 𝐹2𝑘 𝑎𝐹2𝑘−1 + 𝑏𝐹2𝑘 + 𝑎𝐹2𝑘+1 + 𝑏𝐹2𝑘+2

= 𝐹2𝑘 𝐺2𝑘+1 + 𝐺2𝑘+3

Thus we can realize 𝐺𝑖

4𝑘

1

by multiplying the sum 𝐺2𝑘+1 + 𝐺2𝑘+3 with 𝐹2𝑘 .

For instance, we can obtain 𝐺𝑖

4𝑘

1

by multiplying the sum 𝐺11 + 𝐺13 with 55.

REFERENCES

1. Thomas Koshy , “Fibonacci and Lucas Numbers with Applications”, A

Wiley-Interscience Publication

2. Apostol T.M., “Introduction to Analytic Number Theory” , Springer

International Student Edition, Narosa Publishing House (1989).

3. Burton D.M. “Elementary Number Theory” , Tata McGraw-Hill

Edition, Sixth Edition (2006).

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