on the family of diophantine triples { k + 1, 4 k , 9 k + 3}

Periodica Mathematica Hungarica Vol. 58 (1 ), 2009, pp. 59–70DOI: 10.1007/s10998-009-9059-6

ON THE FAMILYOF DIOPHANTINE TRIPLES {k + 1, 4k, 9k + 3}

Bo He1 and Alain Togbe2

[Communicated by Attila Petho]

1Department of Mathematics

Key Laboratory of Numerical Simulation of Sichuan Province

Neijiang Normal University, Neijiang, Sichuan, 641112, P. R. China

E-mail: [email protected]

2Department of Mathematics, Purdue University North Central

1401 S. U.S. 421, Westville, IN 46391 USA

E-mail: [email protected]

(Received June 24, 2008; Accepted July 28, 2008)

Abstract

We prove that if k is a positive integer and d is a positive integer such

that the product of any two distinct elements of the set {k + 1, 4k, 9k + 3, d}

increased by 1 is a perfect square, then d = 144k3 + 192k2 + 76k + 8.

1. Introduction

A set of m distinct positive integers {a1, . . . , am} is called a Diophantine m-

tuple if aiaj + 1 is a perfect square. Diophantus studied sets of positive rationals

with the same property, particularly he found the set of four positive rationals{

116 , 33

16 , 174 , 105

16

}

. But the first Diophantine quadruple was found by Fermat. In

fact, Fermat proved that the set {1, 3, 8, 120} is a Diophantine quadruple. Moreover

Baker and Davenport [2] proved that the set {1, 3, 8, 120} cannot be extended to a

Diophantine quintuple. In general, let n be an integer. A set of m positive integers

{a1, . . . , am} is called a Diophantine m-tuple with the property D(n) or a D(n)−m-

tuple (or a Pn-set of size m), if aiaj + n is a perfect square. Authors also consider

n as a parametric expression. See for example [19].

The problem of extendability of Pn-sets is of big interest. See for examples

[1], [4]–[25], [27]. Several generalizations of the result of Baker and Davenport

are obtained. In 1997, Dujella [6] proved that the Diophantine triples of the form

Mathematics subject classification numbers: 11D09, 11D45, 11B37, 11J68, 11J86.

Key words and phrases: Diophantine m-tuple, Pell equation, Diophantine approximation,linear forms in logarithms.

0031-5303/2009/$20.00 Akademiai Kiado, Budapestc© Akademiai Kiado, Budapest Springer, Dordrecht

60 B. HE and A. TOGBE

{k − 1, k + 1, 4k}, for k ≥ 2, cannot be extended to a Diophantine quintuple.

The Baker–Davenport result corresponds to k = 2. In 1998, Dujella and Petho

[14] proved that the Diophantine pair {1, 3} cannot be extended to a Diophantine

quintuple. In 2008, Fujita [21] obtained a more general result by proving that the

Diophantine pairs {k − 1, k + 1}, for k ≥ 2 cannot be extended to a Diophantine

quintuple. A folklore conjecture is that there does not exist a Diophantine D(1)-

quintuple. In 2004, Dujella [11] proved that there are only finitely many Diophantine

D(1)-quintuples. A stronger version of this conjecture is the following.

Conjecture 1.1. If {a, b, c, d} is a Diophantine quadruple and d >

max{a, b, c}, then d = a + b + c + 2abc + 2√

(ab + 1)(ac + 1)(bc + 1).

A Diophantine quadruple {a, b, c, d} is regular if and only if (a+ b− c− d)2 =

4(ab + 1)(cd + 1). Therefore the quadruple in the above conjecture is regular.

The aim of this paper is to consider the Diophantine triple {k + 1, 4k, 9k + 3}and to prove the following result.

Theorem 1. If k is a positive integer and d is a positive integer such that

the product of any two distinct elements of the set

{k + 1, 4k, 9k + 3, d}

increased by 1 is a perfect square, then d = 144k3 + 192k2 + 76k + 8.

Remark. One can easily check that our quadruple {k + 1, 4k, 9k + 3, d} with

d = 144k3 + 192k2 + 76k + 8 is regular.

The organization of this paper is as follows. In Section 2, we recall some useful

results obtained by Dujella and adapt them to our case. We use a result due to

Bennett [3] on simultaneous approximations of algebraic numbers which are close

to 1 to get an upper bound for k. Finally, in Section 4, we prove Theorem 1 by

means of linear forms in logarithms and the Baker–Davenport reduction method.

2. Preliminaries

Let r, s, t be positive integers defined by

ab + 1 = r2, ac + 1 = s2, bc + 1 = t2. (1)

In order to extend the Diophantine triple {a, b, c} to a Diophantine quadruple

{a, b, c, d}, we have to solve the system

ad + 1 = x2, bd + 1 = y2, cd + 1 = z2. (2)

ON THE FAMILY OF DIOPHANTINE TRIPLES {k + 1, 4k, 9k + 3} 61

Eliminating d, we obtain the following system of Pellian equations:

az2 − cx2 = a − c, (3)

bz2 − cy2 = b − c. (4)

By [9, Lemma 1], there exists a solution (z(i)0 , x

(i)0 ) of (3) such that z = v

(i)m , where

v(i)0 = z

(i)0 , v

(i)1 = sz

(i)0 + cx

(i)0 , v

(i)m+2 = 2sv

(i)m+1 − v(i)

m ,

and |z(i)0 | <

√

c√

c2√

a. Similarly, there exists a solution (z

(i)1 , y

(i)1 ) of (4) such that

z = w(j)n , where

w(i)0 = z

(j)1 , w

(j)1 = tz

(j)1 + cy

(j)1 , w

(j)n+2 = 2tw

(j)n+1 − w(j)

n ,

and |z(j)1 | <

√

c√

c

2√

b.

The initial terms z(i)0 and z

(j)1 are almost completely determined in the fol-

lowing lemma [11, Lemma 8].

Lemma 1.

(a) If the equation v2m = w2n has a solution, then z0 = z1. Furthermore, |z0| = 1

or |z0| = cr − st or |z0| < min{0.869a−5/14c9/14, 0.972b−0.3c0.7}.(b) If the equation v2m+1 = w2n has a solution, then |z0| = t, |z1| = cr − st and

z0z1 < 0.

(c) If the equation v2m = w2n+1 has a solution, then |z0| = cr − st, |z1| = s and

z0z1 < 0.

(d) If the equation v2m+1 = w2n+1 has a solution, then |z0| = t, |z1| = s and

z0z1 > 0.

In the present paper, we have

a = k + 1, b = 4k, c = 9k + 3,

and from (1) we obtain

r = 2k + 1, s = 3k + 2, t = 6k + 1.

One can easily check that

|z0| <

√

c√

c

2√

a<

√

3(9k + 3)

2≤ 4.3

√k

and

|z1| <

√

c√

c

2√

b<

√

2√

3(9k + 3)

4≤ 3.3

√k.


If (b) or (d) holds, this implies |z0| = t = 6k + 1 < 4.3√

k. Thus this is impossible.

If (c) holds, similarly we obtain |z1| = s = 3k+2 < 3.3√

k, which is also impossible.

Therefore we only consider the case (a).

Since cr − st = 1, then the first and the second possibilities in (a) both imply

|z0| = 1. Let us define d0 = (z20 −1)/c in the third possibility of (a). If |z0| > 1, then

d0 < z20/c < a−5/7c9/7/c < c. Thus according to the proof of the above lemma in

[11], {a, b, c, d0} is an irregular Diophantine quadruple. Also by [11, Proposition 1],

if {a, b, c, d} is an irregular Diophantine quadruple with a < b < c < d, then

d > 2.695c3.5a2.5. Therefore we have c > 2.695 · c3.5a2.5. When k ≥ 1, we get a

contradiction.

In other words, we need to solve the system of Pellian equations

(k + 1)z2 − (9k + 3)x2 = −8k − 2, (5)

4kz2 − (9k + 3)y2 = −5k − 3, (6)

with x0 = y1 = 1 and z0 = z1 = ±1. This is equivalent to solving the equation

z = v2m = w2n. (7)

Let λ := z0 = z1, λ ∈ {−1, 1}. We have

v0 = λ, v1 = (9 + 3λ)k + 3 + 2λ, vm+2 = (6k + 4)vm+1 − vm, (8)

w0 = λ, w1 = (9 + 6λ)k + 3 + λ, wn+2 = (12k + 2)wn+1 − wn. (9)

For the relations of indices m and n, we have

Lemma 2. If v2m = w2n, then n ≤ m ≤ 2n.

Proof. By [9, Lemma 3], if vm = wn, then n − 1 ≤ m ≤ 2n + 1. In our even

case, we have 2n − 1 ≤ 2m ≤ 4n + 1. The result is obtained.

In order to get a gap principle between indices m, n and k, we recall the

following lemma:

Lemma 3. We have

v2m ≡ z0 + 2c(az0m2 + sx0m) (mod 8c2),

w2n ≡ z1 + 2c(bz1n2 + ty1n) (mod 8c2).

Proof. These congruences came from items (a) and (c) of [9, Lemma 4].


Using these results, we have

λam2 + sm ≡ λbn2 + tn (mod 4c). (10)

In our case, the congruence becomes

λ(k + 1)m2 + (3k + 2)m ≡ 4λkn2 + (6k + 1)n (mod 36k + 12) .

Multiplying the above congruence by 36λ, we obtain 24m2 + 36λm ≡ −48n2 −36λn (mod 36k + 12). Therefore,

2m2 + 4n2 + 3λm + 3λn ≡ 0 (mod 3k + 1). (11)

If m, n ≥ 2, then

2m2 + 4n2 + 3λm + 3λn ≥ 2m2 + 4n2 − 3m − 3n > 0.

Hence from (11) we obtain

2m2 + 4n2 + 3m + 3n ≥ 2m2 + 4n2 + 3λm + 3λn ≥ 3k + 1.

By Lemma 2, we know that m ≥ n. So we have 6m2 + 6m ≥ 3k + 1 > 3k. Thus we

get

Lemma 4. Assume that v2m = w2n with m, n ≥ 2. Then m ≥√

k/2 − 1.

Now we prove the following result.

Lemma 5. Let x, y, z be positive integer solutions of the system of Pellian

equations (5) and (6) such that z /∈ {1, 36k2 + 30k + 5}. Then

log(2z) > (√

2k − 2) log(6k).

Proof. Using (3) with z0 = ±1, x0 = 1 and v2m = w2n, we have

z =1

2√

a

(

(±√

a +√

c)(s +√

ac)2m + (±√

a −√

c)(s −√

ac)2m)

.

For example, one can refer to the proof of [9, Lemma 1]. Since z /∈ {1, 36k2+30k+5},

then m ≥ 2 and n ≥ 2. We get

2z >1√a(±

√a +

√c)(s +

√ac)2m −

√a +

√c√

a(s +√

ac)2m

≥(

− 1 +

√

c

a

)

(2√

ac)2m − 2√

c√a(2

√ac)2m

> (−1 +√

6)(2√

ac)2m − 1

8a2.5c1.5> (2

√ac)2m .


Hence,

log(2z) > 2m log(2√

ac) > 2m log(2√

(k + 1)(9k + 3)) > 2m log(6k).

From Lemma 4, we obtain the result.

3. Application of a result of Bennett

In this section, we will use the following result of Bennett on simultaneous

approximations of algebraic numbers which are close to 1 to get an upper bound

for k in the system of Pellian equations (3) and (4).

Lemma 6. If ai, pi, q and N are integers for 0 ≤ i ≤ 2, with a0 < a1 < a2,

aj = 0 for some 0 ≤ j ≤ 2, q nonzero and N > M9, where

M = max0≤i≤2

{|ai|},

then we have

max0≤i≤2

{∣

∣

∣

∣

√

1 +ai

N− pi

q

∣

∣

∣

∣

}

> (130Nγ)−1q−µ, (12)

where

µ = 1 +log(33Nγ)

log (1.7N2Π0≤i<j≤2(ai − aj)−2)

and

γ =

(a2 − a0)2(a2 − a1)

2

2a2 − a0 − a1if a2 − a1 ≥ a1 − a0,

(a2 − a0)2(a1 − a0)

2

a1 + a2 − 2a0if a2 − a1 < a1 − a0.

Proof. See [3, Lemma 3.2].

Let us consider the numbers

θ1 =

√

1 +2

3k + 1and θ2 =

√

1 − 1

3k + 1.

The following result gives us some information on the approximations of the alge-

braic numbers θ1 and θ2.

Lemma 7. If k ≥ 171 and (x, y, z) 6= (1, 1, 1), then

∣

∣

∣

∣

θ1 − 6x

2z

∣

∣

∣

∣

< 4z−2 and

∣

∣

∣

∣

θ2 − 3y

2z

∣

∣

∣

∣

< z−2.


Proof. Using (5) and (6), one can verify that

∣

∣

∣

∣

θ1 − 6x

2z

∣

∣

∣

∣

=

√3∣

∣z√

k + 1 − x√

9k + 3∣

∣

z√

3k + 1=

√3(8k + 2)

z√

3k + 1∣

∣z√

k + 1 + x√

9k + 3∣

∣

<

√3(8k + 2)

2z2√

(3k + 1)(k + 1)< 4z−2,

and

∣

∣

∣

∣

θ2 − 3y

2z

∣

∣

∣

∣

=

√3∣

∣

∣z√

4k − y√

9k + 3∣

∣

∣

2z√

3k + 1=

√3(5k + 3)

2z√

3k + 1∣

∣

∣z√

4k + y√

9k + 3∣

∣

∣

<

√3(5k + 3)

4z2√

4k(3k + 1)< z−2.

This completes the proof of the lemma.

The next result gives us the information on d.

Proposition 3.1. If k ≥ 2535 and if the set {k + 1, 4k, 9k + 3, d} is a

Diophantine quadruple, then d has to be 144k3 + 192k2 + 76k + 8.

Proof. If d satisfies the condition, then z2 = cd + 1 = (9k + 3)d + 1. Since

d > 1, we have z 6= 1. And d 6= 144k3 +192k2 +76k+8, we have z 6= 36k2 +30k+5.

Then Lemma 5 implies

log(2z) > (√

2k − 2) log(6k). (13)

Now we apply Lemma 6. We take a0 = −1, a1 = 0, a2 = 2, M = 2, q =

2z, p0 = 3y, p1 = q, p2 = 6x and N = 3k + 1. When k ≥ 171, the condition

N > M9 holds. Therefore, by Lemma 6 and Lemma 7, we have

4z−2 > (130(3k + 1)γ)−1(2z)−µ, (14)

where γ = 7.2 and

µ = 1 +log(33 · 7.2 · (3k + 1))

log(1.7 · (3k + 1)2/36)< 1 +

log(3k + 1) + 5.471

2 log(3k + 1) − 3.052

< 1.5 +3.499

log(3k + 1) − 1.526.

One can check that if k > 1900, then µ < 2. In fact,

2 − µ > 0.5 − 3.499

log(3k + 1) − 1.526.


From (14) we have

(2z)2−µ < 16 · 130 · 7.2(3k + 1).

It follows that

log(2z) < log(14976(3k + 1))/(2 − µ). (15)

Combining (13) and (15), we obtain

√2k − 2 <

log(14976(3k + 1))

(2 − µ) log(6k)<

log(14976(3k + 1))

(0.5 − 3.499log(3k+1)−1.526 ) log(6k)

. (16)

If k ≥ 2535, we get a contradiction.

4. Proof of Theorem 1

In this section, we need to consider the remaining cases, i.e., 1 ≤ k ≤ 2534.

We will use a theorem on lower bounds to linear forms in logarithms to get an upper

bound for m.

Let

α1 = s +√

ac and α2 = t +√

bc.

Solving equations (3) and (4), we have

v2m =1

2√

a

(

(z0

√a + x0

√c)α2m

1 + (z0

√a + x0

√c)α−2m

1

)

and

w2n =1

2√

b

(

(z1

√b + y1

√c)α2n

2 + (z1

√b + y1

√c)α−2n

2

)

respectively. Notice x0 = y1 = 1 and z0 = z1 = λ, λ ∈ {−1, 1}. If z = v2m = w2n

with m, n 6= 0, we have (see [11, formula (60)])

0 < Λ := 2m log α1 − 2n log α2 + log α3 <8ac

3α−4m

1 (17)

where

α3 =

√b(

√c + λ

√a)

√a(

√c + λ

√b)

.

It follows that

log |Λ| < −4m log α1 + log(8ac

3

)

< (2 − 4m) log(2√

ac)

< (2 − 4m) log(3k + 3).(18)

We recall the following result due to Matveev [26].


Lemma 8. Denote by α1, . . . , αl algebraic numbers, not 0 or 1, by log α1,

. . . , log αl their logarithms, by D the degree over Q of the number field K =

Q(α1, . . . , αl), and by b1, . . . , bn rational integers. Define B = max{|b1|, . . . , |bl|},and Ai = max{Dh(αi), | log αi|, 0.16} (1 ≤ i ≤ l), where h(α) denotes the absolute

logarithmic Weil height of α. Assume that the number

Λ = b1 log α1 + · · · + bn log αl

does not vanish. Then

|Λ| ≥ exp{−C(l, κ)D2A1 · · ·Al log(eD) log(eB)},

where κ = 1 if K ⊂ R and κ = 2 otherwise and

C(l, κ) = min

{

1

κ

(

1

2el

)κ

30l+3l3.5, 26l+20

}

.

Now, we apply the lemma with l = 3 and κ = 1 for

Λ = 2m log α1 − 2n log α2 + log α3. (19)

Here we take D = 4, b1 = 2m, b2 = 2n, b3 = 1. Notice that α1 = s +√

ac is a root

of X2 − 2sX + 1 = 0, α2 = t +√

bc is a root of X2 − 2tX + 1 = 0, and α3 is a root

ofa2(c − b)2x4 + 4a2b(c − b)x3 + 2ab(3ab − ac − bc − c2)x2

+ 4a2b(c − b)x + b2(c − a)2 = 0.

Thus we have

h(α1) =1

2log α1, h(α2) =

1

2log α2,

and

h(α3) =1

4

(

log(a2(c − b)2) + 4 log(

√

c/a + λ√

c/b + λ

)

)

<1

2log(ac) +

√

c/a −√

c/b√

c/b − 1< log(

√

(k + 1)(9k + 3)) + 3

< log(k + 1) + log 3 + 3 < log(k + 1) + 4.1 .

Therefore, we take A1 = 2 log(6k + 4), A2 = 2 log(12k + 2), and A3 = 4 log(k + 1) +

16.4. Using Matveev’s result we have

log |Λ| > −8.492 ·1013 · log(6k + 4) · log(12k + 2) ·(log(k + 1) + 4.1) · log(2em). (20)

Combining (18) and (20), we obtain

m

log(2em)< 2.124 · 1013 · log(6k + 4) · log(12k + 2) · (log(k + 1) + 13.1)

log(3k + 3).


By Proposition 3.1, we only need to consider the cases k < 2535. It follows

m

log(2em)< 4.95 · 1015.

This implies m < 2.1 · 1017.

In order to deal with the remaining cases 1 ≤ k ≤ 2534, we will use a Diophan-

tine approximation algorithm, the so-called Baker–Davenport reduction method.

The following lemma is a slight modification of the original version of the Baker–

Davenport reduction method. (See [14, Lemma 5a]).

Lemma 9. Assume that M is a positive integer. Let P/Q be the convergent

of the continued fraction expansion of κ such that Q > 6M and let

η = ‖µ′Q‖ − M · ‖κQ‖,

where ‖ · ‖ denotes the distance from the nearest integer. If η > 0, then there is no

solution of the inequality

0 < mκ − n + µ′ < AB−m

in integers m and n with

log (AQ/η)

log B≤ m ≤ M.

We apply Lemma 9 with

κ =log α1

log α2, µ′ =

log α3

2 log α2, A =

4(k + 1)(3k + 1)

log α2, B = α4

1

and M = 2.1 · 1017.

In one minute all the computations were done. The use of the second con-

vergent was needed in 29 and 32 cases when λ = 1 and −1 respectively. We have

m ≤ 2 except for 1 ≤ k ≤ 9 and k = 16 for λ = ±1. In all these particular cases,

we obtained m ≤ 5. We took again M = 5 in these 20 cases, then we got m ≤ 2. If

m = 2 is a solution of (7), then by Lemma 2 we have 2 ≤ n ≤ 4 and from Lemma 4

we get k ≤ 18. It is easy to check that there is no integer k satisfy the conditions

in (7). Then m must be 1 (m = n = 0 gives the trivial solution d = 0).

Combining this and Proposition 3.1, we have m = n = 1 in equation (7).

When λ = v0 = w0 = −1, v2 = w2 = 36k2+30k+5 gives d = 144k3+192k2+76k+8.

This completes the proof of Theorem 1.


Acknowledgments

The authors express their gratitude to the anonymous referee for constructive

suggestions to improve an earlier draft of this paper. The first author is supported

by the Natural Science Foundation of China (No. 10872085). The second author is

partially supported by Purdue University North Central.

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