Extensions of Some Parametric Families of D(16)-Triples

Let n be an integer. A set of m positive integers is called a D(n)-m-tuple if the product of any two of them increased by n is a perfect square. In this paper, we consider extensions of some parametric families of D(16)-triples. We prove that if {k − 4,k + 4,4k,d}, for k ≥ 5, is a D(16)-quadruple, then d = k3 − 4k. Furthermore, if {k− 4,4k,9k− 12}, for k > 5, is a D(16)-quadruple, then d = 9k3− 48k2 + 76k− 32. But for k = 5, this statement is not valid. Namely, the D(16)-triple {1,20,33} has exactly two extensions to a D(16)quadruple: {1,20,33,105} and {1,20,33,273}.


Introduction
Let n be an integer.A set of m positive integers {a 1 ,a 2 ,...,a m } is called a Diophantine m-tuple with the property D(n) or simply D(n)-m-tuple, if a i a j + n is a perfect square for This problem was first studied by Diophantus for the case n = 1.He found a set of four positive rationals with the property D (1).It was the set {1/16, 33/16,17/4,105/16}.However, the first D(1)-quadruple, {1, 3,8,120}, was found by Fermat.Euler was able to add the fifth positive rational, 777480/8288641, to the Fermat set (see [1, pages 103-104, 232]).Recently, Gibbs [2] found some examples of sets of six positive rationals with the property D (1).The conjecture is that there does not exist a D(1)-quintuple.In 1969, Baker and Davenport [3] proved that the Fermat set cannot be extended to a D(1)-quintuple.More precisely, they proved that if {1, 3,8,d} is a D(1)-quadruple, then d = 120.Dujella gave one generalization to this result in [4], that only extension of D(1)triple {k − 1,k + 1,4k}, for an integer k ≥ 2, to a D(1)-quadruple {k − 1,k + 1,4k,d}, is given by d = 16k 3 − 4k.Fujita (see [5]) obtained a result for the case n = 4, which can be regarded as a generalization of the result from [4].He proved that only extension of D(4)triple {k − 2,k + 2,4k}, for an integer k ≥ 3, to a D(4)-quadruple {k − 2,k + 2,4k,d}, is given by d = 4k 3 − 4k.We will prove a result for the case n = 16, which generalizes results from [4,5].Those results support the strong D(1)-quadruple and D(4)-quadruple conjecture, which state that in the D(1)-quadruple, respectively, D(4)-quadruple {a, b,c,d}, such that a < b < c < d, element d is uniquely determined with a, b, and c.
In the proofs, we will use the same strategy and the methods from [4,5].16)-triples of the form {k − 4,k + 4,4k}

Extension of D(
Eliminating d, we get the following system of Pellian equations: From the theory of Pellian equations, we know that if (z,x) is solution to (2.2), then there exists an integer m ≥ 0 such that where {(z (i) 0 ,x (i) 0 ) : i = 1,...,i 0 } is a finite set of fundamental solutions of (2.2).Indeed, let (z,x) be a solution of (2.2) in positive integers.Consider all pairs of integers (z * ,x * ) of the form k − 4|, we conclude that x * is a positive integer.Among all the pairs (z * ,x * ), we chose the one for which x * is minimal, and we denote it by (z 0 ,x 0 ).Define integers z and x by Alan Filipin 3 where ε = 1, if z 0 ≥ 0, and ε = −1, if z 0 < 0. Then from minimality of x 0 we get which implies, after some calculation, x 2 0 ≤ 3k + 4. Moreover from x 2 0 ≡ 16(mod(k − 4)), we get the following possibilities for x 2 0 : x 2 0 = 16, k + 12, 2k + 8, 3k + 4. Inserting that in (2.2) we get the value for z 2 0 .Then we conclude that only two possibilities are x 0 = 4, z 0 = ±2, or x 2 0 = z 2 0 = 3k + 4, because in other cases fundamental solutions are not integers.At the end it is easy to see that m ≥ 0.
In the exactly same way we conclude that the solutions (z, y) of (2.3) are given by for some integer n ≥ 0, and {(z 1 ) : j = 1,..., j 0 } is finite set of fundamental solutions of (2.3).In this case (omitting the index j), we get only one possibility y 1 = 4, z 1 = ±2, except in the case k = 5.
We have proved the following lemma.

3) are given by
where n is a nonnegative integer.

Congruences.
From (2.4) we conclude that z = v m , for some m ≥ 0, where the sequence (v m ) m≥0 is defined by (2.11) From (2.8) we conclude that z = w n , for some n ≥ 0, where (w n ) n≥0 is defined by (2.12) We have now transformed the system of (2.2) and (2.3) to finitely many Diophantine equations of the form v m = w n .By induction, from (2.11) and (2.12), we get v m ≡ (−1) m z 0 (modk) and w n ≡ z 1 (modk).Then we get z 2 0 = 3k + 4 is possible only in the case k = 7.And that case will be considered at the end of this section.From now we will assume z 0 = ±2, z 1 = ±2.We will consider the equation z = v m = w n for m,n ≥ 6, since for the remaining values of m and n it is easy to check if for some k the equality can hold.
From (2.11) and (2.12) we get the following lemma by induction.
The following lemma can also be proved by induction.

Alan Filipin 5
Proof.Assume v m = w n , for m,n ≥ 6 and m ≤ √ k/3.Then using Lemma 2.4, when we consider the congruence relations, absolute values of both hand sides are less than k 2 , so we actually have the equalities.In the case z 0 = z 1 = 2, we have and m 2 + 2m + n 2 + 2n = k, which is obviously impossible because the left-hand side is less than k (we also use Lemma 2.3).On the same way, we get the contradiction in the remaining three cases.

Large parameters.
In this section we prove that, for k > 2.67 • 10 7 , the equation v m = w n , for n,m ≥ 6, has no solution.First we have to estimate logz, where

.19)
Proof.Let z = v m .We can now consider both cases at the same time, if we define z = |v m |, where (v m ) m∈Z is a sequence defined by , it is not hard to see that for m ≥ 0 we have v m ≥ ϕ m , and Then if z = v m , and m ≥ 6, we have (2.21) The last inequality follows from Lemma 2.5.

.3). Then
Proof.Using (2.2) and (2.3) we get (2.23) Theorem 2.8 [6, Theorem 3.2].Let a i , p i , q, and N be integers for 0 ≤ i ≤ 2 such that a 0 < a 1 < a 2 , a j = 0 for some 0 ≤ j ≤ 2, q = 0, and N > M 9 , where where (2.25) We will now apply Theorem 2.8 to the following numbers: for k > 2.67 • 10 7 > 4 9 , because then the condition of the theorem is satisfied.Then Function on the left-hand side is increasing faster, and for k > 2.67 • 10 7 , the inequality is not satisfied, so we proved the following proposition.Proposition 2.9.If k > 2.67 • 10 7 , then v m = w n has no solution for m,n ≥ 6. Alan Filipin 7

Linear form in logarithms.
In this section we will prepare everything for Baker-Davenport reduction.We want to prove that the statement of Proposition 2.9 holds for k ≤ 2.67 • 10 7 .
Alan Filipin 9 Lemma 2.12.Let M be a positive integer and let p/q be a convergent of the continued fraction expansion of the number κ such that q > 6M.Furthermore, let ε = μq − M κq , where • denotes the distance to the nearest integer.If ε > 0, then the inequality  16)-triples of the form {k − 4,4k,9k − 12}

Lemma 2 .
5 gives us the lower bound of m, depending on k.Lemma 2.5.If v m = w n , and m,n ≥ 6, then m > √ k/3.

Lemma 3 . 5 .k/ 5 . 3 . 3 .Lemma 3 . 6 .Lemma 3 . 7 .
If v m = w n , m,n ≥ 6, then m > √ Large parameters.We now prove that for k > 4 9 , equation v m = w n , for n,m ≥ 6, has no solution.First, we will get the lower bound for log y, where y = v m = w n .Let y = v m = w n , n,m ≥ 6.Then Let x, y, z be positive solutions of the system of (3.2) and(3.3).Thenmax θ 1 − x y , θ 2 − z y < 17y −2 .(3.11)Now we will again apply the same Bennett's theorem.First we see that
Using k < 2.67 • 10 7 , we conclude m < 2 • 10 19 .Now we need one version of Baker-Davenport lemma (see We have implemented this method in Mathematica 5.0, and we got m ≤ 6 for all k < 2.67 • 10 7 .Now we have to see what is happening for m ≤ 6.But it is easy to check, because we get polynomial equations, that it gives us two extensions d = 0, and d = k 3 − 4k.Before we formulate our main result, we have to check what is happening in the case k = 7 for the other fundamental solution.But using the same methods, only it is easier this time, we do not get anything new, actually we get one "extension" d = 3, which will not give us the D(16)-quadruple because in this case k − 4 = 7 − 4 = 3.