NONEXTENDIBILITY OF D(−1)-TRIPLES OF THE FORM {1,10,c}

We prove that there do not exist different positive integers 
 1$" xmlns:mml="http://www.w3.org/1998/Math/MathML"> c , d > 1 such that the product of any two distinct elements of the 
set { 1 , 10 , c , d } diminished by 1 is a perfect square.


Introduction
Let n be an integer.A set of positive integers {a 1 ,a 2 ,...,a m } is said to have the property D(n) if a i a j + n is a perfect square for all 1 ≤ i < j ≤ m.Such a set is called a Diophantine m-tuple (with the property D(n)) or a D(n)-m-tuple.
In fact, this problem was first studied by Diophantus for the case n = 1 and he found a set of four positive rationals with the above property: {1/16, 33/16,17/4,105/16} (see [4]).The first set of four positive integers with the same property was found by Fermat, and it was {1, 3,8,120}.The conjecture is that there does not exist a D(1)-quintuple.In 1969, Baker and Davenport proved that Fermat's set cannot be extended to a D(1)-quintuple, (see [2]).In 2004, Dujella proved that there exists no D(1)-sextuple and there are only finitely many D(1)-quintuples (see [9]).In the case n = −1, the conjecture is that there does not exist a D(−1)-quadruple.This conjecture, for the first time, appeared in [8].
We assume that the D(−1)-triple {a, b,c} can be extended to a D(−1)-quadruple.Then there exist positive integers d, x, y, z such that Eliminating d, we obtain the following system of Pellian equations: 2) The conjecture can thus be written in terms of Pellian equations (see [7]).
Conjecture 1.1.Let {a, b,c} be the set of distinct positive integers with the property that there exist integers r, s, t such that If 1 / ∈ {a, b,c}, then the system of Pellian equations has no solution.If a = 1, then all solutions of system (1.4) are given by (x, y,z) = (0,±r,±s).
For certain triples {a, b,c}, with 1 / ∈ {a, b,c} the validity of Conjecture 1.1 can be verified by simple use of congruences (see [6]).On the other hand, the triples of the form {1, b,c} have one "extension," d = 1.Although we do not count it as a proper extension, its existence implies that such triples extendibility cannot be proved by simple congruence consideration.Recently, the original conjecture if 1 / ∈ {a, b,c} was completely proved by Dujella and Fuchs in [10].There they proved that there exists no D(−1)-quintuple, and if there exists D(−1)-quadruple {a, b,c,d} with a < b < c < d, then a = 1 and b ≥ 5.
In the present paper, we will verify Conjecture 1.1 for all triples of the form {1, 10,c}, and in the end, we will finish the work started by Abu Muriefah and Al-Rashed in [1], proving the same conjecture for all the triples of the form {1, 5,c}.In the first part of our proof we will follow the strategy from [1,7,11].Some of the lemmas are already proved in [10] in a more general context, but in order to keep the paper self-contained, we prefer to give complete proofs everywhere.

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Now from the relation c k = s k 2 + 1, we can form the sequence (c k ) k≥1 , It is easy to check that c k < c l for k < l.
Hence if the triple {1, 10,c} satisfies the property D(−1), then there exists a positive integer k, such that c = c k .Now we can formulate our main result.
Theorem 1.2.Let k be a positive integer and c k as above.All solutions of the system of simultaneous Pellian equations are given by (x, y,z) = (0,±3,± c k − 1).
From this statement, we get immediately the result on the extension of {1, 10,c}.

Preliminaries
Let k be the minimal positive integer, if such exists, for which the statement of Theorem 1.2 is not valid.Then results of Mohanty and Ramasamy (see [14]), and Kedlaya (see [12]) imply that k ≥ 4. For simplicity, we will omit the index k.Now we have , respectively, such that all solutions of (1.9) and (1.10) are given by From (2.1), we conclude that z = v m (i) for some index i and positive integer m, where and from (2.2), we conclude that z = w n ( j) for some index j and positive integer n, where Our system of equations (1.9) and (1.10) is thus transformed to finitely many equations of the form v m (i) = w n ( j) . (2.5) If we choose representatives z 0 (i) + x 0 (i) √ c and z 1 ( j) √ 10 + y 1 ( j) √ c such that |z 0 (i) | and

Application of congruence relations
From (2.3) and (2.4), it follows by induction that So if (2.5) has a solution, we must have 2 , and Now if d 0 > 1, then there exists positive integer l, l < k, such that d 0 = c l .But then the system Thus we proved the following lemma.
For the simplicity, from now on, the superscripts (i) and (j) will be omitted.
The following lemma can be proved easily by induction.
Since we may restrict ourselves to positive solutions of the system of equations (1.9) and (1.10), we may assume that z 0 = z 1 = s.If y 1 = 3, then it is obvious that v l < w l for Alan Filipin 2221 l > 0, and v m = w n , n = 0, implies that m > n.If y 1 = −3, then it is easy to check that v 0 < w 1 .Then v l < w l+1 for l ≥ 0, and thus v m = w n implies that m ≥ n in all cases.Lemma 3.3.If v m = w n , then m and n are even.

Reduction
For completing the proof of Theorem 1.2 for all positive integers k, we must check that for 4 ≤ k ≤ 81, v 2m = w 2n implies that n = m = 0. First, we will prove that where n < 5 • 10 20 , implies n ≤ m ≤ 1.
Having proved that, only one possibility will remain for the solution of (2.5), and that is v 2 = w 2 .However, it is easy to prove that this is not possible.First, we see that v 2 = w 2 is only possible when y 1 = −3.
To finish the proof, we use the reduction method described in [16,Section VI.3] and [17].
2224 Nonextendibility of D(−1)-triples of the form {1, 10,c} If we have the inequality of the type where a i ∈ C are fixed numbers, k 2 , k 3 are positive real constants, and x i are unknown integers with X = max |x i | ≤ X 0 , we can try to reduce the upper bound for X, using the following method.In our case, there is r = 2.
Let L be a lattice generated by the columns of the matrix where K is some large constant and [•] rounds number to the nearest integer.Now using the LLL algorithm (see [13]), we can find reduced basis of L. We do that by using the integral version of the algorithm in the package GP/Pari (see [5]).Let y = (0 Using the properties of reduced basis, it is easy to compute the lower bound k 4 of x − y , where x ∈ L and • denotes the usual (Euclidean) norm of a vector.We can compute that using GP/Pari again.In most cases, if we choose the constant K big enough, we can use the following lemma.Proof of the lemma can be found in [16].In our case, there is an inequality of the type (5.2) with (5.5) Using the described method, after two steps of reduction, in all 78 remaining cases, for 4 ≤ k ≤ 81, we get m ≤ 1, which finishes the proof of Theorem 1.2.

Concluding remarks
We can prove Conjecture 1.1 for all the triples of the form {1, 5,c}, in exactly the same way.This gives a much shorter and neat proof of the result in [1].Since Abu Muriefah and Al-Rashed in [1] have already prepared everything for the use of the described reduction method, we will only give some details here that are different from their paper.

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First, there is no need for three sequences (c k ), because it is well known that all nonnegative solutions of the Pellian equation t 2 − 5s 2 = 4 are given by t = L 2k , s = F 2k (Lucas and Fibonacci numbers).Thus, we only have one sequence c k = F 2k 2 + 1, which leads to significant simplification.Later, when we work with a linear form in logarithms, we get the following.
If This slightly changes the constants which appeared in [1].We get that n < 5 • 10 20 , and finally c = c k < 10 84 , which implies that k ≤ 101.Now we can use the same reduction method, because in this case there is an inequality of type (5.(6.3)By using the reduction method in all cases, for k ≤ 101, after three steps of reduction, we get m ≤ 3. Then to finish our proof, we only need to show that the remaining possibilities, v 1 = w 1 , v 2 = w 2 , v 3 = w 3 , are impossible because we have the relations n ≤ m ≤ n √ 5 and m ≡ n(mod2).However, this is easy to check.