The corresponding elliptic curves of the equations (8) and (10) are in short Weierstrass normal form, whence for a given discriminant D it can be solved by simath.
By (8) and (10) one can compute the coefficients of the elliptic curves in case of the Fibonacci, the Lucas and the Pell sequences. The calculations are summarized inTable 1, as well as all the integer points belonging to them. Every binary recurrence leads to two elliptic equations because of the even and odd suffixes. For the Fibonacci and Lucas sequences D= 5, and for the Pell sequence and its associate sequence D= 8.
Equation Transformed equations All the integer solutions (l, k) Fn= x3
k2 =l3−675l+ 2631150 (15,1620), (−30,1620), (5199,374868), (735,19980), (150,2430), (−129,756) Fn= x3
k2 =l3−675l−2617650 (150,810), (555,12960), (1014,32238), (195,2160), (451,9424), (4011,254016) Ln= x3
k2 =l3−675l−13115250 no solution
Ln= x3
k2 =l3−675l+ 13128750 (375,8100), (−74,3574), (150,4050), (−201,2268), (2391,116964)
Pn= x3
k2 =l3−1728l+ 6746112 (−192,0), (24,2592), (−48,2592), (97,2737) (312,6048), (564,13608), (5208,375840) Pn= x3
k2 =l3−1728l−6690816 (240,2592), (609,14769)
Table 1
The last step is to calculate x and y from the solutions (l, k). By the proof of Theorem 1 it follows that x = 1 +
ql+6D
9D , y = 162Dk in case of the equation (5) and y = 162Dk 2 in case of the associate sequence. Except for some values x and y, they are not integer if x≥3. The exceptions provide all the solutions of the equations (8) and (10). Then the proof of Theorem 2 is complete.
Acknowledgement. The author is grateful to Professor Peth˝o for his valuable remarks.
References
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[3] Cohn, J. H. E., On square Fibonacci numbers, J. London Math. Soc.,39(1964), 537-540.
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Glasgow Math. Assoc., 7(1965), 24-28.
[5] Gebel, L. - Peth˝o, A. - Zimmer, H. G., Computing integral points on elliptic curves, Acta Arithm.68 (1994), 171-192.
[6] London, H. - Finkelstein, R., On Fibonacci and Lucas numbers which are perfect powers, Fib. Quarterly, 7 (1969), 476-481, 487.
[7] McDaniel, W. L., Triangular numbers in the Pell sequence, Fib. Quarterly, 34 (1996), 105-107.
[8] Ming, L., On triangular Fibonacci numbers, Fib. Quarterly,27 (1989), 98-108.
[9] Ming, L., On triangular Lucas numbers, Applications of Fibonacci Numbers, Vol 4., Dordrecht, Netherlands: Kluwer, 1991, 231-240.
[10] Mordell, L. J., Note on the integer solutions of the equationEy2 =Ax3+Bx2+Cx+D, Messenger Math.,51(1922), 169-171.
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London Math. Soc. (2), 21(1923), 415-419.
[12] Peth˝o, A., Full cubes in the Fibonacci sequence, Publ. Math. Debrecen,30(1983), 117-127.
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J´anos Bolyai 60, Sets, Graphs and Numbers Budapest (Hungary), 1991, 561-568.
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[15] SIMATH Manual, Saarbr¨ucken, 1996.
[16] Siegel, C. L. (under the pseudonym X), The integer solutions of the equation y2 = axn+bxn−1+· · ·+k, J. London Math. Soc., 1 (1926), 66-68.
[17] Tallman, M. H., Fib. Quarterly,1 (1963), 47.
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4. Dolgozatok: polinomi´alis-exponenci´alis diofantikus egyenletrendszerek:
diofantikus halmazok
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Clemens Fuchs, Florian Luca, L´aszl´o Szalay
Diophantine Triples with Values in Binary Recurrences
Ann. Sc. Norm. Super. Pisa Cl. Sci., 5 (2008), 579-608.
Diophantine triples with values in binary recurrences Clemens Fuchs, Florian Luca, L´aszl´o Szalay
Abstract
In this paper, we study triples a, b and c of distinct positive integers such thatab+ 1, ac+ 1 andbc+ 1 are all three members of the same binary recurrence sequence.
1 Introduction
A Diophantine m-tuple is a set {a1, . . . , am} of positive integers such that aiaj + 1 is a perfect square (i.e. a square of a number in Z) for all 1 ≤ i < j ≤ m. Finding such sets was already investigated by Diophantus and he found therational quadruple {1/16,33/16,68/16,105/16}. The first quadruple in integers, the set {1,3,8,120}, was found by Fermat. Infinitely many Diophantine quadruples are known and it is conjec-tured that there is no Diophantine quintuple. This was almost proved by Dujella [7], who showed that there can be at most finitely many Diophantine quintuples and all of them are, at least in theory, effectively computable. Several variants of this problem have been studied in the past. For example, Bugeaud and Dujella [2], proved upper bounds for the size m of sets of positive integers with the property that the product of any two distinct elements plus one is a perfect k-th power for fixed k, namely m is bounded by 7 for k = 3, by 5 for k = 4, by 4 for 5 ≤ k ≤ 176, and by 3 for k ≥ 177.
Another variant studied previously is concerned with perfect powers instead of squares or k-th powers for fixed k. The second author proved that the abc-conjecture implies that the size of such sets is bounded by an absolute constant, whereas unconditionally there are bounds depending on the largest element in the set (see [13] and the papers cited therein). For further results on Diophantine m-tuples and its variants, we refer to [8].
In this paper, we treat another variant of this problem. Let r and s be nonzero integers such that ∆ = r2 + 4s 6= 0. Let (un)n≥0 be a binary recurrence sequence of integers satisfying the recurrence
un+2 =run+1+sun for all n≥0.
It is well-known that if we write α and β for the two roots in C of the characteristic equation x2 −rx−s= 0, then there exist constants γ, δ ∈K=Q[α] such that
un=γαn+δβn (1)
holds for all n ≥ 0. We shall assume in what follows that the sequence (un)n≥0 is nondegenerate, which means that γδ 6= 0 and α/β is not root of unity. We shall also make the convention that |α| ≥ |β|. Note that |α|>1.
Here, we look for Diophantine triples with values in the set U = {un : n ≥ 0}, namely sets of three distinct positive integers {a, b, c}, such that ab+ 1, ac+ 1, bc+ 1 are all in U. Clearly, there are always such pairs as e.g. {1, un −1}. Note that if un = 2n+ 1 for alln≥0, then there are infinitely many such triples (namely, takea, b, c to be any distinct powers of two); in this situation, we can even get arbitrarily large sets {a1, . . . , am} with the property that aiaj + 1∈ U for all 1≤i < j ≤m. Our main result is that the above example is representative for the sequences (un)n≥0 with real roots for which there exist infinitely many Diophantine triples with values in U. More precisely we prove the following.
Theorem 1. Assume that (un)n≥0 is a nondegenerate binary recurrence sequence with
∆>0 such that there exist infinitely many sextuples of nonnegative integers (a, b, c;x, y, z)
with 1≤a < b < c such that
ab+ 1 =ux, ac+ 1 =uy, bc+ 1 =uz. (2) Then β ∈ {±1}, δ ∈ {±1}, α, γ ∈ Z. Furthermore, for all but finitely many of the sextuples (a, b, c;x, y, z)as above one has δβz =δβy = 1 and one of the following holds:
(i) δβx= 1. In this case, one of γ or γα is a perfect square;
(ii) δβx=−1. In this case, x∈ {0,1}.
Theorem 1, of course, implies that there are only finitely many triples of positive integers such that the product of any two plus one is inU, except in the cases described (and these cases really occur as we saw above). We mention that the problem can be reformulated as a Diophantine equation of polynomial-exponential type with three independent exponential variables and three additional polynomial variables, namely
(ab+ 1−ux)2+ (ac+ 1−uy)2+ (bc+ 1−uz)2 = 0.
It is well-known that the Subspace theorem is a powerful tool for such problems, e.g. it was also used to classify the solutions to the equation Aux+Buy+Cuz = 0 for fixed A, B, C ∈ Z in [17] (see [18] for a survey on such equations). A new development in applying the Subspace theorem was startet by Corvaja and Zannier (see [22, 23, 10]), and their techniques will also be used in our proof (especially we use [6, 11] and [5]).
We could not prove any finiteness result for the case when ∆ < 0, the reason being
that in this case there is no dominant root in the polynomial-exponential Diophantine equation, which is the main restriction in applying the Subspace theorem with these techniques at present.
For example, it follows for the particular case of the Fibonacci sequence (Fn)n≥0, given by (r, s) = (1,1), F0 = 0 and F1 = 1, that there are at most finitely many triples of positive integers such the product of any two plus one is a Fibonacci number Fn. In the subsequent paper [16] the second and third author show that there is in fact no triple of distinct positive integersa, b and csuch that ab+ 1, ac+ 1 andbc+ 1 are all three Fibonacci numbers.