{F_{2k}, F_{2k+2}, F_{2k+4}, 4F_{2k+1} F_{2k+2}F_{2k+3}}.
Hoggatt and Bergum conjectured that the fourth element in the above set is unique. The conjecture was proved by Dujella [77] in 1999, and this result also implies that if
The main step in the proof of Hoggatt-Bergum conjecture
is a comparison of the
upper bounds for solutions obtained from a theorem of Baker
and Wüstholz with the lower bounds obtained from the
congruence conditions modulo
Motivated by the Hoggatt-Bergum set, several authors considered the question how large Diophantine tuples consisting of Fibonacci numbers can be. He, Togbé and Luca [346] proved that if {F_{2n}, F_{2n+2}, F_{k}} is a Diophantine triple, then k = 2n + 4 or k = 2n - 2 (when n > 1), except when n = 2, in which case k = 1 is also possible. Fujita and Luca [365] proved that there are only finitely many Diophantine quadruples consisting of Fibonacci numbers, and in [393] they proved that there are no such quadruples.
Theorem 4.1: There are no Diophantine quadruples consisting of Fibonacci numbers. |
There are many papers devoted to various generalizations of the result of Hoggatt & Bergum. Let us mention some of the authors: Morgado [19, 27, 39, 50], Horadam [26], Long & Bergum [31], Shannon [32], Dujella [36, 44, 48, 55, 68, 124], Deshpande & Bergum [54], Udrea [59, 93], Beardon & Deshpande [103], Deshpande & Dujella [106], Dujella & Ramasamy [133], Ramasamy [158], Fujita [169], Filipin, He & Togbé [204], Filipin [300], Bacic Djurackovic & Filipin [340], Rihane, Hernane & Togbé [387, 410], Park & Lee [416].
Let the sequence (g_{n}) be defined by
g_{0} = 0, g_{1} = 1, g_{n} = pg_{n-1} - g_{n-2},
where{g_{n}, g_{n+2}, (p ± 2)g_{n+1}, 4g_{n+1}((p ± 2) g_{n+1}^{2} ∓ 1))}
are Diophantine quadruples. For{n, n + 2, 4n + 4, 4(n + 1)(2n + 1)(2n + 3)}.
The proof of this result is bases on the construction of a double sequence x_{i,j}. In the construction, solutions of Pell equation s^{2} - ab t^{2} = 1 were used.
If we have a pair of identities of the form:
4F_{k - 1}F_{k+1} + F_{k}^{2} = L_{k}^{2},
(F_{k}^{2} + F_{k-1}
F_{k+1})^{2} - 4F_{k-1}F_{k+1} F_{k}^{2} = 1
4F_{k-2}F_{k+2} + L_{k}^{2} = 9F_{k}^{2},
(F_{k}^{2} + F_{k-2}F_{k+2})^{2} - 4F_{k-2}F_{k+2} F_{k}^{2} = 1.
{2F_{k-1}, 2F_{k+1}, 2F_{k}^{3} F_{k+1}F_{k+2}, 2F_{k+1}F_{k+2} F_{k+3}(2F_{k+1} ^{2} - F_{k}^{2})}
with the property D(F_{k}^{2}) and the set{2F_{k-2}, 2F_{k+2}, 2F_{k-1} F_{k}L_{k}^{2} L_{k+1}, 2L_{k-1} F_{k} L_{k}^{2} L_{k+1}}
with the property D(L_{k}^{2}) (see [48]). In the same paper, several quadruples were obtained using Morgado identity:F_{k-3}F_{k-2} F_{k-1}F_{k+1} F_{k+2}F_{k+3} + L_{k}^{2} = (F_{k}(2F_{k-1} F_{k+1} - F_{k}^{2}))^{2}.
E.g. the set{F_{k-3}F_{k-2} F_{k+1}, F_{k-1} F_{k+2}F_{k+3}, F_{k}L_{k}^{2}, 4F_{k-1}^{2}F_{k} F_{k+1}^{2} (2F_{k-1}F_{k+1} - F_{k}^{2})}
has the property D(L_{k}^{2}).They also proved in [192] that the only triple of positive integers a < b < c such that ab + 1, ac + 1 and bc + 1 are all members of the Lucas sequence is (a,b,c) = (1,2,3).
Several authors considered other variants of the problem of Diophantus where squares are replaced by members of recursive sequences [177, 266, 266, 293, 309, 310, 311, 321, 327, 334, 343, 354, 376, 396, 402].
1. Introduction
2. Diophantine quintuple conjecture
3. Sets with the property D(n)
5. Rational Diophantine m-tuples
6. Connections with elliptic curves
7. Various generalizations
8. References
Diophantine m-tuples page | Andrej Dujella home page |