{F2k, F2k+2, F2k+4, 4F2k+1 F2k+2F2k+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 {F2n, F2n+2, Fk} 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 (gn) be defined by
g0 = 0, g1 = 1, gn = pgn-1 - gn-2,
where{gn, gn+2, (p ± 2)gn+1, 4gn+1((p ± 2) gn+12 ∓ 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 xi,j. In the construction, solutions of Pell equation s2 - ab t2 = 1 were used.
If we have a pair of identities of the form:
4Fk - 1Fk+1 + Fk2 = Lk2,
(Fk2 + Fk-1
Fk+1)2 - 4Fk-1Fk+1 Fk2 = 1
4Fk-2Fk+2 + Lk2 = 9Fk2,
(Fk2 + Fk-2Fk+2)2 - 4Fk-2Fk+2 Fk2 = 1.
{2Fk-1, 2Fk+1, 2Fk3 Fk+1Fk+2, 2Fk+1Fk+2 Fk+3(2Fk+1 2 - Fk2)}
with the property D(Fk2) and the set{2Fk-2, 2Fk+2, 2Fk-1 FkLk2 Lk+1, 2Lk-1 Fk Lk2 Lk+1}
with the property D(Lk2) (see [48]). In the same paper, several quadruples were obtained using Morgado identity:Fk-3Fk-2 Fk-1Fk+1 Fk+2Fk+3 + Lk2 = (Fk(2Fk-1 Fk+1 - Fk2))2.
E.g. the set{Fk-3Fk-2 Fk+1, Fk-1 Fk+2Fk+3, FkLk2, 4Fk-12Fk Fk+12 (2Fk-1Fk+1 - Fk2)}
has the property D(Lk2).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 |