4. Connections with Fibonacci numbers


4.1. Hoggatt-Bergum conjecture

There are many formulas for Diophantine quadruples with elements represented in terms of Fibonacci numbers. The most popular such quadruple was found in 1977 by Hoggatt & Bergum [8]:

{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 {F2k, F2k+2, F2k+4, d} is a Diophantine quadruple, then d cannot be a Fibonacci number.

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 2F2k F2k+2. This comparison finishes the proof for k > 48. The statement for 2 ≤ k ≤ 48 was proved by a version of the reduction procedure due to Baker & Davenport [3].

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].

Let the sequence (gn) be defined by

g0 = 0,   g1 = 1,   gn = pgn-1 - gn-2,

where p ≥ 2 is an integer. In [55], it was proved that the sets

{gn, gn+2, (p ± 2)gn+1, 4gn+1((p ± 2) gn+12 ∓ 1))}

are Diophantine quadruples. For p = 3 we obtain the Hoggatt-Bergum set, and for p = 2 we obtain a well known polynomial Diophantine quadruple

{n, n + 2, 4n + 4, 4(n + 1)(2n + 1)(2n + 3)}.

4.2. Diophantine quadruples for squares of Fibonacci and Lucas numbers

It was proved in [44] that for any integer n and any set {a, b} with the property D(n2), where ab is not a perfect square, there exist infinitely many sets of the form {a, b, c, d} with the property D(n2).

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: ab + n2 = m2 and s2 - ab t2 = 1, then we can construct the sequence xi,j and obtain infinitely many Diophantine quadruples with the property D(n2). There are several pairs of identities for Fibonacci and Lucas numbers, which have the above form. For example,

                                  4Fk - 1Fk+1 + Fk2 = Lk2,
(Fk2 + Fk-1 Fk+1)2 - 4Fk-1Fk+1 Fk2 = 1

and

                                    4Fk-2Fk+2 + Lk2 = 9Fk2,
(Fk2 + Fk-2Fk+2)2 - 4Fk-2Fk+2 Fk2 = 1.

Applying the construction from [44] to these pairs of identities, we get e.g. the set

{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).

4.3. Fibonacci and Lucas Diophantine triples

In 2008, Luca & Szalay [178] showed that there are no three distinct positive integers a, b, c such that ab + 1, ac + 1, bc + 1 are all three Fibonacci numbers.

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].


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