Fibonacci Diophantine triples

Abstract

Abstract. In this paper, we show that there are no three distinct positive integers a, b, c such that ab + 1, ac + 1, bc + 1 are all three Fibonacci numbers.

Diofantikus számhármasok a Lucas-Lehmer sorozatokban

“…Florian Luca és Szalay László [2] bebizonyították, hogy a Fibonacci sorozathoz nem adható meg diofantikus számhármas, azaz nincsenek olyan 0 < < < egészek, hogy + 1 = , + 1 = és + 1 = volna valamely x, y és z indexekre. Szintén az ő eredményük [3], hogy a Fibonacci-sorozat asszociáltjához, a Lucassorozathoz egyetlen ilyen hármas létezik: = 1 , = 2 és = 3 .…”

Lucas Diophantine Triples

“…According to Theorem 1, there should be only finitely many triples of distinct positive integers {a, b, c} such that ab + 1, ac + 1 and bc + 1 are either all three Fibonacci numbers or all three Lucas numbers. In [12], we showed that there is no such triple for the case of the Fibonacci sequence. In this paper, we deal with the same problem for the case of the Lucas sequence.…”

“…Proof. This is Lemma 6 in [12]. We point out that the particular cases λ = 1 and λ = 3 of Lemma 10 were used in [12].…”

“…In [12], the crucial point of the proof for the case of the Fibonacci sequence was the existence of a factorization of F n −1 (see Lemma 6) in terms of smaller Fibonacci and Lucas numbers. Here, the case of the Lucas numbers is more complicated since such a factorization of L n − 1 exists if and only if n is odd.…”

Balancing Diophantine triples with distance 1

“…(k, ) = (3, 2), (4, 3), (5, 3), (5,4), (6,5), (7,4), (7,5), (7,6), the previous argument provides the upper bounds …”