Abstract
For an integer k≥2, let {Fn(k)}n≥2−k be the k–generalized Fibonacci sequence which starts with 0,…,0,1 (a total of k terms) and for which each term afterwards is the sum of the k preceding terms. In this paper, for an integer d≥2 which is square-free, we show that there is at most one value of the positive integer x participating in the Pell equation x2−dy2=±1, which is a k–generalized Fibonacci number, with a couple of parametric exceptions which we completely characterize. This paper extends previous work from [18] for the case k=2 and [17] for the case k=3.
Highlights
Let d ≥ 2 be a positive integer which is not a perfect square. It is well-known that the Pell equation x2 − dy2 = ±1 (1)
Letting (x1, y1) be the smallest positive solution, all solutions are of the form for some positive integer n, where xn + yn d = (x1 + y1 d)n for all n ≥ 1
They proved that equation (3) has at most one solution (n, m) in positive integers except for d = 2, in which case equation (3) has the three solutions (n, m) = (1, 1), (1, 2), (2, 4)
Summary
Let d ≥ 2 be a positive integer which is not a perfect square. It is well-known that the Pell equation x2 − dy2 = ±1. Where {Fm}m≥0 is the sequence of Fibonacci numbers given by F0 = 0, F1 = 1, and Fm+2 = Fm+1 + Fm for all m ≥ 0 They proved that equation (3) has at most one solution (n, m) in positive integers except for d = 2, in which case equation (3) has the three solutions (n, m) = (1, 1), (1, 2), (2, 4). Where {Tm}m≥0 is the sequence of Tribonacci numbers given by T0 = 0, T1 = 1, T2 = 1, and Tm+3 = Tm+2 + Tm+1 + Tm for all m ≥ 0 They proved that equation (4) has at most one solution (n, m) in positive integers for all d except for d = 2 when equation (4) has the three solutions (n, m) = (1, 1), (1, 2), (3, 5) and when d = 3 case in which equation (4) has the two solutions (n, m) = (1, 3), (2, 5).
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