Powers in products of terms of Pell's and Pell–Lucas Sequences

Abstract

In this paper, we consider the usual Pell and Pell–Lucas sequences. The Pell sequence [Formula: see text] is given by the recurrence un = 2un-1 + un-2 with initial condition u0 = 0, u1 = 1 and its associated Pell–Lucas sequence [Formula: see text] is given by the recurrence vn = 2vn-1 + vn-2 with initial condition v0 = 2, v1 = 2. Let n, d, k, y, m be positive integers with m ≥ 2, y ≥ 2 and gcd (n, d) = 1. We prove that the only solutions of the Diophantine equation unun+d⋯un+(k-1)d = ym are given by u7 = 132 and u1u7 = 132 and the equation vnvn+d⋯vn+(k-1)d = ym has no solution. In fact, we prove a more general result.

On perfect powers that are sum of two balancing numbers

“…In the later case, by [5,Lemma 2.6] we have n ∈ {1, 2, 7} and these values of n, only n = 1 satisfies…”

Pell and Pell-Lucas numbers of the form $-2^a-3^b+5^c$

“…In 1991, Pethő in [10] found all perfect powers in the Pell sequence. In 2015, Bravo, Das, Guzmán and Laishram in [2] found the powers in products of terms of Pell and Pell-Lucas sequences.…”

Perfect Powers in Products of Terms of Elliptic Divisibility Sequences

“…To prove Theorem 1.1, we need to combine several tools, including arithmetic properties of elliptic divisibility sequences, arguments from [3,11] and new variants of bounds, developed in this paper, concerning the greatest prime divisor and the number of prime divisors of blocks of consecutive terms of arithmetic progressions.…”

“…Concerning the general setting, we mention the paper of Luca and Shorey [11], where they gave an effective upper bound for the size of the solutions to the equation when a product of terms from a Lucas sequence or from its companion sequence equals a perfect power. In case of individual recurrences, we refer to Bravo, Das, Guzmán and Laishram [3] who considered the previously mentioned equations with the Pell and Pell-Lucas sequences, listing all solutions. Their proofs also provide a method for Lucas and their companion sequences, in general.…”

“…,(1,1,3,1),(1,1,4,1), (1, 2, 2, 1), (1, 3, 2, 1), (1, 3, 3, 1), (1, 6, 2, 1), (2, 1, 2, 1), (2, 1, 3, 1), (2, 5, 2, 1), (3, 1, 2, 1), (3, 4, 2, 1), (4, 3, 2, 1)and, further, for = 7, we also have the solutions(1,11,2,2),(2,5,3,2),(7,5,2,2).…”