Perfect powers that are sums of squares in a three term arithmetic progression

Abstract

We determine primitive solutions to the equation (x − r) 2 + x 2 + (x + r) 2 = y n for 1 ≤ r ≤ 5, 000, making use of a factorization argument and the Primitive Divisors Theorem due to Bilu, Hanrot and Voutier.

Perfect Powers that are Sums of Squares of an Arithmetic Progression

“…Let 2 ≤ d ≤ 10. Then, Theorems 1.2, 1.3 and results from [10] characterize all primitive solutions to equation (1) with 1 ≤ r ≤ 10 4 , except in the case d = 6, where we have 1 ≤ r ≤ 5000.…”

“…For the proof of Theorem 1.1, it remains to deal with d = 2, 3 and 6 for n ≥ 2, and also with d = 8 for n = 2. The case d = 3 and r ≤ 10 4 has been resolved in [10] and a table of solutions can be found in that paper.…”

“…Lemma 6.2. Let α, β be as in equation (10), with α = (±1+i)/ √ 2 or (±1−i)/ √ 2. Then, the only solutions to equation (7)…”

“…In the proof of Theorem 1.2, we are able to solve the case d = 2, r = 1, n ≥ 3 uniformly with other values of r ≤ 10 4 , using a completely different approach to that taken in [8]. We have also consolidated the results of [10] into Theorem 1.1, where the second author and Koustsianas find all primitive solutions to equation (1) for the case d = 3 and prime exponent n. As in [10] and [11], our main tool is the characterization of primitive divisors in Lehmer sequences due to Bilu, Hanrot and Voutier [6].…”

On the sum of fourth powers in arithmetic progression

“…Also the more general case of equation (1) has been studied before. For the case k = 2 and gcd(x, z) = 1 Koutsianas and Patel [18] used prime divisors of Lehmer sequences to determine all solutions when 1 ≤ y ≤ 5000. Koutsianas [17] further studied this case when y is a prime power p m for specific prime numbers p. The case k = 3 was partially solved by Argáez-García and Patel [1] giving all solutions in case 1 ≤ y ≤ 10 6 using different techniques including the modular method for some Frey curves over the rationals.…”