Perfect Powers that are Sums of Squares of an Arithmetic Progression

Abstract

In this paper, we determine all primitive solutions to the equation (x + r) 2 + (x + 2r) 2 + • • • + (x + dr) 2 = y n for 2 ≤ d ≤ 10 and for 1 ≤ r ≤ 10 4 (except in the case d = 6, where we restrict 1 ≤ r ≤ 5000). We make use of a factorization argument and the Primitive Divisors Theorem due to Bilu, Hanrot and Voutier.

On perfect powers that are sums of cubes of a five term arithmetic progression

“…x 2 + 12r 2 = 12w p 2 . We let x = 6X and obtain (11) 3X 2 + r 2 = w p 2 . We let C 1 = 3, C 2 = r 2 and we see that C 1 C 2 ≡ 0, 3, 4 (mod 8) ≡ 7 (mod 8).…”

“…Thus if (11) has a solution, then we have 5 ≤ p ≤ 13 or p | q − −3 q , where q is some prime q | r and q 6. For fixed values of 1 ≤ r ≤ 10 6 (hence fixed C 1 , C 2 , p), finding the roots of the polynomial (7), leads to solutions (X, w 2 , p).…”

“…Finding perfect powers that are sums of terms in an arithmetic progression has received much interest lately; recent contributions can be found in [1], [2], [3], [4], [5], [6], [10], [11], [14], [15], [17], [18], [19], [21], [23], [24], [26], [27]. This paper aims to demonstrate an application of Theorem 1 in [16] to the problem of finding perfect powers in sums of like powers, thus adding to the diverse range of methodologies currently being used to tackle such problems.…”

The Diophantine equation $(x+1)^k+(x+2)^k+\cdots+(\ell x)^k=y^n$ revisited

“…More recently, Garcia and Patel [9] complemented the work of Cassels, Koutsianas and Zhang by considering the case when k = 3 and showing that equation (3) with n ≥ 5 a prime and 0 < d ≤ 10 6 has only trivial solutions (x, y, n) which satisfy xy = 0. Then, Kundu and Patel [15] determined all primitive solutions to the equation (x + r) 2 + (x + 2r) 2 + • • • + (x + dr) 2 = y n for 2 ≤ d ≤ 10 and for 1 ≤ r ≤ 10 4 (except the case d = 6, where they considered only 1 ≤ r ≤ 5000).…”