Perfect Powers That Are Sums of Consecutive Cubes

Abstract

Abstract. Euler noted the relation 6 3 = 3 3 + 4 3 + 5 3 and asked for other instances of cubes that are sums of consecutive cubes. Similar problems have been studied by Cunningham, Catalan, Gennochi, Lucas, Pagliani, Cassels, Uchiyama, Stroeker and Zhongfeng Zhang. In particular Stroeker determined all squares that can be written as a sum of at most 50 consecutive cubes. We generalize Stroeker's work by determining all perfect powers that are sums of at most 50 consecutive cubes. Our methods include descent, linear forms in two logarithms, and Frey-Hellegouarch curves.

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

“…In the next year, Bennett, Patel and Siksek [3] extended Zhang's result, completely solving equation (2) in the cases k = 5 and k = 6. In 2016, Bennett, Patel and Siksek [4] considered equation (1). They gave the integral solutions to equation (1) using linear forms in logarithms, sieving, and Frey curves when k = 3, 2 ≤ r ≤ 50, x ≥ 1, and n is prime.…”

Perfect powers in an alternating sum of consecutive cubes

“…It is easy to see that for every k and n, (x, y) = (1, 1) is a solution of (1.3). Schäffer ([16]) proved that if k ≥ 1 and n ≥ 2 are fixed, then (1.3) has only finitely many solutions except the following cases (1.4) (k, n) ∈ { (1,2), (3,2), (3,4), (5,2)}.…”

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

“…x k + (x + r) k + · · · + (x + (d − 1)r) k = y n x, y, d, k, r, n ∈ Z, n ≥ 2. This is still a remarkably active field, with recent results due to [ZB13], [Zha14], [Haj15], [BPS16], [BPS17], [Pat17], [Soy17], [BPSS18], [PS17], [Zha17], [Kou17] and [AGP17]. In this paper, we consider the case d = 3 and k = 2, namely the equation…”