Perfect powers in an alternating sum of consecutive cubes

Abstract

In this paper, we consider the problem about finding out perfect powers in an alternating sum of consecutive cubes. More precisely, we completely solve the Diophantine equation (x + 1) 3 − (x + 2) 3 + • • • − (x + 2d) 3 + (x + 2d + 1) 3 = z p , where p is prime and x, d, z are integers with 1 ≤ d ≤ 50.

Perfect powers in sum of three fifth powers

“…Several authors have also studied equations (1.1), (1.2) and its variants using a variety of classical and modern techniques (see e.g. [5,10,14,19,20,23,24,25]).…”