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

TL;DR

This paper proves that certain sums of seven cubes in an arithmetic progression cannot be perfect powers unless trivial solutions occur, using advanced number theory techniques.

Contribution

It extends previous work by analyzing a more complex sum of cubes and applying the Primitive Divisor Theorem for a complete solution.

Findings

Solutions only satisfy xy=0 for the specified range and conditions.

The sum of seven cubes in an arithmetic progression rarely yields perfect powers.

The methodology confirms the non-existence of non-trivial solutions for the equation.

Abstract

We prove that the equation $(x-3r)^3+(x-2r)^3 + (x-r)^3 + x^3 + (x+r)^3 + (x+2r)^3+(x+3r)^3= y^p$ only has solutions which satisfy $xy=0$ for $1\leq r\leq 10^6$ and $p\geq 5$ prime. This article complements the work on the equations $(x-r)^3 + x^3 + (x+r)^3 = y^p$ and $(x-2r)^3 + (x-r)^3 + x^3 + (x+r)^3 + (x+2r)^3= y^p$ . The methodology in this paper makes use of the Primitive Divisor Theorem due to Bilu, Hanrot and Voutier for a complete resolution of the Diophantine equation.