Competition problem — Number theory

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Number theory Difficulty 5.1 AIME, harder Prove it Russia

Do there exist three distinct nonzero integers $a$ , $b$ , $c$ such that $a + b + c = 0$ , and $a^{13} + b^{13} + c^{13}$ is a square of a positive integer?

Существуют ли три попарно различных ненулевых целых числа, сумма которых равна нулю, а сумма тринадцатых степеней которых является квадратом некоторого натурального числа?

Solution

For a natural number $t$ , the triple $3t$ , $-t$ , $-2t$ satisfies all conditions except, possibly, the last one. To ensure that $(3t)^{13} + (-t)^{13} + (-2t)^{13} = t^{13}(3^{13} - 1 - 2^{13})$ is a perfect square, it is sufficient to set, for example, $t = 3^{13} - 1 - 2^{13}$ .

Thus, the triple $3t$ , $-t$ , $-2t$ , where $t = 3^{13} - 1 - 2^{13}$ , satisfies the conditions of the problem.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.