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, 2014

Number theory Difficulty 4.8 AIME Prove it Ireland

Find with proof, all triples of non-negative integers $(x, y, n)$ satisfying
$(x^4 + 1)^3 + (y^4 + 1)^3 = 2014^n.$

Solution

Consider the given equation modulo $13$ . The squares modulo $13$ are $0$ , $\pm 1$ , $\pm 3$ , $\pm 4$ , so the fourth powers modulo $13$ are $0$ , $1$ , $3$ , $9$ . Therefore $x^4 + 1$ must be one of $1$ , $2$ , $4$ , $10$ modulo $13$ . For $z$ congruent to $1$ , $2$ , $4$ , $10$ modulo $13$ , the value of $z^3$ can only be $\pm 1$ or $8$ modulo $13$ .
Since $2014^n \equiv (-1)^n \equiv \pm 1$ modulo $13$ , the equation reduces to
$r + s \equiv \pm 1 \pmod{13}$
where $r$ and $s$ are either $\pm 1$ or $8$ modulo $13$ – this can easily be seen to have no solutions.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.