Competition problem — Number theory

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Number theory Difficulty 5.4 AIME, harder Prove it Vietnam

Let be given a positive integer $n$ and two coprime integers $a, b$ greater than $1$ . Let $p, q$ be two odd divisors greater than $1$ of $a^{6n} + b^{6n}$ . Find the remainder of the division of $p^{6n} + q^{6n}$ by $6 \cdot (12)^n$ .

Solution

The answer follows easily from the following remarks:

a) If $a, b$ are coprime integers greater than $1$ and $p$ is an odd prime divisor of $a^{6n} + b^{6n}$ then $p \equiv 1 \pmod{2^{n+1}}$ .

b) If $x \equiv 1 \pmod{c^k}$ then $x^{c^m} \equiv 1 \pmod{2^{m+k}}$ .

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