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Saudi Arabian Mathematical Competitions, 2012

Number theory Difficulty 5.0 AIME, harder Prove it Saudi Arabia

Find all positive integers $n$ such that $5^{n^2} + 7$ is divisible by $6$ .

Solution

We have $5^{n^2} + 7 = (6 - 1)^{n^2} + 7 \equiv (-1)^{n^2} + 1 \pmod{6}$ . It follows that $5^{n^2} + 7$ is divisible by $6$ if and only if $(-1)^{n^2} + 1 = 0$ . This is equivalent to $n^2$ being odd. Therefore, the possible values of $n$ are all odd positive integers.

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