Find the smallest postive integer $n>1$ such that $2016^n+1$ is a multiple of $n$ .

The answer is 2017.

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Use the factoring identity:

$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...-ab^{n-2}+b^{n-1})$

The equation is $2016^n+1^n$

Substitute the corresponding values of $a$ and $b$ into the identity to yield $2017$ as the first factor and its prime too, so that's the smallest integer.