How to solve following divisibility relation: $$n^2+1 \mid 2^n+1$$ for positive integer $n$?
You might have seen the similar IMO problem. Find all positive integers $n$ such that $\frac{2^n+1}{n^2}$ is an integer.
But I am sure that the extra $1$ has made this problem difficult. $n$ must be even. Let $p$ be the smallest prime divisor of $n^2+1$, then
$$p \mid \gcd \left(2^{p-1}-1, 2^{2n}-1 \right)=4^{\gcd \left((p-1)/2, n \right)}-1$$
Can we say something about $\gcd \left((p-1)/2, n \right)$? Any ideas?
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