Determine all set of integers $n\gt 1$ such that $$\frac{2^n+1}{n^2}\;\;\text{ is a integer. }$$
I found that $n$ is an odd.
Please help by providing me with a hint.
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A reasonable conjecture is that $n=3$ is the only solution, since it does not happen so often that $n$ is a divisor of $2^n+1$, or that $2^n+1$ is not square-free. – Jack D'Aurizio May 18 '17 at 15:48
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This is a really famous one, its the hard cousin of the ubiquitous $n| 2^n-1$. An answer can be found here: https://www.quora.com/If-n-geq-2-divides-2-n-+1-then-3-divides-n-How-can-I-prove-this. All though there are better expositions. – Asinomás May 18 '17 at 15:52