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Determine all integers n > 1 such that \[\frac {2^n + 1}{n^2}\] is an integer.
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For the fraction to be an integer, 2^n+1 should be a multiple of n^2, so: \[2^n+1 = kn^2, k\]
\[k \in \mathbb{R} \]
ie \[n^2|2^n+1 \implies n\in\mathbb{N}: 2^n\equiv-1\pmod{n^2} \]
I think you can write: \[2^n + 1 = kn^2\] Differentiate for n: \[2^n log2 = 2kn\] Differentiate again: \[2^n (log2)^2 = 2k\] It could be possible to find n now. But you should wait from another response, because I am not sure about the answer.
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