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Find all positive integer $n$ such that there exists $m$ with $2^n-1|m^2+17^2$.

I have tried to mod $2^n-1$ and use the fact that $2^n \equiv 1 \pmod{2^n-1}$. I have also tried to factorize $m^2+17^2$ into $(17+mi)(17-mi)$ and use the arithmetic of $\mathbb{Z}[i]$ but it didn't work out, please help, thank you so much

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  • $\begingroup$ Why on earth would anyone vote to close this question??? $\endgroup$ – barak manos Mar 25 '15 at 13:31
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For $n\geqslant2$, $2^n-1\equiv3\pmod4$ which means $2^n-1$ has at least one prime divisor $p\equiv3\pmod4$. Then $m^2+17^2\equiv0\pmod p$. As $p\neq17$, let $a$ be an inverse of $17$ modulo $p$. Then $(am)^2\equiv-1\pmod p$, contradicting $p\equiv3\pmod4$.

It remains to check $n=1$, which clearly satisfies the condition.

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