I believe this is a well known problem of Euler, but I was unable to find its solution anywhere. If $2^{2^m}+1$ is divisible by some prime $p$, then $p\equiv 1\pmod{2^{m+2}}$. Thanks in advance.
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reminds me of http://youtu.be/tJZpdXWm4Gg?t=1m15s – Alexander Grothendieck Dec 08 '13 at 23:29
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For every integer of the form $N=a^4+1$, the residue $2$ is a square modulo $N$, with $r=a^3-a$ being a square root, i.e. $r^2\equiv 2 \pmod{N}$. This applies to $m>1$ and $N=2^{2^m}+1$, so $2^{2^m} \equiv r^{2^{m+1}}\equiv -1\pmod{N}$. This congruence also holds for every prime divisor $p$ of $N$, thus $r$ has order $2^{m+2}\pmod{p}$, which by Fermat's little theorem implies that $p-1$ must be divisible by $2^{m+2}$.
ccorn
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I know it is in Will Dunham's book, "Journey Through Genius":http://books.google.com/books?id=_IbWAAAAMAAJ&focus=searchwithinvolume&q=2%5E%7B2n%7D%2B1+is+prime , where it is done by cases, but I cannot remember the actual page number.
user99680
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