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Let $p$ be a prime divisor of the Fermat number $F_n = 2^{2^n} + 1$. Prove that $p$ must have the form $2^{n+1}k + 1$.

Thank you in advance.

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  • $\begingroup$ $1k$? $\space \space$ $\endgroup$ – D_S Oct 24 '16 at 3:38
  • $\begingroup$ (2^(n+1))*k + 1 $\endgroup$ – Zed1 Oct 24 '16 at 3:43
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This is not very hard.

From Fermat's little theorem, $$p\mid 2^{p-1} -1,$$ also $$p\mid 2^{2^n} + 1 \mid 2^{2^{n+1}} -1,$$ so $$p \mid gcd(2^{p-1} -1, 2^{2^{n+1}}-1) = 2^{gcd(p-1, 2^{n+1})}-1.$$

On the other hand, $$gcd(p-1, 2^{n+1})=2^m$$ for some positive integer $m$(as $p$ is odd). If $m<n+1$, then $$p\mid 2^{2^m}-1 \mid 2^{2^n} -1= F_n -2,$$ which is impossible. So $m=n+1$, and $2^{n+1} \mid p-1$.

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