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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$1k$? $\space \space$ – D_S Oct 24 '16 at 03:38
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(2^(n+1))*k + 1 – Zed1 Oct 24 '16 at 03:43
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Or, possibly a better duplicate target. – Jyrki Lahtonen Jul 10 '21 at 10:16
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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 $2^{n+1} \mid p-1$.
Sammy Black
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S. Y
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We have that for $n=0$ as well as $n=1$. For $n\ge2$, I thought of it related to this question – J. W. Tanner May 26 '20 at 00:54
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You are right. I was thinking around these lines. Don’t quite remember though. – S. Y May 26 '20 at 00:57