So, I encountered this issue in the proof of the constructability of a regular n-gon, where it is shown that the n-gon is constructable iff it is a product of $2^k$ and distinct primes $p_i$ such that $p_i-1 = 2^k$. I understand that, however, I don't understand how we can follow that if $p - 1 = 2^k$ $p$ must be a Fermat prime, i.e. $p = 2^{2^s} + 1$. It is clear that if $k$ would divide $p-1$, $k$ would be a power of two, and my textbook claims that it is elementary to see that $k$ must divide $p-1$ - however, I don't know how I could prove that. What am I missing?
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The easiest way to show that a prime of the form $2^k+1$ must be a Fermat prime is to assume that $k$ has an odd prime divisor , say $q$ , then with $r:=2^{\frac{k}{q}}$ we have $2^k=r^q\equiv (-1)^q=-1\mod (r+1)$ , hence because of $1<r+1<2^k+1$ , $r+1$ is a nontrivial divisor of $2^k+1$ , hence $2^k+1$ is not prime. I am confused about the claim that $k$ must divide $p-1$ and also of $p-1=2^kp$. This does not really make sense. – Peter Sep 27 '22 at 07:47
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The largest Fermat prime is with high probability $F_4=2^{16}+1=65537$. The next possible Fermat prime is $F_{33}$. The probability of a further Fermat prime is less than $1:10^9$ , but nevertheless there could be another Fermat prime , we cannot even rule out infinite many of them. – Peter Sep 27 '22 at 07:56