Fermat primes 17 and 257 appear a lot in the prime composition of numbers of the form $a^{2^n}+1$. For example, $11^8+1$ is divisible by 17 and $11^{32}+1$ is divisible by 257. I have verified the following statement for 3, 5, 17 and 257 but can't prove it. I would be grateful if someone could provide a hint.
Let $F = 2^{2^k} + 1$ be a Fermat prime.
Then for $n=0, 1, ..., 2^k-1, x^{2^n} + 1 \equiv 0 \mod F$ has exactly $2^n$ solutions.
For example, when k=2 we have:
$(n=0) \hspace{10pt} x^1 + 1 \equiv 0 \mod 17$ has one solution (x=16).
$(n=1) \hspace{10pt} x^2 + 1 \equiv 0 \mod 17$ has two solutions (x=4 and 13).
$(n=2) \hspace{10pt} x^4 + 1 \equiv 0 \mod 17$ has four solutions (x=2, 8, 9 and 15).
$(n=3) \hspace{10pt} x^8 + 1 \equiv 0 \mod 17$ has eight solutions (x=3, 5, 6, 7, 10, 11, 12 and 14).