Show that: $2^{n+ 1} \mid q- 1$ & $2^{n+ 2} \mid q- 1$ For every $n \in \mathbb{N}$, let $F_{n}= 2^{2^{n}}+ 1$. For each $n \in \mathbb{N}$, let $q$ be a prime divisor of $F_{n}$, show that: $2^{n+ 1} \mid q- 1$. Furthurmore, if $n\geqq 2$, show even more that: $2^{n+ 2} \mid q- 1$
I can't find the relation of the 
hypothesis & the expression. To be honest, progress I've achieved on the problem is beneath mention (something like Euler's theorem or Fermat's theorem). I need to the help! Thanks!
 A: For the first argument, it is really only a matter of Fermat's little theorem. We know that $q\ | \ 2^{2^n} + 1 \ | \ 2^{2^{n+1}}-1$. What is the smallest $o \in \mathbb N$ such that $2^o \equiv 1 \pmod q$? From the information given we see that $o \ | \ 2^{n+1}$. So $o = 2^{n'}$ for some $n' \leq n+1$. Can it be strictly smaller, that is, $n' < n+1$? No, because if it happens $q \ | \ 2^{n'-1} + 1$. This cannot happen since we know that Fermat primes are mutually relatively prime.
This means that the $o = 2^{n+1}$ is the smallest positive integer such that $2^o \equiv 1 \pmod q$. But we know from Fermat's little theorem that $o \ | \ q-1$, hence the first argument is proved.
For the second argument, we need to find a number $b$ such that $o' = 2o = 2^{n+2}$ is the smallest number such that $2^{o'} \equiv 1 \pmod q$. This is somehow equivalent to finding the "square root" of $2$ modulo $q$. Does the square root exist? You need to look at the Legendre's symbol, that is, $\left(\frac 2q\right)$. The theorem is that $\left(\frac 2q\right) = (-1)^{(q^2-1)/8}$. If $8 \ | \ q$, that is guaranteed when $n \geq 2$, we know that $\left(\frac 2q\right) = 1$, so the square root exists, and we can apply similar arguments we used to prove the first part.
