I want to show that $p$ is a Fermat prime $\iff$ every quadratic non-residue of $p$ is also a primitive root mod $p$
These are some facts that I know:
$F_n = 2^{2^n} + 1$
Every prime divisor $p$ of $F_n$ is of the form $2^{k + 1}k + 1$
$\mathrm{ord}_{F_n}(2) = \mathrm{ord}_{p}(2) = 2^{n + 1}$
This is what I wrote for the ($\Rightarrow$) direction and I want to make sure it is correct:
$p = F_n = 2^{2^n} + 1$ is prime, $\phi(p) = p - 1 = 2^{2^n}$
let $a$ be a quadratic non-reside mod $p$ so $x^2 \equiv_p a$ has no solutions. Suppose $ord_p(a) = h$. Since $h \ | \phi(p) = 2^{2^n}$ then $h = 2^k$ for some $k \leq 2^{n}$
Then $(x^2)^h \equiv_p x^{2h} \equiv_p a^h \equiv_p 1$ which is the same as $x^{2^{k + 1}} \equiv_p 1$ has no solutions therefore $2^{k + 1}\nmid 2^{2^n}$ We also know $2^k \mid 2^{2^n}$ so from these two we can conclude that $2^k = 2^{2^n}$ so the order of $a$ is $\phi(p)$ hence it is a primitive root mod $p$.
I don't know what to do for the other direction. Hints would be appreciated.
We will use the fact that the set of quadratic non-residues mod $p$ is the same as the set of primitive roots mod $p$. Which means $\phi(\phi(p)) = \phi(p - 1) = \frac{p - 1}{2}$ Hence $2 \mid p - 1$ so we can write $p = 2^km$ where $m$ is odd. Since $2^k$ and $m$ are relatively prime and the Euler Phi function is multiplicative we have $\phi(p - 1) = \phi(2^k)\phi(m) = 2^{k - 1}\phi(m) = \frac{p - 1}{2} = 2^{k - 1}m$ thus $m = \phi(m)$ which means $m = 1$.
So now we have $p - 1 = 2^k \rightarrow p = 2^k + 1$. Now suppose $k = 2^st$ where $t$ is odd. Now we can write $p = (2^{2^s})^t + 1$. If $t > 1$ then $p$ could be factorised since $t$ is odd therefore $t = 1$ so now we have $p = 2^{2^s} + 1 = F_s$.