Let $p=2^{2^k}+1$ be a prime where $k\ge1$. Prove that the set of quadratic non-residues mod $p$ is the same as the set of primitive roots mod $p$. Use this to show that $7$ is a primitive root mod $p$.
I've already shown the theorem to be true. The second part asks to use the first part to show the result which leads me to think that I have to show $7$ is a quadratic non-residue mod $p$ then use the first part to imply that it must be a primitive root.
To show $7$ to be a quadratic non-residue for $k\ge1$ is to show that the Legendre symbol $\left(\frac{7}{p}\right) = -1$. Now, $$\left(\frac{7}{p}\right)=\left(\frac{p}{7}\right)(-1)^{\left(\frac{7-1}{2}\right)\left(\frac{p-1}{2}\right)} = \left(\frac{p}{7}\right)(-1)^{3(2^{(2^k)-1})} = \left(\frac{p}{7}\right)$$ since $2^{2^k-1}$ is even (as $k\ge1$).
Then it suffices to know $p$ mod $7$ to determine the Legendre symbol. Since $\left(\frac{p}{7}\right) = -1$ when $p\equiv 3,5,6$ mod $7$, I suspect I somehow have to show that $p$ must be congruent to those values but I don't know how to do that. Although, trivially, $p\not\equiv 1$ mod $7$ otherwise, $7|2^{2^k}$ which is not possible.
Unfortunately, I don't know where to go from here.
Any guidance would be appreciated. However, assuming I’ve taken the right approach, I would prefer a constructive hint to a full blown solution, as I think I may be able to work it out on my own, given a nudge in the right direction.
Thank you for taking the time.
Powers of 2 are congruent to $1, 2, $ or $4$ modulo $7$ according as the power is congruent to $0, 1$ or $2$ modulo $3$ (as $3$ is the order of $2$ modulo $7$). As $3\nmid 2^k,$ $p\equiv 3\ or\ 5 \pmod 7.$