$2^{4n+1} \equiv 1 \pmod{8n+7}$, this has been bugging me

Question

Here is the question:

Suppose that $p$ is an odd prime. The law of quadratic reciprocity says that $x^2\equiv 2\pmod p$ has a solution. if $p\equiv1 \text{ or } 7 \pmod 8$. Prove that $2^{4n+3}\equiv1 \pmod{8n+1}$. $8n+1$ is a prime

I honestly don't know where to start, I tried starting it with Fermat's Theorem because it looked similar to that, $a^{p-1} = 1 \pmod p$, but it didn't lead me anywhere, or I missed something.

Your help will be appreciated!

Answer 1

If $2$ is a quadratic residue $\pmod p$, then $\exists a\in\Bbb Z/p\Bbb Z$ s.t. $a^2\equiv 2\pmod p$.

In the case $p=8n+7$ where $p$ is prime, $$2^{4n+3}\equiv a^{8n+6}\equiv a^{\phi(p)}\equiv 1\pmod p$$

Answer 2

$\rm mod\ p\!:\ 0\not\equiv a \equiv x^2 \Rightarrow\ a^{(p-1)/2}\equiv x^{p-1}\equiv 1\ $ by little Fermat (and conversely: Euler's criterion)