There are infinitely many primes $p \equiv 7 \ (\text{mod} \ 8).$
I have just started to learn elementary number theory. May I know if my proof to the a/m problem is correct? Please advise, thank you.
Proof: Suppose there are finitely many such primes. Let $N = (p_1p_2...p_k)^2-2,$ where $p1,...,p_k \equiv 7\ (\text{mod} \ 8).$
Let $p$ be a prime divisor of $N.$ Then, $(p_1...p_k)^2 \equiv 2 \ (\text{mod} \ p).$ Hence, $2$ is quadratic residue modulo $p$ and $\Big(\frac{2}{p}\Big)=1 \implies p \equiv 1 \ \text{or} \ 7 \ (\text{mod} \ 8).$
If all prime divisors of $N$ are $\equiv 1 \ (\text{mod} \ 8),$ then $N \equiv 1 (\text{mod} \ 8).$
On other hand, since $(p_1)^2,...,(p_k)^2 \equiv 1 \ (\text{mod} \ 8), $ it follows that $N=(p_1...p_k)^2 -2 \equiv -1 \equiv 7 \ (\text{mod} \ 8).$ Hence, there exists a prime divisor $q \equiv 7\ (\text{mod} \ 8)$
Then, $q \mid (p_1...p_k)^2$ and $q \mid N \implies q \mid 2.$ (Contradiction).
