Show there are infinitely many primes that are equivalent to $1 \pmod{8}.$
I've tried using proof by contradiction:
How do I finish this off?
A one line answer would be to use Dirichlet's theorem for primes in arithmetic progression (Wiki link: https://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions). The statement of this theorem says that if $a$ and $b$ are integers with $gcd$ 1, then there are infinitely many primes in the arithmetic progression $\{a+bn:n\in\mathbb{Z}\}$. In your question $a=1$ and $b=8$ which are coprime. So this theorem gives your result. But this is like killing an ant with a tank!
Your proof can be completed by showing that the only prime divisors of $M=(2P)^4+1$ are of the form $8k+1.$ The proof to this fact is given in Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$. Now suppose $p$ is an odd prime dividing $M$ ($2\nmid M$ as $M$ is odd) then $p\equiv 1(\bmod 8).$ But since there are only $n$ such primes by hypothesis, it must be one out of these $n$ many primes. But if $$ M=(2P)^4+1\equiv 0(\bmod p)\\ \implies (2P)^4\equiv -1(\bmod p) $$ which is a contradiction since $P\equiv 0(\bmod p).$ Thus there are no divisors of $M$ and hence $M$ is a prime with $M\equiv 1(\bmod 8)$ which contradicts the fact that there are only $n$ such primes. Thus there must be infinitely many such primes.
Note: This type of "Euclidean elementary proof" can be given only for specific cases. Ram murty has discussed in https://mast.queensu.ca/~murty/murty-thain2.pdf the extent to which Euclidean proof can be given.
