I am doing revision for my number theory exam and I am stuck on the following question.
Let $x$ be an even integer. Show that every prime divisor $p$ of $x^4 + 1$ satisfies $\big(\frac{-1}{p}\big)$ = $\big(\frac{2}{p}\big) = 1$ where $\big(\frac{a}{p}\big)$ denotes the Legendre Symbol. Deduce that there are infinitely many primes $p \equiv 1 \pmod{8}$.
Hint: Observe that $x^4 + 1 = (x^2 + 1)^2 - 2x^2$
So as $x$ is even this means $x^4 + 1$ is odd, hence $p$ has to be an odd prime. Don't know where to go from here. How does the hint help?
Any advice would be very much appreciated!