Legendre Symbol Problem 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!

If $p$ divides $x^4+1$, then it divides $(x^2+1)^2 - 2x^2$, hence $(x^2+1)^2\equiv 2x^2\pmod{p}$. 
Can you conclude from this that $\left(\frac{2}{p}\right)=1$?
And if $p$ divides $x^4+1$, then $x^4\equiv -1\pmod{p}$; can you conclude that $\left(\frac{-1}{p}\right) = 1$?
Since $\left(\frac{2}{p}\right) = 1$ if and only if $p\equiv \pm 1\pmod{8}$, and $\left(\frac{-1}{p}\right)=1$ if and only if $p\equiv 1\pmod{4}$, can you see why this forces $p\equiv 1\pmod{8}$? And why you can now conclude that there are infinitely many such primes?
