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!
 A: 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?
A: This does not use the hint, so does not answer your question. But as part compensation, we prove a stronger result. Let $p$ be an odd prime that divides $x^{2^n}+1$. Then $x^{2^n}\equiv -1 \pmod{p}$. It follows that $x^{2^{n+1}}\equiv 1\pmod p$, and therefore the order of $x$ modulo $p$ divides $2^{n+1}$. But since $x^{2^n}\not\equiv 1\pmod{p}$, the order of $x$ is not $\le 2^n$. So the order of $x$ is exactly $2^{n+1}$, and therefore $2^{n+1}$ divides $p-1$, that is, $p\equiv 1 \pmod{2^{n+1}}$.
Now let $k$ be a large positive integer,  let $K=(k!)^{2^n}+1$, and let $p$ a (necessarily odd) prime divisor of $K$. It is clear that $p>k$. By the result above, $p \equiv 1\pmod{2^{n+1}}$.  Thus $p$ is prime $>k$ which is congruent to $1$ modulo $2^{n+1}$. So for every positive integer $n$, there are arbitrarily large primes congruent to $1$ modulo $2^{n+1}$. 
A: Alternate proof that if an odd prime $p|n^4+1$ for some $n\in\mathbf{Z}$, then $p\equiv 1\pmod{8}$. We have $n^4\equiv -1\pmod{p}\implies n^8\equiv 1\pmod{p}$, so $\text{ord}_p(n)|8$. Note that $\text{ord}_p(n)\nleq 4$, so $\text{ord}_p(n)=8$. Additionally, by FLT $n^{p-1}\equiv 1\pmod{p}$, so $\text{ord}_p(n)|n-1\implies 8|p-1$, as desired. 
This directly implies that $2$ is a quadratic residue in $\mathbf{Z}/p\mathbf{Z}$, so $\left(\frac{2}{p}\right)=1$.
