Show that a prime divisor to $x^4-x^2+1$ has to satisfy $p=1 \pmod{12}$ Suppose that $x$ solves $x^4-x^2+1= 0 \pmod p$. Show that $p=1 \pmod {12}$. Following a hint I have rewritten the equation as $(x^2-1)^2=-3 \mod p$ and $(2x^2-1)^2=-x^2 \pmod p$. The first equation gives that $p=1 \pmod 3$ by using quadratic reciprocity and noting that $1$ is the only applicable quadratic residue. However, I am not able to show that $p=1 \pmod 4$. Since this is a condition in some formulations of quadratic reciprocity I guess that it should be used here as well. This is exercise 13 chapter 5 from the book by Ireland and Rosen if it is any help. Thankful for any hints!

Observe that $p$ cannot divide $x$, so that $x$ is invertible modulo $p$. Rewrite the second rewritten congruence as
$$\left( \frac{2x^2-1}{x} \right)^2 \equiv -1 \pmod{p} .$$
This shows that -1 is a quadratic residue modulo $p$, implying that $p\equiv 1\pmod{4}$.
