# 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!

-
Maybe you could rewrite the second rewritten congruence as $\left( \frac{2x^2-1}{x}\right)^2 \equiv -1 \pmod{p}$ (clearly p cannot divide x), thereby showing that -1 is a quadratic residue. –  Daan Michiels May 6 '12 at 17:42
Thank you! If you post this as an answer I will gladly accept. –  user3620 May 6 '12 at 19:40

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 $p\equiv 1\pmod{4}$.
This is the minimal polynomial $f(x)$ of a primitive twelfth root of unity. It has no root mod $2$ or $3$. Therefore if $f$ has a root in $\mathbb{F}_p$ then it is a primitive twelfth root of unity since $f$ is a factor of $X^{12}-1$ and the latter has no double roots in $\mathbb{F}_p$ (the gcd of $X^{12}-1$ and $12 X^{11}$ is $1$ in $\mathbb{F}_p$). Therefore $12$ divides the order of $\mathbb{F}_p^{\ast}$. In other words $p \equiv 1 \pmod{12}$.