Prove that $Φ_{12}(x) = x^4 − x^2 + 1$ is reducible over $\mathbb{F}_{p}$ for every prime $p$. As the title explains, I'm trying to solve the following problem. 

Prove that $Φ_{12}(x) = x^4 − x^2 + 1$ is reducible over $\mathbb{F}_{p}$ for every prime $p$.

I've shown it's reducible over $p=2$ and $p=3$, and I've followed a hint I'm given to show for $p>3$ that $p^2-1$ is divisible by $12$.
I have no idea where to go from here - I can't see how divisibility by $12$ is supposed to help.
I should note that at this point in our course we've only just been introduced to the cyclotomic polynomials, so I don't think we're supposed to use any results about them that aren't basic.
I'd really appreciate any help you could give me.
 A: Think first about what happens over $\Bbb{Q}$. To get the splitting field of $\Phi_{12}(x)$ you need to adjoin the fourth root of unity $i=\sqrt{-1}$ as well as the third root of unity $\omega=(-1+i\sqrt3)/2$. In other words, we get the splitting field $\Bbb{Q}(\sqrt3,\sqrt{-1})$ by adjoining two square roots.
Over a finite field we then have the following consequence of the uniqueness of the field $\Bbb{F}_{p^2}$. Namely, $\sqrt{3}$ is either in $\Bbb{F}_p$ or generates the field $\Bbb{F}_{p^2}$. All according to whether $3$ is a quadratic residue or not.
And. The same holds for $\sqrt{-1}$. Either it is already in the prime field, or it will be an element of $\Bbb{F}_{p^2}$.

Irrespective of the details of adjoining those two square roots we see that $\Phi_{12}(x)$ splits into linear factors over $\Bbb{F}_{p^2}$. Therefore it will have quadratic factors at worst over $\Bbb{F}_p$. In particular, it won't be irreducible.


Alternatives.


*

*As Arthur suspected, we can use the fact that $12\mid p^2-1$ for any prime $p>3$. Given that the multiplicative group of $\Bbb{F}_{p^2}$ is cyclic of order $p^2-1$, this means that the roots of $\Phi_{12}(x)$ are all in $\Bbb{F}_{p^2}$. Consequently the minimal polynomials of those roots are quadratic at worst.

*We can write down a factorization using simple trigonometry, if one of $i$, $\sqrt{3}$ or $\sqrt{-3}$ is available in $\Bbb{F}_p$. Guess what, because $-3=(-1)\cdot3$, basic properties of quadratic residues tell us that at least one of those square roots will be available irrespective of choice of $p>3$. The cases $p=2,3$ are easy to handle anyway.

A: A simple answer by using Frobenius:
Suppose $p\neq2,3$ is prime. Let $\zeta$ be a root of $\Phi_{12}$ in $\overline{\mathbb F_p}$. Then we know that the other three roots are $\zeta^5, \zeta^7, \zeta^{11}$.
Now the Frobenius $x\mapsto x^p$ acts on the set of roots. If we identify these four roots with the multiplicative group $G = (\mathbb Z/12\mathbb Z)^\times$, then the Frobenius action just maps $a$ to $pa$. This means that the orbits of this action are exactly the quotient of $G$ by the subgroup generated by $p$.
Since $G$ is not cyclic, the Frobenius never acts transitively on $G$. Hence the polynomial $\Phi_{12}$ is always reducible.

This automatically generalizes to all $\Phi_n$ such that $(\mathbb Z/n\mathbb Z)^\times$ is not cyclic, i.e. $n$ is not of the form $p^r$ or $2p^r$.
