Let $p(x) = x^2 -ax + 1 \in \mathbb{Z}_q[x] $ where $q > 3$ is a prime. For what values of $a \in \mathbb{Z}_q $ is $p(x) $ irreducible, i.e. has no roots in $\mathbb{Z}_q $?
For instance, consider $ \mathbb{F}_4 = \{ 0,1, \alpha, \alpha + 1 \}$. The polynomials $p(x) = x^2 - \alpha x + 1 $ and $q(x) = x^2 - ( \alpha + 1 )x +1 $ are irreducible, but not when $a = 0$ or $a = 1$.
But if $\mathbb{F}_q = \mathbb{Z}_5 = \{0,1,2,3,4 \}$ then $p(x) = x^2 + x + 1$ is irreducible, i.e when $a = 1$.
I do not necessarily need to limit the field to $\mathbb{Z}_p$, it can be $\mathbb{F}_{p^n}$, but it is perhaps easier to begin in that end. I was wondering if there is any results involving this and what the intuition is behind it?
Best regards