Computing the number of irreducible polynomials in a field

Question

Let p be an odd prime. Compute the number of irreducible polynomials of the form $f(x) = x^2 + x + a$ over $\mathbb{Z}_p$

I know there are $p$ total possibilities for that polynomial and that a degree 2 polynomial is reducible in $\mathbb{Z}_p$ iff it has a root in $\mathbb{Z}_p$ and for $f$ to be reducible that would mean it is of the form $f(x) = (x-b)(x-c)$ where $b,c \in \mathbb{Z}_p$. Mulptiplying these out yields $x^2 + (-b-c)x + bc$ so we need $bc=a$ and $b+c=-1=p-1$. How do I count the number of possible values for $b$ and $c$? I think the number of possible combinations for $b$ and $c$ to meet the condition $b+c=-1$ is $p$ (not sure though), but I can't figure out which of those $p$ choices will yield $bc=a$

Answer 1

$x^2+x+a$ is irreducible over $\mathbb{F}_p$ iff its discriminant $1-4a$ is a quadratic non-residue $\!\!\pmod{p}$.
There are $\color{blue}{\frac{p-1}{2}}$ quadratic non-residues and $4$ is invertible $\!\!\pmod{p}$ since $p$ is odd.

Answer 2

if $x^2+x+a$ is reducible, then it has roots in $\mathbb{Z}_p$, thus $a=-b^2-b$ for some $b \in \mathbb{Z}_p$

Now note that for the map $s:\mathbb{Z}_p \to \mathbb{Z}_p$ defined by $$s(c) =-x^2-x $$ We can say $x^+x+a$ is irreducible if and only if $a \not\in s(\mathbb{Z}_p)$

$$-x^2-x=-y^2-y \iff y-x=x^2-y^2$$

and this implies $(-1)(x-y)=(x-y)(x+y)$

Thus $x+y+1=0$

Thus we have the following solutions

So $\{0,p-1\}, \{1,p-2\} \dots \{\frac{p-1}{2},\frac{p-1}{2}\}$ are the only pairs where $s(x)$ assumes same values and hence we have $\frac{p+1}{2}$ elements in $s(\mathbb{Z}_p)$. Thus for $p-\frac{p+1}{2}=\frac{p-1}{2}$ values of $a$, the polynomial $x^2+x+a$ is irreducible.