How many irreducible monic quadratic polynomials are there in $\mathbb{F}_p[X]$?

Can some of you help me with my homework?

I had to count the irreducible, quadratic, monic polynomials in $\mathbb{F}_p[X]$ for arbitrary $p$.

I will show you what I tried myself.

Research effort

First of all the set $\{X^ 2 + bX +c : b, c \in \mathbb{F}_p \}$ contains $p^2$ elemenst. We need to substract the amount of elements that is reducible. These elements are exactly the set $A = \{(X+c)(X+d) : c, d \in \mathbb{F}_p \}$. There are $p(p-1)$ pairs of $(a,b) \in \mathbb{F}_p \times \mathbb{F}_p$ such that $a \neq b$, and there are $p$ pairs such that $a=b$.

When $a$ and $b$ are unequal, whe know that $a$ and $b$ can be swapped without changing anything to the polynomial $(X+a)(X+b)$, so I obtained that $|A| \leq p +(p-1)- \frac{1}{2}(p-1)=p-\frac{1}{2}(p-1)$.

In other to find out these are the elements, or that there are still less, I thought I had to assume that $(X+c)(X+d) = (X+\gamma)(X+\delta)$, to deduce some information about the realation between $(c,d)$ and $(\gamma, \delta)$, but this led to taking square roots in $\mathbb{F}_p$, and I am not familiar with that.

Could you please provide me some more information?

Monic reducible degree two polynomials have a root, so they are of the form $(X-a)(X-b)$.

There are $p$ polynomials of the form $(X+a)^2$ and $\binom{p}{2}$ of the form $(X+a)(X+b)$ with $a\ne b$ (as you computed). So $$ p^2 - p - \binom{p}{2}=\frac{p(p-1)}{2} $$ is the correct number.

Examples. For $p=2$ the reducible polynomials are $X^2$, $X(X+1)$ and $(X+1)^2$; so you get $4-3=1$, indeed the only one is $X^2+X+1$.

For $p=3$ we have $X^2$, $(X+1)^2$, $(X+2)^2$, $X(X+1)$, $X(X+2)$, $(X+1)(X+2)$. The three irreducible polynomials are $X^2+1$, $X^2+X+2$, $X^2+2X+2$.

Hint: the product of all monic irreducible polynomials of degree dividing $n$ with coefficients in $\mathbf F_p$ is $x^{p^n}-x$.

So, with $n=2$, you just have to exclude the linear monic polynomials...