Finding the number of irreducible quadratics in $\Bbb Z_p[x]$, where $p$ is a prime The problem is to find the number of irreducible quadratics in $\Bbb Z_p[x]$, where $p$ is a prime number.
To solve this, I wish to find first the number of reducible quadratics of the form $x^2+ax+b$, then the number of reducible quadratics, and subtract this from the total number of quadtratics.
I know that each reducible quadratic that is of the form $x^2 +ax+b$ is a product $(x+c)(x+d)$ for $c,d\in \Bbb Z_p$.
So, I don't know what to do next. Please help me solve this problem.
Thanks for the help!
 A: MJ. Rivo, your idea is correct. Here is how to do the calculation.
The total number of monic quadratic polynomials in $\Bbb Z_p[x]$ is
$p^2$.
To see this, just note there is a bijection associating each monic quadratic polynomials $x^2 +ax+b$ to $(a,b) \in \Bbb Z_p[x] \times \Bbb Z_p[x]$.
On the other hand, the total number of reducible monic quadratic polynomials in $\Bbb Z_p[x]$ is
$$\frac{p(p+1)}{2}.$$
To see this, just note that each reducible monic quadratic polynomial bijectively corresponds to a product $(x+c)(x+d)$ for $c,d\in \Bbb Z_p$. How many such products we have?  Since multiplication is commutative, the order does not matter. So we get $\frac{p(p-1)}{2}$ (for $c\neq d$) plus $p$ (for $c= d$).  The number then is $\frac{p(p+1)}{2}$.
So, the total number of irreducible monic quadratic polynomials in $\Bbb Z_p[x]$ is
$$p^2-\frac{p(p+1)}{2}=\frac{p(p-1)}{2}.$$
So the answer for irreducible monic quadratic polynomials is $\frac{p(p-1)}{2}$.
If you want the number of all irreducible quadratic polynomials (not necessarily monic), just note that each irreducible quadratic polynomials  bijectively corresponds to $e(x^2 +ax+b)$ where $e \in \Bbb Z_p \setminus \{0\}$ and $x^2 +ax+b$ is an irreducible monic quadratic polynomials in $\Bbb Z_p[x]$. So the total number of all irreducible quadratic polynomials in $\Bbb Z_p[x]$ is
$$\frac{p(p-1)^2}{2}$$
