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!
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 $$\frac{p(p-1)^2}{2}$$