Number of irreducible polynomials of degree $2$ over $\Bbb F_{p^n}$ What is the number of irreducible polynomials, $p(x)$, of degree $2$ over $\Bbb F_q$, $q=p^n$?
 A: Note that a polynomial of degree $2$ is "reducible" iff its roots are in the field. Let us count the reducible polynomials. First count the monic ones, for which the coefficient of $x^2$ is $1$ (then you can multiply your answer by $q-1$ to allow the coefficient of $x^2$ to vary).
So how many monic reducible polynomials of degree $2$ are there?  If the roots are to be distinct, they can be chosen in $\binom{q}{2}$ ways.  And there are in addition $q$ monic polynomials that have a double root.  Now put all these facts together.  Subtract your answer from the number $(q-1)(q)(q)$ of degree $2$ polynomials.
A: Here is a sketch of a solution that proceeds directly, instead of counting the complement.  It has the serious disadvantage that it does not work in characteristic $2$. Also, it uses more machinery than the counting the complement approach, but perhaps it has some merit.
Look at the quadratic $ax^2+bx+c$.  The coefficient $a$ can be chosen in $q-1$ ways.  We will show that for every such choice, and every one of the $q$ possibilities for $b$, there are $(q-1)/2$ choices for $c$ that yield an irreducible quadratic.
The idea is that (for characteristic $\ne 2$) a suitably interpreted version of the quadratic formula gives the zeros, and that in particular the quadratic has no zero in the field precisely when the discriminant $b^2-4ac$ is not a square in the field.
So let $a$ be a fixed non-zero element, and let $b$ be arbitrary but fixed. As $c$ ranges over all the elements of the field, so does $b^2-4ac$.  But exactly $(q-1)/2$ elements of the field are non-squares. This completes the argument.
