One can show using zeta functions that the number of (monic)square-free polynomials of degree $n$ over a finite field $\Bbb F_q$ is $q^n-q^{n-1}$. For instance, this is done here.
The answer is simply enough to suggest a very nice combinatorial interpretation - the number of all monic polynomials is $q^n$ and the number of monic square-full polynomials is $q^{n-1}$.
Is there a combinatorial way to show the second statement about the number of monic squarefull polynomials is $q^{n-1}$? This is equivalent to the statement that there are q^{n-1} solutions to the discriminant being zero but the discriminant is not linear.
However, the discriminant is in general irreducible($2\not\mid q)$ and perhaps it is true that any irreducible polynomial in $n$ variables over $\Bbb F_q$ has $q^{n-1}$ solutions?
Update: A little calculation seems to show that the number of monic polynomials of degree $n$ that have no $k$-powers dividing them is $q^n - q^{n-k+1}$. For example, $k=2$ corresponds to the original square-free case.
One can generalize the condition that a polynomial $p(x)$ has a square factor if and only if $\gcd(p(x),p'(x)) \neq 1$ and for $k-$free, the criterion is that $\gcd(p(x),p^{(1)}(x),\dots,p^{(k-1)}(x))\neq 1$.
The corresponding discriminant condition would be that $p(x),p^{(1)}(x),\dots,p^{(k-1)}(x)$ all have $0$ discriminant.
I guess an optimal answer would explain all of these at once.
