Number of elements of "$\mathrm{SL}_n(\mathbb{F}_p^\times)$" mod $p$ How many elements of $\mathrm{SL}_n(\mathbb{F}_p)$ have all nonzero entries? Just the answer mod $p$ would be fine as well. This seems like it should be easy/in the literature but I couldn't find it.
 A: I will sharpen my earlier comment a little. The number of such matrices is divisible by $(p-1)^{2n-2}$ (though it may be zero). Let $T$ be the group of invertible diagonal matrices, and let $\mathcal{M}$ denote the set of invertible matrices of full rank ( though not necessarily determinant 1).
Then $M \to TMU^{-1}$ defines an action of $T \times T$ on $\mathcal{M},$ and the only pairs $(T,U)$ which fix any element are those with $T = U$ both scalar. Hence $(T \times T)/{\rm diag} (S)$ acts semi-regularly on $\mathcal{M},$ where $S$ is the subgroup of $T$ consisting of scalars. Thus $|\mathcal{M}|$ is divisible by $(p-1)^{2n-1}$, and hence, as user74230 notes in comments, the number of matrices in $\mathcal{M}$ which have determinant $1$ is a multiple of $(p-1)^{2n-2}.$ Also, the number of such matrices is divisible by $\frac{n!}{2},$ since $A_{n}$ acts by right multiplication (in its representation by permutation matrices) on such matrices, and no non-identity element has any fixed point.
Actually, I realise that this implies that when $n \geq p,$ the number of such matrices is divisible by $p.$ For when $p =2$ the number is zero, and when $p$ is odd and $n \geq p,$ then $|A_{n}|$ is divisible by $p^{\frac{n - \sigma_{p}(n)}{p-1}}$, where $\sigma_{p}(n)$ is the sum of the digits in the base $p$ expansion of $n.$ 
A: Mod $p$ it's $(-1)^{n+1} n!$.
Let's compute the number of points with determinant $1$ and all entries nonzero by inclusion-exclusion, modulo $p$. For each set of entries, we get a term for matrices in $SL _n$ with those entries $0$. This is an affine hypersurface of degree $n$ in some affine space. By Warning's theorem the number of points is a multiple of $p$ unless the number of variables is at most $n$. But the number of variables is the number of nonzero entries. A matrix with $\leq n$ nonzero entries that is invertible is a permutation matrix times a diagonal matrix. We can easily count the contribution if these. It is $(-1)^{n^2-n} (p-1)^{n-1} n!$. Mod $p$ we get the stated claim.
A: Here's another argument obtaining the mod $p$ count. It's enough to do for $p$ odd. The desired count is $\sum_{A\in \mathrm{SL}_n(\mathbb{F}_p)} \prod_{i,j=1}^n A_{ij}^{p-1}\cdot \det(A)^{p-1}$. Changing "$\mathrm{SL}$" to "$\mathrm{GL}$" (even to all $n\times n$ matrices) changes the sum by a factor of $-1$. So it suffices to calculate $\sum_{A_{ij}\in \mathbb{F}_p} \prod_{i,j=1}^n A_{ij}^{p-1}\cdot \det(A)^{p-1}$.
Now, for $e_{ij}\geq 0$, $\sum_{A_{ij}\in \mathbb{F}_p} \prod_{i,j=1}^n A_{ij}^{e_{ij}} = 0$ unless all $e_{ij}$ are divisible by $p-1$, in which case it is $(-1)^{n^2} = (-1)^n$. Therefore the cross-terms in $\det(A)^{p-1}$ all die, and the only terms that can contribute are the $(p-1)$-st powers of terms in $\det(A)$. There are $n!$ of these, and they all carry a sign of $(\pm 1)^{p-1} = 1$. Therefore the count for $\mathrm{GL}_n$ is $(-1)^n n!$, so the count for $\mathrm{SL}_n$ is $(-1)^{n+1} n!$.
A: EDIT: There's an error in this approach, it doesn't work.  Computing some examples seems to show the answer is pretty complicated...
(All I have to add to what's been said is an ad-hoc argument for why this number is divisible by $p$.  If you let $f(n, k)$ be the number of $n \times n$ matrices with nonzero entries having rank $k$, then you can easily write down the recurrence (letting $m=p-1$ for simplicity of notation)
$$f(n, k)=m^{2k}f(n-1, k)+(2m^{n+k-1}-m^{2k-1}-m^{2k-2})f(n-1, k-1)+m^{2k-3}(m^{n-k+1}-1)^2f(n-1, k-2)$$
You can then show by induction that $f(n, n)$ and $f(n, n-1)$ are both divisible by $p$ for $n \ge 3$.  It's true for $n=3$, because $f(3, 2)=m^5(m+1)^2(m-1)$ and $f(3, 3)=m^6(m+1)(m-1)^2$ are both divisible by $m+1$.  Then since $m=p-1 \equiv -1\pmod{p}$, you can just use the above equation $\pmod{p}$ to write
$$f(n, n-1) \equiv f(n-1, n-1)+2f(n-1, n-2)$$
$$f(n, n) \equiv -2f(n-1, n-1)-4f(n-1, n-2)$$
which are both zero by hypothesis.
(I did also turn the above recurrence into a generating function identity, but the result seems very hard to solve.  It's been a long time since I used generating functions, so I don't know if it's hard just because of that.))
