TITLE: Matrices over Finite Prime Fields
QUESTION [5 upvotes]: Let $p$ be an odd prime and $\mathbb Z_p$ be the prime field of order $p$. Consider the matrix ring $R=M_n(\mathbb Z_p)$. Is there any method to count the solutions of the equation (in the ring $R$)
$$X^2=I.$$
Where $I$ is the identity matrix?

REPLY [2 votes]: Because $p$ is different from $2$, the solutions are the same as direct sum decompositions $\mathbb{Z}_p^{\oplus n} = E_{+1} \oplus E_{-1}$ as $\mathbb{Z}_p$-vector spaces.  You can index these by the dimensions, say $r$ and $n-r$.  For each, the number of solutions is
$$
\frac{(p^n-1)(p^n-p) \cdot \dots \cdot (p^n-p^{r-1})}{(p^r-1)(p^r-p) \cdot \dots \cdot (p^r-p^{r-1})} \cdot p^{r(n-r)}
$$
So the final answer is the sum over $r$ from $0$ to $n$ of this factor.

REPLY [2 votes]: There are $\displaystyle \sum_{i=0}^n \frac{N(n)}{N(i)N(n-i)}$ solutions to your problem, where 
$N(r)$ is the number of elements in $GL_r(\mathbb{F}_p)$.
Any such $X$ is invertible, and $X\in GL_n(\mathbb{F}_p)$ is a solution if and only if it is conjugate in $GL_n(\mathbb{F}_p)$ to a diagonal matrix with the first $i$ entries equal to $1$ 
and last $n-i$ entries equal to $-1$. This explains the number of summands. Now $g\in GL_n(\mathbb{F}_p)$  commutes with such diagonal matrix if and only if it preserves both eigenspaces. Hence, there are $\frac{N(n)}{N(i)N(n-i)}$ elements in the conjugacy class. 
The formula can be made even more explicit by substituting a formula for $N(r)$.