Involutory matrix $2 \times 2$ I want to find out how many  $2 \times 2$ involutory matrices are there over $\mathbb Z_{26}$.
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Is there any formula to calculate this? 
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Thanks for your help.

Let $p$ be an odd prime, and $A=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)\in M_2(\mathbb Z/p\mathbb Z)$. 
Then $A^2=I$ is equivalent to $a^2+bc=1,$ $d^2+bc=1,$ $b(a+d)=0,$ $c(a+d)=0.$ 
If $a+d\ne0$, then $b=c=0$, so $a^2=d^2=1$, and there are $2$ solutions. 
If $a+d=0$, then consider the following cases:
(i) $a=0$. Then $bc=1$, that is, $c=b^{-1}$ and find $p-1$ solutions.
(ii) $a=1$. Then $bc=0$, and we have $2p-1$ solutions.
(iii) $a=-1$. Similarly, we get $2p-1$ solutions.
(iv) $a\ne0,\pm1$. Then $c=(1-a^2)b^{-1}$, so there are $(p-1)(p-3)$ solutions.
Summing up we get $p^2+p+2$ matrices $2\times 2$ over $\mathbb Z/p\mathbb Z$.
