Is $A^{q+2}=A^2$ in $M_2(\mathbb{Z}/p\mathbb{Z})$? I'm wondering, why is it that for $q=(p^2-1)(p^2-p)$, that $A^{q+2}=A^2$ for any $A\in M_2(\mathbb{Z}/p\mathbb{Z})$?
It's not hard to see that $GL_2(\mathbb{Z}/p\mathbb{Z})$ has order $(p^2-1)(p^2-p)$, and so $A^q=1$ if $A\in GL_2(\mathbb{Z}/p\mathbb{Z})$, and so the equation holds in that case. But if $A$ is not invertible, why does the equality still hold?

If $A$ is not invertible, then its characteristic polynomial is either $x^2$ or $x(x-a)$ for some $a\in\mathbb{Z}/p\mathbb{Z}$. In the former case, by the Cayley-Hamilton Theorem we have $A^2 = 0$, hence $A^{q+2}=A^2$. In the latter case, the matrix is similar to a diagonal matrix, with $0$ in one diagonal and $a$ in the other. So, up to conjugation, we have
$$A^{q+2}=\left(\begin{array}{cc}
0 & 0\\
0 & a\end{array}\right)^{q+2} = \left(\begin{array}{cc}
0 & 0\\
0 & a^{q+2}
\end{array}\right).$$
But $a^{p} = a$. Since $q = p^4 -p^3-p^2 + p$, we have
$$a^{q} = \frac{a^{p^4}a^p}{a^{p^3}a^{p^2}} = 1$$
so $a^{q+2} = a^2$, hence $A^{q+2}=A^2$. 
