Which power of an integer matrix is identity modulo $p^\alpha$? I've read this question about identity power of an integer matrix. 
But how about power of a matrix modulo $p^\alpha$.
$$A^m \equiv I \pmod{p^\alpha} $$
How can I find the minimal $m$ that the above equation hold?
Or how to prove that such $m$ does not exist? 
 A: We assume that $p$ is a prime. If $\alpha >1$, then $\mathbb{Z}/p^{\alpha}\mathbb{Z}$ is a ring and not a field. Then the probem is difficult. For instance consider the equation $A^3=I_4$ over $\mathbb{Z}/4\mathbb{Z}$; I think that the problem is feasible but I do not know how to do.
If $\alpha=1$, then $K=\mathbb{Z}/p\mathbb{Z}$ is a field and it's easier...
Case 1. $p$ does not divide $m$, then $x^m-1$ has only simple roots. We decompose $x^m-1=p_1(x)\cdots p_k(x)$ in irreducible over $K$. The minimal polynomial $m(x)$ of $A$ is a product of some of the $p_i$, for instance $m(x)=p_1(x)\cdots p_r(x)$. Finally, $A$ is similar over $K$ to $diag(C_1,\cdots,C_1,\cdots ,C_r,\cdots C_r)$, where $C_i$ is the companion matrix of $p_i$.
Case 2. $m=pq$. Then $x^m-1=(x^q-1)^p$ and it suffices to decompose $x^q-1$.
Method. STEP 1.Calculate the minimal polynomial $m(x)$ of the matrix $A$ and $d=\det(A)$. If $d=0$, then there is no $m$. Otherwise, let $s$ be the minimum of $t>0$ s.t. $d^t=1$; if $m$ exists, then $s$ divides $m$. Find $s$.
STEP 2. for $u=1,2,\cdots$, calculate the remainder of the division of  $x^{us}-1$ by $m(x)$. You stop when $m(x)$ divides $x^{us}-1$.
Example: $p=7$,$A=\begin{pmatrix}1&2&2&2&2\\1&1&0&1&0\\2&3&4&0&2\\2&2&3&2&5\\2&3&6&2&1\end{pmatrix}$, $\det(A)=3$, $s=6$. We obtain $m=300$.
It remains to obtain a stopping test.
