Matrix Equation $A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}$ How can I solve in $\mathcal{M}_{2}(\mathbb{Z})$ the equation $$A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}?$$
I try to use $$A^2-Tr(A)A+detA\cdot I_2=O_2$$ but I don't still obtain anything. 
thanks. 

Let $t=\operatorname{trace}(A)$ and $d=\det(A)$. By Cayley-Hamilton theorem, $A^2 = tA - dI$ and hence
$$
\pmatrix{-7&-9\\ 3&2} = A^3 - 3A = tA^2 - (d+3)A = (t^2-d-3)A - tdI.\tag{1}
$$
Taking traces on both sides, we get $-5 = (t^2-d-3)t - 2td$. Hence $t$ divides $5$, i.e. $t=1,-1,5,-5$, and we can express $d$ in terms of $t$ as $d = \frac13\left(t^2 + \frac5t\right) - 1$. Since $d$ must be an integer, $(t,d)=(1,1)$ or $(-5,7)$. However, by inspecting the $(2,1)$-th entries on both sides of $(1)$, we see that $(t^2-d-3)\mid3$. Therefore $(t,d)=(1,1)$. So, from $(1)$ we obtain
$$
\pmatrix{-7&-9\\ 3&2} = -3A - I\quad\Longrightarrow\quad A=\pmatrix{2&3\\ -1&-1}.
$$
