TITLE: Matrix Equation $A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}$
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: Here's an approach that works even if the left hand side is a more complicated expression of the form $AP(A)$, with an arbitrary polynomial $P$ (with integer coefficients).
Let $M$ denote the matrix on the right hand side.  The equality implies that $A$ and $M$ commute.  If we write this as
$$\pmatrix{a & b\\ c & d}\pmatrix{-7 & -9\\ 3 & 2}=\pmatrix{-7 & -9\\ 3 & 2}\pmatrix{a & b\\ c & d}$$
we find that
$$b=-3c\text{ and }d=a+3c$$
so
$$A=\pmatrix{a & -3c\\c & a+3c}$$
with
$$\det(A)=a^2+3ac+3c^2={(2a-3c)^2+3c^2\over4}$$
Since we're working with integers, $AP(A)=M$ implies $\det(A)$ divides $\det(M)=13$, which limits the possibilites to integer solutions to
$$(2a-3c)^2+3c^2=4\text{ or } 52$$
of which there are clearly a (fairly small) finite number.
More generally, if
$$M=\pmatrix{p & q\\ r & s}$$
this approach shows that $rb=qc$, $rd=ra+(s-p)c$, and
$$r(ps-qr)=k(ra^2+(s-p)ac-qc^2)$$
where the inequality that makes the pertinent quadratic form on the right hand side positive definite (thereby limiting the number of possible solutions) is
$$(p+s)^2-4(ps-qr)\lt0$$
which is to say, $M$ should have complex eigenvalues.  In sum, if $M$ has complex eigenvalues, then the commutativity condition $AM=MA$ and the divisibility condition $\det(A)|\det(M)$ all by themselves limit the possibilities for $A$ to a finite set, regardless of the polynomial $P(A)$ on the left hand side.