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How can I solve in $\mathcal{M}_{2}(\mathbb{Z})$ the equation $$B^3+C^3=\begin{pmatrix}-1 & 1\\ 0 & -2\end{pmatrix}?$$

I try to use $$X^2-Tr(X)X+det X\cdot I_2=0_2$$ but there are $B$ and $C$, I don't still obtain anything.

thanks.

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I don't know any systematic way to solve the equation, but I think the easiest solution is given by: $$ \pmatrix{-1&1\\ 0&-2}=\pmatrix{0&1\\ 0&-1}+(-I)=\pmatrix{0&1\\ 0&-1}^3+(-I)^3. $$ While the rank-one matrix $B=\pmatrix{0&1\\ 0&-1}$ is the unique integer cube root of itself, $-I$ has infinitely many integer cube roots. More specifically, you may replace $C=-I$ by $C=U\pmatrix{0&-1\\ 1&1}U^{-1}$ for any unimodular matrix $U$.

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The easiest part. We assume that $B,C$ are upper-triangular.

Then $b_{1,1}^3+c_{1,1}^3=-1$; according to Fermat (up to order) $b_{1,1}=-1,c_{1,1}=0$.

And $b_{2,2}^3+c_{2,2}^3=-2$, that implies $b_{2,2}=c_{2,2}=-1$ (see $(a+1)^3-a^3$ when $a>0$).

By identification, we conclude that there is an integer $p$ s.t. (up to order)

$B=\begin{pmatrix}-1&p\\0&-1\end{pmatrix},C=\begin{pmatrix}0&1-3p\\0&-1\end{pmatrix}$.

EDIT. There are solutions $(B,C)$ in $M_2(\mathbb{R})$ s.t. $B^3,C^3$ are not upper-triangular; for example

$B\approx \begin{pmatrix}0&-2.0107\\1&-1.8503\end{pmatrix},C\approx\begin{pmatrix}1.1426&-2.7184\\1&0\end{pmatrix}$.

Thus the question is: can we find such matrices in $M_2(\mathbb{Z})$ ?

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