# Matrix Equation $B^3+C^3=\begin{pmatrix}-1 & 1\\ 0 & -2\end{pmatrix}$

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.

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 $$U$$.

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 $$M_2(\mathbb{Z})$$ ?