<s>[INST] $A^2+B^2=AB$ and $BA-AB$ is non-singular The question is:

Are there square matrices $A,B$ over $\mathbb{C}$ s.t. $A^2+B^2=AB$ and $BA-AB$ is non-singular?

From $A^2+B^2=AB$ one could obtain $A^3+B^3=0$. Can we get something from this?
Edit: 
$$A^2+B^2=AB\implies\\
A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\
A^3+(A^2+B^2)B=A^2B \implies \\ 
A^3+A^2B+B^3=A^2B \implies \\
A^3+B^3=0
$$
 [/INST] The answer is yes. We shall construct a concrete example below. First, suppose $A$ and $B$ are invertible and let $C=AB^{-1}$. From $A^2+B^2=AB$ we get $C + A^{-1}C^{-1}A = I$ and $A^{-1}CA=(I-C)^{-1}$. Therefore the spectrum of $C$ is invariant under the transformation $z\mapsto \frac1{1-z}$.
Let $w$ be the cubic root of unity. If $A^2+B^2=AB$, then $(A+wB)(A+w^2B)=w(BA-AB)$. So, if we want $BA-AB$ to be invertible, neither $-w$ nor $-w^2$ can be an eigenvalue of $C$. As $-w$ and $-w^2$ are fixed points of $z\mapsto\frac1{1-z}$, the spectrum of $C$ has to be splittable into three cycles of the form $\{z,\ \frac1{1-z},\ \ 1-\frac1z\}$, where $z\ne-w,-w^2$, and for every feasible pair of invertible matrices $A,B$, their dimensions must be divisible by 3.
Now we can construct an example on $M_3(\mathbb C)$. Take $z=2$ and 
$C=\operatorname{diag}(2, -1, \frac12)$. So $(I-C)^{-1}=\operatorname{diag}(-1, \frac12, 2)$. To make $A^{-1}CA=(I-C)^{-1}$ or $A^2+B^2=AB$, we pick $A=\pmatrix{0&0&1\\ 1&0&0\\ 0&1&0}$ and $B=C^{-1}A=\pmatrix{0&0&\frac12\\ -1&0&0\\ 0&2&0}$. One can readily verify that $BA-AB=\pmatrix{0&-\frac32&0\\ 0&0&-\frac32\\ 3&0&0}$ is invertible.
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