A commutator that dictates the dimension

Let $A$ and $B$ be real $n \times n$ matrices such that $A^2+B^2=A B$ . Prove that if $B A-A B$ is an invertible matrix then $n$ is divisible by 3.

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Set $M=A+\omega B$ , where $\omega=-\frac{1}{2}+i \frac{\sqrt{3}}{2}$ . We have
$\begin{aligned} M \bar{M} & =(A+\omega B)(A+\bar{\omega} B)=A^2+\omega B A+\bar{\omega} A B+B^2 \\ & =A B+\omega B A+\bar{\omega} A B=\omega(B A-A B), \end{aligned}$ because $\bar{\omega}+1=-\omega$ . Since $\det(M \bar{M})=\det M . \det \bar{M}$ is a real number and $\operatorname{det} \omega(B A-A B)=\omega^n \operatorname{det}(B A-A B)$ and $\operatorname{det}(B A-A B) \neq 0$ , then $\omega^n$ is a real number. This is possible only when $n$ is divisible by 3 .

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