$A^2$ $B=A^2-B$ then $AB=BA$

Question

If for $2$ real $n$ by $n$ matrices we have $A^2B=A^2-B$ then prove that the two matrices commute.

This is a problem from a competition.

I've tried several manipulations but none of them work.

Can't come up with a counter example, as well.

Answer 1

As Sangchul Lee says in the comments, putting everything to one side allows us to add $I$ to both sides and factor as $I=(I+A^2)(I-B)$, telling us $I+A^2$ and $I-B$ are invertible, and $B=I-(I+A^2)^{-1}$.

Obviously to determine $B$ commutes with $A$, it suffices to see $(I+A^2)^{-1}$ commutes with $A$.

Prove that whenever $X$ is invertible, $X$ and $A$ commute if and only if $X^{-1}$ and $A$ commute. Then you can apply this principle with $X=I+A^2$.