IMC 2023 · Day 1 · P2

Hint: Factorize $B^3 - ABC$.

Note that $B^3 = A^2 B$, from which it follows that $$A^2 B - ABC = 2I \ \Longrightarrow \ A (AB - BC) = 2I.$$ Similarly, using that $B^3 = B C^2$, we find that $$B C^2 - ABC = 2I \ \Longrightarrow \ (BC - AB) C = 2I.$$ It follows that $A$ is a left-inverse of $(AB - BC)/2$, whereas $-C$ is a right inverse. Hence $A = -C$ and as such, it must hold that $ABA = 2I - B^3$. It follows that $ABA$ must commute with $B$, and so it follows that $(AB)^2 = (BA)^2$. Now we compute that $$(AB - BA)(AB + BA) = (AB)^2 + A B^2 A - B A^2 B - (BA)^2 = (AB)^2 + A^4 - B^4 - (AB)^2 = 0.$$ However, we noted before that the matrix $AB - BC = AB + BA$ must be invertible. As such, it must follow that $AB = BA$. We conclude that $ABA = A^2 B = B^3$ and so it readily follows that $B^3 = I$. Finally, $A^6 = B^6 = (B^3)^2 = I^2 = I$, completing the proof.