Prove that no 2 orthogonal matrices satisfy this equation $A^2-B^2=AB$ This question came up in my linear algebra finals, and I couldn't prove it. Could anyone help me?

Question: Prove that no $2$ orthogonal matrices satisfy this condition:  $A^2-B^2=AB$

Attempt: Assume $A$ and $B$ are orthogonal. Mutiplying $A^T, B^T$ to both sides, we have:
$$
A-A^TB^2=B \\
AB^T-A^TB=I
$$
Taking transpose of both sides:
$$
(AB^T-A^TB)^T=I\\
BA^T-B^TA=I
$$
Again multiplying $B, A$ to both sides:
$$
B^2A^T-A=B \\
B^2-A^2=BA
$$
So combining, I have $AB=-BA$, but after this I'm stuck. I thought of the fact that $2x2$ orthogonal matrices represent a rotation in the $\mathbb{R}^2$ space, and so they commute, but in general rotations do not commute with reflections and an $nxn$ orthogonal matrices do not commute. 
Hints appreciated.

I start from $AB=-BA$. This implies that $ABA^T = -B$ and $AA B A^T A^T = - A B A^T = - (-B) = B$.
This means that $A A B  = B A A$. In a similar manner, $B B A = A B B$.
Now from $A A - B B = A B$, we find that  $A A B - B B B = A B B$.
Apply the equality $AB=-BA$ to it, we obtain $B A A - B B B = B B A$. This gives $A A - B B = B A = A B$.
But $B A = - A B$ and so, $A B = -A B$ or $A B= 0$, a contradiction.
