Could anybody solve it??

1)Suppose $det(AB-BA) \neq 0$.From hypothesis we have $(A-B)^2=AB-BA$ ,so A-B is invertible.Also from the initial condition we have $A(A-B)=(A-B)B$,SO $A=(A-B)*B*(A-B)^{-1}$.Thus,we have $A-B=(A-B)*B*(A-B)^{-1}-B$ and multiplying this by $(A-B)^{-1}$ we conclude that $I_n=B(A-B)^{-1}-(A-B)^{-1}*B$.From trace we get that n=0,false. 2)It s well known that rank(A-B)=1 implies $(A-B)^2=tr(A-B)*(A-B)$.From hypothesis we have also $(A-B)^2=AB-BA$ ,so $tr(A-B)*(A-B)=AB-BA$ and ,from trace again,we have that $tr(tr(A-B)*(A-B))=0$,so A=B or trA=trB which means that $tr(A-B)=0$,so AB=BA

Nice solution!! is it also correct? If $A \in M_n(R)$ $rankA=1$ implies $A^2=trA \cdot A$? Here $R$ is the set of real numbers

a possible argument is $ A=b c^T $, where $b,c$ column vectors. Then $ A^2 = b c^T b c^T = b \cdot tr(A) \cdot c^T = tr(A) \cdot b c^T = tr(A) \cdot A $.

This is my problem which was offered at the Romanian National Olympiad 2014 for xi-th grade !