THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) From $A^3=B^3$ we get $\det A=\det B=d$ and by Hamilton-Cayley theorem we obtain $A^2=aA-d{I}_2$, $B^2=bB-d{I}_2$, where $a=\mathrm{tr}A$, $b=\mathrm{tr}B$. As a consequence one can write $A^3=a{A}^2-dA=(a^2-d)A-ad{I}_2$ and $B^3=(b^2-d)B-bd{I}_2$, so $(a^2-d)A-ad{I}_2=(b^2-d)B-bd{I}_2$ ().

By right and left multiplication with $B$ we get $(a^2-d)BA-adB=(b^2-d)B^2-bdB=(a^2-d)AB-adB$, so $d=a^2$. In the same way, $d=b^2$, and by (), we deduce $d=0$ or $a=b$. If $d=0$, then $a=b=0$, so $A^2=B^2=O_2$, from where $A^n=B^n$, for any $n\ge 2$. So $\text{tr }A=a=0=b=\text{tr }B$ and $\text{tr }{A}^n=\text{tr }{B}^n$, for all $n\ge 2$. If $d\ne 0$, then $a=b$ and $C^2=aC-a^2{I}_2$, $C^3=-a^3{I}_2$, so $C^{3k}=(-a^3{)}^k{I}_2$, $C^{3k-2}=(-a^3{)}^{k-1}C$, $C^{3k-1}=(-a^3{)}^{k-1}(aC-a^2{I}_2)$, $\forall C\in \{A,B\},\forall k\ge 1$.

b) For $A=$, $B=$, we have $AB=$, $BA=$, $A^n=$, $B^n=$ and $\text{tr }{A}^n=\text{tr }{B}^n=2,\forall n\ge 1$.