74th Romanian Mathematical Olympiad

Let us denote $d=\det (X)$ and $t=\text{tr}(X)$. From the assumption $X^{2023}=X^{2022}$, we obtain $d^{2023}=d^{2022}$. Therefore, $d\in \{0,1\}$.

If $d=1$, then $X$ is invertible. Then $X^{2022}$ is also invertible. We find $X=I_2$. So, the relation $X^3=X^2$ is verified.

Assume now $d=0$. The Cayley-Hamilton theorem implies $X^2=tX$. So we obtain $X^{n+1}=t^{n}X,\forall n\in {\mathbb{N}}^{\ast }$. Hence $t^{2022}X=t^{2021}X$. We get $t=0$ or $t=1$ or $X=O_2$.

If $t=0$, then $X^2=O_2$, so $X^3=X^2=O_2$.

If $t=1$, then $X^2=X$, so $X^3=X^2$.

If $X=O_2$, then clearly $X^3=X^2=O_2$.