67th Romanian Mathematical Olympiad

If $M\in {\mathcal{M}}_2(\mathbb{R})$, we have $$\det (M-x{I}_2)=x^2-\mathrm{tr}(M)x+\det (M),\forall x\in \mathbb{R}.(1)$$ Let $A\in {\mathcal{M}}_2(\mathbb{R})$ be a matrix satisfying the hypothesis. From (1) we obtain $$\mathrm{tr}(A^{2014})=\mathrm{tr}(A^{2016})=0.(2)$$ Using the Cayley-Hamilton theorem, we obtain $$A^2-\mathrm{tr}(A)A+\det (A)I_2=O_2.\text{\hspace{1em}(3)}$$ We analyze several cases.

If $\mathrm{tr}(A)=0$, then (3) implies $A^2=-\det (A)I_2$, whence $A^{2014}=(-\det (A){)}^{1007}{I}_2$. From (2) it results $\det (A)=0$.

If $\det (A)=0$, then this time (3) implies $A^2=\mathrm{tr}(A)A$. Inductively, $A^{n+1}={\mathrm{tr}}^n(A)A$, $\forall n\in {\mathbb{N}}^{\ast }$. It follows that $\mathrm{tr}(A^n)={\mathrm{tr}}^n(A)$, $\forall n\in {\mathbb{N}}^{\ast }$. As a special case, using (2), $0=\mathrm{tr}(A^{2014})={\mathrm{tr}}^{2014}(A)$, whence $\mathrm{tr}(A)=0$.

If $\mathrm{tr}(A)\ne 0$ and $\det (A)\ne 0$. From (3), $A^{2016}-\mathrm{tr}(A)A^{2015}+\det (A)A^{2014}=O_2$. Then $\mathrm{tr}(A^{2016})-\mathrm{tr}(A)\mathrm{tr}(A^{2015})+\det (A)\mathrm{tr}(A^{2014})=0$. Using (2) and our assumption, we get $\mathrm{tr}(A^{2015})=0$. From (3), it follows that $$\mathrm{tr}(A^n)=\frac{1}{\det (A)}\left[\mathrm{tr}(A)\mathrm{tr}(A^{n+1})-\mathrm{tr}(A^{n+2})\right],n\in {\mathbb{N}}^{\ast }.$$ Thus, starting with $\mathrm{tr}(A^{2014})=\mathrm{tr}(A^{2015})=0$, we obtain recursively $\mathrm{tr}(A^{2013})=0$, $\mathrm{tr}(A^{2012})=0$, ..., $\mathrm{tr}(A)=0$, a contradiction. It follows that $\mathrm{tr}(A)=0$ and $\det (A)=0$.

Finally, from (3) it follows that $A^2=O_2$, hence $A^n=O_2$, $\forall n\ge 2$. Then, for $n\ge 2$, we have $\det (A^n-I_2)=\det (-I_2)=1=\det (I_2)=\det (A^n+I_2)$. For $n=1$, from (1), $\det (A-I_2)=1=\det (A+I_2)$.