IMC 2021 · Day 2 · P5

Hint: The determinant is the product of the eigenvalues.

Let $B_m$ be the corresponding matrix $B$ depending on $m$: $$2021 B_m = A^m + B_m^2.$$ For $m = 1$, we obtain $A = 2021 B_1 - B_1^2$. Since $B_1$ is real and symmetric, so is $A$. Thus $A$ is diagonalizable and all eigenvalues of $A$ are real.

Now fix a positive integer $m$ and let $\lambda$ be any real eigenvalue of $A$. Considering the diagonal form of both $A$ and $B_m$, we know that there exists a real eigenvalue $\mu$ of $B_m$ such that $$2021 \mu = \lambda^m + \mu^2 \quad \Rightarrow \quad \mu^2 - 2021 \mu + \lambda^m = 0.$$ The last equation is a second degree equation with a real root. Therefore, the discriminant is non-negative: $$2021^2 - 4 \lambda^m \ge 0 \quad \Rightarrow \quad \lambda^m \le \frac{2021^2}{4}.$$ If $|\lambda| > 1$, letting $m$ even sufficiently large we reach a contradiction. Thus $|\lambda| \le 1$.

Finally, since $\det A$ is the product of the eigenvalues of $A$ and each of them has absolute value less then or equal to 1, we get $|\det A| \le 1$ as desired.

Different solution can be found in paper s2002