IMC 2018 · Day 1 · P3

We will show that the only such number is $a = 0$.

Let $A = \begin{pmatrix} a & -a & -1 & 0 \\ a & -a & 0 & -1 \\ 1 & 0 & a & -a \\ 0 & 1 & a & -a \end{pmatrix}$ and suppose that $A = B^2$. It is easy to compute the characteristic polynomial of $A$, which is $$p_A(x) = \det(A - xI) = (x^2 + 1)^2.$$ By the Cayley-Hamilton theorem we have $p_A(B^2) = p_A(A) = 0$.

Let $\mu_B(x)$ be the minimal polynomial of $B$. The minimal polynomial divides all polynomials that vanish at $B$; in particular $\mu_B(x)$ must be a divisor of the polynomial $p_A(x^2) = (x^4 + 1)^2$. The polynomial $\mu_B(x)$ has rational coefficients and degree at most 4. On the other hand, the polynomial $x^4 + 1$, being the 8th cyclotomic polynomial, is irreducible in $\mathbb{Q}[x]$. Hence the only possibility for $\mu_B$ is $\mu_B(x) = x^4 + 1$. Therefore, $$\tag{1} A^2 + I = \mu_B(B) = 0.$$ Since we have $$A^2 + I = \begin{pmatrix} 0 & 0 & -2a & 2a \\ 0 & 0 & -2a & 2a \\ 2a & -2a & 0 & 0 \\ 2a & -2a & 0 & 0 \end{pmatrix},$$ the relation (1) forces $a = 0$.

In case $a = 0$ we have $$A = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}^2,$$ hence $a = 0$ satisfies the condition.