International Mathematics Competition
for University Students
2018

IMC2018: Day 1, Problem 3

Problem 3. Determine all rational numbers \(\displaystyle a\) for which the matrix

\(\displaystyle \begin{pmatrix} a & -a & -1 & 0 \\ a & -a & 0 & -1 \\ 1 & 0 & a & -a \\ 0 & 1 & a & -a \end{pmatrix} \)

is the square of a matrix with all rational entries.

(Proposed by Daniël Kroes, University of California, San Diego)

Solution. We will show that the only such number is \(\displaystyle a=0\).

Let \(\displaystyle A = \begin{pmatrix} a & -a & -1 & 0 \\ a & -a & 0 & -1 \\ 1 & 0 & a & -a \\ 0 & 1 & a & -a \end{pmatrix}\) and suppose that \(\displaystyle A=B^2\). It is easy to compute the characteristic polynomial of \(\displaystyle A\), which is

\(\displaystyle p_A(x)=\det(A-xI)=(x^2+1)^2. \)

By the Cayley-Hamilton theorem we have \(\displaystyle p_A(B^2)=p_A(A)=0\).

Let \(\displaystyle \mu_B(x)\) be the minimal polynomial of \(\displaystyle B\). The minimal polynomial divides all polynomials that vanish at \(\displaystyle B\); in particular \(\displaystyle \mu_B(x)\) must be a divisor of the polynomial \(\displaystyle p_A(x^2)=(x^4+1)^2\). The polynomial \(\displaystyle \mu_B(x)\) has rational coefficients and degree at most \(\displaystyle 4\). On the other hand, the polynomial \(\displaystyle x^4+1\), being the \(\displaystyle 8\)th cyclotomic polynomial, is irreducible in \(\displaystyle \mathbb{Q}[x]\). Hence the only possibility for \(\displaystyle \mu_B\) is \(\displaystyle \mu_B(x)=x^4+1\). Therefore,

\(\displaystyle A^2+I = \mu_B(B) = 0. \tag1 \)

Since we have

\(\displaystyle A^2 + I = \begin{pmatrix} 0 & 0 & -2a & 2a \\ 0 & 0 & -2a & 2a \\ 2a & -2a & 0 & 0 \\ 2a & -2a & 0 & 0 \end{pmatrix}, \)

the relation (1) forces \(\displaystyle a=0\).

In case \(\displaystyle a=0\) we have

\(\displaystyle A = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}^2, \)

hence \(\displaystyle a=0\) satisfies the condition.