### International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria

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### 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 DaniĆ«l 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.