IMC 2008 · Day 2 · P11

Call a square matrix of type (B), if it is of the form $$\begin{pmatrix} 0 & b_{12} & 0 & \dots & b_{1,2k-2} & 0 \\ b_{21} & 0 & b_{23} & \dots & 0 & b_{2,2k-1} \\ 0 & b_{32} & 0 & \dots & b_{3,2k-2} & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ b_{2k-2,1} & 0 & b_{2k-2,3} & \dots & 0 & b_{2k-2,2k-1} \\ 0 & b_{2k-1,2} & 0 & \dots & b_{2k-1,2k-2} & 0 \end{pmatrix}.$$ Note that every matrix of this form has determinant zero, because it has $k$ columns spanning a vector space of dimension at most $k - 1$.

Call a square matrix of type (C), if it is of the form $$C' = \begin{pmatrix} 0 & c_{11} & 0 & c_{12} & \dots & 0 & c_{1,k} \\ c_{11} & 0 & c_{12} & 0 & \dots & c_{1,k} & 0 \\ 0 & c_{21} & 0 & c_{22} & \dots & 0 & c_{2,k} \\ c_{21} & 0 & c_{22} & 0 & \dots & c_{2,k} & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & c_{k,1} & 0 & c_{k,2} & \dots & 0 & c_{k,k} \\ c_{k,1} & 0 & c_{k,2} & 0 & \dots & c_{k,k} & 0 \end{pmatrix}.$$ By permutations of rows and columns, we see that $$|\det C'| = \left| \det \begin{pmatrix} 0 & C \\ C & 0 \end{pmatrix} \right| = |\det C|^2,$$ where $C$ denotes the $k \times k$-matrix with coefficients $c_{i,j}$. Therefore, the determinant of any matrix of type (C) is a perfect square (up to a sign).

Now let $X'$ be the matrix obtained from $A$ by replacing the first row by $\begin{pmatrix} 1 & 0 & 0 & \dots & 0 \end{pmatrix}$, and let $Y$ be the matrix obtained from $A$ by replacing the entry $a_{11}$ by 0. By multi-linearity of the determinant, $\det(A) = \det(X') + \det(Y)$. Note that $X'$ can be written as $$X' = \begin{pmatrix} 1 & 0 \\ v & X \end{pmatrix}$$ for some $(n-1) \times (n-1)$-matrix $X$ and some column vector $v$. Then $\det(A) = \det(X) + \det(Y)$. Now consider two cases. If $n$ is odd, then $X$ is of type (C), and $Y$ is of type (B). Therefore, $|\det(A)| = |\det(X)|$ is a perfect square. If $n$ is even, then $X$ is of type (B), and $Y$ is of type (C); hence $|\det(A)| = |\det(Y)|$ is a perfect square.

The set of primes can be replaced by any subset of $\{2\} \cup \{3, 5, 7, 9, 11, \dots\}$.