If $B$ is an integer matrix, does $\mathrm{det}(B^TB)$ has to be a perfect square?

Question

Suppose $B \in \mathbb{Z}^{m \times n}$ be an integer matrix of rank $n$. I want to prove that $\mathrm{det}(B^TB)$ is a perfect square.

Of course if $m=n$, $\mathrm{det}(B^TB) = \mathrm{det}(B)^2$ and we are done since $B$ is an integer matrix, implying that $\mathrm{det}(B)$ is an integer.

However, how about the case of rectangular matrices, i.e. when $m>n$?

In particular this would help me to prove that the parallelopiped having the columns of $B$ as sides have integral volume (which is $\sqrt{\mathrm{det}(B^TB)}$).

Thanks in advance.

Answer 1

It is not true if $m\neq n$. For example, if $B = \begin{pmatrix}1 \\ 2\end{pmatrix}$ then $B$ is an integer matrix of rank one with $$B^TB = \begin{pmatrix}1 & 2\end{pmatrix} \begin{pmatrix}1\\2\end{pmatrix} = 1^2 + 2^2 = 5$$ So $\det(B^TB) = 5$, which is not a perfect square.

Even if $n > 1$, the result still does not hold; if $$B = \begin{pmatrix}1&2\\3&4\\5&6\end{pmatrix}$$ then $$B^TB = \begin{pmatrix}35 & 44\\44 & 56\end{pmatrix}$$ so $\det(B^TB) = 24$, which is not perfect.