I already ask this but now its "for all"
Prove or disprove:
$A \in M(3,\mathbb{Z})$ has a square root with integer entries if and only if $XAX^{-1} \in M(3,\mathbb{Z})$ has a square root with integer entries, for all invertible $X \in M(3,\mathbb{R})$
Consider the matrices $$A=\begin{pmatrix} 0 & 0 & 9 \\ 9 & 0 & 0 \\ 0 & 9 & 0 \\ \end{pmatrix} \quad$$
and
$$B=\begin{pmatrix} 0 & 3 & 0 \\ 0 & 0 & 3 \\ 3 & 0 & 0 \\ \end{pmatrix} \quad$$
Observe that $A,B \in \Bbb M_3 (\Bbb Z)$ with $A=B^2.$ So $A$ has a square root with integer entries.
Let us consider the matrix
$$X=\begin{pmatrix} \frac 1 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \quad$$
Then you will easily find that
$$XAX^{-1} = \begin{pmatrix} 0 & 0 & 3 \\ 81 & 0 & 0 \\ 0 & 3 & 0 \\ \end{pmatrix} \quad$$
Then $XAX^{-1}$ is a matrix with integer entries. But there doesn't exist any $C \in \Bbb M(3,\Bbb Z)$ such that $XAX^{-1} = C^2.$
For a proof of the above assertion see this link.
