Let
$ B^2 = \begin{bmatrix} -2&0&0 \\ -1&-4&-1\\ 2&4&0\\ \end{bmatrix}^2 = \begin{bmatrix} 4&0&0\\ 4&12&4\\ -8&-16&-4\\ \end{bmatrix} = A similar to \begin{bmatrix} 4&0&0\\ 0&0&-16\\ 0&1&8\\ \end{bmatrix} = rcf(A) $
Now, as far as i know there is a fact that if $A,B$ in $M(3,\Bbb Z)$ are similar, then, $A$ has square root in $M(3,\Bbb Z) \iff B$ has a square root in $M(3,\Bbb Z) $
Therefore, since $A$ has a square root over $\Bbb Z$, and $A \sim rfc(A)$ . rcf of $A$ must have a square root over $\Bbb Z$.
However , using other research on computing square root of a matrix, I come up with
$\begin{bmatrix} 2&0&0\\ 0&1&-4\\ 0&1/4&3\\ \end{bmatrix}$
That is not in $M(3,\Bbb Z)$, maybe becuase it is just one of its square roots.
But, by the fact, I am certain that $rcf(A)$ has a square root over $\Bbb Z$.
I tried a desperate move letting $a,b,c,d,e,f,g,h,i$ be integers, and
$ \begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i\\ \end{bmatrix}^2 = \begin{bmatrix} 4&0&0\\ 0&0&-16\\ 0&1&8\\ \end{bmatrix}$
Then I tried to evaluate $a,b,c,d,e,f,g,h,i$ . But it does not work. I got stuck. Whatever I do.
Please help me with this huhuhu.
