TITLE: Does $A$ admit a square root with integer entries?
QUESTION [8 upvotes]: Let us consider the matrix
$$ A =
 \left( \begin{array}{crc} 0 & 0 & 3 \\
 81 & 0 & 0 \\ 0 & 3 & 0 \end{array}
 \right).$$
Does the matrix $A$ admit a square root each of whose entry is an integer? Please help me in this regard.
Thank you very much.
EDIT $:$
I have tried by taking determinant of $A$ but that is not working since $\det (A) =729,$ which is a perfect square. I don't know what other tools should I require to solve this. I think I can also use the trace argument here. If $B$ was the square root of $A$ then I also found that trace of $B$ is $2\sqrt 2 \cos \left (\frac {3 \pi} {8} \right) + 3,$ which is not an integer. Because $\lambda$ is an eigen value of $A$ iff $\sqrt {\lambda}$ is an eigen value of $B.$ So such a $B$ cannot be found. 
Please check my argument above whether it holds good or not.

REPLY [3 votes]: $A$ is a pseudo-permutation. Thus $A$ has $3$ distinct eigenvalues and its commutant is $C(A)=span(I,A,A^2)$. If $B^2=A$, then $B\in C(A)$ and
$B=aI+bA+cA^2=\begin{pmatrix}a&9c&3b\\81b&a&243c\\243c&3b&a\end{pmatrix}$. We want that $a,3b,9c \in\mathbb{Z}$.
$B^2-A=0$ is equivalent to $a^2+1458bc=0,2ac+b^2=0,2ab+9\times(81c^2)-1=0$.
If $a\not= 0$, then $bc<0,ac<0,ab<0$, that is contradictory.
If $a=0$, then $b=0$ and $9\times(81c^2)=1$, that is contradictory (because $81c^2\in\mathbb{Z}$).
Conclusion. No solutions.