Does $A$ admit a square root with integer entries? 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.
 A: $A$ 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.
A: Using the similarity matrix $X=\begin{pmatrix}1&0&0\\0&9&0\\0&0&3\end{pmatrix}$, $A$ is similar to the circulant matrix $A'=\begin{pmatrix}0&0&9\\9&0&0\\0&9&0\end{pmatrix}$. This $A'$ has an easily found square root $B'=\begin{pmatrix}0&3&0\\0&0&3\\3&0&0\end{pmatrix}$. Now, reverse the similarity, and
$B=\begin{pmatrix}0&\frac13&0\\0&0&9\\9&0&0\end{pmatrix}$ is a square root of $A$.
So there's a rational square root - but that's not an integer matrix. We need to look deeper. $A'$ can be diagonalized by the Fourier matrix; its eigenvalues are $9w^2$ for each cube root $\omega$ of unity, with eigenvectors $\begin{pmatrix}1&\omega&\omega^2\end{pmatrix}^T$. With this complete set of eigenvectors with different eigenvalues, any square root $B'$ of $A'$ must share those same eigenvectors, and its eigenvalues must be square roots $\pm 3\omega$ of the corresponding eigenvalues of $A'$. That gives us eight possibilities.
First, if we take $+$ signs everywhere or $-$ signs everywhere, we get $\begin{pmatrix}0&3&0\\0&0&3\\3&0&0\end{pmatrix}$ or $\begin{pmatrix}0&-3&0\\0&0&-3\\-3&0&0\end{pmatrix}$. Undoing the similarity, these each lead to square roots of $A$ with one non-integer entry as found before.
Taking a $-$ sign for the eigenvalue $1$ and $+$ signs for the other two eigenvalues leads to $\begin{pmatrix}-2&1&-2\\-2&-2&1\\1&-2&-2\end{pmatrix}$. Reverse the similarity and we get a square root $B=\begin{pmatrix}-2&\frac19&-\frac23\\-18&-2&3\\3&-\frac23&-2\end{pmatrix}$. Of course, taking a $+$ sign at $1$ and a $-$ sign at the other two would lead to the negative of this matrix. (See also @user's comments on the question)
The other four choices come from taking different signs for the two non-real roots. The trace of the square roots we get out of these have nonzero imaginary part $\pm 3\sqrt{3}$, so they can't be rational.
So then, $A$ has four rational square roots. None of them have integer entries, and thus the answer is no.
Your argument, on the other hand, fails; it's an attempt to prove that there aren't any rational square roots, when in fact there are.
