$A\in \mathbb{R}^{n\times n}$ has eigenvalues in $\mathbb{Z}$ with at least 3 different eigenvalues. $\det(A)^n = 5^4$, find $A$'s eigenvalues $\newcommand{\adj}{\text{adj}}$
The question as it appeared in the first place: $A\in \mathbb{R}^{n\times n}$ such that all $A$'s eigenvalues are in $\mathbb{Z}$ and $A$ has at least 3 different eigenvalues.
Let $B$ be the matrix results in making the next row operations upon $A$ : $R_1 \leftrightarrow R_2$ , $R_2 \rightarrow 5R_2$ , $R_4 \rightarrow R_4-16R_2$.
Given that $\det((B\cdot \adj(A))^{-1}) = -\frac{1}{5^5}$ Find $A's$ eigenvalues.
So: $$\begin{align}\det((B\cdot \adj(A))^{-1}) = -\frac{1}{5^5} &\Rightarrow \det((B\cdot \adj(A))) = -5^5 \\&\Rightarrow \det(B)\cdot \det(A)^{n-1}= -5^5 \\&\Rightarrow -1\cdot 5\cdot \det(A)\cdot \det(A)^{n-1} = -5^5 \\&\Rightarrow \det(A)^n = 5^4\end{align}$$
This is where I got to the question represented in the title.
I don't know anything else about $A$, I got few examples for $A$ which satisfies the conditions: $\text{diag}(1,-1,-1,5), \text{diag}(1,1,-1,-5)...$ 
I don't see how do the conditions determine $A's$ eigenvalues.

Since the eigenvalues are all integers, their product $\det(A)$ is an integer.  What are the divisors of $5^4$?
EDIT: Use the fact that $n \ge 4$.
