For $A \in \mathcal{M}_3(\mathbb{C})$, does $\mathrm{tr}(A^2) = \mathrm{tr}(A^3) \in \mathbb{Q}$ imply $\mathrm{tr}(A^4) \in \mathbb{Q}$? Let $A \in \mathcal{M}_3(\mathbb{C})$ such that $\mathrm{tr}(A^2) = \mathrm{tr}(A^3) \in \mathbb{Q},$ where $\mathrm{tr}(A)$ is the trace of $A.$ It is possible to prove that $\mathrm{tr}(A^4) \in \mathbb{Q}?$ 

We fix some $t\in \Bbb C$. Let $a,b,c$ be the complex roots of the equation 
$$
X^3-6tX^2+18t^2 X -36 t^3\ .
$$
Let $e_n$ be the elementary symmetric polynomial of degree $n$ in $a,b,c$.
Let $p_n$ be the Newton symmetric polynomial of degree $n$ in $a,b,c$.
By Vieta,
$$\left\{
\begin{aligned} 
e_1 &= a+b+c = 6t\ ,\\
e_2 &= ab+bc+ca = 18t^2\ ,\\
e_3 &= abc = 36t^3\ .
\end{aligned} 
\right.
$$
Let $A$ be the matrix with diagonal entries $a,b,c$. Then, using the Newton identities,
$$
\begin{aligned} 
\operatorname{Trace}(A^2)
&=a^2+b^2+c^2\\
&=p_2\\
&=e_1p_1-2e_2\\
&=6t\cdot 6t-2\cdot 18t^2=0\ ,
\\
\operatorname{Trace}(A^3)
&=a^3+b^3+c^3\\
&=p_3\\
&=e_1p_2-e_2p_1+3e_3\\
&=0-18t^2\cdot 6t+3\cdot 36t^3=0\ ,
\\
\operatorname{Trace}(A^4)
&=a^4+b^4+c^4\\
&=p_4\\
&=e_1p_3-e_2p_2+e_3p_1-4e_4\\
&=0-0+36t^3\cdot 6t-0\\
&=216t^4\ .
\end{aligned} 
$$
It is clear that we can chose $t$, such that the last expression is not rational. This gives and explicit counterexample.

Later edit: The general family of cubics with roots $a,b,c$, so that $p_1=p_1(a,b,c)=e_1=e_1(a,b,c)=s$ and $p_2=p_3=p_2(a,b,c)
=p_3(a,b,c)=q$, thus parametrized by $s,q$, is obtained by solving
$$
\begin{aligned}
e_1&=p_1=s\ ,\\
2e_2 &=e_1p_1-p_2=s^2-q\ ,\\
3e_3 &=e_2p_1-e_1p_2+p_3=\frac 12(s^2-q)s-sq+q\ ,\\
\end{aligned}
$$
and $p_4$ is obtained algebraically as $p_4=e_1p_3-e_2p_2+e_3p_1-4e_4$. (Here, $e_4=0$.) The question "It is still possible to find a counterexample so that $s$ (trace of $A$) is a natural (or even rational) number?" has a negative answer. If both $s,q$ are rational, than $p_4$ is rational, as an algebraic expression in $s,q$.
