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Let $A = \begin{pmatrix} 3&1 \\ 0&1 \end{pmatrix}$ and $B = \begin{pmatrix} 1&0\\ 1&2 \end{pmatrix}$. I want to show that the only elements of the semigroup generated by $A$ and $B$ that have integer eigenvalues are elements of the form $A^n$ and $B^n$, $n \in \mathbb{N}$. I have tried every way that I can think of. Is it possible that a problem like this is undecidable?

Let $$A = \begin{pmatrix} 3&1 \\ 0&1 \end{pmatrix}$$ and $$B = \begin{pmatrix} 1&0\\ 1&2 \end{pmatrix}$$ I want to show that the only elements of the semigroup generated by $A$ and $B$ that have integer eigenvalues are elements of the form $A^n$ and $B^n$, where $n \in \mathbb{N}$. I have tried every way that I can think of. Is it possible that a problem like this is undecidable?

Let \[ A= \left [\begin{matrix} 3 & 1 \\ 0 & 1 \end{matrix} \right ] \] and $$ B=\left [{\begin{matrix} 1 & 0 \\ 1 & 2 \\ \end{matrix}} \right ] $$

I want to show that the only elements of the semigroup generated by $A$ and $B$ that have integer eigenvalues are elements of the form $A^n$ and $B^n$, $n \in \mathbb{N}$. I have tried every way that I can think of. Is it possible that a problem like this is undecidable?

Let $A = \begin{pmatrix} 3&1 \\ 0&1 \end{pmatrix}$ and $B = \begin{pmatrix} 1&0\\ 1&2 \end{pmatrix}$. I want to show that the only elements of the semigroup generated by $A$ and $B$ that have integer eigenvalues are elements of the form $A^n$ and $B^n$, $n \in \mathbb{N}$. I have tried every way that I can think of. Is it possible that a problem like this is undecidable?