Let $A = \begin{pmatrix} 3&1 \\ 0&1 \end{pmatrix}$ and $B = \begin{pmatrix} 1&0\ &0\\ 1&2 \end{pmatrix}$. 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? 3 deleted 100 characters in body; added 2 characters in body Let $A= A = \left [\begin{matrix} begin{pmatrix} 3& 1 \&1 \ 0& 1 \end{matrix} \right ] &1$end{pmatrix}$ and $$B=\left [{\begin{matrix} 1 & 0 \B = \begin{pmatrix} 1& 2 \\ \end{matrix}} &0\ 1&2 \right ]$$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?          2 fixed matrices Let $A= \left [{\begin{array}{cc} \begin{matrix} 3 & 1 \\ 0 & 1 \ \end{array}} end{matrix} \right ]$ and [ $$B=\left [{\begin{array}{cc} {\begin{matrix} 1 & 0 \\ 1 & 2 \\ end{array}} \end{matrix}} \right ] ]$$ 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?