Take the 2-minute tour ×  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?  share|improve this question
Can't seem to get the latex right, but these are two by two
matrices. –  Hej May 15 '10 at 22:54 Markdown was munching your
backslashes. –  S. Carnahan May 15 '10 at 22:59 It could, in
principle, be undecidable, but I very much doubt that. Without
any further arguments, it sounds rather like defeatism to me at
that point. –  Xandi Tuni May 15 '10 at 23:07 Heuristically, the
2^n words of length n in your semigroup are going to have
determinant around 6^n, so I guess the "probability" of the
discriminant being a square is something like 6^{-n/2}; so I
guess it seems reasonable to me that yo