eigenvalues of integral matrices Is it possible that a $3$-by-$3$ matrix with integer values and determinant 1 has a real eigenvalue with algebraic multiplicity 2, that is not equal to $\pm 1$? 
Doing some elementary computations one can rephrase the question as follows. Do there exist integers $k$ and $m$ such that $a=\frac{1}{3}(k\pm \sqrt{k^2-3m})$ and  $b=\frac{1}{3}(k\mp 2\sqrt{k^2-3m})$ are real numbers and $a^2b=1$ but $a,b \neq \pm 1$?
 A: As in Ewan's proof, write
$$x = \frac{2 a^3+1}{a^2} \quad y = \frac{a^3+2}{a}.$$
Notice that
$$a = \frac{xy-9}{2 (x^2-3y)}. \quad (\ast)$$
So, if $x$ and $y$ are integers then either $a$ is rational or else $xy=9$ and $x^2=3y$. The latter case implies $x (x^2/3) = 9$, so $x=y=3$ and $a=1$, which we already ruled out. 
So we focus on the case where $a$ is rational. Write $a=p/q$ in lowest terms. So
$$x = \frac{2p^3+q^3}{p^2 q} \quad y=\frac{p^3+2 q^3}{p q^2}.$$
From the formula for $x$, any prime that divides $p$ also divides $q$; from the formula $y$, any prime that divides $q$ also divides $p$. Since $p/q$ is in lowest terms, we get $a = \pm 1$. 
So, where did $(\ast)$ come from? Since the map $a \mapsto \left( \frac{2a^3+1}{a^2}, \frac{a^3+2}{a} \right)$ is generically injective, the subfield of $k(a)$  generated by $(2a^3+1)/a^2$ and $(a^3+2)/a$ should be all of $k(a)$. So such a formula should exist, and I just needed to find it. To find it, I used that $a$ is double root of $t^3-x t^2+y t - 1$, so it is also a root of $3 t^2 - 2x t + y$. Therefore, $a$ is a root of $(t^3-x t^2+y t - 1) - (t/3 - x/9) (3 t^2 - 2x t + y) = (2y/3 - 2x^2/9) t +(xy/9 -1)$. I solved this equation for $t$.
