ds.dynamical systems - Spectrum of a generic integral matrix.

Yes, a generic integer matrix has no more than two eigenvalues of the same norm. More precisely, I will show that matrices with more than two eigenvalues of the same norm lie on a algebraic hypersurface in $mathrm{Mat}_{n times n}(mathbb{R})$. Hence, the number of such matrices with integer entries of size $leq N$ is $O(N^{n^2-1})$.

Let $P$ be the vector space of monic, degree $n$ real polynomials. Since the map "characteristic polynomial", from $mathrm{Mat}_{n times n}(mathbb{R})$ to $P$ is a surjective polynomial map, the preimage of any algebraic hypersurface is algebraic.
Thus, it is enough to show that, in $P$, the polynomials with more than two roots of the same norm lie on a hypersurface. Here are two proofs, one conceptual and one constructive.

Conceptual: Map $mathbb{R}^3 times mathbb{R}^{n-4} to P$ by
$$phi: (a,b,r) times (c_1, c_2, ldots, c_{n-4}) mapsto (t^2 + at +r)(t^2 + bt +r) (t^{n-4} + c_1 t^{n-5} + cdots + c_{n-4}).$$

The polynomials of interest lie in the image of $phi$. Since the domain of $phi$ has dimension $n-1$, the Zariski closure of this image must have dimension $leq n-1$, and thus must lie in a hyperplane.

Constructive: Let $r_1$, $r_2$, ..., $r_n$ be the roots of $f$. Let
$$F := prod_{i,j,k,l mbox{distinct}} (r_i r_j - r_k r_l).$$
Note that $F$ is zero for any polynomial in $mathbb{R}[t]$ with three roots of the same norm. Since $F$ is symmetric, it can be written as a polynomial in the coefficients of $f$. This gives a nontrivial polynomial condition which is obeyed by those $f$ which have roots of the sort which interest you.