Is the set of certain polynomials finite or infinite? Let us consider the set of all polynomials with the following properties:
i) all coefficients are integer;
ii) the leading coefficient equals one;
iii) all zeros are real and simple and belonging to $[-1.99,1.99] $.
Is this set finite or infinite?
 A: It is finite. This follows by combining two separate results:


*

*If $p \in \mathbb{Z}[x]$ is monic integer having all its complex roots lying in $[-2,2]$, then all these roots are of the form $2\cos(2\pi q)$ with $q \in \mathbb{Q}$.
For the proof of this, write $p = \prod_{i=1}^d (x - 2\cos(2\pi \, t_i))$ and observe that the sequence of degree $d$ monic polynomials $p_n := \prod_{i=1}^d(x - 2\cos(2\pi n t_i))$ is also in $\mathbb{Z}[x]$ and that, of course, it has bounded coefficients (in terms of $d$ alone, independently of $n$). Thus there must be two $p_n = p_{n'}$ with $n \neq n'$; considering their roots, we get the claim with $q(n-n') \in \mathbb{Z}$.


*For a given $d \in \mathbb{N}$, the monic polynomial $P_d$ with simple roots at the set $\{2\cos(2\pi m/d) \mid (m,d) = 1\}$ is in $\mathbb{Z}[x]$ and irreducible.
This follows from the corresponding irreducibility property of the cyclotomic polynomial $\Phi_d(z)$, which can be written as $z^{\deg{P_d}} P_d(z+1/z)$.
Combining these two statements, you can see that your set consists precisely of the products of pairwise different minimal polynomials of those $2\cos(2\pi / d)$ that do not exceed $1.99$. There are only finitely many such $d$.
