Polynomial , and accumulation points of a sequence Hy , I would like to have some help on the following problem .
let $p \in \mathbb{R}[X]$ and for  a positive integer $n$ , let  $u_{n}=\cos(\pi p(n))$
suppose that   the set of accumulation points of $(u_{n})$  is finite . 
Prove that : 
$p(X)-p(0) \in \mathbb{Q}[X]$  .  
it's feels like it is related to weyl theorem . 
 A: You could use equidistribution results like Weyl's, Corollary 6 here; note rational coefficients in the polynomial can be ignored since it would suffice to show equidistribution modulo $1/q$ to contradict the finiteness of the accumulation points. But equidistribution is overkill.
We will show that if $p\in\mathbb R[X]$ satisfies $p(X)-p(0)\not\in\mathbb Q[X]$ then $p$ has infinitely many accumulation points mod $1.$
Assume by induction that this proposition is true for polynomials of smaller degree. By multiplying by a suitable integer we can assume that the top coefficient $c_n$ of $p$ is irrational.
If $p$ is linear then it is of the form $c_1n+c_0$ with $c_1$ irrational; this set is dense mod $1$ (I assume you know this), so has infinitely many accumulation points.
Otherwise, the degree of $p$ is greater than $1.$ The polynomial defined by $q(n)=p(n+1)-p(n)$ is non-constant and has the same top coefficient but smaller degree, so by the induction hypothesis must have infinitely many accumulation points mod $1.$ That implies that $p$ has infinitely many accumulation points mod $1,$ because the accumulation points of $q$ are differences of accumulation points of $p.$
