Problem Statement
Let $x_1,x_2,…,x_m$ be real numbers such that the set
$$ A = \lbrace \sum_{i=1}^m cos(n\pi x_i) | n \in \mathbb{N} \rbrace $$
is finite. Prove that all the $x_i$ are rational numbers.
Idea
(borrow from the proof that $\lbrace n\alpha \rbrace = n\alpha - \lfloor n\alpha \rfloor$ is dense on $[0,1]$)
Let $k$ be an integer. Consider the tuple
$$ (\lbrace \dfrac{t x_1}{2} \rbrace,\lbrace \dfrac{t x_2}{2} \rbrace,…,\lbrace \dfrac{t x_m}{2} \rbrace)$$
Each fractional part $\lbrace \dfrac{t x_i}{2} \rbrace$ belongs to exactly one interval of the set
$$ K = \lbrace \lbrack 0, \dfrac{1}{k}\rparen, \lbrack \dfrac{1}{k}, \dfrac{2}{k}\rparen, … , \lbrack \dfrac{k-1}{k}, 1\rparen \rbrace $$
By pigeonhole principle, dividing infinite tuples of the above to an finite set of $K \times K \times…\times K$ ($m$ times) means there exists 2 integers $p < q$ such that p and q falls into the same hypercube, that is their distance in each dimesion is less than $\frac{1}{k}$:
$$ \| \{ \dfrac{qx_i}{2}\} -\{ \dfrac{px_i}{2}\}\| < \dfrac{1}{k}$$ $$ \Leftrightarrow 0 \leq \lbrace (q-p)\dfrac{x_i}{2} \rbrace < \dfrac{1}{k}\text{ or } \dfrac{k-1}{k} \leq \lbrace (q-p)\dfrac{x_i}{2} \rbrace < 1$$
This implies, by choosing arbitarily large $k$, there exists an integer $t \geq 1$ such that $t \pi x_i$ gets arbitarily close to some $2 m \pi | m \in \mathbb{Z}$ points for every $i$. This means $cos(t\pi x_i)$ gets arbitarily close to 1 for every $i$ as well.
As the set A is finite, then the value $n$ (which is achieved when $cos(t\pi x_i) = 1$ for every i) must be in A.
Then the conclusion that all $x_i$ are rational numbers comes naturally.
Aftermath
This proof differs from one from the book that utilize the Chebyshev polynomial’s recurrence. In general,this method can be extended for any continous, periodic function as well.