I want to show that $\forall n \in \mathbb{Z}, P(n) \in \mathbb{Z}$, assuming I have both $e^{2 i \pi P(n)} \underset{n \to +\infty}{\longrightarrow} 1$ and $P \in \mathbb{R}[X]$.
My idea was to start doing it by mathematical induction, I write $P = \alpha X^N + Q$ where $\deg Q \leq N-1$. So, if I show that $\alpha \in \mathbb{Z}$, by mathematical induction all coefficients of $P$ will be in $\mathbb{Z}$ and then we'll have the desired result (initialisation is easy).
But I dont't find a way to exploit this track. Any help is welcome.
Following @Kelenner's hint, if we do it by mathematical induction, we remark that $e^{2 i \pi P(n-1)} \underset{n \to +\infty}{\longrightarrow} 1$ then we also have that $e^{2 i \pi (P(n) -P(n-1))} \underset{n \to +\infty}{\longrightarrow} 1$. As $Q = P(X) - P(X-1)$ is of degree $< \deg P$, one has $Q(n) \in\mathbb{Z}$ for all $n \in \mathbb{Z}$.
As $P(n) - P(0)= \sum_{k = 1}^n P(k) - P(k-1) = \sum_{k = 1}^n Q(k)$, we have $\forall n \in \mathbb{Z}, P(n) - P(0) \in \mathbb{Z}$.
Then: $e^{2 i \pi P(0)} = e^{2 i \pi P(n)} e^{- 2 i \pi (P(n) - P(0))}\underset{n \to +\infty} {\longrightarrow} 1 \times 1 = 1$ then $P(0) \in \mathbb{Z}$.
Then, for all $n \in \mathbb{Z}$, $P(n) = P(0) + Q(1) + \dots + Q(n) \in \mathbb{Z}$.