Polynomial such that $e^{2i\pi P(n)} \rightarrow 1$ here is a  problem i’ve been having quite a lot of trouble with .
Let $P$ be a polynomial such that the sequence $e^{2i\pi P(n)}$ converges to 1 $(i^2=-1).$
Show that $\forall n ,P(n)$ is an integer.
Taking the imaginary part we are left with some $P$ polynomial such that
$$\sin(2\pi P(n))\rightarrow0$$
But a half angle factorization also gives  that: $$\sin(\pi P(n))\rightarrow 0$$
And we wish to force P to , i guess even if it’s not equivalent, to have only integers coefficients.
May you help me please.
 A: Write $f(x)= \exp(2\pi ix)$. If $P(n)=a_1n+a_0$ is linear, you could use  $$f(a_1)=f(P(n+1)-P(n))= \frac{f(P(n+1))}{f(P(n))} \to 1 \; \text{as} \; n \to \infty$$
to deduce that $a_1$ is an integer, whence so is $a_0$.  This approach generalizes, but one needs to use the representation of a polynomial in terms of binomials.
Write a degree $d$ polynomial as
$P(x)=\sum_{k=o}^d a_k {x\choose k}$ (Every polynomial has a unique representation as such a combination of binomial coefficients.)
Claim:    The assumption $f(P(n)) \to 1$ as $ n \to \infty$ implies that all $a_j$ in the representation $P(x)=\sum_{k=o}^d a_k {x\choose k}$ are integers.
(This then  implies that $P(m)$ is an integer for all integer $m$.)
Proof By induction on $d$. We just need the induction step. Given $P$ that satisfies the assumption,
$$Q(x):=P(x+1)-P(x)=\sum_{k=1}^d a_k {x \choose k-1} $$
also satisfies $f(Q(n)) \to 1$ as $n \to \infty$, so by the induction hypothesis,
$a_1, \ldots , a_d$ are integers. It then   follows that $f(a_0)=1$ so $a_0$ is also an integer.
A: (Some years ago, I saw a proof.)
We use Mathematical Induction to prove the following statement:
Let $P(x)$ is a polynomial with real coefficients. If $\lim_{n\to\infty} \mathrm{e}^{2\pi \mathrm{i}P(n)} = 1$, then $P(n)$ is an integer
for any $n\in \mathbb{N}_{> 0}$.
When $\mathrm{deg}(P) = 0$, the statement is true.
Assume that the statement is true for $\mathrm{deg}(P) = k$ ($k\ge 0$).
Let us prove that the statement is also true for $\mathrm{deg}(P) = k + 1$.
We have $\lim_{n\to \infty} \mathrm{e}^{2\pi \mathrm{i}(P(n) - P(n-1))} = 1$.
Let $Q(x) = P(x) - P(x-1)$. Then $\mathrm{deg}(Q) = k$. By the induction hypothesis, $Q(n)$ is an integer for any $n\in \mathbb{N}_{> 0}$.
Since $P(n) = Q(n) + Q(n-1) + \cdots + Q(2) + P(1)$, we have
$\lim_{n\to\infty} \mathrm{e}^{2\pi \mathrm{i}P(n)} = \lim_{n\to\infty} \mathrm{e}^{2\pi \mathrm{i}P(1)} = 1$ and thus $P(1)$ is an integer.
Thus, $P(n)$ is an integer for any $ n\in \mathbb{N}_{> 0}$.
We are done.
