Polynomial such that $e^{2i\pi P(n)} \rightarrow 1$

Question

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.

Answer 1

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.

Answer 2

(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.