If $f(n) \in \mathbb{Z}$ for an infinite number of $n \in \mathbb{Z}$, then $f \in \mathbb{Q}[x]$.

Question

Do you think that the following statement is true? Do you have any idea about the proof?

Let $\; f(x) \in \mathbb{C}[x]$ be a polynomial. If $f(n) \in \mathbb{Z}$ for an infinite number of $n \in \mathbb{Z}$, then $f \in \mathbb{Q}[x]$.

Answer 1

The following uses too much machinery: I am being lazy.

Suppose that $P(x)=w_0+w_1x+\cdots +w_{n-1}x^{n-1}$, where the $w_i$ are complex numbers.

Let $b_1$ to $b_n$ be distinct integers such that $P(b_i)=m_i$, where $m_i$ is an integer.

This yields the system of $n$ linear equations $$w_0+b_iw_1+b_i^2w_2+\cdots +b_i^{n-1}w_{n-1}=m_i$$ ($i=1,2,\dots, n$).

We will show that the above system has unique rational solutions $w_i$. This is straightforward, as long as we prove that the $n\times n$ matrix with $i$-th row $$1\quad b_i \quad b_i^2 \quad b_i^3\quad \cdots\quad b_i^{n-1}$$ is invertible.

But that matrix is the well known Vandermonde matrix, and it is invertible.