The complex rational function is a polynomial if it takes integer value when variable is an integer

Question

Question: Let $R(x)=\dfrac{f(x)}{g(x)}$ is a rational function with the coefficients of $f(x),g(x)$ all in $\mathbb{C}$. Suppose $R(n)\in\mathbb{Z}$ for all $n\in\mathbb{Z}$. Now my quesion is: Is $R(x)$ a polynomial?

The following is my idea:

First, we can simplify it into real case: $\text{Re}(R)(x):=\dfrac{R(x)+\overline{R(x)}}{2}$, $\text{Im}(R)(x):=\dfrac{R(x)-\overline{R(x)}}{2i}$ are real value functions and $R(x)=\text{Re}(R)(x)+i\text{Im}(R)(x)$. Since for all $n\in\mathbb{Z}$, $\text{Im}(R)(n)=0$, i.e., $\text{Im}(R)$ has infinite many roots, hence $\text{Im}(R)(x)\equiv0$. Thus, we only need to focus on the real case.

Second, if $\deg(f)<\deg(g)$, it is clear that when $n\to\infty$, $0<|R(n)|<1$, which is not an integer, contradiction!

Third, if $\deg(f)=\deg(g)$, since $\lim\limits_{n\to\infty}\dfrac{f(x)}{g(x)}=\lambda$, where $\lambda$ is the proportion of leading coefficient of $f(x)$ and $g(x)$, the set $\{R(n):n\in\mathbb{Z}\}$ is a finite subset of $\mathbb{Z}$. So there is a $m\in\mathbb{Z}$ such that there are infinite many $n\in\mathbb{Z}$ such that $R(n)=m$. Thus $f(x)=mg(x)$ has infinite many roots, which follows that $f(x)\equiv mg(x)$, so $R(x)=m$ is a polynomial.

However, I am struggling at the case when $\deg(f)>\deg(g)$. If you write $\dfrac{f(x)}{g(x)}=p(x)+\dfrac{r(x)}{g(x)}$, where $\deg(r)<\deg(g)$, even when $n\to\infty$, $\dfrac{r(n)}{g(n)}\to 0$, I don't know $p(n)$ at all, since it is not necessarily integer.

Answer 1

Say, $R(x)$ is a rational function with $\deg(numerator)-\deg(denominator)=k>0$. Now look at its finite difference derivative: $R(x+1)-R(x)$. It is still a rational function, it also takes integer values at integer arguments, and its degree is lower than $k$. Go ahead, and after a few steps you will reach the situation with $\deg(f)<\deg(g)$ which you know how to treat. That derivative is a polynomial, hence so is the original $R(x)$.

Answer 2

In order to prove this, we first introduce Weyl's equidistribution theorem (see https://terrytao.wordpress.com/tag/weyls-equidistribution-theorem/). By this theorem, for any real polynomial $F(x)$ with at least one coefficient is irrational, $\{F(n)\}_{n\in\mathbb{N}} \mod 1$ equidistributes in $[0,1]$; otherwise, $\{F(n)\}_{n\in\mathbb{N}}\mod 1$ is periodic in $[0,1]$.

Now go back to the problem. For $\dfrac{f(x)}{g(x)}$ with $\deg(f)<\deg(g)$, $f(x)\equiv0$. For $\deg(f)\ge\deg(g)$, we can write it as $\dfrac{f(x)}{g(x)}=p(x)+\dfrac{r(x)}{g(x)}$ with $\deg(r)<\deg(g)$ and $r\not 0$. When $x\to \infty$, $\dfrac{r(x)}{g(x)}\to 0$. However, since $\{p(n)\}_{n\in\mathbb{N}}\mod 1$ is equidistributed or periodic in $[0,1]$, when $n$ is large, $p(x)+\dfrac{r(x)}{g(x)}$ is not an integer!

So we have $r(x)=0$ and all the coeffiecients of $p(x)$ are rational.