If $f((\text{deg}(f) + 1$) consecutive $\mathbb{Z}\text{s}) \subset \mathbb{Z}$ then $f(\mathbb{Z}) \subseteq \mathbb{Z}$

Question

Question:

Let $f:\mathbb{R}\to\mathbb{R}$ be a polynomial of degree $n$. Suppose there exists an integer $m\in\mathbb{Z}$ such that $f(m)\in\mathbb{Z}$, $f(m+1)\in\mathbb{Z}$, ..., $f(m+n)\in\mathbb{Z}$ (i.e. there exists $n+1$ consecutive integers all satisfying $f(x)\in\mathbb{Z}$). Prove that $\forall x\in\mathbb{Z}, f(x)\in\mathbb{Z}$.

Context:

This is a problem I thought of on my own when I was preparing for a contest next month. This problem at first glance "felt" like one of those classic and old problem with lots of solutions and thoughts around it. Surprisingly, I could find neither solutions nor questions regarding the same problem.

After spending a while on the problem, I came up with a finite difference solution. The solution is as follows:

My solution:

Let $P(n)$ denote the statement we want to prove for a function of degree $n$. We shall prove that $P(n)$ is true for all $n$ by induction.

Base case: When $n=0$, $f(x)=c$ where $c$ is a constant. Since $f(m)\in\mathbb{Z}$, we have $c\in\mathbb{Z}$, which means that $\forall x\in\mathbb{Z}, f(x)=c\in\mathbb{Z}$. Hence, $P(0)$ is true.

Induction hypothesis: $P(k)$ is true where $k\in\mathbb{Z}^+$.

Inductive step: We claim that $P(k+1)$ is true. We have the condition that there exists an integer $m\in\mathbb{Z}$ such that $f(m)\in\mathbb{Z}$, $f(m+1)\in\mathbb{Z}$, ..., $f(m+k+1)\in\mathbb{Z}$.

Consider a function $g(x)=f(x)-f(x-1)$ (i.e. first finite difference). Since $f(x)$ has degree $k+1$, we have that $g(x)$ has degree $k$. From the conditions given we also know that $$g(m+1)=f(m+1)-f(m)\in\mathbb{Z} \\ g(m+2)=f(m+2)-f(m+1)\in\mathbb{Z} \\ \dots \\ g(m+k+1)=f(m+k+1)-f(m+k)\in\mathbb{Z}$$

Hence, we have $k+1$ consecutive integers all satisfying $g(x)\in\mathbb{Z}$. By the induction hypothesis, this means that $\forall x\in\mathbb{Z}, g(x)\in\mathbb{Z}$. Then, we have $$f(m+k+2)=f(m+k+1)+g(m+k+2)\in\mathbb{Z} \\ f(m+k+3)=f(m+k+2)+g(m+k+3)\in\mathbb{Z} \\ \dots$$ and $$f(m-1)=f(m)-g(m)\in\mathbb{Z} \\ f(m-2)=f(m-1)-g(m-1)\in\mathbb{Z} \\ \dots$$

It can now be easily proven by induction that $\forall x\in\mathbb{Z}, f(x)\in\mathbb{Z}$. This completes the induction, $Q.E.D.$

First I'd like to ask you to verify if my proof is correct (and the spaces for improvements). Second, I'd also like to ask for more insightful and/or elegant proofs (or any generalisation from this). Any help is appreciated!

Answer 1

WLOG we may assume $m = 0$.

I will use the Lagrange basis polynomials to prove this. By unicity of the interpolating polynomial, we have $f(x) = \sum_{j=0}^n y_{j} L^{(n)}_j(x)$ where $y_j = f(j)$ and $$L^{(n)}_j(x) = \frac{x(x-1)\cdots\widehat{(x-j)}\cdots(x-n)}{j(j-1)\cdots\widehat{(j-j)}\cdots(j-n)}$$ where the hat denotes omission of that factor, for each $j \in \{0, \ldots, n\}$. Further, we know that for each $i \in \{0, \ldots, n\}$: $L^{(n)}_j(i) = \delta_{ij}.$

It follows that we really only have to show the statement for the Lagrange polynomials $L^{(n)}_j$, $j \in \{0, \ldots, n\}$.

For $j = 0$, this boils down to showing that the product of $n$ consecutive integers is divisible by $n$ factorial, see e.g. The product of $n$ consecutive integers is divisible by $n$ factorial .

Also the case $j > 0$ uses the same fact, since if $x$ is an integer, $$L^{(n)}_j(x) = \frac{\overbrace{x(x-1)\cdots(x-(j-1))}^{\overset{\Large \text{$j$ consecutive integers}}{\text{are divisible by $j!$}}}\cdot \widehat{(x-j)} \cdot \overbrace{(x-(j+1)) \cdots(x-n)}^{\overset{\Large \text{$n-j$ consecutive integers}}{\text{are divisible by $(n-j)!$}}}} {j(j-1)\cdots 1 \cdot \widehat{(j-j)}\cdot{(-1) \cdots(-(n-j))}}.$$