Is the Lagrange polynomial integer-valued for points with consecutive integer x-values?

Question

What I'm really wondering is, does Lagrange polynomial interpolation have an answer for every question of "what's the next integer in this sequence"? Does it define an infinite integer sequence to extend every finite integer sequence?

Let $S$ be a finite set of points, at least two of them, all having consecutive integer $x$-coordinates and integer $y$ coordinates.

(Perhaps a restatement is helpful? Let $a, b\in \Re$ with $2 < b - a$. Let $s:\mathbb{Z}\to\mathbb{Z}$. Let $S$ be such that $\langle n, s(n)\rangle\in S$ if, and only if, $n\in [a, b]\cap \mathbb{Z}$.)

Is the Lagrange polynomial through $S$ necessarily an integer-valued polynomial?

I quickly see that it is for two points.

I suppose this is elementary. I didn't get this problem statement from anywhere though. I simply wondered.

Answer 1

For the sake of having an answer here, I'll give a proof.

Case 1: $p$ interpolates the points $(0,1)$, $(1,0),\dots , (n,0)$. Then $p(x) = \frac{(x-1) \cdots(x-n)}{n!}$. Plugging an integer for $x$, we get the product of $n$ consecutive integers in the numerator. Such a product is divisible by $n!$, as is witnessed by the fact that binomial coefficients are integers.

Case 2: $p$ interpolates the points $(-m,0),\dots, (-1,0)$, $(0,1)$, $(1,0),\dots , (n,0)$. Such $p$ is the product of two polynomials from Step 1, in one of which $x$ is replaced by $-x$. Thus, $p$ is integer-valued.

General case: The Lagrange interpolating polynomial is a linear combination of integer translates of polynomials from Case 2, with integer coefficients.

Answer 2

Yes, here is how to write the Lagrange interpolation polynomial for consecutive integers explicitly in terms of binomial coefficients. For $x_0, x_1, \dots, x_n$ we have $$ L_n(x)=\sum_{k=0}^{n} f(x_k) l_k(x) $$ where $$ l_k(x)=\frac{x-x_0}{x_i-x_0}\cdots \frac{x-x_{i-1}}{x_i-x_{i-1}} \frac{x-x_{i+1}}{x_i-x_{i+1}}\cdots \frac{x-x_{n}}{x_i-x_{n}}. $$ But for consecutive integer values $x_i=x_0+i$ and integer $m$ we have $$ l_k(m)=\frac{m-x_0}{k}\cdots \frac{m-x_0-k+1}{1} \frac{m-x_0-k-1}{-1}\cdots \frac{m-x_0-n}{k-n}\\ =\frac{(m-x_0)!}{(m-x_0-n-1)!(m-x_0-k)}\frac{1}{k!(n-k)!(-1)^{n-k}}\\ =(-1)^{n-k}\binom{m-x_0}{k}\binom{m-x_0-k-1}{n-k}, $$ and so $l_k(m)$ is clearly an integer. So if $f(x_0),f(x_1),\dots f(x_n)$ are also integers, then $L_n(m)=\sum_{k=0}^{n} f(x_k) l_k(m)$ is sum of integers and hence also an integer.