polynomial of degree $n$ with real coefficient takes integral values on a certain set $n+1$ consecutive integers , then $f(x)$ is integer-valued

Question

Show that if $f(x)$ is a polynomial of $deg n$ with real coefficient ,which takes integral values on a certain set $n+1$ consecutive integers , then $f(x)$ is integer-valued.

How to prove this I thought of this like $P(x) =\sum_{i=0}^{\infty}a_{i}x^{i}$ where $a_{n}=\frac{P^{n}(0)}{n!}$ Your answers are really nice. But , I don't know or I want answer from strong principle of induction.

Answer 1

Along the same lines as GEdgar’s answer: use the fact that if $f(x)$ has degree $n>1$ then $g(x) := f(x+1) - f(x)$ has degree $n-1$ (this is easy to see by looking at the leading term, but ultimately we’re still appealing to binomial expansion). Now note that if $f$ has integer values at $n+1$ consecutive integers, then $g$ has integer values at $n$ consecutive integers. Induct from the fact that the claim is trivial for the degree $0$ case (constant polynomials).

Answer 2

Hint.
If $f(x)$ has degree $n$, show that it satisfies the recurrence equation $$ \sum_{k=0}^{n+1} (-1)^k\binom{n+1}{k} f(x-k) = 0 $$ Use this, together with induction (forward and backward).

Answer 3

►Let $P_{n+1}(x)=A_0+A_1x+A_2x^2+\cdots+A_{n+1}x^{n+1}$ where $A_i\in\mathbb R$ for $i=0,1,...,n+1$ and $a_1,a_2,...,a_{n+2}$ a set of $n+2$ consecutive integers.We have the equation $$\begin{pmatrix}1&a_1&a_1^2&\cdots&a_1^{n+1}\\1&a_2&a_2^2&\cdots&a_2^{n+1}\\\vdots\\1&a_{n+2}&a_{n+2}^2&\cdots&a_{n+2}^{n+1}\end{pmatrix}*\begin{pmatrix} A_0\\A_1\\\vdots\\A_{n+1}\end{pmatrix}=\begin{pmatrix} m_1\\m_2\\\vdots\\m_{n+1}\end{pmatrix}$$ where each $m_i$ is integer.This leads to a Vandermonde's determinant (VD) of order $n+2$ which like as for a proof of it (there are several ones) can be reduced to a VD of order $n+1$. We do this to go to a polynomial $P_n$ of degree $n$ so we can apply induction.

We have $$\det\begin{pmatrix}1&a_1&a_1^2&\cdots&a_1^{n+1}\\1&a_2&a_2^2&\cdots&a_2^{n+1}\\\vdots\\1&a_{n+2}&a_{n+2}^2&\cdots&a_{n+2}^{n+1}\end{pmatrix}=\det\begin{pmatrix}1&0&0&\cdots&0\\1&a_2-a_1&a_2(a_2-a_1)&\cdots&a_2^n(a_2-a_1)\\\vdots\\1&a_{n+2}-a_1&a_{n+2}(a_{n+2}-a_1)&\cdots&a_{n+2}^n(a_{n+2}-a_1)\end{pmatrix}=$$ $$(n+1)!\det\begin{pmatrix}a_2&a_2^2&\cdots&a_2^{n+1}\\\vdots\\a_{n+2}&a_{n+2}^2&\cdots&a_{n+2}^{n+1}\end{pmatrix} $$ ►Now for $n=1$ we have $P_1(x)=A_0+A_1x$ so if $$A_0+A_1a=m_1\\A_0+A_1(a_1+1)=m_2$$ we have $$A_1=m_2-m_1\in\mathbb Z\\A_0=m_1-A_1a\in\mathbb Z$$ ►Suposse it is true for $P_n(x)$ and prove that this is so for $P_{n+1}(x)$.

Note that $$\det\begin{pmatrix}a_2&a_2^2&\cdots&a_2^{n+1}\\\vdots\\a_{n+2}&a_{n+2}^2&\cdots&a_{n+2}^{n+1}\end{pmatrix}=a_2a_3\cdots a_{n+2}\det\begin{pmatrix}1&a_2&\cdots&a_2^n\\\vdots\\1&a_{n+2}&\cdots&a_{n+2}^n\end{pmatrix} $$ and you are done.