0
$\begingroup$

If $f(x)$ is a polynomial such that if $y \in\mathbb{Z}$, then $f(y)\in\mathbb{Z}$, show that there exist $n$ such that $n!f(x)\in\mathbb{Z}[x]$

$\endgroup$

closed as off-topic by Watson, Joey Zou, Claude Leibovici, JMP, ervx Aug 6 '16 at 13:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Watson, Joey Zou, Claude Leibovici, JMP, ervx
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Welcome to Math SE!!To get better answers to your questions you must try to give some information regarding your background knowledge on the subject,your attempt to solve it and detail your question as much as possible.If you do these,you will get better answers and have a nice time in our site.Thanks!!(From a Review)... $\endgroup$ – tatan Jul 22 '16 at 13:35
  • 1
    $\begingroup$ What's $n$? Obviously if you choose $n$ large enough you can clear all coefficients' denominators. Is perhaps $n$ the degree of your polynomial? $\endgroup$ – Gregory Grant Jul 22 '16 at 13:36
  • $\begingroup$ And where if $f$ taken from? $\Bbb Q [x]$, $\Bbb R[x]$, $\Bbb C[x]$? $\endgroup$ – Alex M. Jul 22 '16 at 13:37
4
$\begingroup$

A well known result is that any integer-valued polynomial is the sum of integer multiples of $\binom{x}{k} =\dfrac{x(x-1)...(x-k+1)}{k!} $.

https://en.wikipedia.org/wiki/Integer-valued_polynomial

By multiplying by the largest $k!$, we get your statement.

$\endgroup$
2
$\begingroup$

Newton's interpolation formula is $$ f(n) = d_0 \binom{n}{0} + d_1 \binom{n}{1} + d_2 \binom{n}{2} + d_3 \binom{n}{3} +\cdots $$ where $d_i$ are the numbers in the first column of the repeated differences array.

This proves that a polynomial takes integral values at integers iff it is an integer linear combination of the binomial polynomials. See Integer-valued polynomial.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.