A monic integer-valued polynomial in $\mathbb{Q}[x]$ must be in $\mathbb{Z}[x]$

Question

Is it true that if $f(x) \in \mathbb{Q}[x]$ is a $\textbf{monic}$ polynomial such that $f(k) \in \mathbb{Z}$ for all (sufficiently large) $k\in \mathbb{Z}$ then $f(x) \in \mathbb{Z}[x]$?

I am aware of non-monic counterexamples such as $\binom{x}{n}$ but I cannot see how to proceed in the monic case.

Any help appreciated!

Answer 1

You can turn a non-monic counterexample into a monic counterexample by adding a sufficiently high power of $x$. For example, $\binom{x}{n}$ becomes $x^{n+1}+\binom{x}{n}$.