# Rational functions and polynomials with infinitely many integer values

Let $f\in \mathbb{C}(x)$ be a rational function. Assume that we have an infinite collection $\{p_n\}_{n\in \mathbb{N}}$ of positive integers such that for every $n$ it holds that $f(p_n)\in\mathbb{N}$. Is it necessarily true that $f$ is a polynomial? What if we assume that $p_n=n$ for every $n$?

• For your second question, that should follow by the generalized Schinzel's hypothesis. – Jason Starr Mar 8 '17 at 13:01
• What is the generalized Schnizel's Hypothesis? – Ehud Meir Mar 8 '17 at 13:03
• Actually, I think this is elementary. Write $f(x) = g(x)/h(x)$, where both $g$ and $h$ have degree $\leq d$. In the $2(d+1)$-dimensional vector space of pairs $(g,h)$ of such polynomials, consider the linear equations $a_nh(p_n)-g(p_n)=0$ for specified $a_n\in \mathbb{Q}$. Since a rational function has at most finitely many zeroes, this collection of countably many linear equations specifies a $1$-dimensional subspace. So finitely many of these linear equations suffice. Since the equations have rational coefficients, this $1$-dimensional space is defined over $\mathbb{Q}$ . . . – Jason Starr Mar 8 '17 at 13:26
• . . . So assume that $g(x)$ and $h(x)$ have rational coefficients. By the division algorithm in $\mathbb{Q}[x]$, $g(x) = h(x)q(x) + r(x)$. Thus, $f(x) = q(x) + (r(x)/h(x))$, where $q(x) = \tilde{q}(x)/N$ for some $\tilde{q}(x)\in \mathbb{Z}[x]$ and $N\in \mathbb{Z}_{>0}$, and where $r(x)$ has degree less than $h(x)$. But then there exists $P>0$ such that for all $x>P$, the absolute value of $r(x)/h(x)$ is less than $1/N$. So, for $p_n>P$, $q(p_n)$ is a fraction with denominator $N$, yet $r(p_n)/h(p_n)$ has absolute value less than $1/N$. Since the sum is an integer, $r(x)$ equals $0$. – Jason Starr Mar 8 '17 at 13:30
• I missed that. Many thanks for this! – Ehud Meir Mar 8 '17 at 13:39