# Rational function over Q

Assume that $f$ is a rational function with rational coefficients. Suppose moreover that $f$ takes integer values at the integers. Then in fact $f$ is a polynomial with rational coefficients.

## 2 thoughts on “Rational function over Q”

1. nachodarago says:

The main ingredient to prove this is the following:
Let $F\in\mathbb{Z}[X]$ a polynomial with integer coefficients and consider the set $S = \{p\text{ prime such that } p\mid F(n)\text{ for some }n\in\mathbb{Z}\}$. Then $S$ is an infinite set.

The proof of this fact is easy. An sketch of a proof is the following: suppose $S$ is finite and enumerate $p_1,\ldots,p_r$ its elements. Consider $F((p_1\cdots p_r)^\ell)$ for sufficiently large $\ell$ such that $F((p_1\cdots p_r)^\ell)$ is greater than the independent coefficient of $F$ and also greater the valuation of each prime $p_i$ of the independent coefficient. This will give the desired contradiction.

Now, suppose $f$ is a rational function which is not a polynomial, then there exist $g,h\in\mathbb{Q}[X]$ such that $f=g/h$. Moreover, we can suppose that $g,h$ are coprime and thus by Euclid algorithm, there are polynomials $A,B\in\mathbb{Q}[X]$ such that $Ag+Bh=1$. By clearing denominators we obtain polynomials $\tilde{A},\tilde{B},\tilde{g},\tilde{h}\in\mathbb{Z}[X]$ such that $\tilde{A}\tilde{g}+\tilde{B}\tilde{h}=m$ for a certain integer $m$ and $\tilde{g}/\tilde{h}$ is an integral multiple of $g/h$. Since $h(n)\mid g(n)$ for every $n\in\mathbb{Z}$ we must have that every prime divisor of $\tilde{h}(n)$ divides $m$. But we saw that this set must be infinite. This is a contradiction and we are done.

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2. Let $r= f/g$ be a rational function and suppose that it takes integer values and $g$ has degree greater than $f$. Then $r=0$. Indeed, $f/g\to 0$, so $f$ must have infinitely many roots, and $f=0$. Taking now $f/g$ we can assume $g$ has degree at most that of $f$. Writing $f/g = A + S/g$ with $A,S$ rational polynomials and $S$ of degree less than $g$, we can assume by multiplying by a common denominator that $A$ is a polynomial in integers. Then $S/g$ takes integral values at the integers, and we conclude that it is zero by the above.

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