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A polynomial with integer coefficients maps integers to integers. A rational function (which is a quotient of two polynomials) tends to map integers to rational numbers. Other than the trivial case where the denominator is a factor of the numerator, are there any rational functions that map all integers to integers? What is known about them and how can they be found?

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Let $r(z) = \frac{p(z)}{q(z)}$ such that $r:\mathbb{Z}\rightarrow\mathbb{Z}$ and $p,q$ polynomials. Then $q(z)|p(z)$ for all $z \in \mathbb{Z}$ and $\deg p \geq \deg q$ (check what happens at $\infty$).

You can check here, for example, how the division between polynomials over a commutative ring (like $\mathbb{Z}$) works.

So if $q$ is monic, then you can divide the polynomials, so $p(z)=f(z)q(z)+g(z)$ with $f,g$ polynomials and $\deg g < \deg q$. Then if $g \not \equiv 0$, $r(z)=f(z)+\frac{g(z)}{q(z)}$, so $\frac{g(z)}{q(z)}$ is also an "integer" rational function, which is absurd. So $g\equiv0$ and $r$ is a integer polynomial.

If $q$ is non-monic, you can apply the theorem of the linked answer: call $q_0$ the leading coefficient of $q$, then $q_0^kp(z) = f(z)q(z) + g(z)$. Consider the integer rational function $q_0^kr(z) = \frac{q_0^kp(z)}{q(z)} = f(z)+ \frac{g(z)}{q(z)}$. As above, this implies that $g(z) \equiv 0$. So $r(z)=\frac{p(z)}{q(z)}=\frac{q_0^kp(z)}{q_0^kq(z)}=\frac{f(z)q(z)}{q_0^kq(z)}=\frac{f(z)}{q_0^k}$. Then again $r$ is a integer polynomial.

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