# Rational input, integer output

Standard

Consider the following polynomial equation [source: Berkeley Problems in Mathematics, problem 6.13.10]:

$f(t) = 3t^3 + 10t^2 - 3t$

Let’s try to figure out the rational values of $t$ for which $f(t)$ is an integer. Clearly, if $t\in\mathbb{Z}$ then $f(t)$ is an integer. So let’s consider the case when $t=m/n$ where $\gcd(m,n)=1$ and $m\neq \pm 1$. Substituting this value of $t$ we get:

$\displaystyle{f\left(\frac{m}{n}\right) = \frac{3m^3}{n^3} + \frac{10m^2}{n^2} - \frac{3m}{n}= \frac{m(3m^2+10mn-3n^2)}{n^3}=k \in \mathbb{Z}}$

Since, $n^3\mid (3m^2+10mn-3n^2)$ we conclude that $n\mid 3$. Also it’s clear that $m\mid k$. Hence, $n=\pm 3$ and we just need to find the possible values of $m$.

For $n=3$ we get:

$\displaystyle{f\left(\frac{m}{3}\right) = \frac{m(m^2+10m-9)}{9}=k \in \mathbb{Z}}$

Hence we have $9\mid (m^2+10m)$. Since $\gcd(m,n)=\gcd(m,3)=1$, we have $9\mid (m+10)$, that is, $m\equiv 8\pmod 9$.

Similarly, for $m=-3$ we get $n\equiv 1 \pmod 9$. Hence we conclude that the non-integer values of $t$ which lead to integer output are:

$\displaystyle{t = 3\ell+ \frac{8}{3}, -3\ell-\frac{1}{3}}$ for all $\ell\in\mathbb{Z}$