Ideas and methods in deciding solvability of rational expression equals integer I would like to know if there are results concerning the solvability - or even the solution - of equations of the form
$$
R(t)=z,
$$
where $t$ and $z$ are both unknown, $t\in \mathbb{Q}$, $z\in\mathbb{Z}$ and $R(t)$ is rational function in the variable $t$. For example,
$$
\dfrac{t+4}{t^2-3t+8}=z.
$$
Of course, I am not interested in the solution of this special case, but looking for methods or ideas which can be used if one comes across a similar equation. If there are no general results (like if this and that are true, then there cannot be a solution, etc...), then what can be said if we have more assumptions on $R(t)$, like
$$
R(t)=\dfrac{p(t)}{q(t)},
$$
where $\deg p$ and $\deg q$ are bounded or if we suppose that $t$ is also an integer, etc..? Any comment, answer will be welcomed!
 A: Okay, so for me it seems I found a partial answer. As a rational function of $t$ we can always write
$$
R(t)=\dfrac{p(t)}{q(t)},
$$
where $q(t)$ is nonzero. We can also suppose that $p(t)$ and $q(t)$ have integer coefficients. If not, then we can mutliply both by the less common multiple of the denominators in all of their coefficients. This does not change the range of $R(t)$ but gives integer coefficients. Now, if $\deg p(t)\geq \deg q(t)$ then we can use division with remainder and rewrite $R(t)$ a
$$
R(t)=g(t)+\dfrac{r(t)}{q(t)}.
$$
Now, if $t$ is an integer, then $g(t)$ is an integer. Hence, it remains to show that when $r(t)/q(t)$ can be integer. We know that $\deg q>\deg r$, that is, if $|t|$ is great enough, we have
$$
|q(t)|>|r(t)|\implies R(t)\not\in\mathbb{Z}.
$$
Thus, we have only finitely many possibilities for $t$ and in specific cases one can check them all.
It still remains open question for me what to do with the general case, when $t$ is a rational but not an integer.
