Let $f, g \in \mathbb{Z}[x]$ be coprime polynomials. I am interested in an upper bound for $$ N(B) = \# \{ x \in [-B, B] \cap \mathbb{Z}: f(x)\mid g(x) \}. $$ I assume there must be something known about this quantity... If someone could provide me a reference it would be appreciated. Thank you
ps I assume $\deg f > 0$.
The assumption that $f(X)$ and $g(X)$ are relatively prime means that there is a positive integer $R_{f,g}=\operatorname{Resultant}(f,g)$ and polynomials $a(X)$ and $b(X)$ in $\mathbb Z[X]$ so that $$ a(X)f(X) + b(X)g(X) = R_{f,g}. $$ Hence if $x\in\mathbb Z$ satisfies $f(x)\mid g(x)$, then $f(x)\mid R$. Since $\bigl|f(x)\bigr|\to\infty$ as $|x|\to\infty$, this shows that your quantity $N(B)$ is bounded as $B\to\infty$. A relatively trivial bound is simply $$ N(B) \le \#\Bigl\{x\in\mathbb Z : \bigl| f(x) \bigr| \le R_{f,g} \Bigr\}. $$ One can bound $R_{f,g}$ in terms of the coefficients of $f$ and $g$ and their degrees, if you want a more explicit bound.
The condition $f(x) | g(x), x \in \mathbb{Z}$ can be interpreted as the equation
$$\displaystyle g(x) = yf(x), x,y \in \mathbb{Z}.$$
This evidently defines an algebraic plane curve, of degree $d = \max\{\deg f, \deg g\}$. One can apply the determinant method to get a good bound which is best possible in general. The version due to Bombieri and Pila does not seem to work well with skew boxes if I recall, but the affine version due to Heath-Brown does work. In this case, we suppose that $|x| \asymp B$ and hence $|y| \ll \frac{|g(x)|}{|f(x)|} \ll B^{\deg g - \deg f} = B^k$, say. We may assume that $k \geq 1$, otherwise the result is easier. Put $\deg g = d_1, \deg f = d_2$, and
$$\displaystyle V = \exp \left(\frac{(\log B)(\log B^{k})}{\log(B^{d_1})} \right) = B^{\frac{k}{d_1}}.$$
Then it is a theorem of Heath-Brown (see equation (3) in Browning's paper: Power-free values of polynomials) that the number of solutions is bounded by
$$\displaystyle O_\varepsilon \left(B^{\frac{k}{d_1} + \varepsilon} \right).$$
Evidently the exponent is less than one for $\varepsilon$ sufficiently small.
