Given a polynomial $P=a\_3z^3+a\_2z^2+a\_1z+1, z >0$ with non-negative integer coefficients $a\_1, a\_2, a\_3\ne 0$, it appears if $P$ is **not** factorizable then there are **finitely** many positive integers $x, z$ such that $xz+1 \mid P(z)$, $xz+1<P(z)$. If $a\_2=a\_1=0$, the claim is true. [The Diophantine equation $ (xz+1)(yz+1)=az^{3}+1$ has no solutions in positive integers with $z > a^2+2a$](https://mathoverflow.net/questions/386989/the-diophantine-equation-xz1yz1-az31-has-no-solutions-in-positive-i). However the proof for the general case doesn't follow directly from the proof for the case $P=a\_3z^3+1$. Also for a particular triple $(a\_1, a\_2, a\_3)$, what's the minimum value of $z$ such that $xz+1$ is not a proper divisor of $P(z)$ for all $x>1$?


My thoughts:
If $P$ is factorizable then we can find integers $b\_1, b\_2, b\_3$ such that $a\_3z^3+a\_2z^2+a\_1z+1=(b\_1z+1)(b\_2z^2+b\_3z+1)$. Expanding and comparing coefficients we get $b\_1+b\_3=a\_1$, $b\_1b\_3+b\_2=a\_2$, and $b\_1b\_2=a\_3$. Since $P$ is assumed non-factorizable, we will have to use this result somewhere in the proof.



Is $xz+1 $ a proper divisor of $a_3z^3+a_2z^2+a_1z+1$ finitely often?
The conjecture is true. That is, if the integral cubic polynomial
$$P(Z)=a\_3 Z^3+a\_2 Z^2+a\_1 Z+1$$
is irreducible in $\mathbb{Z}[Z]$ (hence also in $\mathbb{Q}[Z]$ by [Gauss's lemma](https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial))), then there are only finitely many positive integer solutions of the equation
$$(xz+1)(yz+1)=P(z).$$


**1.** First we consider the case when $x\mid a\_3$ or $y\mid a\_3$. By symmetry, it suffices to deal with the case $x\mid a\_3$. We fix $x$ for this section. By long division, we get an integral quadratic polynomial $Q\in\mathbb{Z}[Z]$ and a nonzero integer $r\in\mathbb{Z}$ such that
$$a\_3^2 P(Z)=(xZ+1)Q(Z)+r.$$
If $xz+1\mid P(z)$, then $xz+1\mid r$, hence there are finitely many possibilities for $z$ (and also for $y$).


**2.** Now we consider the case when $x\nmid a\_3$ and $y\nmid a\_3$. We rewrite the original equation as
$$tz=x+y-a\_1\qquad\text{where}\qquad t:=a\_3z+a\_2-xy.$$
Here $t$ is an integer. If $t\leq 0$, then $x+y\leq a\_1$, which leads to finitely many triples $(x,y,z)$. So let us focus on the case $t>0$. We use an identity inspired by the OP's earlier post:
\begin{align\*}
(tx-a\_3)(ty-a\_3)&=t^2 xy-a\_3 t(x+y)+a\_3^2\\
&=t^2(a\_3z+a\_2-t)-a\_3 t(tz+a\_1)+a\_3^2\\
&=-t^3+a\_2 t^2-a\_1 a\_3 t+a\_3^2.
\end{align\*}
We conclude that $t\leq 3\max(|a\_1|,|a\_2|,|a\_3|)$, for otherwise the LHS is positive, while the RHS is negative. Moreover, the factors on the LHS are nonzero integers by $x\nmid a\_3$ and $y\nmid a\_3$. So there are finitely many possibilities for the factors on the LHS (namely they are integral divisors of the finitely many possible values of the RHS), hence also for the triple $(x,y,z)$.


