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 $z > a^2+2a$. 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.

onlyfinitely many such integers. \\ Also, TeX note: using`|`

with manual spacing doesn't work well; prefer`\mid`

. Compare, for example, the spacing of $a \not | \ b$`a \not | \ b`

to $a \nmid b$`a \nmid b`

. I have edited accordingly (but not to make the other changes I suggested). $\endgroup$8more comments