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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.

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    $\begingroup$ Why do you want to prove that and how do you know it's true? $\endgroup$ May 5, 2021 at 20:03
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    $\begingroup$ Questions in the imperative voice ("Prove that …") are usually not well received; I would suggest editing it. I'm also pretty sure that you mean that there are only finitely 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). $\endgroup$
    – LSpice
    May 5, 2021 at 20:26
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    $\begingroup$ Because it has application to primality testing. If this is true which is most likely the case beyond doubt then if $z$ is prime, we can prove whether $P(z)$ is prime by checking if $b^{P(z) - 1} \equiv 1 ($mod $P(z))$ and $b^{(P(z) - 1) /p} \not\equiv 1 ($mod $P(z)) $. There would be no need to obtain another factor of $P(z) - 1$ required in Pocklington's test $\endgroup$
    – ASP
    May 5, 2021 at 20:28
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    $\begingroup$ It's most likely true because I have examined a number of cases by experiment . Also I already proved the case when $a_2=a_1=0$ which gives me more confidence $\endgroup$
    – ASP
    May 5, 2021 at 20:31
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    $\begingroup$ Perhaps I don't understand what you're asking. You always have $P(z) = x z + 1$ where $x = a_3 z^2 + a_2 z + a_1$. That makes infinitely many solutions. $\endgroup$ May 5, 2021 at 21:03

1 Answer 1

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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), 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 $(x,y,z)$.

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  • $\begingroup$ For the above result to have application in primality testing, we need to explicitly determine a value $z_{min}$ as function of $a_3, a_2, a_1$ so that for all $z >z_{min}$, $xz+1$ is not a proper divisor of $P(z) $. From the last inequality of Case 2 of your proof, all solutions $(x, y, z)$ have $z<a_3^2 +2a_3+a_2m^2 - a_1a_3-1$ where $m=3$max$(|a_1|,|a_2|,|a_3|$). However the upper bound of $z$ in Case 1 is not clear. $\endgroup$
    – ASP
    May 6, 2021 at 12:40
  • $\begingroup$ Currently I am researching on the question : Given a prime $p$ and a positive integer $a$, Is there an efficient way of determining whether $sp+1$ is a proper divisor of $ap+1$ or at least can we find some forms of integers $a$ such that it can be efficiently determined whether $sp+1$ is a proper divisor of $ap+1$. I have found a few such integers $a$, I'll share the idea in another post to see if the idea can be generalized. $\endgroup$
    – ASP
    May 6, 2021 at 12:42
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    $\begingroup$ @DavidJones: In Case 1, the nonzero integer $r$ can be explicitly determined as a polynomial of $(a_1,a_2,a_3,x)$ by running the division algorithm for $a_3^2 P(Z)$ and $xZ+1$. As $x$ is a divisor of $a_3$, we get an explicit upper bound for $r$ in terms of $(a_1,a_2,a_3)$, which is then an upper bound for $z$ as well. $\endgroup$
    – GH from MO
    May 6, 2021 at 14:44

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