Are there infinitely many positive integer solutions to $(xz+1)(yz+1)=P(z)$? Let $P(z)\equiv 1($ mod $ \ z) $ be a polynomial of degree $n>3$ with integer coefficients. Are there infinitely many positive integers $x, y, z$ such that $(xz+1)(yz+1)=P(z)$?
If $P(z) = a_nz^n+1$, it has be proven that the Diophantine equation has infinitely many solutions in positive integers $x, y, z$ Prove that the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many solutions in positive integers..
From experiment, it appears the assertion is true for all polynomials $P(z)\equiv 1$(mod$ \ z)$ of degree $n>3$. How do we go about proving this?
Note if $n=3$, it has been shown that the Diophantine equation has a finite number of solutions
https://mathoverflow.net/questions/392002/is-xz1-a-proper-divisor-of-a-3z3a-2z2a-1z1-finitely-often/392018#392018
 A: your first recipe, 5,20, 51, 104..  works
$$ z = n^3 + 2n^2 + 2n   $$
$$x = n+1 $$
$$ 1+xz = n^2 + 3 n^3 + 4 n^2 +2n+1 $$
$$ y = n^5 +3n^4 +5n^3 + 4 n^2 + 1 $$
$$  1 + yz = n^8 +5n^7 +13n^6 +20n^5 + 19n^4 + 10n^3 + 2 n^2 + 1$$
$$  ( 1+xz)(1+yz) = n^{12} + 8n^{11} + 32n^{10} + 81n^9 + 142n^8 + 178n^7 + 161n^6 + 104n^5 + 48n^4 + 17n^3 + 6n^2 + 2n + 1 $$
which is the same as $z^4 + z^3 + z^2 + z + 1.$
A: A partial answer :
Let $P(z) = z^4 + z ^3+z^2+z+1$.
Define three sequences as follows ;
$x_1 = 4$, $x_2=27$,
$x_n=7x_{n-1}-x_{n-2}-1$
$y_1 = x_2$, $y_2=x_3$,
$y_n=7y_{n-1}-y_{n-2}-1$
$z_1 = 10$, $z_2=70$,
$z_n=7z_{n-1}-z_{n-2}+1$
For all $n<10^5$, I have verified that $(x_n, y_n, z_n)$ is a solution of the Diophantine equation $(xz+1)(yz+1)=P(z)$.
A natural question: Is $(x_n, y_n, z_n) $ always a solution of the given equation?
This partial solution could serve as a starting point to proving the assertion in the question.
A: From a maple output,  the terms of the sequence $Z= 5, 20, 51, 104, 185, 300, 455, 656,909$ are all solutions in $z$ for the Diophantine equation $(xz+1)(yz+1)=P(z)$ with $P(z) =z^4 + z^3 +z^2 +z+1$.
Computing the difference table for the sequence $Z$, the differences converge to 6 at the third level.
It appears every term generated by extending sequence $Z$ using the difference table is a solution to the given Diophantine equation
A: A general observation:
Let $P(z) = z^4 +z^3+z^2+z+1$. Suppose $(x, y, z) =(x_1,y_1,z_1), (x_2,y_2,z_2), (x_3,y_3,z_3)$
are distinct positive integer solutions to the Diophantine equation $(xz+1)(yz+1)=P(z)$ with $x_2=y_1$ and $x_3=y_2$.  Let $C$ be the ceiling of $z_3/z_2$ and $r = z_3 - Cz_2 +z_1$. Define a sequence $Z$ as follows:
$Z_1=z_1$, $ Z_2=z_2$, $Z_n= CZ_{n-1}-Z_{n-2}+r$,  $n \ge 3$.
It appears $Z_n$ is a solution in $z$ of the Diophantine equation for all $n \ge 3 $.
A: I figured out all solutions in positive integers $\ x, y , z \ $  to $(xz+1)(yz+1)=z^4+z^3+z^2+z+1$.
First, define some sequences as follows:
$r_m = m^2+m-1, \ \ \ m=1, 2, \ldots $
$q_m = (r_m+2)^2-2,\ \ \ m=1, 2, \ldots $
$a_m = m+1, \ \ \ m=1,2,\ldots $
$b_m = m^5 + 3m^4 + 5m^3 + 4m^2 +m, \ \ \ m=1,2,\ldots $
$c_m = m^3 + 2m^2 +2m, \ \ \ m=1,2,\ldots $
$e_m = m^3 + m^2 +m+1, \ \ \ m=1,2,\ldots $
$g_m = m^5 + 2m^4 + 3m^3 + 3m^2 +m, \ \ \ m=1,2,\ldots $
And for a particular $m$, define sequences $A,B,C,E,F,G$ as follows;
$A_1 = a_m$,  $A_2 = b_m$, $A_n = q_mA_{n-1} - A_{n-2} - r_m, \ \ \ n = 3, 4 , \dots$
$B_n = A_{n+1}, \ \ \ n = 1, 2, \ldots $
$C_1 = c_m$,  $C_2 = q_mC_1+r_m$, $C_n = q_mC_{n-1} - C_{n-2} + r_m, \ \ \ n = 3, 4 , \dots$
$E_1 = e_m$,  $E_2 = q_mE_1-r_m$, $E_n = q_mE_{n-1} - E_{n-2} - r_m, \ \ \ n = 3, 4 , \dots$
$F_n = E_{n+1}, \ \ \ n = 1, 2, \ldots $
$G_1 = g_m$,  $G_2 = q_mG_1+ r_m - m$, $G_n = q_mG_{n-1} - G_{n-2} + r_m, \ \ \ n = 3, 4 , \dots$
All positive integer solutions $(x,y,z)$ are given by $(A_n, B_n, C_n)$, $(E_n, F_n, G_n), n = 1, 2, \dots $.
NB. This is simply an observation. A proof is required to show that $(A_n, B_n, C_n)$, $(E_n, F_n, G_n)$ are indeed solutions for every $n$ and that these are the only solutions in positive integers.
