Are there infinitely many positive integer solutions to $(xz+1)(yz+1)=P(z)$?

Question

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

Answer 1

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 $z^4 + z^3 + z^2 + z + 1.$

Answer 2

I figured out, eventually, the bottleneck in my program, now it is quick enough to investigate $z$ bigger then 10,000.

This includes the smallest $z$ that has two pair of your divisors, that being $z = 11660 \; , \; \; \; $ with $P(z) = 31 \cdot 41 \cdot 61 \cdot 211 \cdot 4591 \cdot 246131 $

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 z: 5  x:  2 y:  14  p(z) =781  xz+1:  11
 z: 10  x:  4 y:  27  p(z) =11111  xz+1:  41
 z: 20  x:  3 y:  138  p(z) =168421  xz+1:  61
 z: 36  x:  14 y:  95  p(z) =1727605  xz+1:  505
 z: 51  x:  4 y:  660  p(z) =6900505  xz+1:  205
 z: 70  x:  27 y:  184  p(z) =24357971  xz+1:  1891
 z: 102  x:  15 y:  700  p(z) =109314931  xz+1:  1531
 z: 104  x:  5 y:  2180  p(z) =118121641  xz+1:  521
 z: 185  x:  6 y:  5730  p(z) =1177716661  xz+1:  1111
 z: 248  x:  95 y:  650  p(z) =3798056761  xz+1:  23561
 z: 300  x:  7 y:  12894  p(z) =8127090301  xz+1:  2101
 z: 455  x:  8 y:  25928  p(z) =42953754481  xz+1:  3641
 z: 481  x:  184 y:  1260  p(z) =53639428805  xz+1:  88505
 z: 516  x:  40 y:  6669  p(z) =71029912405  xz+1:  20641
 z: 656  x:  9 y:  47880  p(z) =185471804305  xz+1:  5905
 z: 909  x:  10 y:  82710  p(z) =683492207581  xz+1:  9091
 z: 945  x:  138 y:  6478  p(z) =798338453221  xz+1:  130411 
 z: 1220  x:  11 y:  135410  p(z) =2217151897621  xz+1:  13421
 z: 1595  x:  12 y:  212124  p(z) =6476123466121  xz+1:  19141
 z: 1701  x:  650 y:  4454  p(z) =8376693917005  xz+1:  1105651
 z: 1780  x:  85 y:  37296  p(z) =10044401482181  xz+1:  151301 
 z: 2040  x:  13 y:  320268  p(z) =17327408387641  xz+1:  26521
 z: 2561  x:  14 y:  468650  p(z) =43033624601605  xz+1:  35855 
 z: 3164  x:  15 y:  667590  p(z) =100249723211821  xz+1:  47461
 z: 3298  x:  1260 y:  8635  p(z) =118340747834111  xz+1:  4155481
 z: 3855  x:  16 y:  929040  p(z) =220907368166881  xz+1:  61681 
 z: 4640  x:  17 y:  1266704  p(z) =463623595038241  xz+1:  78881
 z: 4797  x:  700 y:  32880  p(z) =529626147427261  xz+1:  3357901
 z: 4830  x:  156 y:  149575  p(z) =544350277130731  xz+1:  753481  
 z: 5525  x:  18 y:  1696158  p(z) =931982466249901  xz+1:  99451  
 z: 6516  x:  19 y:  2234970  p(z) =1802980203022405  xz+1:  123805  
 z: 7619  x:  20 y:  2902820  p(z) =3370147427418361  xz+1:  152381  
 z: 8528  x:  660 y:  110205  p(z) =5289805397731921  xz+1:  5628481
 z: 8840  x:  21 y:  3721620  p(z) =6107425684618441  xz+1:  185641 
 z: 10185  x:  22 y:  4715634  p(z) =10761846073176661  xz+1:  224071
 z: 11130  x:  259 y:  478332  p(z) =15346865227395031  xz+1:  2882671
 z: 11660  x:  23 y:  5911598  p(z) =18485510549623261  xz+1:  268181
 z: 11660  x:  4454 y:  30527  p(z) =18485510549623261  xz+1:  51933641
 z: 13271  x:  24 y:  7338840  p(z) =31020394955386705  xz+1:  318505

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Answer 3

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.

Answer 4

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

Answer 5

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 $.

Answer 6

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.