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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

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  • $\begingroup$ By $P(z)\equiv1\pmod{z}$ do you mean that $P(0)=1$? $\endgroup$
    – Servaes
    Commented Aug 17, 2021 at 11:24
  • $\begingroup$ Also, what have you tried? What kind of experiment have you done? Have you found any general results? Have you tried (or been able) to reduce the problem to monic or irreducible $P(z)$? $\endgroup$
    – Servaes
    Commented Aug 17, 2021 at 12:09
  • $\begingroup$ By $P(z) \equiv 1$ (mod $\ z) $, I mean $P(z)$ is of the form $a_nz^n+a_{n-1}z^{n-1}+\cdots + a_1z+1$ $\endgroup$
    – ASP
    Commented Aug 17, 2021 at 12:54
  • $\begingroup$ Finally got some results $\endgroup$
    – ASP
    Commented Sep 1, 2021 at 19:17
  • $\begingroup$ 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 $\endgroup$
    – ASP
    Commented Sep 2, 2021 at 16:25

6 Answers 6

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

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  • $\begingroup$ The proof is simple, direct and clear. There's still a lot to find out. The family of solutions proven does not exhaust all solutions. Is it possible to construct all solutions for $z$ using some formulae? Can this proof be adapted for at least all polynomials of the form $P(z) =z^n + z^{n-1}+ \cdots + z^2+z+1$ i.e with all coefficients equal to 1? $\endgroup$
    – ASP
    Commented Sep 2, 2021 at 20:21
  • $\begingroup$ I don't know. You might try degrees 6 and 10 as a prime minus 1 (degree) should be, well, somewhat predictable; irreducible for one thing. $\endgroup$
    – Will Jagy
    Commented Sep 2, 2021 at 20:31
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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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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.

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  • $\begingroup$ for each of your $x,y,z,$ taking three constants, say $a_z,b_z,c_z$ we find $$ z_n = a_z (\frac{7 + 3 \sqrt 5}{2})^n + b_z (\frac{7 - 3 \sqrt 5}{2})^n + c_z $$ where $c_z$ should be rational, most likely $\pm1/5.$ Same for $x,y$ with different $a,b,c$ $\endgroup$
    – Will Jagy
    Commented Sep 2, 2021 at 19:09
  • $\begingroup$ Let me use this closed form to prove that $(x_n, y_n, z_n) $ is always a solution. $\endgroup$
    – ASP
    Commented Sep 2, 2021 at 20:11
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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

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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 $.

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  • 1
    $\begingroup$ turns out $C= (r+2)^2 - 2.$ Also $X_n = C X_{n-1} - X_{n-2} - r $ $\endgroup$
    – Will Jagy
    Commented Sep 5, 2021 at 3:08
  • $\begingroup$ Also $r = n^2 + n - 1, n = 1, 2, 3, \cdots $ $\endgroup$
    – ASP
    Commented Sep 5, 2021 at 9:18
  • $\begingroup$ Every solution in $z$ is a term of a sequence of the form $Z_n = CZ_{n-1} - Z_{n-2}+r$ $\endgroup$
    – ASP
    Commented Sep 5, 2021 at 17:30
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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.

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  • $\begingroup$ If all solutions are given by the sequences A, B, C, E, F, G then its actually possible to reduce the Diophantine equation $(xz+1)(yz+1)=z^4 +z^3 +z^2 +z+1 $ from three unknowns $x, y, z$ to two variables $m, n, $ $\endgroup$
    – ASP
    Commented Sep 7, 2021 at 6:51
  • $\begingroup$ fixed a few typos in the answer $\endgroup$
    – ASP
    Commented Sep 7, 2021 at 9:50
  • $\begingroup$ anyone verified these solutions? $\endgroup$
    – ASP
    Commented Sep 7, 2021 at 17:57

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