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Let $P(z) = z^4 +z^3 +z^2 +z+1$ where $z$ is a positive integer.

While working with the diophantine equation $(xz+1)(yz+1)=P(z)$, I was able to construct a seemingly infinite and complete solution set to the diophantine equation in positive integers $(x, y, z)$.

First, define functions $r$ and $q$ of a positive integer $m$ as follows:

\begin{align} r(m) &= m^2+m-1 \\ q(m) &= (r(m) +2)^2-2 \\ \end{align}

And for a particular positive integer $m$, define sequences $A, B, C, D$ as follows;

\begin{align} A_1 &= m+1\\ A_2 &= m^5 + 3m^4 + 5m^3 + 4m^2 +m\\ A_n &= q(m)A_{n-1} - A_{n-2} - r(m)\\ \end{align}

\begin{align} B_1 &= m^3 + 2m^2 +2m\\ B_2 &= q(m)B_1+r(m)\\ B_n &= q(m)B_{n-1} - B_{n-2} + r(m) \\ \end{align}

\begin{align} C_1 &= m^3 + m^2 +m+1 \\ C_2 &= q(m)C_1-r(m)\\ C_n &= q(m)C_{n-1} - C_{n-2} - r(m)\\ \end{align}

\begin{align} D_1 &= m^5 + 2m^4 + 3m^3 + 3m^2 +m\\ D_2 &= q(m)D_1+ r(m)- m\\ D_n &= q(m)D_{n-1} - D_{n-2} + r(m) \\ \end{align}

It appears all positive integer solutions $(x,y,z)$ are given by $(A_n, A_{n+1} , B_n)$, $(C_n, C_{n+1}, D_n), n = 1, 2, \dots $.

How does one go about proving that $(A_n, A_{n+1} , B_n)$, $(C_n, C_{n+1}, D_n)$ are indeed solutions for all $n$ and showing that these are the only solutions? My approach would be to obtain closed formulae for $A_n, B_n, C_n$ and $ D_n$ then substitute into $(xz+1)(yz+1)=P(z)$ and check if the $LHS=RHS$. However, the closed forms of these sequences are too bulky. And also this method does not prove whether these are the only solutions.

If sequences $A, B, C, D $ cover all solutions in positive integers $x, y, z$ then it becomes much more efficient to determine if $P(z) $ possesses any proper factor of the form $xz+1$ for a particular $z$ using these sequences as compared to trial divisions on $P(z) $ or prime factorizing $P(z) $.

Also, is it possible to construct all solutions for irreducible polynomials $P(z) = z^n+z^{n-1}+ \cdots + z^2+z+1$ of degree $n>4?$

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  • $\begingroup$ Regarding your last question: if $n+1$ is not prime then $z^n + z^{n-1}... +1$ will not be irreducible(and will in fact factor into non-trivial cyclotomic polynomials), so the behavior of general solutions to this will likely be not great. Note also that there's a preprint on Arxiv by Sean Bibby, Pieter Vyncke and myself which looks at a related sort of Diophantine equation arxiv.org/abs/1908.09420 . What we call a $\sigma_{4,1}$ quasisolution there is similar to what you are looking at but with the left hand sides +1s replaced with -1s. Similar techniques may be of use/interest. $\endgroup$
    – JoshuaZ
    Commented Sep 9, 2021 at 10:57
  • $\begingroup$ You should consult the paper of D. Badziahin (Finding special factors of values of polynomials at integer points, Int. J. of Numb. Theor., V. 13(1), 2017), where slightly more general problem was considered, i.e., the existence, for a given polynomial $P$, divisors $d$ of $P(n)$ such that $d\equiv 1\pmod{n}$. $\endgroup$
    – Maciej Ulas
    Commented Sep 9, 2021 at 14:52
  • $\begingroup$ The paper is great. It considers a more general polynomial $P(z) = z ^4 +c_1z^3 +c_2z^2 +c_3z+1$. Just wondering, are there any methods that can decide whether $P(z) $ has a proper divisor $d \equiv 1 $ (mod $z) $ where $P(z) $ is a polynomial of degree $n>4$. If such methods exist, then a faster primality test exists for numbers $P(z) $ (faster than the current $N-1 $ tests). $\endgroup$
    – ASP
    Commented Sep 9, 2021 at 17:35
  • $\begingroup$ arxiv.org/abs/2005.02327 (General purpose primality test, Theorem 2.2) $\endgroup$
    – ASP
    Commented Sep 9, 2021 at 17:44
  • $\begingroup$ @Maciej Ulas, the paper you cited by D. Badziahin has been really very useful in my research. The author has classified all divisors of $P(x) =x^4 +c_1x^3 +c_2x^2 +c_3x+1$ that are congruent to $1$ modulo $x$. Before I attempt to extend the ideas of this paper with my supervisor, are there any other papers building on this research or similar papers that classify divisors of numbers $P(n) $ where $P$ is a polynomial of degree greater than $4$? $\endgroup$
    – ASP
    Commented Oct 19, 2021 at 16:28

2 Answers 2

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@David Jones. Since you are not certain if the parametric solution you provide is a general solution I can confirm that it is not. There is a certain numerical solution to your equation & is given below.

(x,y,z)=(6478,138,945)

Your parametric solution does not satisfy the above numerical solution. Hence there is another parametric solution out there for the above solution. So having said this there is no point looking for a proof.

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    $\begingroup$ There is still need for a proof! The given equation $(xz+1)(yz+1)=z^4+z^3 +z^2+z+1$ is symmetrical in $x$ and $y$ i.e if $(x, y, z) =(a, b, c) $ is a solution then $(b, a, c) $ is also a solution. Therefore, to find all solutions, it suffices to find those with $x \le y$. That said, the parametric solution given above appears to cover all solutions $(x, y, z) $ with $x \le y$. Taking $m=2$ and $n=2$, $(A_2, A_3, B_2) = (138, 6478, 945)$. $\endgroup$
    – ASP
    Commented Sep 15, 2021 at 7:59
  • $\begingroup$ have you found any numerical solutions that do not satisfy the parametric solution with $x \le y$ $\endgroup$
    – ASP
    Commented Sep 16, 2021 at 19:01
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"OP" wants to know of a numerical solution which does not satisfy his parametric solution. It is given below.

$(x,y,z)=(138,49331482518,2609165)$

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  • $\begingroup$ Your solution is obtained by taking $m=137$ and $n=1$; $ \ \ (A_1, A_2, B_1)=(138, 49331482518, 2609165)$ $\endgroup$
    – ASP
    Commented Sep 19, 2021 at 17:57
  • $\begingroup$ @Samuel: As the OP shows, this answer is mistaken. Please delete it. $\endgroup$
    – Alex M.
    Commented Sep 21, 2021 at 7:26

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