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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} - F_{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 $n>4?$

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 $n>4?$

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 $A,B,E,F$ 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} E_1 &= m^3 + m^2 +m+1 \\ E_2 &= q(m)E_1-r(m)\\ E_n &= q(m)E_{n-1} - E_{n-2} - r(m) \end{align}

\begin{align} F_1 &= m^5 + 2m^4 + 3m^3 + 3m^2 +m\\ F_2 &= q(m)F_1+ r(m)- m\\ F_n &= q(m)F_{n-1} - F_{n-2} + r(m) \\ \end{align}

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

How does one go about proving that $(A_n, A_{n+1} , B_n)$, $(E_n, E_{n+1}, F_n)$ are indeed solutions for all $n$ and showing that these are the only solutions? My approach would be to obtain closed formulae for $A_n, B_n, E_n$ and $ F_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, E, F $ 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 $n>4?$

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} - F_{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 $n>4?$

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, we define some functions of a positive integer $m$ as follows:

\begin{align} r(m) &= m^2+m-1 \\ q(m) &= (r(m) +2)^2-2 \\ a(m) &= m+1 \\ b(m) & = m^5 + 3m^4 + 5m^3 + 4m^2 +m\\ c(m) &= m^3 + 2m^2 +2m \\ e(m) &= m^3 + m^2 +m+1 \\ f(m) &= m^5 + 2m^4 + 3m^3 + 3m^2 +m\\ \end{align}

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

\begin{align} A_1 &= a(m)\\ A_2 &= b(m)\\ A_n &= q(m)A_{n-1} - A_{n-2} - r(m)\\ \end{align}

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

\begin{align} E_1 &= e(m)\\ E_2 &= q(m)E_1-r(m)\\ E_n &= q(m)E_{n-1} - E_{n-2} - r(m) \end{align}

\begin{align} F_1 &= f(m)\\ F_2 &= q(m)F_1+ r(m)- m\\ F_n &= q(m)F_{n-1} - F_{n-2} + r(m) \\ \end{align}

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

How does one go about proving that $(A_n, A_{n+1} , B_n)$, $(E_n, E_{n+1}, F_n)$ are indeed solutions for all $n$ and showing that these are the only solutions? My approach would be to obtain closed formulae for $A_n, B_n, E_n$ and $ F_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, E, F $ 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 $n>4?$

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 $A,B,E,F$ 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} E_1 &= m^3 + m^2 +m+1 \\ E_2 &= q(m)E_1-r(m)\\ E_n &= q(m)E_{n-1} - E_{n-2} - r(m) \end{align}

\begin{align} F_1 &= m^5 + 2m^4 + 3m^3 + 3m^2 +m\\ F_2 &= q(m)F_1+ r(m)- m\\ F_n &= q(m)F_{n-1} - F_{n-2} + r(m) \\ \end{align}

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

How does one go about proving that $(A_n, A_{n+1} , B_n)$, $(E_n, E_{n+1}, F_n)$ are indeed solutions for all $n$ and showing that these are the only solutions? My approach would be to obtain closed formulae for $A_n, B_n, E_n$ and $ F_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, E, F $ 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 $n>4?$