Timeline for Positive integer solutions to the diophantine equation $(xz+1)(yz+1)=z^4+z^3 +z^2 +z+1$
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 16, 2021 at 19:01 | comment | added | ASP | have you found any numerical solutions that do not satisfy the parametric solution with $x \le y$ | |
| Sep 15, 2021 at 7:59 | comment | added | ASP | 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 $n=2$, $(A_2, A_3, B_2) = (138, 6478, 945)$. | |
| S Sep 13, 2021 at 22:11 | review | First answers | |||
| Sep 13, 2021 at 23:20 | |||||
| S Sep 13, 2021 at 22:11 | history | answered | Jamie | CC BY-SA 4.0 |