Timeline for What is the smallest unsolved Diophantine equation?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 20, 2022 at 4:01 | comment | added | A. J. | For the equation x^2 y = y^2 + z^2 + 1, x^2 = z^2 + 2 and y=z^2 + 1. I doubt that there are integer solutions to this problem. | |
| Aug 31, 2020 at 18:12 | comment | added | Victor Ostrik | @JoshuaZ Assume that $y=4k+1$ and $x^2-y=4l+1$. Then $x^2=4k+4l+2$ which is impossible. Similarly assuming $y=4k+1$ and $x^2-y=2(4l+1)$ or vice versa we get that $x^2$ is congruent to 3 modulo 4. | |
| Aug 31, 2020 at 17:59 | comment | added | JoshuaZ | @VictorOstrik Sorry, how are you getting that $y(x^2-y)$ must have a 3 mod 4 factor? | |
| Aug 31, 2020 at 17:07 | comment | added | Victor Ostrik | Equation $x^2y=y^2+z^2+1$ has no integral solutions: rewrite it as $y(x^2-y)=z^2+1$. Thus we want to factorize $z^2+1$ into two factors sum of which is a square (in particular, both factors are positive). But all prime factors of $z^2+1$ are of the form $4k+1$ or 2 appearing with exponent at most 1, so all positive factors are congruent to 1 or 2 modulo 4. This is a contradiction. | |
| Aug 31, 2020 at 13:46 | history | answered | Bogdan | CC BY-SA 4.0 |