Timeline for What are the solutions of this Diophantine equation?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 5, 2019 at 22:09 | comment | added | Gerry Myerson | No, McEonald, I'm just providing an alternative form that might, or might not, be of any use. | |
| Sep 5, 2019 at 18:55 | history | became hot network question | |||
| Sep 5, 2019 at 17:30 | comment | added | Emil Jeřábek | @SteveHuntsman According to the paper linked by user44191, any other solution must have $|x|,|y|,|z|>\frac1310^{16}$. | |
| Sep 5, 2019 at 17:08 | answer | added | user44191 | timeline score: 17 | |
| Sep 5, 2019 at 17:03 | comment | added | user44191 | $(x + \frac{1}{3})^3 + (y + \frac{1}{3})^3 + (z + \frac{1}{3})^3 = \frac{1}{9}$ may also be useful. Or $(3x + 1)^3 + (3y + 1)^3 + (3z + 1)^3 = 3$. | |
| Sep 5, 2019 at 13:39 | comment | added | Steve Huntsman | According to MATLAB, not in $\{-100,\dots,100\}^3$. | |
| Sep 5, 2019 at 13:29 | comment | added | McRonald | @GerryMyerson, by ''also known as'', are you saying the question is a known open problem that can be expressed in the form that you wrote ? If yes, what is the name of the conjecture ? Any references ? | |
| Sep 5, 2019 at 11:38 | comment | added | Gerry Myerson | Also known as $x(x+1)^3+y(y+1)^3+z(z+1)^3=x^4+y^4+z^4$. | |
| Sep 5, 2019 at 9:50 | review | Close votes | |||
| Sep 6, 2019 at 20:27 | |||||
| Sep 5, 2019 at 9:00 | review | First posts | |||
| Sep 5, 2019 at 10:09 | |||||
| Sep 5, 2019 at 8:59 | history | asked | McRonald | CC BY-SA 4.0 |