Timeline for Are there any solutions to the diophantine equation $x^n-2y^n=1$ with $x>1$ and $n>2$?
Current License: CC BY-SA 3.0
9 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Dec 13, 2016 at 22:21 | vote | accept | Sophie | ||
| Dec 13, 2016 at 21:35 | answer | added | user102449 | timeline score: 2 | |
| Dec 13, 2016 at 15:19 | answer | added | KConrad | timeline score: 20 | |
| Dec 13, 2016 at 14:55 | comment | added | WhatsUp | For $n = 4$ it is also easy to see that there is no non-trivial solution, by writing the equation as $(x^2 + 1)(x^2-1)=2y^4$. | |
| Dec 13, 2016 at 14:48 | comment | added | WhatsUp | For $n = 3$ this should follow from the fact that the Mordell-Weil group of the curve $x^3 - 2y^3 = 1$ over $\mathbb{Q}$ is of rank $0$. | |
| Dec 13, 2016 at 14:17 | review | First posts | |||
| Dec 13, 2016 at 14:26 | |||||
| Dec 13, 2016 at 14:11 | history | asked | Sophie | CC BY-SA 3.0 |