Timeline for Awfully sophisticated proof for simple facts
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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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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| Mar 4, 2012 at 12:37 | comment | added | Zsbán Ambrus | If such a proof works for n = 4, then it's a better answer for this question than the n = 3 one, because the simplest proof for n = 4 is much simpler than the simplest proof for n = 3. | |
| Jun 5, 2011 at 17:07 | comment | added | James Weigandt | @paul Monsky: Yes, but the curve $x^4 + y^4 = z^2$ is an elliptic of conductor 32. | |
| Oct 18, 2010 at 16:39 | comment | added | Cam McLeman | @Pete: Nice point. It's like arguing that sending email is frivolous overkill since a carrier pigeon could do the same job with much less technology. | |
| Oct 18, 2010 at 4:41 | comment | added | paul Monsky | @adrian But isn't the Jacobian of the Fermat quartic isogenous to a product of 3 elliptic curves, each of analytic rank 0? | |
| Oct 18, 2010 at 1:48 | comment | added | Adrian Barquero-Sanchez | @muad Are you sure something similar works for $n = 4$? I mean, the genus of the Fermat curve $x^n + y^n = z^n$ is $\frac{(n-1)(n-2)}{2}$ so when $n = 4$ the genus of the curve is 3. | |
| Oct 17, 2010 at 18:03 | comment | added | Pete L. Clark | This is actually a nice answer, because it treats $x^3 + y^3 = z^3$ like what it is -- a rational elliptic curve -- and proceeds to find all rational points on it in the way which is easiest given the current level of technology. | |
| Oct 17, 2010 at 16:32 | comment | added | Martin Brandenburg | 1+ for "Math Underflow" (and your answer). | |
| Oct 17, 2010 at 15:44 | history | answered | muad | CC BY-SA 2.5 |