Timeline for The elliptic curve for $x_1^9+x_2^9+\dots+x_6^9 = y_1^9+y_2^9+\dots+y_6^9$
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
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| Aug 19, 2023 at 10:39 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Trimmed post for brevity
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 10, 2015 at 7:07 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
added 6 characters in body
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| Nov 10, 2015 at 6:53 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Added incidental fact.
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| Nov 10, 2015 at 6:16 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Better flow.
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| Nov 10, 2015 at 6:06 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
More relevant details.
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| Nov 10, 2015 at 6:01 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
More relevant details.
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| Jan 18, 2015 at 18:41 | vote | accept | Tito Piezas III | ||
| Jan 18, 2015 at 18:38 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
More details.
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| Jan 14, 2015 at 13:30 | comment | added | individ | System $(1),(2)$ - has a simple solution. artofproblemsolving.com/blog/109106 | |
| Jan 14, 2015 at 13:11 | answer | added | Allan MacLeod | timeline score: 9 | |
| Jan 14, 2015 at 4:32 | comment | added | individ | I think it is necessary to substitute $(1)$ in $(2),(3)$ . Then you need to reformulate the problem as the solution to this system of 2 equations. If you want to make it so, and then we will solve the system. | |
| Jan 14, 2015 at 0:46 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
More details.
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| Jan 13, 2015 at 19:37 | comment | added | Jesper Petersen | If I am understanding the link between the diophantine equation of the title and the family of curves $D$ correctly, then while there is an abundance of curves of zero rank, it might be of interest to look for a subfamily of $D$ of higher rank. | |
| Jan 13, 2015 at 18:35 | comment | added | Tito Piezas III | @JeremyRouse: I think you mean $x = 0, y = \pm c_2$. I cobbled up some basic Mathematica code and found that $$n =1/2,\, 1/8,\, 5/3,\, 5/4,\, 5/13,\, 7/9,\, 11/2,\, 11/16,\,17/9\,\dots$$ will do. Quite a lot, it seems | |
| Jan 13, 2015 at 17:56 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Clarify.
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| Jan 13, 2015 at 17:52 | comment | added | Jeremy Rouse | I didn't find any other than the fairly silly $x = 0$, $y = \pm c_{1}$. I think it's quite likely that your elliptic surface has rank zero, given that it has many rank zero specializations. | |
| Jan 13, 2015 at 17:36 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Clarify.
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| Jan 13, 2015 at 17:27 | comment | added | Tito Piezas III | @JeremyRouse: Yes, did you find some? | |
| Jan 13, 2015 at 17:04 | comment | added | Jeremy Rouse | Would solutions where $x$ and $y$ are rational functions in $n$ be admissible? | |
| Jan 13, 2015 at 16:12 | comment | added | Tito Piezas III | @LaurentMoret-Bailly: A polynomial solution to $(1)$ is simply to express $x,y$ as polynomials $p_1(n), p_2(n)$ in terms of $n$. Or express $x,y,n$ as polynomials $p_1(m),p_2(m),p_3(m)$ in terms of some variable $m$. For example, the simplest is $x = -\frac{4(n^3-9n)}{7(n^2+3)}$ but this makes $y=0$. | |
| Jan 13, 2015 at 8:12 | comment | added | GH from MO | I don't understand the question either. Please clarify. | |
| Jan 13, 2015 at 7:26 | comment | added | Laurent Moret-Bailly | I don't understand the question, but the upvotes suggest I may be dumb. | |
| Jan 13, 2015 at 1:31 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Stream-lined.
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| Jan 13, 2015 at 1:19 | history | edited | Tito Piezas III | CC BY-SA 3.0 |
Details.
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| Jan 13, 2015 at 1:08 | history | asked | Tito Piezas III | CC BY-SA 3.0 |