Timeline for Nontrivial solutions for $\sum x_i = \sum x_i^3 = 0$
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2015 at 8:41 | history | edited | Alexey Ustinov |
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| Oct 24, 2015 at 6:08 | answer | added | Alexey Ustinov | timeline score: 3 | |
| Apr 21, 2015 at 3:49 | comment | added | Noam D. Elkies | $N=4$ is not possible even over $\bf C$. The 1st and 3rd elementary symmetric functions of the $x_i$ would vanish, so they would be roots of a polynomial $x^4 + s_2 x^2 + s_4 = 0$, and would thus split into two $\{x,-x\}$ pairs. | |
| Apr 21, 2015 at 0:16 | vote | accept | CommunityBot | ||
| Apr 21, 2015 at 0:11 | answer | added | Michael Renardy | timeline score: 6 | |
| Apr 21, 2015 at 0:08 | answer | added | R.P. | timeline score: 14 | |
| Apr 20, 2015 at 23:52 | comment | added | Zack Wolske | Then the minimum is $N=0$. | |
| Apr 20, 2015 at 23:49 | comment | added | user22139 | not sure I understand. $\{x, -x\}$ is a solution to the system with $N=2$. That's the kind of solutions I don't want. | |
| Apr 20, 2015 at 23:49 | answer | added | Gerry Myerson | timeline score: 12 | |
| Apr 20, 2015 at 23:48 | comment | added | Zack Wolske | The "fundamental" condition is not necessary. A minimal solution won't include both $x$ and $-x$. | |
| Apr 20, 2015 at 23:46 | history | edited | user22139 | CC BY-SA 3.0 |
deleted 4 characters in body
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| Apr 20, 2015 at 23:46 | comment | added | user22139 | I'll edit to address it, thanks. To the question you ask, yes, intended meaning is "if $x$ belongs to the set, then $-x$ does not" | |
| Apr 20, 2015 at 23:45 | comment | added | Yemon Choi | Minor nitpick: the double use of $i$ in the line after the displayed equation really makes me wince, and the formulation looks odd: are you sure you mean to say "if $x_i$ belongs to the set $\{x_1,\dots, x_n\}$, then ... " ? | |
| Apr 20, 2015 at 23:42 | history | asked | user22139 | CC BY-SA 3.0 |