Timeline for solutions to special diophantine equations
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2014 at 13:41 | vote | accept | mohi | ||
| Oct 12, 2014 at 10:15 | history | edited | Alexey Ustinov | CC BY-SA 3.0 |
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| Oct 12, 2014 at 8:28 | history | edited | Alexey Ustinov | CC BY-SA 3.0 |
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| Oct 12, 2014 at 8:24 | comment | added | Alexey Ustinov | If you need a parametrization only then you can change variables $v\to -v$, $z\to -z$. If you need more then you can use Rogovskaya's observations. | |
| Oct 12, 2014 at 5:41 | comment | added | mohi | $a_1+a_2+a_3=0$ would be correct if $x+y+z=u+v+w$ here we have $x+y+v=u+w+z$ | |
| Oct 12, 2014 at 5:37 | comment | added | mohi | Here $a_1+b_2=b_3$ | |
| Oct 12, 2014 at 5:33 | comment | added | Alexey Ustinov | $a_1+a_2+a_3=0$ is the second (linear) equation of the given system. | |
| Oct 12, 2014 at 5:29 | comment | added | mohi | Thanks but I'm not sure I agree with the parameterization. Please note that $a_1+a_2+a_3\neq 0$ as here $x+y+v=u+w+z$ | |
| Oct 12, 2014 at 3:38 | history | edited | Alexey Ustinov | CC BY-SA 3.0 |
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| Oct 12, 2014 at 3:30 | history | edited | Alexey Ustinov | CC BY-SA 3.0 |
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| Oct 12, 2014 at 3:24 | history | answered | Alexey Ustinov | CC BY-SA 3.0 |