Timeline for Does this quadratic form over a large field represent 1?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 1, 2013 at 15:02 | comment | added | Noam D. Elkies | Once you have one solution $(v_1,v_2,v_3,v_4)$, you can parametrize the others by the line joining any other solution $(w_1,w_2,w_3,w_4)$ to $(v_1,v_2,v_3,v_4)$. Often in such problems this gets you from a trivial or spurious solution to one that's relevant to your intended application. For starters it will probably give you a better handle on whether "useful" solutions exist. | |
| Sep 1, 2013 at 13:15 | comment | added | Neil Strickland | Indeed, this gives a solution, thank you. Unfortunately I learn from it that my formulation of the problem was not optimal. Your solution gives $u_2=2/y_1$, and the denominator of $y_1$ prevents it from being useful for the intended application. I will have to think in more detail about which denominators would be harmless. Anyway, this at least gives me more insight into the structure of the problem. | |
| Sep 1, 2013 at 13:06 | vote | accept | Neil Strickland | ||
| Sep 1, 2013 at 7:00 | comment | added | Laurent Moret-Bailly | How could I miss that? | |
| Aug 31, 2013 at 21:44 | comment | added | Noam D. Elkies | Wait, isn't that a solution? We're looking to solve $b_1^{\phantom2} v_1^2 + b_2^{\phantom2} v_2^2 + b_3^{\phantom2} v_3^2 = b_4^{\phantom2} v_4^2$ and the displayed identity says $(v_1,v_2,v_3,v_4) = (5y_2-2, 4, 0, 2y_1)$ works. | |
| Aug 31, 2013 at 20:09 | history | answered | Laurent Moret-Bailly | CC BY-SA 3.0 |