Timeline for 3 square theorem
Current License: CC BY-SA 2.5
10 events
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Jul 16, 2021 at 17:13 | history | edited | YCor |
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Jan 5, 2011 at 18:04 | vote | accept | t.k | ||
Jan 5, 2011 at 18:04 | vote | accept | t.k | ||
Jan 5, 2011 at 18:04 | |||||
Dec 31, 2010 at 10:22 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
deleted 3 characters in body
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Dec 30, 2010 at 3:29 | answer | added | Will Jagy | timeline score: 7 | |
Dec 29, 2010 at 10:44 | comment | added | Qiaochu Yuan | Do you know a proof which doesn't? (Not rhetorical; I only know one proof.) | |
Dec 29, 2010 at 9:54 | answer | added | Pete L. Clark | timeline score: 6 | |
Dec 29, 2010 at 8:22 | comment | added | Alex B. | Google does. Here is a concrete application towards the 3-square theorem: to prove that $M\equiv 3\pmod{8}$ is a sum of three integer squares, apply the Hasse principle to the form $x^2 + y^2 + z^2 - Mw^2$ and thereby show that this form represents 0 over $\mathbb{Q}$ (it does so over all completions). Using Cassels's (sic) lemma completes the proof of the theorem for $M\equiv 3\pmod{8}$. | |
Dec 29, 2010 at 8:13 | comment | added | Kevin Buzzard | Assuming you mean the standard classification of which integers are the sum of three integer squares, the only other thing you need is the local-global principle, which is true in this situation (Hasse-Minkowski). See for example Serre's book "A course in arithmetic". The reason you get the funny $4^m(8n+7)$ business is the local condition at 2; all other local conditions are satisfied. | |
Dec 29, 2010 at 8:04 | history | asked | t.k | CC BY-SA 2.5 |