Timeline for Sum of squares modulo a prime
Current License: CC BY-SA 3.0
12 events
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| Jul 6, 2011 at 14:41 | comment | added | Noam D. Elkies | This idea can also be used to get at a few memorable cases of higher reciprocity. For example: if $p \equiv 1 \bmod 3$, and you already know that the number of solutions of $x^3 + y^3 = 1 \bmod p$ is $p - 2 - a$ where $4p = a^2 + 27 b^2$ for some integers $a,b$, then it follows at once that $2$ is a cubic residue iff $a$ is even, that is, iff $p$ itself can be written as $a^2 + 27b^2$. [NB the count is $p-2-a$, not the familiar $p+1-a$, because for this purpose we must exclude the three points at infinity.] | |
| Jul 6, 2011 at 6:41 | comment | added | LSpice | To save the agony of scrolling, the answer by KConrad that Qiaochu references is mathoverflow.net/questions/1420/… (I think). Perhaps it is pleasant that the comment numbered 12345 is a number-theoretic one. :-) | |
| Jul 6, 2011 at 5:01 | comment | added | Qiaochu Yuan | More precisely, see the discussion after Proposition 8.6.1 in Ireland and Rosen. But there does not seem to be a more specific citation there. | |
| Jul 6, 2011 at 4:46 | comment | added | Qiaochu Yuan | @Noam: I believe this is the proof mentioned by KConrad in his answer to this MO question: mathoverflow.net/questions/1420?sort=votes#sort-top . He says it is due to V. Lebesgue and can be found in Ireland and Rosen. | |
| Jul 6, 2011 at 3:28 | comment | added | GH from MO | Actually Gauss in Disquisitiones initially regards 0 a quadratic residue, but later he excludes it for convenience. I included zero only to make the formula simpler (without zero I would have needed two lines for n=2k and n=2k+1, or alternately a notation for n mod 2). | |
| Jul 6, 2011 at 3:17 | comment | added | Noam D. Elkies | :-) If no reference turns up here I'll post a reference request as a question... It worked quite well for my one previous M.O. query (on the integral for $\frac{22}{7} - \pi$). | |
| Jul 6, 2011 at 3:15 | comment | added | Wadim Zudilin | Noam, your last remark is more interesting than the question! I have not seen this proof of quadratic reciprocity before. | |
| Jul 6, 2011 at 0:01 | comment | added | Noam D. Elkies | For quadratic reciprocity: let $n$ be an odd prime prime $l \neq p$, and $N$ the number of solutions of $\sum_{i=1}^l a_i^2 = 1 \bmod p$. Cyclic permutation of the coordinates has two fixed points or none according as $N \equiv 2$ or $0 \bmod l$. From the formula for $N$ it soon follows that $(l/p) \equiv {p^*}^{(l-1)/2} \bmod l$ where $p^* = \gamma(1)^2 = p$ or $-p$ according as $p \equiv 1$ or $-1 \bmod 4$. By Legendre's formula this means $(l/p) = (p^*/l)$. Exercise: modify this to determine $(2/p)$ from the count of solutions of $x_1^2 + x_2^2 \equiv 1 \bmod p$. [Original source?] | |
| Jul 5, 2011 at 23:55 | comment | added | Noam D. Elkies | While 0 is indeed a square, it does not count as a "quadratic residue" (nor as a "quadratic nonresidue" — though this term should really have been "nonquadratic residue"). Properly 0 should count as half square and half non-square (think about the number of square roots), and then the limit would really be 1/2. | |
| Jul 5, 2011 at 23:12 | comment | added | Gerhard Paseman | From your answer I infer 0 does not count as a quadratic residue mod p. (I definitely agree it is not as interesting as a quadratic residue, but should it be really ostracized from the set of squares?) Gerhard "Supports The Rights of Zero" Paseman, 2011.07.05 | |
| Jul 5, 2011 at 22:58 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
added 106 characters in body
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| Jul 5, 2011 at 22:51 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |