Timeline for What's the "best" proof of quadratic reciprocity?
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2022 at 12:08 | comment | added | Arsenii Sagdeev | @Oblomov once you have the general case of QR, you can derive the supplementary low for $p=2$ in two steps. 1) Extend QR to the similar statement about the Jacobi symbols of two odd arguments. 2) Replace $2$ with $-(p-2)$ (which is odd) and apply the QR of the Jacobi symbols. Though this fix still fits for undergrads, it isn't as simple as I'd like it to be. I found this solution in Leo Goldmakher's note on Rousseau's proof. | |
| Jul 24, 2022 at 12:02 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added 48 characters in body
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| Oct 20, 2021 at 13:17 | comment | added | Emil Jeřábek | @NoahSnyder I tried that, to no avail. (That is, more precisely, I tried to compute the product of elements of $(\mathbb Z/8p\mathbb Z)^\times/\{\pm1,4p\pm1\}$ in two different ways, but they just turned out to be equal without yielding any nontrivial information. But perhaps there is a better way to arrange the things.) | |
| Oct 9, 2021 at 17:12 | comment | added | Noah Snyder | @Oblomov: I wonder if you could look at a quarter of the elements of $(Z/8pZ)^\times$ instead of half? | |
| Oct 9, 2021 at 16:32 | comment | added | LSpice | @MRB's post mentioned by @NoahSnyder. | |
| Oct 9, 2021 at 16:31 | history | edited | LSpice | CC BY-SA 4.0 |
Finishing TeXing while this is on the front page
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| Oct 9, 2021 at 15:34 | history | edited | Noah Snyder | CC BY-SA 4.0 |
added 17 characters in body
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| Oct 1, 2021 at 15:33 | comment | added | Oblomov | I was wondering what is the natural adaptation of this proof to the $p=2$ case? I tried to mimic the proof replacing $Z/(pq)^\times$ by $Z/(8p)^\times$ but it failed. Any idea? | |
| Mar 4, 2019 at 21:37 | comment | added | Álvaro Lozano-Robledo | Thanks for noticing... It's here stacky.net/files/115/RousseauQR.pdf | |
| Mar 4, 2019 at 19:50 | comment | added | KConrad | @ÁlvaroLozano-Robledo that tinyurl link you gave in your comment isn't working properly anymore. | |
| Apr 1, 2018 at 21:21 | comment | added | Fedor Petrov | Note that if $M$ is a finite set, $|M|=n$, then the sign of a permutation $\pi$ of $M$ equals $\prod_{x<y} (f(\pi(x))-f(\pi(y)))/(f(x)-f(y))$, where $f$ is arbitrary injection from $M$ to any field (with characteristics not 2) and '$<$' is an arbitrary linear order on $M$. And the main property of the sign is that this does not depend on the chosen injection and linear order. Now applying this to Zolotarev's proof we exclude the notion of the sign and I feel (did not check) that we should get Rousseau's proof for the natural choices of orders and injections. | |
| Jan 21, 2018 at 2:49 | comment | added | Alexey Ustinov | Direct link to the file | |
| S Apr 10, 2015 at 8:44 | history | suggested | Yous | CC BY-SA 3.0 |
Correct the exponent of latter element of second product
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| Apr 10, 2015 at 8:35 | review | Suggested edits | |||
| S Apr 10, 2015 at 8:44 | |||||
| Apr 20, 2013 at 22:01 | history | edited | Douglas Zare | CC BY-SA 3.0 |
typeset Legendre symbol
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| Apr 20, 2013 at 21:55 | comment | added | Douglas Zare | @Daniel Miller: That edit wasn't correct. It's not $a^p=a \mod p$, it's $a^{\frac{p-1}{2}} = (\frac{a}{p})$ where the RHS is the Legendre symbol. | |
| Apr 20, 2013 at 21:43 | history | edited | Daniel Miller | CC BY-SA 3.0 |
cleaned up last bit of TeX
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| Jun 12, 2012 at 13:04 | history | edited | darij grinberg | CC BY-SA 3.0 |
latex doesn't understand ()
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| Jul 25, 2011 at 21:05 | comment | added | Noah Snyder | Thanks! I put in the latter URL. While I was at it I also texed my answer (remarkably enough, this answer was from the days when MO didn't have tex support). | |
| Jul 25, 2011 at 21:04 | history | edited | Noah Snyder | CC BY-SA 3.0 |
added 136 characters in body
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| Jul 25, 2011 at 17:43 | comment | added | Álvaro Lozano-Robledo | The link you provided to MR1125443 won't work for people outside of the Columbia network. I think this link ams.org/mathscinet/pdf/1125443.pdf should work for anyone with a mathscinet subscription. Or even better, point it here tinyurl.com/3z62fdw where a free PDF copy of Rousseau's paper is available. | |
| Oct 26, 2009 at 15:08 | vote | accept | Ben Webster♦ | ||
| Oct 22, 2009 at 2:38 | history | edited | Noah Snyder | CC BY-SA 2.5 |
fixed typos in formulas
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| Oct 20, 2009 at 19:56 | comment | added | Noah Snyder | See MRB's post below for a link to a rediscovery of this proof by Tim Kunisky. | |
| Oct 20, 2009 at 19:37 | history | edited | Noah Snyder | CC BY-SA 2.5 |
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| Oct 20, 2009 at 19:31 | history | answered | Noah Snyder | CC BY-SA 2.5 |