Talk:Jacobi symbol

message: Do all the brackets here represent the Legendre symbol?
If so, it should say so!
::::Johnbibby 18:44, 10 October 2006 (UTC)
replies: text: BrunoX 00:06, 23 November 2006 (UTC)
thread_title: Question

message: is it just me or rule #10 is wrong since it should be stated that m,n must be relativly prime? —The preceding unsigned comment was added by 217.132.13.228 (talk • contribs) 06:15, 21 February 2007  (UTC)
:It's OK. If m and n share a factor, then both sides of the equation are zero. -- EJ 10:09, 21 February 2007 (UTC)
thread_title: #10?

message: I rewrote the page, added references, explained in more detail just what Jacobi added to teh calculateon of (a|p) Virginia-American (talk) 21:38, 2 March 2008 (UTC)
thread_title: rewrite

message: I just cut huge rambling sections of this page, including all the calculations. My objection is that the whole point of the Jacobi symbol is that it is polynomial time computable, and the calculations shown used factorization.  —Preceding unsigned comment added by 18.51.7.209 (talk) 00:47, 28 October 2008 (UTC)
replies: text: And twice your mass removal has been reverted. There is no consensus here on this talk page for your removal of a huge portion of the article. Discuss it here before you remove it again, please. Ariel♥Gold 00:55, 28 October 2008 (UTC)
replies: text: Kingofaustria  —Preceding undated comment was added at 04:25, 28 October 2008 (UTC)

text: The problem is, even if it is a good, "atypical" lecture, it still doesn't read like an encyclopedia article should.  While it is good to include information on how it can be calculated easily, it shouldn't be done in that style, and it shouldn't cloud the actual definition of the symbol, as it was. CountingPine (talk) 09:50, 4 November 2008 (UTC)
thread_title: October 28

message: Rule #6 is incorrect, quadratic reciprocity should be something like..
If both n and m are odd, then (n/m) = (m/n) unless both n and m are congruent to 3 mod 4. In this case, (n/m) = -(m/n).  —Preceding unsigned comment added by 97.77.52.94 (talk) 02:48, 9 March 2011 (UTC)
replies: text: That is, in fact, exactly what it says (see the piecewise part on the righthand side). They also have to be positive coprime. Dcoetzee 15:22, 23 March 2013 (UTC)
thread_title:  Rule #6 

message: It said on the image caption that any non-zero residue m mod n must have (m|n)=1. I initially couldn't figure out how to prove this, then I found a counterexample: m=10 is not congruent to 0 mod n=15, and it is congruent to 5^2=25 mod 15, so it is a non-zero residue. However, (10|15)=(10|3)(10|5)=1*0=0, not 1. Thus, I'm removing the claim. Any objections? David815 (talk) 15:29, 22 February 2015 (UTC)
thread_title:  Removing Claim Made in Image Caption 

message: So 9 has no quadratic residues aside from 0? 2² = 4 doesn't make 4 a quadratic residue? Am I missing something? Quite misleading...
2804:14D:BAA1:3E5:51B2:48F8:7C8A:685A (talk) 00:16, 5 May 2016 (UTC)
thread_title:  Table 

message: The worked problem begins with "Given that 9907 is prime," but that is unnecessary.  The calculation will work merely given that 9907 is odd and positive. Colin McLarty (talk) 13:20, 14 January 2017 (UTC)
thread_title:  Example of calculations 

message: I'd like to remove the first one. Any problems?
James in dc (talk) 02:49, 7 December 2023 (UTC)
thread_title:  Why are there two tables? 

message: I have rewritten this article
the main changes are
*  removed the table from the beginning
*  the sections have been reordered (table and history at end)
*  the relation between Legendre and Jacobi symbols has been expanded
*  the computer programs have been desk-checked
*  removed Legendre symbol calculation (but say they are impractical)
*  removed (broken) link to calculator - I couldn't find a replacement that prints out the steps  
Please visit my sandbox User:James in dc/sandbox/Jacobi symbol and comment. I will replace the article in a week or so, after I finish with the notes.
James in dc (talk) 23:13, 18 December 2023 (UTC)
thread_title:  Rewrite