Talk:Wilson's theorem

message: ::In mathematics, Wilson's Theorem states that p is a prime number just if
::::(p-1)!\ \equiv\ -1\ (\mbox{mod}\ p)
::(see factorial and modular arithmetic for the notation). There is a converse result.
"Just if" could mean "only if".  It could also mean "precisely if", i.e. "if and only if".  Given this vagueness, what does the last sentence mean, about the "converse result". Michael Hardy 22:35, 7 April 2006 (UTC)
replies: text: And now I've deleted the part about the converse result, since that makes no sense when it says "if and only if". Michael Hardy 20:45, 1 May 2006 (UTC)
thread_title: "just if"?

message: It appears that al-Haytham discovered a version of the CRT rather than Wilson's theorem as such; also, if he did apply it to primes rather than congruences (do we have another source on this?), it would be described in modern terms as a conjecture, as he didn't have a proof.  However I think that it's probably best to mention him and let the reader decide the significance rather than delete it.  How do others feel?
CRGreathouse (t | c) 23:23, 28 July 2007 (UTC)
thread_title: Ibn al-Haytham

message: Do they use results that in fact come from Wilson's theorem? Should a proof that does not use group theory be presented?--Jrm2007 (talk) 07:41, 18 July 2008 (UTC)
replies: text: The proofs are not circular. For the first proof, the only non-trivial dependence is the fact that multiplication modulo p forms a group (i.e., the fact that inverses are present). This follows trivially from p being prime. Misof (talk) 22:13, 13 December 2009 (UTC)
thread_title:  Are the proofs presented in the article circular? 

message: I think the equation
 log n/log q ≤ n/q − 1 
does not hold for n = 2 q
SettingMoon (talk) 20:21, 15 August 2008 (UTC)
thread_title:  A question about the ‘Converse‘ part 

message: Isn't this actually the contrapositive of the forward direction?  I.e., not the converse (can an "if and only if" statement have a converse, since that covers both directions?)
tacotank10 (talk) 07:12, 27 February 2009 (UTC)
replies: text: Agreed. I edited the section to make it clear that this is not the converse of the statement (which does not make much sense for an implication). This is a proper generalization. I also provided a much simpler proof than the one with logarithms. Misof (talk) 22:08, 13 December 2009 (UTC)
thread_title:  converse section... 

message: The "simple generalisation" is misstated - since it is not in the form of a congruence, the "−1" should be replaced by "n − 1".  If it is restated as congruence i.e. 
:
\forall n\in\mathbb{N}:
(n-1)! \equiv 
\begin{cases}
-1 \pmod n & \gets n \text{ is prime}
\\
2 \pmod n & \gets n=4
\\
0 \pmod n & \text{otherwise}
\end{cases}
or as I suggested as
:
\forall n\in\mathbb{N}:
(n-1)! ~\bmod~ n = 
\begin{cases}
n-1 & \gets n \text{ is prime}
\\
2 & \gets n=4
\\
0 & \text{otherwise}
\end{cases}
it breaks down at n = 1 either way, as the top and  bottom cases collapse into one. Quondum (talk) 20:17, 5 August 2011 (UTC)
replies: text: I have now modified the article to match the first fork above, though the exclusion of n=1 is clumsily stated. Quondum (talk) 18:34, 11 August 2011 (UTC)
thread_title:  A simple generalization - a flaw 

message: The article currently says,
:This further generalizes to the fact that in any finite abelian group, either the product of all elements is the identity, or there is precisely one element a of order 2. In the latter case, the product of all elements equals a.
While this is strictly true, as stated the possibility that there is precisely one element a of order 2 and that the product of all elements is still identity is not discounted. This actually cannot occur. That is, "or" has been used when "xor" was probably meant (and is in fact true). Applying the fundamental theorem of finite abelian groups and doing some mundane calculations, one finds that the group contains a unique element of order 2 if and only if the Sylow 2-subgroup is cyclic (of order > 1); if it is cyclic, the product of the group's elements is the unique order 2 element, and otherwise the product is the identity.
The statement is unsourced right now anyway, but perhaps a source can be found for this stronger version. I haven't edited the article since it's OR. 208.107.152.253 (talk) 08:42, 19 March 2012 (UTC)
replies: text: I would think this follows immediately from the fact that the order of the identity element is never 2.  When I first read this I disambiguated to 'xor' on this basis.  I think that 'xor' is also implied adequately by "either ... or".  — Quondum☏✎ 15:49, 19 March 2012 (UTC)
replies: text: Oh yes, of course the 'xor' can be quickly deduced from the 'or' case (though even that is OR). The Sylow 2-subgroup information I mentioned is really the content of my refinement. I don't agree about "either ... or" implying 'xor'; usually "either ... or ... but not both" is used for 'xor'. 208.107.152.253 (talk) 09:20, 20 March 2012 (UTC)
replies: text: Perhaps you can simply reword it to make it clearer that the cases are exclusive – I doubt you'll get much objection unless you make it too verbose.  If you have sources for any alternative arguments (and I'm not going to pretend that I'm familiar with Sylow p-subgroups) of merit in this context, you could add these as a footnote.  — Quondum☏✎ 12:30, 20 March 2012 (UTC)
replies: text: "either ... or else"? AlexFekken (talk) 12:30, 21 March 2012 (UTC)
replies: text: I've simply added "(but not both)" to the page. As for my refinement, unfortunately I couldn't even find the original statement in my algebra books. It's almost perfect for an exercise problem.... 208.107.152.253 (talk) 15:14, 21 March 2012 (UTC)
thread_title:  Finite abelian group generalization 

message: I'm comparatively well educated in math, but I had a problem with the "fancy" notation too.  It's jargon no matter how "standard" it is.  Jargon is specifically disfavored by WP policy and is unencyclopedic.  There's no such thing as a "technical article" in WP, WP is not a repository for math papers.  Rather, it's our duty to be clear to lay readers.  Being fancy and pedantic can never be justified as "compromise".  Any math notation beyond maybe middle school or high school should at least link to something that explains what it means.  For example, "(m mod n)" is explained, but "(mod n)" by itself is not.  And, what's that "triple equals" sign, eh?  I remember something like that as "is defined as", but that makes no sense here.    98.217.10.176 (talk) 16:28, 19 August 2019 (UTC)
replies: text: The relevant link is to modular arithmetic; it appears earlier in the same sentence as the equation does, and explicitly indicates that it's the link you should click on to understand the notation.  Moreover, the preceding sentence already explains the statement in a non-jargon, non-notation way.  There are problems with accessibility of some mathematics articles on Wikipedia, but the first two sentences of this article are not a good example of that problem. (Possibly it is also worth noting that this status quo resulted from the earlier discussion 18 months ago.) --JBL (talk) 16:32, 19 August 2019 (UTC)
thread_title: Yes.  It's inscrutable jargon.

message: Regarding the generalization of Wilson's theorem stated in the Wilson prime article, it seems to be quite important to mention it on this page too. What do you think ?
Lancelot Chardonnet (talk) 14:31, 26 December 2020 (UTC)
thread_title:  Adding more generalizations 

message: — Assignment last updated by HUco2024 (talk) 16:42, 12 December 2022 (UTC)
thread_title: Wiki Education assignment: Number Theory I

message: note 7. "Gauss, DA, art. 78" is ungoogleable and should contain a link 128.237.82.1 (talk) 05:41, 8 March 2024 (UTC)
replies: text: It's to one of the two translations of Disquisitiones Arithmeticae found in the references section of this article. --JBL (talk) 17:47, 8 March 2024 (UTC)
thread_title:  cryptic reference 

message: Hello people. I have noticed also that (p-2)! \equiv_p 1 for prime p, and not for composite p. Is this true generally and is it worth discussing?
63.40.186.52 (talk) 22:08, 13 April 2024 (UTC)
replies: text: This is an essentially trivial consequence of Wilson's theorem (becauce p - 1 = -1 is a unit modulo p) and is mentioned in passing in the third proof.  To discuss it at greater length, we should have sources to base the discussion on.  --JBL (talk) 21:18, 15 April 2024 (UTC)
replies: text: Hello. I only brought it up because its marginally simpler.  You have literally one fewer multiplication on the left hand side. On the right hand side you have a simple 1, not a p-1. There is no -1 in modulus, so even the necessary subtraction is avoided.  I know its trivial and insignificant, but its an observation nonetheless. It might actually be a useful fact to someone implementing Wilson at the very edge of integer sizes that their hardware can handle. 63.40.186.52 (talk) 01:18, 17 April 2024 (UTC)
thread_title:  variant