Talk:Carmichael function

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I'm afraid that Lower Bound in Theorem 1 makes very little sense. Are you sure that everything is perfectly OK?

And why this page is - well - extremele similar to that on answers.com? http://www.answers.com/topic/carmichael-function

The answers.com link refs wikipedia as the source so their info came from here rather than the other way around. I do find the article very difficult to follow though so a rewrite might be a good idea. Maddog Battie 11:59, 11 October 2006 (UTC)Reply

And then there is the section that is called "Average and Typical value", but with no explanation of what these really mean, and what is the distinction between the two.
In many mathematical, statistical, and scientific, and engineering applications, the average value and the typical value are one and the same. (So, there would be no reason to mention both of them.) On the other hand, there are situations where the average value exists, but some other value (such as possibly some median value) is more "typical".
Also, there are some cases, such as where we have a Cauchy-distributed random variable, where the average value does not exist mathematically, but there is a "typical" value that can be defined. It might be a median, or a value of maximum likelihood, or to get more technical, it could be the "Cauchy Principal Value" of the average. Note that the Cauchy Principal Value of something is not just for Cauchy random variables, but it is a more general concept that can be applied to lots of infinite sums and indefinite integrals.72.146.44.141 (talk) 15:58, 6 October 2008 (UTC)Reply

I think the recursive definition of the function is incomplete. What to do with the cases p = 2^1 and p = 2^2 ? The example sequence (1,1,2,2,...) indicates these ones fall in the general p^(k-1)*(p-1) case. I'm no mathematician, could a math guy (or girl) fix this? —Preceding unsigned comment added by Rkomatsu (talk • contribs) 11:15, 10 March 2010 (UTC)Reply

In the case of ${\displaystyle p=2}$, let ${\displaystyle a}$ be coprime with respect to ${\displaystyle p}$; that is, ${\displaystyle a=1+2n}$. One can immediately obtain ${\displaystyle a^{1}\equiv 1{\text{ mod 2}}}$, where the Euler's totient function of ${\displaystyle 2}$, ${\displaystyle \psi (2)}$, is equal to ${\displaystyle 1}$. Thus, ${\displaystyle a^{\psi (2)}\equiv 1{\text{ mod 2}}}$. Likewise, in the case of ${\displaystyle p=4}$, let ${\displaystyle a}$ be coprime with respect to ${\displaystyle p}$; that is, either ${\displaystyle a=1+4n}$ or ${\displaystyle a=3+4n}$. In the first case, ${\displaystyle a^{1}\equiv 1{\text{ mod 4}}}$. In the second case, ${\displaystyle a^{1}\equiv 3{\text{ mod 4}}}$, and ${\displaystyle a^{2}\equiv 9\equiv 1{\text{ mod 4}}}$, where the Euler's totient function of ${\displaystyle 4}$, ${\displaystyle \psi (4)}$, is equal to ${\displaystyle 2}$. Therefore, ${\displaystyle a^{\psi (4)}\equiv 1{\text{ mod 4}}}$. D4nn0v (talk) 04:24, 15 January 2018 (UTC)Reply

In the phrase "for every integer a that is both coprime to and smaller than n." the limitation "smaller than n." doesn't seem to be necessary. a^m mod n = 1 is valid for any integer a coprime to n, according to at least one other source. I ask a math expert to verify this, I don't know whether this is a mistake or simply a more recent result not included in the original definition of the Carmichael function. Rkomatsu (talk) 14:18, 10 March 2010 (UTC)Reply

You are right, the statements

${\displaystyle a^{m}\equiv 1{\pmod {n}}}$ for every integer a that is coprime to n

and

${\displaystyle a^{m}\equiv 1{\pmod {n}}}$ for every integer a that is both coprime to and smaller than n

and

${\displaystyle a^{m}\equiv 1{\pmod {n}}}$ for every positive integer a that is both coprime to and smaller than n

are equivalent. AxelBoldt (talk) 13:13, 19 March 2018 (UTC)Reply

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Values for 15 and 16 are shown in bold indicating they are different when the actual values shown seem to match. --N8 08:06, 31 August 2018 (UTC)Reply

Nevermind. I see what I did. --N8 03:24, 1 September 2018 (UTC)Reply

In the first paragraph of numerical examples, after 2^2=4\ne 1 (mod 5), the fact that 3^2=9\ne 1 (mod 5) is irrelevant, it should read 2^3=8\ne 1 (mod 5), as we want to show that the 4th is the minimal power, as 2nd and 3rd powers do not result in 1. LaurV (talk) 09:41, 14 August 2024 (UTC)Reply

That ${\displaystyle 3^{2}=9\not \equiv 1{\pmod {5}}}$ is in parentheses suggests to me that it was considered additional reinforcement, not necessary to make the point, given that exponent 2 was already ruled out by the example ${\displaystyle 2^{2}=4\not \equiv 1{\pmod {5}}}$. I suspect that exponent 3 was not mentioned since it's ruled out on the general principle that the exponent of a group must divide the order of the group, which is ${\displaystyle \varphi (5)=4}$ in this example. Might be worth adding a remark about that to the introduction. Will Orrick (talk) 16:53, 14 August 2024 (UTC)Reply

I have made some changes, following up on my previous comment. Let me know if you think anything important or clarifying has been omitted. Will Orrick (talk) 16:10, 16 August 2024 (UTC)Reply