20131229, 12:03  #1 
Dec 2011
100100_{2} Posts 
Carmichael Numbers
Any constructive comments on the following statement, favorable or otherwise, will be appreciated.
If (a , m) = 1 and a^{m1} = 1 (mod. m) then m is either prime or a Carmichael number. I believe the proof can be accomplished in 3 steps, supposing m is not prime: 1. m is square free. 2. If p is any prime dividing m then (p  1) divides (m  1). 3. There are at least 3 distinct primes dividing m. 
20131229, 15:20  #2  
Mar 2006
2×3^{2}×29 Posts 
Quote:
For all a, 1 < a < n, If (a,m) = 1 and... Here is a counterexample with a specific a and m, that break conditions 1 and 2 of your proposed proof: a = 2 m = 5654273717 (a,m) = 1 a^(m1) == 1 (mod m) m is not prime m is not a carmichael 1. m is not square free m = 1093^2 * 4733 2. none of the primes p dividing m (1093, 4733) have (p1)(m1) But, if you start your condition with: For all a, 1 < a < n, Then you are defining Carmichael numbers, or primes. So then, yes, it would definitely be true that the statement identifies only Carmichael numbers or primes. 

20131229, 15:39  #3  
"Bob Silverman"
Nov 2003
North of Boston
1D3B_{16} Posts 
Quote:


20131229, 17:46  #4  
Dec 2011
2^{2}·3^{2} Posts 
Quote:
(1093 is a Weiferich prime.) Then I believe the statement to be true. Last fiddled with by Stan on 20131229 at 17:54 

20131229, 20:22  #5 
Dec 2011
2^{2}×3^{2} Posts 
Sorry, incorrect spelling of Wieferich.
Last fiddled with by Stan on 20131229 at 20:25 
20131230, 14:22  #6  
"Bob Silverman"
Nov 2003
North of Boston
1110100111011_{2} Posts 
Quote:
Definitions are neither true nor false. They are not subject to verification. 

20131230, 19:07  #7  
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
2^{4}×719 Posts 
Quote:
In my view, an axiom is a special case of a definition. By definition (!) axioms are true. Therefore, and again in my view, at least definitions are true. 

20131230, 22:33  #8 
"Jane Sullivan"
Jan 2011
Beckenham, UK
3^{2}·5·7 Posts 

20131231, 01:47  #9  
"Bob Silverman"
Nov 2003
North of Boston
7·1,069 Posts 
Quote:
An Axiom makes an unproven (and unprovable) assumption that a certain condition is true. Quote:
does not constitute a testable condition. In the instance under discussion, one simply attaches the label "Carmichael Number" to a set of numbers. It makes no assertion about the set itself. Saying "A Carmichael Number is a number such that ....." simply attaches a label to a set of numbers. Last fiddled with by R.D. Silverman on 20131231 at 01:48 Reason: dropped sentence 

20131231, 01:51  #10 
"Bob Silverman"
Nov 2003
North of Boston
7×1,069 Posts 
Yes. In Euclidean geometry the parallel axiom is assumed to be true.
An assumption that parallel lines satisfy a different assumption(s) simply leads to a different geometry. The fact that one can get a different system of mathematics from a different axiom does not make the parallel axiom untrue in Euclidean geometry. 
20131231, 03:05  #11 
∂^{2}ω=0
Sep 2002
Repรบblica de California
2DE1_{16} Posts 
What about an Ansatz  is that just an axiom with Sauerkraut?

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