20131229, 12:03  #1 
Dec 2011
24_{16} 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
541 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
2^{3}·3·313 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
2^{3}×3×313 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
11,719 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
7·47 Posts 

20131231, 01:47  #9  
"Bob Silverman"
Nov 2003
North of Boston
1D58_{16} 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
1110101011000_{2} 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
5×2,351 Posts 
What about an Ansatz  is that just an axiom with Sauerkraut?

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Carmichael numbers and Devaraj numbers  devarajkandadai  Number Theory Discussion Group  0  20170709 05:07 
Carmichael numbers (M) something  devarajkandadai  Miscellaneous Math  2  20130908 16:54 
Carmichael Numbers  devarajkandadai  Miscellaneous Math  0  20060804 03:06 
Carmichael Numbers II  devarajkandadai  Math  1  20040916 06:06 
Carmichael Numbers  devarajkandadai  Math  0  20040819 03:12 