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 2013-12-29, 12:03 #1 Stan   Dec 2011 22·32 Posts Carmichael Numbers Any constructive comments on the following statement, favorable or otherwise, will be appreciated. If (a , m) = 1 and am-1 = 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.
2013-12-29, 15:20   #2
WraithX

Mar 2006

5·101 Posts

Quote:
 Originally Posted by Stan Any constructive comments on the following statement, favorable or otherwise, will be appreciated. If (a , m) = 1 and am-1 = 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.
I believe you might want to start your condition by saying:
For all a, 1 < a < n, If (a,m) = 1 and...

Here is a counter-example with a specific a and m, that break conditions 1 and 2 of your proposed proof:
a = 2
m = 5654273717
(a,m) = 1
a^(m-1) == 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 (p-1)|(m-1)

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.

2013-12-29, 15:39   #3
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by WraithX I believe you might want to start your condition by saying: 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.
Note that this is not a proof, but a DEFINITION.

2013-12-29, 17:46   #4
Stan

Dec 2011

448 Posts

Quote:
 Originally Posted by WraithX I believe you might want to start your condition by saying: For all a, 1 < a < n, If (a,m) = 1 and... Here is a counter-example with a specific a and m, that break conditions 1 and 2 of your proposed proof: a = 2 m = 5654273717 (a,m) = 1 a^(m-1) == 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 (p-1)|(m-1) 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.
Many thanks WraithX for the counter-example. I need to amend the original condition to exclude a Weiferich prime being a factor of m.
(1093 is a Weiferich prime.)
Then I believe the statement to be true.

Last fiddled with by Stan on 2013-12-29 at 17:54

2013-12-29, 20:22   #5
Stan

Dec 2011

22×32 Posts

Quote:
 Originally Posted by Stan Many thanks WraithX for the counter-example. I need to amend the original condition to exclude a Wieferich prime being a factor of m. (1093 is a Wieferich prime.) Then I believe the statement to be true.
Sorry, incorrect spelling of Wieferich.

Last fiddled with by Stan on 2013-12-29 at 20:25

2013-12-30, 14:22   #6
R.D. Silverman

Nov 2003

1D2416 Posts

Quote:
 Originally Posted by Stan Many thanks WraithX for the counter-example. I need to amend the original condition to exclude a Weiferich prime being a factor of m. (1093 is a Weiferich prime.) Then I believe the statement to be true.
Learn to read. I already told you that this is a definition.
Definitions are neither true nor false.

They are not subject to verification.

2013-12-30, 19:07   #7
xilman
Bamboozled!

"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across

71·157 Posts

Quote:
 Originally Posted by R.D. Silverman Learn to read. I already told you that this is a definition. Definitions are neither true nor false. They are not subject to verification.
How do you (personally, not generally) distinguish between a definition and an axiom?

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.

2013-12-30, 22:33   #8
BudgieJane

"Jane Sullivan"
Jan 2011
Beckenham, UK

283 Posts

Quote:
 Originally Posted by xilman By definition (!) axioms are true.
Including the parallel axiom?

2013-12-31, 01:47   #9
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by xilman How do you (personally, not generally) distinguish between a definition and an axiom?
A definition attaches a name (or label) to an object.

An Axiom makes an unproven (and unprovable) assumption
that a certain condition is true.

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.
Sorry, but wrong. Definitions are neither true nor false. A definition
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 2013-12-31 at 01:48 Reason: dropped sentence

2013-12-31, 01:51   #10
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by BudgieJane Including the parallel axiom?
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.

 2013-12-31, 03:05 #11 ewmayer ∂2ω=0     Sep 2002 Repรบblica de California 2DAE16 Posts What about an Ansatz - is that just an axiom with Sauerkraut?

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