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Old 2012-01-23, 19:49   #12
Stan
 
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Originally Posted by ccorn View Post
Well then: 5*phi(5) | phi(66) and 5 | (66-1), but 25 does not divide 66.
Once again you are correct, but I have not been sufficiently explicit since all the primes in the sequence (2^ni)-1; 0 < i < 5, are congruent to 3 mod.4
So I require p*phi(p) | phi(m) and p*phi(p) | (m - 1) where p is a prime congruent to 3 mod.4 and m = 2^p - 1.

Last fiddled with by Stan on 2012-01-23 at 20:33
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Old 2012-01-23, 20:08   #13
ccorn
 
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Once again you are correct, but I have not been sufficiently explicit since all the primes in the sequence (2^ni)-1; 0 < i < 5, are congruent to 3 mod.4
So I require p*phi(p) | phi(m) and p | (m - 1) where p is a prime congruent to 3 mod.4 and m = 2^p - 1.
If you require p | (m - 1), you cannot conclude p^2 | m anyway...

P.S.: You have presented no reasoning why from such phi(a) | phi(b) you could possibly conclude a | b. In fact, many coprime numbers share the same totient function value, so you cannot easily carry divisibiliy properties across applications of the totient function. With very strong conditions on a or b this might be possible (e. g. if a and b are increasing powers of 2 or 6), but you would still need to provide a proof why such a rather singular case should happen.

P.P.S.: This is the second of the two alleged non sequiturs, the first one deserves attention too.

Last fiddled with by ccorn on 2012-01-23 at 20:38
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Old 2012-01-23, 20:45   #14
Stan
 
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If you require p | (m - 1), you cannot conclude p^2 | m anyway...

P.S.: You have presented no reasoning why from such phi(a) | phi(b) you could possibly conclude a | b. In fact, many coprime numbers share the same totient function value, so you cannot easily carry divisibiliy properties across applications of the totient function. With very strong conditions on a or b this might be possible (e. g. if a and b are increasing powers of 2 or 6), but you would still need to provide a proof why such a rather singular case should happen.
I take it that you have read the PDF I asked to be checked for errors.
I do not assume phi(a) | phi(b) implies a | b but I have discovered a situation where p*phi(p) | phi(m) and p*phi(p) | (m-1) from which I propose that m is a prime.
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Old 2012-01-23, 21:20   #15
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I take it that you have read the PDF I asked to be checked for errors.
I do not assume phi(a) | phi(b) implies a | b but I have discovered a situation where p*phi(p) | phi(m) and p*phi(p) | (m-1) from which I propose that m is a prime.
Aha. For what reason? For the time being, take p = 5, m=341 as a counterexample. There may well be a large gcd(phi(m),m-1).
Also, consider Batalov's second counterexample.
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Old 2012-01-23, 23:34   #16
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Aha. For what reason? For the time being, take p = 5, m=341 as a counterexample. There may well be a large gcd(phi(m),m-1).
Also, consider Batalov's second counterexample.
So I require p*phi(p) | phi(m) and p*phi(p) | (m - 1) where p is a prime congruent to 3 mod.4 and m = 2^p - 1. In fact, p = 2^(2^q - 1) - 1
where q is a prime of the same form.
In Batalov's second counterexample, 19 is not 2^(a prime) - 1.
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Old 2012-01-23, 23:53   #17
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19 is not 2^(a prime) - 1.
Is your starting element 2 a 2^(a prime) - 1?
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Old 2012-01-24, 07:22   #18
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So I require p*phi(p) | phi(m) and p*phi(p) | (m - 1) where p is a prime congruent to 3 mod.4 and m = 2^p - 1. In fact, p = 2^(2^q - 1) - 1
where q is a prime of the same form.
In Batalov's second counterexample, 19 is not 2^(a prime) - 1.
Where is the logical link (the reasoning) between all those requirements and your desired conclusion?

The current state, as I perceive it, is that you list a number of true properties (mostly laid down in propositions 1 and 2) and then claim what you like.

For example, you essentially claim that gcd(phi(m),m-1) = m-1 (where m=n5) but you provide only a divisor of the gcd. (And only a divisor of that divisor is actually granted, due to the first non sequitur.) You will agree that this is not a proof.
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Old 2012-01-24, 14:15   #19
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Quote:
Originally Posted by Stan View Post
So I require p*phi(p) | phi(m) and p*phi(p) | (m - 1) where p is a prime congruent to 3 mod.4 and m = 2^p - 1. In fact, p = 2^(2^q - 1) - 1
where q is a prime of the same form.
In Batalov's second counterexample, 19 is not 2^(a prime) - 1.
2^(2^q-1)-1 means p is a double Mersenne if m=2^p-1 then you are talking about triple Mersennes ?

2^3-1 = 2^(2^2-1)-1
2^7-1=2^(2^3-1)-1 = 2^(2^(2^2-1)-1)-1

so we can assume that you speak of all the terms > M7=MM3=MMM2 correct ?

of course M127=MM7=MMM3=MMMM2

Last fiddled with by science_man_88 on 2012-01-24 at 14:29
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Old 2012-01-24, 14:46   #20
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2^(2^q-1)-1 means p is a double Mersenne if m=2^p-1 then you are talking about triple Mersennes ?

2^3-1 = 2^(2^2-1)-1
2^7-1=2^(2^3-1)-1 = 2^(2^(2^2-1)-1)-1

so we can assume that you speak of all the terms > M7=MM3=MMM2 correct ?

of course M127=MM7=MMM3=MMMM2
oh and how could I forget , say p= 3 mod 4 2*p+1 = (3*2+1) mod 4 = 7 mod 4 = 3 mod 4 so every mersenne >=3 works out as a possible p.
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Old 2012-01-30, 20:03   #21
Stan
 
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Post Carmichael Numbers

Is it true that all Carmichael numbers are congruent to 1 modulo 4?
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Old 2012-01-30, 20:30   #22
R.D. Silverman
 
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Is it true that all Carmichael numbers are congruent to 1 modulo 4?
Don't be lazy.

A simple Google search will reveal the answer. (hint: Wikipedia)

One might ask: what motivates this question? do you have some reason to
suspect that it is true?
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