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 2010-05-03, 14:04 #12 Gammatester     Mar 2009 468 Posts Although English is not my native langage, I think your Proposition 1: Code: If M_p is not prime, when this is true: phi(M_p) = 0 mod p^2 should read Code: If M_p is not prime, then this is true: phi(M_p) = 0 mod p^2. But M_4 = 15 is a counterexample because phi(15) mod 16 = 8. Or am I missing something here, e.g. that p should be prime?
2010-05-03, 14:25   #13
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

22×1,877 Posts

Quote:
 Originally Posted by Gammatester Although English is not my native langage, I think your Proposition 1: Code: If M_p is not prime, when this is true: phi(M_p) = 0 mod p^2 should read Code: If M_p is not prime, then this is true: phi(M_p) = 0 mod p^2. But M_4 = 15 is a counterexample because phi(15) mod 16 = 8. Or am I missing something here, e.g. that p should be prime?
If p is prime, the conjecture is true and TRIVIAL.

Given N = 2^p-1, and N is composite then N is the product of at least two
primes, each of which is 1 mod p. phi(N) will be divisible by p^k, where
k is the number of distinct prime factors of N.

This does not merit calling it a 'conjecture'. It is an elementary homework
problem that one might assign to a beginning number theory class.

2010-05-07, 06:08   #14
sascha77

Jan 2010
germany

2×13 Posts

Quote:
 Originally Posted by R.D. Silverman If p is prime, the conjecture is true and TRIVIAL. Given N = 2^p-1, and N is composite then N is the product of at least two primes, each of which is 1 mod p. phi(N) will be divisible by p^k, where k is the number of distinct prime factors of N. This does not merit calling it a 'conjecture'. It is an elementary homework problem that one might assign to a beginning number theory class.
Hello Silverman, Yes i mean that p must be Prime. M_p is the abbreviation of Mersenne-number with p prime. Thank you, that you showed me, that Proposition 1 is trivial. But it was not the conjecture , the conjecture is in top of the pdf. (Conjecture 1) And I showed that proposition1 is true, because i need this for example in proposition5. And I wasnt shure if everybody see the Proposition1 is trivial. So I simply showed that this is true. I have question to you: Let M_p be a Mersenne-PRIME Number (M_p is Prime). When it has the form : M_p = 2^p-1 = 2*k*p+1 , with k is natural number. Can you show that k==0 (mod p) is always false. Perhaps it is also trivial, I do not know. Sascha

2010-05-07, 11:09   #15
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

11101010101002 Posts

Quote:
 Originally Posted by sascha77 Hello Silverman, Yes i mean that p must be Prime. M_p is the abbreviation of Mersenne-number with p prime. Thank you, that you showed me, that Proposition 1 is trivial. But it was not the conjecture , the conjecture is in top of the pdf. (Conjecture 1) And I showed that proposition1 is true, because i need this for example in proposition5. And I wasnt shure if everybody see the Proposition1 is trivial. So I simply showed that this is true. I have question to you: Let M_p be a Mersenne-PRIME Number (M_p is Prime). When it has the form : M_p = 2^p-1 = 2*k*p+1 , with k is natural number. Can you show that k==0 (mod p) is always false. Perhaps it is also trivial, I do not know. Sascha
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 2010-05-08, 00:33 #16 cheesehead     "Richard B. Woods" Aug 2002 Wisconsin USA 22·3·641 Posts Also, Mathworld (http://mathworld.wolfram.com/) is your mathematical friend, usually. http://mathworld.wolfram.com/search/...erich&x=10&y=9

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