20120123, 19:49  #12  
Dec 2011
2^{2}·3^{2} Posts 
Quote:
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 20120123 at 20:33 

20120123, 20:08  #13  
Apr 2010
5×31 Posts 
Quote:
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 20120123 at 20:38 

20120123, 20:45  #14  
Dec 2011
100100_{2} Posts 
Quote:
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)  (m1) from which I propose that m is a prime. 

20120123, 21:20  #15  
Apr 2010
5·31 Posts 
Quote:
Also, consider Batalov's second counterexample. 

20120123, 23:34  #16  
Dec 2011
44_{8} Posts 
Quote:
where q is a prime of the same form. In Batalov's second counterexample, 19 is not 2^(a prime)  1. 

20120123, 23:53  #17 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
19·23^{2} Posts 

20120124, 07:22  #18  
Apr 2010
5·31 Posts 
Quote:
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),m1) = m1 (where m=n_{5}) 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. 

20120124, 14:15  #19  
"Forget I exist"
Jul 2009
Dartmouth NS
2·3·23·61 Posts 
Quote:
2^31 = 2^(2^21)1 2^71=2^(2^31)1 = 2^(2^(2^21)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 20120124 at 14:29 

20120124, 14:46  #20  
"Forget I exist"
Jul 2009
Dartmouth NS
2·3·23·61 Posts 
Quote:


20120130, 20:03  #21 
Dec 2011
2^{2}·3^{2} Posts 
Carmichael Numbers
Is it true that all Carmichael numbers are congruent to 1 modulo 4?

20120130, 20:30  #22 
"Bob Silverman"
Nov 2003
North of Boston
1110101010010_{2} Posts 

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