20110704, 16:28  #1 
Apr 2010
Over the rainbow
976_{16} Posts 
Mersenne, another question
First of all, sorry, i don't know howto Latex.
Let p1 and p2 be random prime Let M(p1) and M(p2) be composite mersenne Let C be a number, factor of M(p1) or M(p2)>0 As we know, factor of M(P) are in the form of (2*k*P)+1 Does a couple (M(p1),M(p2)) exist where the factor of p2 is aqual to (((2*k*p1)*p2) +1)*C Last fiddled with by firejuggler on 20110704 at 16:29 
20110704, 16:43  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Let p1 and p2 be random prime Let M_{p1} and M_{p2} be composite mersenne Let C be a number, factor of M_{p1} or M_{p2}>0 As we know, factor of M_{P} are in the form of (2*k*P)+1 Does a couple (M_{p1},M_{p2}) exist where the factor of p2 is equal to (((2*k*p1)*p2) +1)*C and this didn't need anything but sub tags. and a underline. 

20110704, 16:51  #3  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20110704 at 16:59 

20110704, 17:07  #4  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20110704, 18:08  #5 
Dec 2010
Monticello
5·359 Posts 
I expect (warning: naively) such a number to exist....you might want to look among all the factored Mersenne numbers to see if there is a couple like that.

20110704, 19:26  #6  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts 
Quote:
(2k*p_{1}*p_{2}+1)*(2kP+1) =2Pk*2k*p_{1}*p_{2}+2kP+2k*p_{1}*p_{2}+1 =4Pk^{2}*p_{1}*p_{2}+2k*p_{1}*p_{2}+2kP+1 =(2*p_{1}*p_{2})(2Pk^{2}+k)+2kP+1 I don't think this can be a factor of p_{2} unless P=p_{2}, since otherwise it wouldn't necessarily be of the form 2kp_{2}+1. That means that C is a factor of p_{2}, not p_{1}, and so p_{1} hardly enters into the equation. But I think you're really asking this: Is there a number of the form 2*k*p_{1}*p_{2}+1 that divides both M(p_{1}) and M(p_{2})? I'm with Christenson: my gut says yes, and I don't know of anything preventing the possibility, but I don't know of any specific examples. I'm going to try some brute force searching real quick to see if I can find any examples... 

20110704, 20:25  #7  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts 
Quote:
Edit: Will Edgington's list of Mersenne factors contains no prime twice. I find it unlikely this is a coincidence, as the list starts out quite dense: it includes 143 of the 168 primes below 1000. Edit 2: Just found on http://www.garlic.com/~wedgingt/mersenne.html, "Lemma 1: A prime number divides at most one primeexponent Mersenne" (with proof) So the answer to "Is there a number of the form 2*k*p1*p2+1 that divides both M(p1) and M(p2)?" is no. I never noticed this fact before, though I have visited W.E.'s pages in the past. Last fiddled with by MiniGeek on 20110704 at 21:00 

20110704, 20:27  #8 
Apr 2010
Over the rainbow
2422_{10} Posts 
what i was meaning is that
M(p1) has a factor of (2*k*p1)+1 and M(p2) has a factor of (2*k*p1*p2)+1 I was using P as a 'general case the C i used is a nonspecific cofactor 
20110704, 21:59  #9  
Nov 2003
2^{2}×5×373 Posts 
Quote:
problem in a first year number theory class. 

20110704, 22:09  #10  
Nov 2003
2^{2}·5·373 Posts 
Quote:
Ask: what is GCD(2^a1, 2^b1)???? When are you people going to stop prattling and making meaningless "conjectures" and pick up and READ some number theory books???? The question under discussion is well within the scope of any 1st year number theory book. 

20110704, 22:30  #11 
Apr 2010
Over the rainbow
2×7×173 Posts 
why do you think i put it under 'Miscellaneous Math Threads' ? i knew i was risking your anger would it be under ' Math'

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