 mersenneforum.org Mersenne, another question
 Register FAQ Search Today's Posts Mark Forums Read  2011-07-04, 16:28 #1 firejuggler   Apr 2010 Over the rainbow 22×641 Posts Mersenne, another question First of all, sorry, i don't know how-to 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 2011-07-04 at 16:29   2011-07-04, 16:43   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20C016 Posts Quote:
 Originally Posted by firejuggler First of all, sorry, i don't know how-to 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
ctan.org helps and so does the tex tags.

Let p1 and p2 be random prime
Let Mp1 and Mp2 be composite mersenne
Let C be a number, factor of Mp1 or Mp2>0
As we know, factor of MP are in the form of (2*k*P)+1
Does a couple (Mp1,Mp2) 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.   2011-07-04, 16:51   #3
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts Quote:
 Originally Posted by science_man_88 ctan.org helps and so does the tex tags. Let p1 and p2 be random prime Let Mp1 and Mp2 be composite mersenne Let C be a number, factor of Mp1 or Mp2>0 As we know, factor of MP are in the form of (2*k*P)+1 Does a couple (Mp1,Mp2) 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.
now to answer the question to be of the form 2*t*p2+1 t = k*p1 okay so this then becomes a matter of the mod 3 remainder of t to determine what p2 could be. because for p2=6n-1 t = 1 or 0 mod 3 will work to give a 6n+1 Mersenne number, 6n+1 t= 2 or 0 mod 3 works. oddly enough I was looking at these relations last night by hand for other reasons.

Last fiddled with by science_man_88 on 2011-07-04 at 16:59   2011-07-04, 17:07   #4
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

838410 Posts Quote:
 Originally Posted by science_man_88 now to answer the question to be of the form 2*t*p2+1 t = k*p1 okay so this then becomes a matter of the mod 3 remainder of t to determine what p2 could be. because for p2=6n-1 t = 1 or 0 mod 3 will work to give a 6n+1 Mersenne number, 6n+1 t= 2 or 0 mod 3 works. oddly enough I was looking at these relations last night by hand for other reasons.
okay that wasn't quite an answer I know , but I'm not a number theory person, R.D. Silverman likely has a clue or more extra to help.   2011-07-04, 18:08 #5 Christenson   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.   2011-07-04, 19:26   #6
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

10000101010112 Posts Quote:
 Originally Posted by firejuggler First of all, sorry, i don't know how-to 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
Not sure what you're trying to get at with (((2*k*p1)*p2) +1)*C. Where C is a factor of M(px), let P=px. That final expression is equivalent to:

(2k*p1*p2+1)*(2kP+1)
=2Pk*2k*p1*p2+2kP+2k*p1*p2+1
=4Pk2*p1*p2+2k*p1*p2+2kP+1
=(2*p1*p2)(2Pk2+k)+2kP+1

I don't think this can be a factor of p2 unless P=p2, since otherwise it wouldn't necessarily be of the form 2kp2+1. That means that C is a factor of p2, not p1, and so p1 hardly enters into the equation.

But I think you're really asking this:
Is there a number of the form 2*k*p1*p2+1 that divides both M(p1) and M(p2)?
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...   2011-07-04, 20:25   #7
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17·251 Posts Quote:
 Originally Posted by Mini-Geek 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...
After searching all combinations where p1 < p2 < 500, k < 10000, and up to p < 1000 with k < 1000, and finding no results, I'm no longer so sure this happens. If it does, it would seem to be rather rare. Of course, it's possible that I've done something wrong with the code. The speed (or lack thereof) makes me not want to go farther with this particular implementation (very simple Java - doesn't even check if the potential factor is +-1 mod 8 or prime, just does a BigInteger.mod on everything). It's finding enough factors that I don't have much reason to doubt it, but you never know.
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 prime-exponent 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 Mini-Geek on 2011-07-04 at 21:00   2011-07-04, 20:27 #8 firejuggler   Apr 2010 Over the rainbow 256410 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 non-specific cofactor   2011-07-04, 21:59   #9
R.D. Silverman

Nov 2003

746010 Posts Quote:
 Originally Posted by Mini-Geek But I think you're really asking this: Is there a number of the form 2*k*p1*p2+1 that divides both M(p1) and M(p2)? I'm with Christenson: my gut says yes, and I don't know of anything preventing the possibility, ...
I do. It is an elementary exercize that might be given as a homework
problem in a first year number theory class.   2011-07-04, 22:09   #10
R.D. Silverman

Nov 2003

22·5·373 Posts Quote:
 Originally Posted by R.D. Silverman I do. It is an elementary exercize that might be given as a homework problem in a first year number theory class.
I will even give a hint:

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.   2011-07-04, 22:30 #11 firejuggler   Apr 2010 Over the rainbow 1010000001002 Posts why do you think i put it under 'Miscellaneous Math Threads' ? i knew i was risking your anger would it be under ' Math'   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post paulunderwood Miscellaneous Math 1 2016-01-24 01:41 ShiningArcanine Math 21 2012-04-27 01:38 LaurV Math 11 2012-03-16 12:10 rong123 Marin's Mersenne-aries 7 2007-11-09 00:34 stpascu Factoring 1 2006-10-16 16:31

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