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 2006-03-25, 08:13 #1 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 641 Posts search for MMM127 small factors? Would the allure of MM127's possible primality be diminished if a Catalan-Mersenne number was found to be composite? Has anyone searched for small factors of the CM numbers greater than MM127? mild tangent: I'm not aware if attempting to find small factors of these monstrous numbers is even feasible, so, if I'm talking out of my arse, leave your inflammatory comments at the door. Frankly, I'm too busy with my other math classes to go learning computational mathematics from scratch and buying \$70 books to do so, nor is the question compelling enough for me to go through that much toil. If you want to make toast but you're out of bread, you don't go out and plant wheat.
 2006-03-25, 12:44 #2 alpertron     Aug 2002 Buenos Aires, Argentina 33×72 Posts Travis, It's not a bad idea. Since MM127 is almost surely composite, MMM127 factors should not have the form 2*k*MM127+1. My program to compute factors of googolplexplex could be adapted to perform trial division by small factors (say less than 10^12). If MMM127 has such a small factor, we are sure that MM127 is not prime.
 2006-03-25, 13:09 #3 alpertron     Aug 2002 Buenos Aires, Argentina 33×72 Posts What I wrote in my previous post is not correct. If MM127 is equal to the product of primes a*b*c*..*z, then after dividing MMM127 by 2^a-1, 2^b-1, etc. (which are very large numbers whose factors have the form 2*k*a+1, 2*k*b+1, which are out of reach) we get the primitive factor. The factors of this primitive factor have the form 2*k*MM127+1. So if we cannot find a factor of MM127 we also cannot expect to find a factor of MMM127. Last fiddled with by alpertron on 2006-03-25 at 13:10
2006-03-25, 15:50   #4
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by alpertron What I wrote in my previous post is not correct. If MM127 is equal to the product of primes a*b*c*..*z, then after dividing MMM127 by 2^a-1, 2^b-1, etc. (which are very large numbers whose factors have the form 2*k*a+1, 2*k*b+1, which are out of reach) we get the primitive factor. The factors of this primitive factor have the form 2*k*MM127+1. So if we cannot find a factor of MM127 we also cannot expect to find a factor of MMM127.
This is correct.

 2006-03-25, 17:05 #5 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 641 Posts Ahh, okay. Thanks to both :)
 2006-05-02, 02:23 #6 David John Hill Jr     Jun 2003 Pa.,U.S.A. C416 Posts one assurence numerically speaking If one could show x|(x-1)! evenly and with multiples of 5, you will be quaranteed of finding factors . So also for m^x 127.
2006-05-02, 16:28   #7
ewmayer
2ω=0

Sep 2002
República de California

101100111111102 Posts

Quote:
 Originally Posted by alpertron The factors of this primitive factor have the form 2*k*MM127+1.
Good luck doing arithmetic modulo numbers of this size.

 2006-06-11, 15:38 #8 ixfd64 Bemusing Prompter     "Danny" Dec 2002 California 2×1,151 Posts I think that testing one trial factor of MMM127 (2k * MM127 +1) takes roughly the same amount of computation as LL-testing MM127.

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