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Old 2006-03-25, 08:13   #1
Orgasmic Troll
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Default 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.
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Old 2006-03-25, 12:44   #2
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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.
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Old 2006-03-25, 13:09   #3
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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
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Old 2006-03-25, 15:50   #4
R.D. Silverman
 
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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.
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Old 2006-03-25, 17:05   #5
Orgasmic Troll
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Ahh, okay.

Thanks to both :)
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Old 2006-05-02, 02:23   #6
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Default 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.
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Old 2006-05-02, 16:28   #7
ewmayer
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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.
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Old 2006-06-11, 15:38   #8
ixfd64
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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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