20120808, 04:49  #1 
May 2004
100111100_{2} Posts 
is 2^976005417520179872111 prime?
Is 2^p1 a Mersenne prime? Here p =
97600541752017987211 a prime number. 
20120808, 05:52  #2 
May 2004
2^{2}×79 Posts 
is2^p1 prime?

20120808, 05:57  #3 
May 2004
2^{2}×79 Posts 
is 2^p1 prime?

20120808, 06:14  #4 
Jul 2003
So Cal
2·3·347 Posts 
No, M3299 is not a factor of M97600541752017987211. Try again.

20120808, 07:21  #5 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
22411_{8} Posts 
How did you arrive at that number?
There are no factors through 2^97. 
20120808, 11:41  #6 
Aug 2002
Buenos Aires, Argentina
5·269 Posts 
That's right. Since gcd(2^{a}1,2^{b}1) = 2^{gcd(a,b)}1 (see lemma 2 at Will Edgington Mersenne Prime which is credited to Donald Knuth), 2^{p}1 cannot be a multiple of 2^{q}1 when both p and q are prime.

20120808, 13:11  #7 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20120809, 03:18  #8 
May 2004
2^{2}·79 Posts 
is 2^p1 prime?
Agreed; that was a hasty conclusion; on recheck both are nonfactors.

20120813, 05:41  #9 
May 2004
2^{2}×79 Posts 
is 2^p  1 prime?
Whether 2^P  1 is prime or not, all factors of 2^P  1 seem to have the shape P*k + 1 ( here P is any prime and k belongs to N ). Hence the factors of 2^p  1 must also have the shape
97600541752017987210*k + 1. 
20120813, 05:54  #10 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
16065_{8} Posts 
It's a well known fact that all Mersenne factors are of the form 2*p*k+1. So any factors must be 2*97600541752017987210*k + 1. Someone even mentioned this a few posts up.

20120813, 09:35  #11 
Apr 2010
Over the rainbow
43·59 Posts 

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