20140309, 00:06  #1 
Dec 2010
2·37 Posts 
Modular restrictions on factors of Mersenne numbers
Could anybody please help me understand why any factor of Mp must be of the form 2*k*p+1?
I understand that any factor q divides 2^(q1)1, due to Fermat's little theorem, and I understand that (apparently) p divides q1, which easily leads to the conclusion. I don't understand how FLT leads to p dividing q1. Is anybody out there knowledgeable and kind enough to help me understand this piece of the Mersenne puzzle? 
20140309, 00:33  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010011010111_{2} Posts 
There's this page at UTM. Maybe someone will explain that in easier terms?
___________________________________________ Interestingly (similarly),
Last fiddled with by Batalov on 20140309 at 00:43 Reason: GQ and EQ's 
20140309, 00:45  #3  
Dec 2010
2×37 Posts 
Quote:


20140309, 01:20  #5  
Nov 2003
2^{2}×5×373 Posts 
Quote:
by itself. Instead, look up Lagrange's Theorem. Definition: Within a group G, the order of an element a is the smallest integer x, such that a^x = e, where e is the group identity. 

20140309, 04:34  #6 
May 2013
East. Always East.
11·157 Posts 
2kp+1 or +/ 1 (mod 8). The second one is important. 2^61 = 63 = 7*9.
Now whether or not this proof is good for composite exponents is unknown to me. Last fiddled with by TheMawn on 20140309 at 04:34 
20140309, 12:51  #7 
Dec 2010
2·37 Posts 
Now I get it. Thank you. The order is the smallest exponent such that the congruency is one. 2^p is congruent to one, and p is prime, so p is the order. It's the smallest possible exponent. Therefore, q1 is divisible by p, and the rest is simple algebra.
Beautiful. TheMawn, if the exponent is composite, then 2^c is congruent to one, but c may not be the order as c is not prime. If d is a factor of c, 2^d could be congruent to one, and d could be the order. While d would divide q1, c might not. So for composite exponents, I suppose one of the factors of the composite would likely replace the "p" value in "2kp+1". As for the +/ 1 (mod 8) rule, I don't understand quadratic residues yet. I'll get there though! Last fiddled with by siegert81 on 20140309 at 12:59 
20140309, 13:58  #8  
Nov 2003
1D24_{16} Posts 
Quote:
Quote:
Exercize for the student: determine what he said that was wrong. 

20140309, 16:03  #9 
Dec 2010
2×37 Posts 
Actually, I *DO* get it, but I used the letter p instead of q, on accident.
If p is a prime factor of Mq, then 2^q is congruent to 1 (mod p). Any power of 2 that is congruent to 1 (mod p) must be a multiple of the order of 2 (mod p). Since q is prime, q is the order of 2 (mod p). Due to FLT, 2^(p1) is congruent to 1 (mod p). As a result, p1 is divisible by q, and p=2kq+1. Last fiddled with by siegert81 on 20140309 at 16:08 
20140310, 12:51  #10 
Nov 2003
2^{2}·5·373 Posts 

20140311, 04:16  #11 
Aug 2010
Kansas
547 Posts 
Quoted for posterity's sake. Your proofreading comment becomes rather ironic given your grammatical error in the above quote.

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