20100503, 11:04  #1 
Sep 2009
2^{2}·3^{2} Posts 
Number of Factors for a Mersenne Number
Dear All,
Sometimes I see this expression: Let Mq=u*v = (2aq+1)(2bq+1) where u and v are primes and Mq a mersenne number. (Like in Sascha Pfallers Mersenne conjecture). This is true if there are only 2 prime factors of Mq. If we have more than 2 factors then (2bq+1) is not a prime but a product of primes. then (2bq+1)= (2cq+1) (2dq+1) (2eq+1) (2fq+1) .... Am I right or am I missing something? Is there a way to detect if Mq has only 2 factors without trial factoring or any other method? 
20100503, 11:29  #2  
"Bob Silverman"
Nov 2003
North of Boston
5^{2}·13·23 Posts 
Quote:
Quote:
the number of prime factors is to factor the number. Last fiddled with by R.D. Silverman on 20100503 at 11:29 Reason: typo 

20100503, 11:45  #3  
Sep 2009
2^{2}×3^{2} Posts 
Quote:
I assume that there is no way to know if Mq has more than 2 factors besides factoring... 

20100503, 12:25  #4  
Sep 2009
2^{2}·3^{2} Posts 
Quote:
Is there a known rule for (2bq+1) as (2bq+1)= (2cq+1) (2dq+1) (2eq+1) (2fq+1) .... is not a prime and all the prime divisors are congruent to + 1 mod 8. 

20100503, 12:39  #5 
"Bob Silverman"
Nov 2003
North of Boston
5^{2}·13·23 Posts 

20100503, 12:42  #6  
Sep 2009
2^{2}·3^{2} Posts 
Quote:
(2bq+1)= (2cq+1) (2dq+1) (2eq+1) (2fq+1) .... is not a prime and all the prime divisors are congruent to + 1 mod 8. Should I replace the word rule with property? 

20100503, 12:54  #7  
"Bob Silverman"
Nov 2003
North of Boston
5^{2}·13·23 Posts 
Quote:
arithmetic? Trivially if a = b = 1 mod 8, then ab = 1 mod 8 as well. If a = 1 mod 8 and b = 1 mod 8 then ab = 1 mod 8........ Why is this a mystery? 

20100503, 12:58  #8 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10266_{8} Posts 
It's already known that any and all factors q (whether prime, as proven here, or composite, as trivially proven at the end of this post given the previous proof concerning prime factors) of a Mersenne number (2^p1 with p prime) must be of the form q=2kp+1 and q=+1 mod 8.
This is quite useful for trial factoring, but I don't see how what you're doing is helpful... Proof that all composite factors also follow the rules for prime factors: (where 2ap+1 and 2bp+1 are prime factors, show that their product can be expressed as 2kp+1 by determining k) (2ap+1)*(2bp+1)=2kp+1 4abp^2+2bp+2ap+1=2kp+1 4abp^2+2bp+2ap=2kp 2abp^2+bp+ap=kp 2abp+b+a=k And, two numbers p and q that are + 1 mod 8 will always multiply to + 1 mod 8. (e.g. when p and q are both equal to 1 mod 8, their product is equal to (1)(1) mod 8, which is 1 mod 8) Last fiddled with by MiniGeek on 20100503 at 13:05 
20100503, 13:23  #9  
Sep 2009
2^{2}·3^{2} Posts 
Quote:
Quote:
2kp+1 in your example is equal to + 1 mod 8. Is it equal to a value modulus other than 8? 

20100503, 13:33  #10  
Einyen
Dec 2003
Denmark
6470_{8} Posts 
Quote:
Yes, they are equal to one of these 16 numbers mod 120: +1, +7, +17, +23, +31, +41, +47, +49 

20100503, 13:36  #11 
Sep 2009
2^{2}·3^{2} Posts 

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