20110904, 16:00  #1 
"James Heinrich"
May 2004
exNorthern Ontario
2^{2}×13×73 Posts 
Can two Mersenne numbers share a factor?
Is it possible for two composite Mersenne numbers to have a common factor?
In the 33,415,185 factors I've looked at there are no duplicates, but I don't know if this is by chance or guaranteed? 
20110904, 16:02  #2 
Jun 2003
2^{2}·3·449 Posts 
GCD(2^a1,2^b1)=2^GCD(a,b)1. So, for prime exponent mersennes, the answer is no.

20110905, 14:32  #3 
"Åke Tilander"
Apr 2011
Sandviken, Sweden
1000110110_{2} Posts 
Are all primes ±1 mod 8 factors?
... and then you could ask yourself if all primes of the form ±1 mod 8 are a factor to a specific prime exponent Mersenne number each?
Last fiddled with by aketilander on 20110905 at 14:48 
20110905, 17:13  #4 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
1010001100001_{2} Posts 
Mersenne factors are of the form (2k^p)+1.
Where p is the mersenne prime and k is any whole, positive number. That would suggest that cannot be duplicated. 
20110905, 17:20  #5 
"Bob Silverman"
Nov 2003
North of Boston
2^{3}×3×311 Posts 

20110905, 17:46  #6 
"Forget I exist"
Jul 2009
Dumbassville
3·2,797 Posts 

20110905, 17:47  #7  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
2·3·23·31 Posts 
That's 2kp+1.
Quote:
As axn already correctly stated, no two primeexponent Mersenne numbers share a factor. It's just not suggested by 2kp+1. Last fiddled with by MiniGeek on 20110905 at 17:49 

20110906, 02:59  #8  
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
1010001100001_{2} Posts 
Oops. Yes that's what I meant. I was in too much of a hurry this AM.
Quote:


20110906, 03:38  #9 
Jun 2003
2^{2}×3×449 Posts 
Even if k is always an even number (which it needn't be), 2kp1p2+1 can be rewritten as 2Kp1+1 or 2Kp2+1 (where K=kp1 or K=kp2 will also be an even number, since even * odd = even)

20110906, 08:20  #10 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2·3·1,669 Posts 
Actually... if we start making gcd's, then a stronger affirmation could be true, isn't is? like "two mersenne numbers with coprime exponents can not share the same factor". Or am I wrong?
For example, assuming k divides 2^{a}1 and 2^{b}1, with a and b not necessarily prime, a>b, then it divides their difference too, which should be 2^{a}12^{b}+1, or 2^{b}(2^{ab}1). As k is odd, it can't divide 2^b, so it divides the parenthesis. Repeating the process with 2^{ab}1 and any one from the initially 2 mersenne involved, we could see that k divides 2^11=1, when gcd(a,b) is 1, as we already recognize the "gcding" process for the exponents. That is, 2^{15}1 and 2^{28}1 (substitute 15 and 28 with any 2 coprimes) can't have common factors, even if 15 and 28 are not primes. Particularly, the affirmation is true for primes, as they are always coprime. If this reasoning is right and I am not missing something, then we can answer yes for the general case: two mersenne numbers never share a common factor. Except of course in the obvious case when their exponent share a common factor. In this case both 2^{pa}1 and 2^{pb}1 are divisible (algebrically) by 2^{p}1. Last fiddled with by LaurV on 20110906 at 08:40 
20110906, 11:53  #11 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
4278_{10} Posts 
Yep, this is right, implied by the formula GCD(2^a1,2^b1)=2^GCD(a,b)1: If a and b are coprime, GCD(a,b)=1, so GCD(2^a1,2^b1) is also 1.

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