Hi, The _only_ incidence of 2^p-1 & 2^p+1 both being prime is p=2 yielding the prime pair (3, 5).
Here's a proof by induction: Consider the difference between the second successor of two consecutive Mersenne numbers with odd exponents: (2^(n+2)+1) - (2^n+1) = 2^(n+2) - 2^n = 2^n * (2^2 - 1) = 2^n * (4 - 1) which is clearly divisible by 3. Now 2^1 + 1 = 3 is divisible by 3, therefore 2^p+1 is divisible by 3 for _every_ odd p (irrespective of whether or not 2^p-1 is a Mersenne prime). However I believe there is at least one known Mersenne prime 2^p-1 for which it is not known whether (2^p+1)/3 is prime or composite. Regards Brian Beesley On Saturday 05 April 2003 18:47, Bjoern Hoffmann wrote: > Hi, > > I wondered if someone already have checked if the last mersenne > numbers +2 are double primes? > > like 3+5, 5+7, 9+11, 11+13 or 9 = 3^2 (well, usually) > > 824 633 702 441 > and > 824 633 702 443 > > regards > Bjoern > > _________________________________________________________________________ > Unsubscribe & list info -- http://www.ndatech.com/mersenne/signup.htm > Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers _________________________________________________________________________ Unsubscribe & list info -- http://www.ndatech.com/mersenne/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
