Quote:
Originally Posted by a1call
Aren't all Mersennenumbers with Primeexponents, Fermat'sProbablePrime in base 2^n?

That is true.
For any N if you consider the b bases for that N is a Fermat pseudoprime then these bases form a group in Z_N.
For Mersenne numbers this means that 2^n is such a base, because mp is a Fermat pseudoprime for base=2, fortunately
these means only p such bases, because 2^p==2^0 mod mp.
In an elementary way without group:
you need: (2^n)^(2^p1)==2^n mod (2^p1)
but we have: 2^p=a*p+2
hence: (2^n)^(2^p1)==2^(n*(a*p+1))==2^n mod (2^p1)
what we needed.
ps. this is the reason why we are using base=3 for Fermat testing the Mersenne numbers, base=2,4,8 etc is "bad". But you shouldn't fix base=3 to all numbers, because for other type of numbers: N could be a (trivial) pseudoprime for base=3, and you need to choose another base.