Re: Fermat conjecture; please post it
> theorem: iff F(n)== -1 mod (2^(n-1) +1), then F(n) is prime!did you mean "if" ?
But it's still wrong for n=16 (and n=36) :
F(16) = -1 (mod 2^15+1)
but F(16) is known to be composite,
as Peter wrote, it is divisible by 825753601.
> so, if (r +1)*b = 2^(2^n)+2, then b must equal 2; so, r +1=I think here you use that
2^(2^n-1)+1 has no factor of the form 2^h+1,
but this is not true in general...
For example with n=16 you have
2^65535+1 which is divisible by 2^3+1 = 9.
[Sorry, in my previous mail I misread the definition of r vs 2^h, please ignore...]