Hi Jeff
> >prime, unless we find a factor. Interestingly enough, when we find the
next
> >Mersenne prime, searching for a factor of M(M(p)) might allow us to find
an
> >even bigger prime. If for example, 6*M(p)+1 divides M(M(p)), then it
must
> >be prime!
> Which one must be prime? 6*M(p)+1, or M(M(p))?
> And why? Enquiring minds, and all.... Thanks!
Just to clarify... and make sure my own thinking isn't muddled on this
issue.
Let q be a factor of M(p), p a prime. We know all factors of M(p) are of the
form 2kp+1. So all factors of q must be of that form too. It follows then
that if q<(2p+1)^2, q is necessarily prime.
On the surface it looks like a great shortcut in primality testing, taking
as long as it would take to test a number the size of p to test a number
that could be as large as 4p^2. In practice though you'd have to be very
lucky to find such a q, and failure does not necessarily prove
compositeness. Perhaps though there's a lucky prime or two to be discovered
because of it :)
Chris
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