On 28 Apr 2001, at 0:57, [EMAIL PROTECTED] wrote:
> > Question (deep) - if we did discover a factor of 2^(2^727-1)-1,
> > would that help us to find a factor of 2^727-1 ?
The reason for asking this question was twofold: firstly, to find out
whether it _might_ be possible to know a factor of 2^c-1 without
knowing a factor of c, and secondly, if this statement is not true,
(in other words, if finding a factor of M(M(p)) automatically leads
to the derivation of a factor of M(p)) to possibly find a factor of
the recalcitrant number M(727) by attempting to factor M(M(727)),
which probably hasn't had much effort expended on it. Yet.
>
> I am skeptical too. Show us how the factors
>
> 131009 of M_(M11) = 2^(2^11 - 1) - 1
> 724639 of M_(M11)
> 285212639 of M_(M23)
>
> lead to factorizations of M11 and M23. Why don't the factors
>
> 231733529 of M_(M17)
> 62914441 of M_(M19)
>
> lead to similar factorizations of M17 and M19?
The "obvious" answer here is that 11 and 23 are "3 mod 4" Sophie
Germain primes, whereas neither 17 nor 19 are. I tinkered around a
bit with the algebra but wasn't able to come up with a formal
justification for this distinction.
The point of course is that there is a formal proof that, if a prime
p is congruent to 3 modulo 4 and 2p+1 is also prime, then 2^p-1 is
divisible by 2p+1 - which makes searching for a factor of M(p) by
trying to factor M(M(p)) somewhat pointless!
>
> With c = 2^727 - 1, 2^751 - 1, 2^809 - 1, 2^997 - 1, 2^1051 - 1, I
> looked for factors 2*k*c + 1 of 2^c - 1, but found none with k <=
> 20000. I invite those with a special search program for M_(M127)
> factors to search these further.
None of 727, 751, 809, 997 & 1051 are "3 mod 4" S-G primes (else they
wouldn't be on the "factors wanted" lists!)
Regards
Brian Beesley
1775*2^332181+1 is prime! (100000 digits) Discovered 22-Apr-2001
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