Brian J. Beesley [EMAIL PROTECTED] wrote,
in response to my earlier message
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
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
On 26 Apr 2001, Brian J. Beasley wrote
On 26 Apr 2001, at 6:34, Hans Riesel wrote:
Hi everybody,
If 2^p-1 is known to be composite with no factor known, then so is
2^(2^p-1)-1.
Much as I hate to nitpick a far better mathematician than myself,
this is seems to be an
On 26 Apr 2001, at 6:34, Hans Riesel wrote:
Hi everybody,
If 2^p-1 is known to be composite with no factor known, then so is
2^(2^p-1)-1.
Much as I hate to nitpick a far better mathematician than myself,
this is seems to be an overstatement.
It is certainly true that, for composite c,
On Thu, 26 Apr 2001 06:34:07 +0200 (MET DST), Hans Riesel wrote:
Hi everybody,
If 2^p-1 is known to be composite with no factor known, then so is
2^(2^p-1)-1.
For that matter, the same argument can be made with regard to
2^(2^R-1)-2 for some RSA factoring challange number R, and probably
Hans Riesel wrote: Hi everybody,
If 2^p-1 is known to be composite with no factor known, then so is
2^(2^p-1)-1.
Hans Riesel
It has been a long time since I have seen a more elegant
argument ender than this. Thanks Hans! spike
