# Re: General factoring result for k^m = q mod N

*From*: Mark Murray <w.h.oami@xxxxxxxxxxx>*Date*: Wed, 16 Jun 2010 19:01:39 +0100

On 13/06/2010 15:36, JSH wrote:

On Jun 13, 3:21 am, Mark Murray<w.h.o...@xxxxxxxxxxx> wrote:On 13/06/2010 03:27, JSH wrote:

> Yet I'm the one who found it, over 200 years since Gauss introduced

> "mod" in 1801.

1) Chinese remainder theorem.

2) Modular exponentiation.

Interesting, chased the link to Wikipedia for modular exponentiation

and that got me to wondering my result could be used to find e.

But your extensive "research" never got you to

3) Discrete logarithm.

http://en.wikipedia.org/wiki/Discrete_logarithm

Given c = b^e mod m, where c, b and m are known, yeah, it seems to me

that is should, potentially, maybe be possible using my result to

figure out e. But maybe not. I decided to stop thinking on it after

a point. Kind of overwhelming. So the rest may not be valid, but I

have to toss it out there anyway for national security reasons, as the

"unknown" is not good. It's bad.

Cool. Well guess that breaks something in encryption. NSA should

start looking for a new method, fast.

Not until you tell them how to factor fast enough.

M

--

Mark "No Nickname" Murray

Notable nebbish, extreme generalist.

.

**References**:**General factoring result for k^m = q mod N***From:*JSH

**Re: General factoring result for k^m = q mod N***From:*Mark Murray

**Re: General factoring result for k^m = q mod N***From:*JSH

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