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From: Mark Murray on 21 Jul 2010 02:43
On 21/07/2010 03:02, JSH wrote:
> You're on your on.
>
> I am walking away.

Next time (and there /will/ be a next time), concede with good
grace.

M
--
Mark "No Nickname" Murray
Notable nebbish, extreme generalist.
From: José Carlos Santos on 21 Jul 2010 04:42
On 21-07-2010 3:02, JSH wrote:

> I have no problem with wandering off for YEARS, and am in the process
> of doing just that, while the paper at the Annals may offer a chance
> of something happening,

In your dreams.

> I am walking away.

As much as a boomerang.

Best regards,

Jose Carlos Santos
From: JSH on 21 Jul 2010 10:19
On Jul 21, 1:42 am, José Carlos Santos <jcsan...(a)fc.up.pt> wrote:
> On 21-07-2010 3:02, JSH wrote:
>
> > I have no problem with wandering off for YEARS, and am in the process
> > of doing just that, while the paper at the Annals may offer a chance
> > of something happening,
>
> In your dreams.
>
> > I am walking away.
>
> As much as a boomerang.

I'm letting this matter drop as k^m = q mod N is encryption relevant.
The equation ITSELF regardless of any of my claims around it.

My point is that I can walk away for a couple of years from THIS
RESULT not that I've stopped posting.

For instance there is still currently discussion about twin primes--a
much safer result.

Sometimes I get the impression that some of you are simply just not
that bright.

I'm still leaving open the door to argue about prime gaps or anything
else--not encryption related--which for me leaves open just about all
of number theory.


James Harris
From: MichaelW on 21 Jul 2010 23:47
On Jul 22, 10:03 am, JSH <jst...(a)gmail.com> wrote:
>
> There is no accepted method for doing that, while my approach isn't
> being accepted.
>
> If there is, demonstrate it with k^m = 52 mod 103.
>

52^1 = 52 mod 103.

What's my prize?

Joking aside I assume you meant to specify one of k or m. Also you
might want to use larger numbers as two digit numbers are too easy to
do with methods that don't work as efficiently for larger values.

Regards, Michael W.
From: Tim Little on 22 Jul 2010 00:22
On 2010-07-22, JSH <jstevh(a)gmail.com> wrote:
> Give a technique for finding m, when k^m = q mod N, with k, q and N
> known, by integer factorization in reply

Sure. Since you appear to ignore almost all links that don't go to
Wikipedia or Mathworld, try:
http://en.wikipedia.org/wiki/Index_calculus_algorithm
and
http://en.wikipedia.org/wiki/Pohlig-Hellman_algorithm

Both of these involve factorizations to solve the discrete log
problem. Feel free to ask if you don't understand anything there.

Note that these methods are massive overkill for the trivial problems
you post here. It would be like using the Space Shuttle to go next
door instead of into space. They are useful for computer solution of
much bigger problems.


There is also a website with Java applet and source code that solves
discrete logarithms using the Pohlig-Hellman algorithm:

http://www.alpertron.com.ar/DILOG.HTM

Not that you will look at it, as you are no doubt scared of what you
might find there. I mention it only since other people reading this
thread might be interested.


> If there is, demonstrate it with k^m = 52 mod 103.

You haven't specified k, so m can be just about anything. Feel free
to provide a value of k though.


- Tim
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