Re: Factoring with cubic equations?

From: lapin des pyrenees (moc_at_com)
Date: 10/06/03 

Date: Sun, 5 Oct 2003 19:30:52 -0400

"Simon Johnson" <Ckwop@hotmail.com> wrote in message
news:f5668ae7.0310050718.6f75490b@posting.google.com...
> "lapin des pyrenees" <moc@com> wrote in message
news:<vT2dnd0IK4_jWOOiU-KYjA@golden.net>...
> > the reason we use f(x)=x^2 is not because it's the "quickest non-linear
> > polynomial ", the real reason is because we are too dumb to derive a
quicker
> > polynomial!
> >
> > btw, I derived a way to generate all the squares used in the QS method.
> >
> > Note: I did not say that: given a number N I can tell you the squares
x^2
> > and y^2. All I am saying is that there exist a way to generate all those
> > squares independently of the corresponding number N to which they
relate.
> >
> > If this result is important,by that I mean ( as a question ) will it
make
> > the search for the squares related to a given number faster if we knew
where
> > to look?
> > if it is faster, I will release the way it's done.
> >
> > le lapin des vosges
> >
>
> I conjecture (someone correct me if i'm wrong) that the problem of
> finding a polynomial that colides quicker than f(x) = x^2 mod n is
> roughly equal to the amount of effort to brute-force that collision
> with f(x) = x^2 mod n.
>

the nice thing about " I conjecture" is that it's easy and you don't have to
prove a thing!

> Finding your polynomials is probaby harder than using GNFS on big
> numbers.
>
how can you say it's harder if you don't even know how " my method " works.
what is the basis or your judgment? this is about science not politics. how
about some hard work on your part that demonstrate something useful instead
of conjecturing...

I conjecture that you don't understand how the method works! prove me wrong!

le lapin des alpes maritimes
vive la france

> Simon.

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