Re: Factoring with cubic equations?

From: Simon Johnson (Ckwop_at_hotmail.com)
Date: 10/05/03 

Date: 5 Oct 2003 08:18:02 -0700

"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.

Finding your polynomials is probaby harder than using GNFS on big
numbers.

Simon.

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