| Re: JSH: Newest factoring algorithm |
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Group: alt.math · Group Profile
Author: JSHJSH Date: Jun 10, 2008 21:40
On Jun 8, 7:22 pm, JSH gmail.com> wrote:
> After my latest miss-steps I went back, looked everything over again
> and finally noticed something that was bizarrely, simply obvious and
> staring me in the face.
>
> Here's the factoring algorithm that results.
>
> Given an odd target composite T, for a test odd prime p, iterate
> through non-zero integers k until
>
> T - k^2 mod p
>
> is a quadratic residue. If no k will work, discard that prime p and
> try another. If one does work then get á from
>
> á^2 = (T - k^2)(k^2)^{-1} mod p.
>
> Then you have residues of your factors modulo p from
>
> f_1 = ák mod p
>
> and
>
> f_2 = á^{-1}(1 + á^2)k mod p
>
> where f_1*f_2 = T.
>
I think there's a subtlety that is probably escaping most of you here.
Normally with f_1*f_2 = T mod p, you have, how many solutions for f_1
mod p and f_2 mod p?
But somehow, someway with these weird equations you have FEWER
solutions.
How is that possible?
___JSH
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