# Re: Surrogate factoring, update

From: Marcel Martin (mm_at_ellipsa.no.sp.am.net)
Date: 07/11/04

```Date: Sun, 11 Jul 2004 08:31:16 +0200

```

James Harris a écrit :
>
> Jean-Luc Cooke <jlcooke@engsoc.org> wrote in message news:<ccp66s\$73c\$1@driftwood.ccs.carleton.ca>...
> > James,
> >
> > Keep looking into this. You'll learn a lot. But honestly, this is just
> > a variation of Fermat's Difference of Squares factorization:
>
> Actually in many ways it is. It's where it's different that things
> get interesting.
>
>
> > pq = n
> >
> > M = mean of p and q
> > E = error from mean
> >
> > M = (p+q)/2
> > p = M+E
> > q = M-E
> >
> > n^2 = M^2 - E^2
> >
> > Using quadratic recipcosity you can eliminate many possible values, but
> > then end result is no net improvment in factorization problem.
>
> That's just one side of what I gave in my original post.
> [...]
> No, I didn't just come up with the standard congruence of squares.
> Here are the equations pulled out:
>
> (jk - Tk + T)(jk + Tk + T) = T^4
>
> k = (-jT +/- T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2)
>
> and
>
> j = (-T +/- T sqrt(k^2 + T^2))/k

Your method is essentially no more than a Fermat method.
Take both sides of (jk - Tk + T)(jk + Tk + T) = T^4 modulo T.
It comes (jk)^2 = 0 mod T and, since, T has no square factors
we can write jk = 0 mod T, i.e., jk = aT for some a.

(aT - kT + T)(aT + kT + T) = T^4
(a + 1 - k)(a + 1 + k)T^2 = T^4
(a+1)^2 - k^2 = T^2

Which is equivalent to something like n^2 = M^2 - E^2.
It leads nowhere, this is just a (uselessly complicated) Fermat's
method.

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