Re: Surrogate factoring, update
From: Marcel Martin (mm_at_ellipsa.no.sp.am.net)
Date: 07/11/04
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Date: Sun, 11 Jul 2004 08:31:16 +0200
James Harris a écrit :
>
> JeanLuc 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 = ME
> >
> > 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.
 mm http://www.ellipsa.net/ mm@ellipsa.no.sp.am.net ( suppress no.sp.am. )
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