# 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 :

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