Re: Critiquing surrogate factoring
From: C. Bond (cbond_at_ix.netcom.com)
Date: 03/30/05
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Date: Wed, 30 Mar 2005 20:38:08 GMT
C. Bond wrote:
> jstevh@msn.com wrote:
>
> > The idea is simple enough, factor one number and use its factorization
> > to get the factorization of another. The point being taking a number
> > that is hard to factor, and yet, factoring it, by factoring an easier
> > number.
> >
> > I thought it might help to try and write the the gist of it in a
> > theorem.
> >
> > Surrogate Factoring Theorem:
> >
> > Given M, a target natural number to be factored, and j, an integer
> > chosen such that j^2>M^2, a rational factor b_2 of M is given by
> >
> > b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2
> >
> > where T = M^2 - j^2, and f_1 is a rational factor of T, and where Az is
> > given by
>
> If j^2>M^2, as required above, then T is negative. Is that what you want?
Since James has apparently chosen not to answer this question, perhaps he will
answer another. Does the statement "f_1 is a rational factor of T" mean that
f_1 is *any* rational factor of T? Or does it mean that f_1 is a particular
rational factor T whose value is to be determined by some means? By the way,
for completeness, what is your definition of a "rational factor"?
-- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com
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