Re: Checking my logic, surrogate factoring
jstevh_at_msn.com
Date: 03/05/05
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Date: 5 Mar 2005 04:59:17 -0800
jstevh@msn.com wrote:
> I have one thread started with the correct perfect algorithm for
> surrogate factoring, and I will continue to reply in it and monitor
it
> closely.
>
> However I still want to start another thread to talk out the
solution,
> because I feel like it.
>
> yx^2 + Ax - M^2 = 0
>
> and
>
> yz^2 + Az - j^2 = 0
>
> with T = M^2 - j^2
>
> and I have easily after using
>
> y = (M^2 - Ax)/x^2 = (j^2 - Az)/z^2
>
> z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
>
> where wrapped up in the square root is Ax, but I have z and x naked
> outside, but related as a ratio.
>
> That is,
>
> z/x = (-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
>
> and I can get integer Ax by iterating through the factors f_1 and f_2
> such that
>
> f_1 f_2 = Tj^2.
>
> That's kind of the old news.
>
> But the important thing that I didn't put into proper perspective is
in
>
> y = (M^2 - Ax)/x^2 = (j^2 - Az)/z^2
>
> (M^2 - Ax)/x^2 = (j^2 - Az)/z^2
>
> so if the numerator of x, as x is often a fraction, has prime factors
> in common with M, then some of those will divide up into the
numerator.
>
> But with z, they're blocked as j is coprime to M.
>
> So necessarily there is an asymmetry between x and z in terms of
having
> prime factors of M.
>
Yes, if x has M itself as a factor, otherwise A may simply divide off
prime factors of x in common with M, and that allows x and z to match.
That was the possibility I was ignoring, which invalidates the
argument.
I noticed that after replying to a sci.crypt poster who tried my idea
and it didn't work, as I calculated with some of his numbers and found
that Ax was coprime to M.
That left me focusing on solving for x, A or y, and so I began probing
that area, which lead to an easy answer (so it may be wrong, sigh)
which involves getting two linear equations relating y and A^2, which I
just realized can't work as in both cases y is simply related to A^2 as
a ratio.
Oh well, back to research...
James Harris
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