Re: I was right, surrogate factoring proof
From: David Kastrup (dak_at_gnu.org)
Date: 02/13/05
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Date: Sun, 13 Feb 2005 19:41:39 +0100
jstevh@msn.com writes:
> Well, let me give just a bit of mathematics, and the theorem this
> time, followed by proof, in line with the convention one poster
> keeps bringing up.
>
> Given a natural number M, a natural number j can be chosen such that
> the factorization of M^2 - j^2 and j will factor M.
A "factorization" does not "factor M". And "can be chosen" does not
mean that it is efficient to do so. Given a natural number M, a
natural number j can be chosen such that j will factor M, by choosing
j to be one of M's factors.
That's not earth shaking.
> It can be shown that with non-zero integer A, and rationals x and z,
>
> Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
Now you are pulling a lot of unrelated and undefined variables out of
your arse and state meaningless relations.
> and
>
> Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
>
> where A is an integer such that Ax and Az are both integers.
>
> Proof complete.
Proof of what? That it is possible to factor M? Of course it is.
That's not the question. The question is how to do it, efficiently.
-- David Kastrup, Kriemhildstr. 15, 44793 Bochum
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