Re: I was right, surrogate factoring proof

jstevh_at_msn.com
Date: 02/13/05 

Date: 13 Feb 2005 09:09:42 -0800

ođin wrote:
> > Last weekend I came across a simpler way to describe surrogate
> > factoring with a couple of quadratics and solutions for x and z,
which
> > I promptly programmed--and it didn't seem to work.
>
> No.. it did not work. You were certain that it was simple and proven,
but
> you were wrong. And all who doubted you were, according to you,
liars.
>

Mathematical proof is what matters.

Here it's easy. If you refuse to accept mathematical proof then there
is no room for discussion.

Here is the proof.

Given

yx^2 + Ax - M^2 = 0

yz^2 + Az - j^2 = 0

and

T = M^2 - j^2

it follows that

x = z(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

and

z = x(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

and, you can multiply both equations by A, to

get

Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

where for any rational x and z there must be an integer A such that
both Ax and Az are integers, so looping through *all* possible integer
Ax and Ay as required by the square roots MUST factor M.

So, I can't be wrong, or, algebra is wrong.

If you disagree, show a false step.

The math is not hard, or do you disagree?

I'm asking you to do more than just blindly criticize, but to show you
are actually using rational thought.

Can you do math, or are you just someone disagreeing to be
disagreeable?

James Harris

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