SF: Two Az's

jstevh_at_msn.com
Date: 04/24/05 

Date: 24 Apr 2005 14:24:14 -0700

It turns out that the 50% factoring percentage can be seen a different
way, which goes to focusing on what I call Az, in the SFT (which you
can see further down in this post).

As if you have

b_1 b_2 = M^2

with rational b_1 and b_2, you can express those rationals using
integers as

b_1 = (g_1 d_2)/d_1

and

b_2 = (g_2 d_1)/d_2

where g_1, g_2, d_1 and d_2 are integers, and

g_1 g_2 = M^2

and then you find from an equation in the derivation of the SFT that

Az = (-f_1 g_1 d_2^2 - f_2 g_2 d_1^2 + 2M^2 d_1 d_2 )/d_1 d_2

where f_1 f_2 = T.

So if g_1 and g_2 are trivial then you have one value for Az, but
notice, you can use the same d_1 and d_2, with a non-trivial value,
which will give you a second value for Az.

For every trivial Az value there is a second non-trivial value.

What's intriguing here is that mathematically there isn't a lot of
indication that the mathematics cares either way.

What posters who disagree with me are basically arguing is that for
some reason g_1 and g_2 will be picked mathematically as trivial
factors, more often than not.

There's just no mathematical reason that I can see from that equation,
and knowing that given d_1 and d_2, you can use g_1 and g_2 that are
trivial or non-trivial, depending on your mood, it hardly makes sense
that the mathematics would tend to go one way.

But from a social perspective, if you accept what is mathematically
true, then it's this really big deal.

Part of my point is that math society is corrupt. It claims that
what's mathematically true is all it cares about, but here, with
something where the truth is embarrassing to math society, and
potentially dangerous to the world's economy, they fudge.

Mathematicians lie.

And they will lie stupidly, as here, there really isn't a lot of room
with the SFT to lie in an intelligent way, except to people who are not
mathematically sophisticated.

Here's the SFT.

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 Az is given
by

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

and Ax is given by

Ax = +/- (k_1 + k_2) + 2j^2

where k_1 k_2 = -Tj^2, and k_1 and k_2 are rationals.

James Harris

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