Surrogate factoring ideas to ponder

jstevh_at_msn.com
Date: 03/22/05 

Date: 21 Mar 2005 15:26:37 -0800

Part of the reason I find myself still focusing on a sign variation on
my earlier quadratics is that I can easily get to a surprising
relation.

The quadratic I have now is

yz^2 - Az + j^2 = 0

where you can add T to both sides to get

yz^2 - Az + M^2 = T

as T = M^2 - j^2,

and j is some integer you choose, while M is the number you're trying
to factor.

One thing you can do trivially is consider

(y/A^2)A^2 z^2 - Az + j^2 = 0

and solve for Az, and importantly, those solutions for Az are the same
with

yz^2 - Az + M^2 = T

as adding T to both sides doesn't change the solutions.

However, my next step is to factor so that I have

(a_1 z + b_1)(a_2 z + b_1) = T, so that I can get

a_1 z + b_1 = f_1

and

a_2 z + b_2 = f_2

where f_1 f_2 = T, and notice that

b_1 b_2 = M^2, and now I can solve to get

Az = f_1 b_2 - f_2 b_1 - 2M^2

which is rather remarkable.

As now I have the factors of M related to the solution for my earlier
quadratic, but importantly, b_1 and b_2 can be rationals.

The way I normally do the analysis I keep f_1 and f_2 integer factors
of T, and that still leaves b_1 and b_2 rational factors of M^2.

So far though I have just that equation relating factors of T and M to
Az, as clearly on its own Az has no relation to either as it comes from

(y/A^2)A^2 z^2 - Az + j^2 = 0

where you pick j.

The big picture then is that what I have is a way to find a rational Az
based on factors of T and M, but no reason at this point to think I can
easily go the other way, as Az can be rational, so I have the full set
of rationals, and yes, sure there are solutions that give me factors of
M, but it's an infinite set.

There are a lot of possibilities.

But from my other work I know I can get values for Az that *are*
related to M through the factorizaton of T, and that raises a BIG
question as for each Az I get THAT way I can solve

Az = f_1 b_2 - f_2 b_1 - 2M^2

to find b_1 or b_2, as I talked about in a previous post.

But now then, how does the mathematics pick what rational factors of M
to use?

If it's random, then there's just as good a chance that I'll get a
single prime factor of M in the numerator of b_1 or b_2, as not.

If it's not random, then what drives it? What would block those
factors?

Here the crucial point, possibly the subtle point, is that rational
factors of M MUST be selected, and some mechanism must decide that
selection.

But what is it?

I've noticed some posters wish to dismiss surrogate factoring as a
random process, where you just take a lot of gcd's and can get lucky,
but here I'm showing that randomness would actually mean the *opposite*
of what they suppose.

In failed algorithms, something interfered to block, but what was it?

James Harris

Relevant Pages

  • JSH: Factoring
    ... and there is this remarkable relationship that shows a connection ... but that STILL leaves the huge set of rationals for b_1 and b_2 to ... And if you know anything about factoring, you know that I have Az setup ... I think that these quadratics may be beyond most of you. ...
    (sci.math)
  • Surrogate factoring ideas to ponder
    ... Part of the reason I find myself still focusing on a sign variation on ... my earlier quadratics is that I can easily get to a surprising ... quadratic, but importantly, b_1 and b_2 can be rationals. ... selection. ...
    (sci.math)
  • Simplifying discussion on surrogate factoring
    ... factoring by focusing on a single quadratic, ... Focusing on the single quadratics ... actually talking about finding rationals such that you get the ... If you can solve either of the conics, ...
    (sci.crypt)
  • Simplifying discussion on surrogate factoring
    ... factoring by focusing on a single quadratic, ... Focusing on the single quadratics ... actually talking about finding rationals such that you get the ... If you can solve either of the conics, ...
    (sci.math)
  • Re: BigNum -- Floating Point
    ... but that's not what makes their digits such a convoluted mess. ... The reason is because they each have something called a ... represent rationals, then you get division as a property of rational numbers. ... Paul Hsieh ...
    (comp.programming)