Re: Factoring problem, solved

From: ScottD (bubba_at_aol.com)
Date: 01/19/05 

Date: Wed, 19 Jan 2005 06:04:09 GMT

Job well done James. Congratulations on solving
the pesky factoring problem. I would have done it
my self but I had to work late tonight. A couple
of months from now we will be joking about how
cave men used factoring difficulty as a measure
of encryption security, he he he he he.

By the way, refresh my memory on flt, I was writing
down the answer when that stupid mechanical pencil
ran out of led. A cute exercise for nephew.
-- bubba

<jstevh@msn.com> wrote in message news:1106108880.869064.289660@c13g2000cwb.googlegroups.com...
>I am increasingly certain that I've solved the factoring problem.
>
> That is a bold claim to add to previous bold claims of mine, but here
> in THIS post I'll give the underlying theory as it's very basic, and
> direct you to a site where you can download a rough, and somewhat
> flawed, but still good enough to show the idea, prototype factoring
> program, which implements some of the theory.
>
> I'll explain why I say some of the theory in just a bit.
>
> My work boils down to the analysis of a system of equations:
>
> a_1 x + b_1 = f_1
> a_2 x + b_2 = f_2
>
> a_1 b_2 + a_2 b_2 = A
>
> a_1 a_2 x^2 + Ax + b_1 b_2 = f_1 f_2
>
> where to match the variables in my initial paper
>
> y = a_1 a_2, T = f_1 f_2, and j^2 = - b_1 b_2.
>
> (Here b_1 b_2 is a negative number so j is still an integer.)
>
> I have a rather complex looking solution in the paper, but remarkably
> you can encompass it by solving the first three equations for key
> variables:
>
> a_1 = A(b_1 - f_1)/(b_1 f_2 + b_2 f_1 - 2b_1 b_2)
>
> and
>
> x = (b_1 f_2 + b_2 f_1 - 2b_1 b_2)/A
>
> which shows that you can actually factor simply by cycling through the
> factors of j and T, while my focus in the program is on cycling
> through the factors of T.
>
> All together you then get the factors of your target M, as
>
> M^2 = j^2 + T, so
>
> a_1 a_2 x^2 + Ax + b_1 b_2 = f_1 f_2
>
> is
>
> a_1 a_2 x^2 + Ax - M^2 = 0
>
> so x is a factor of M.
>
> The idea may seem pathetically simple, but that is because you are
> programmed to believe that the factoring problem is hard. It is not.
>
> I have just shown you in a few lines how to factor an arbitrary
> non-zero positive integer M in polynomial time.
>
> Many of you are programmed to believe that mathematics must be abstruse
> and difficult to work. You are wrong. The factoring solution I've
> presented here while looking extremely simple can work and is simple
> because the PROBLEM is simple.
>
> I have demonstration code, which is rough as I threw something
> together, to test out my own theory, and hopefully to be more
> convincing. That code implements a VERSION of my own theory, as I
> don't factor both j and T, but only T, as j gets background factored by
> the mathematics.
>
> If you watch it factoring and spitting out primes (though it also
> mislabels some as prime and can't factor some easy numbers...rough
> version remember) then you can at least see that something is happening
> here worth investigating further.
>
> In the hopes of convincing *someone* willing to help I have put put up
> a Yahoo! Group with my paper and some rudimentary code:
>
> http://groups.yahoo.com/group/sufactor/
>
> It is CRITICAL that the information be taken seriously as soon as
> possible so that some other method can be used for Internet security,
> and quickly implemented.
>
> Time is a factor.
>
>
> James Harris
>

Relevant Pages

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