Re: Surrogate Factoring Solution

guenther.vonKnakspott_at_gmx.de
Date: 03/10/05 

Date: 9 Mar 2005 22:28:25 -0800

jstevh@msn.com wrote:
> My apologies up front, as I have the solution now, so it makes sense
to
> start a new thread, though I wonder about the sense of posting an
> actual solution.
>
> It turns out that solving for A and y is key to that solution, as
> remember
>
> yx^2 + Ax - M^2 = 0
>
> and
>
> yz^2 + Az - j^2 = 0,
>
> so
>
> A = (z^2 M^2 - x^2 j^2)/xz(z-x)
>
> and
>
> y = -(zM^2 - xj^2)/zx(z-x)
>
> and focusing again on y/A^2, which I have an algorithm for
calculating,
> I have
>
> y/A^2 = -xz(z-x)(zM^2 - xj^2)/(z^2 M^2 - x^2 j^2)^2
>
> and focusing on the denominator, I wish to know what to multiply
>
> 8j^2 M^2(y/A^2)
>
> times so that it is an integer, so assuming x has a single prime
factor
> of M that I'll call g_1, so x = g_1, and using g_2 = M/g_1, I have
>
> y/A^2 = -g_1^2 z(z - g_1)(zg_2^2 - j^2)/g_1^4(z^2 g_2^2 - j^2)^2
>
> which is
>
> y/A^2 = -z(z - g_1)(zg_2^2 - j^2)/g_1^2(z^2 g_2^2 - j^2)^2
>
> and now focusing on
>
> z^2 g_2^2 - j^2, I can just let
>
> n = z^2 g_2^2 - j^2, and solve for z, so
>
> z = sqrt(n + j^2)/g_2^2
>
> and notice you need to get an integer z, but you don't know what g_2
> is, or you wouldn't need to try and factor, but, you do know that M^2
> will handle that g_2^2, so you need
>
> n + j^2 = k^2 M^4
>
> where k is some arbitary natural.
>
> The simplest solution then is n = M^4 - j^2, which you will have to
> factor.
>
> If it's hard to factor you can iterate k, and get another n.
>
> You use that n with
>
> sqrt((8j^2 M^2 (y/A^2) + n^2(2j^2 + T))^2 - n^4 T^2)
>
> so the algorithm is as follows.
>
> Given a target natural M that you wish to factor, you must first
> select a non-zero integer j, such that j is coprime to M, and any
such
> j will work, but you wish to pick one that is large with respect to
M,
> like
>
> j = M - 1
>
> as that will make T smaller, but most importantly you want to pick j
> such that T is easy to factor.
>
> Next you calculate T, where
>
> T = M^2 - j^2
>
> and you factor T.
>
> Next you find n, where simplest is to use
>
> n = M^4 - j^2
>
> and factor it, but if it's hard to factor you can find another n
using
>
> n = k^2 M^4 - j^2
>
> using k an arbitrary natural number.
>
> Now you consider integers f_1 and f_2, such that
>
> f_1 f_2 = n^4 T^2
>
> and you need to consider all such solutions, including seemingly
> trivial ones like where f_1 = n^4 T^2, or f_1 = 1, or -1.
>
> Then you calculate a number I call y/A^2, where the numerator num is
> given by
>
> num = +/-(f_1 + f_2) - 2n^2(2j^2 + T)
>
> and take the gcd with M, and for at least one f_1 and f_2, you will
> have a single prime factor of M.
>
> And that is the easy answer, which is actually rather easy to derive.
>
> I did just figure it all out, but I've had a strong feeling a simple
> answer was there.
>
> And deep down I'd hoped that someone else might find it first,
> especially if it was someone who could have helped me take all of
this
> in-house to a friendly government agency.
>
> I did try the NSA by direct contact a couple of months ago, but in
> reply I was told that they got many claims of important mathematical
> results, and they weren't interested.
>
> And now the full answer is just out there.
>
> Well, my dad did call me paranoid, and maybe he's right. I can keep
> telling myself all kinds of scary things, but knowledge is valuable,
> and I have to believe that the truth is good.
>
> I believe that the truth is the best.
>
>
> James Harris

http://www.rsasecurity.com/rsalabs/node.asp?id=2093

Either you're wrong again, or those math people bastards are still
ignoring your major results. Then, maybe you should just go and try to
get some help with that delusional mind of yours.

Quantcast