Re: Proof factoring solution is closed form

jstevh_at_msn.com
Date: 02/09/05 

Date: 9 Feb 2005 04:24:30 -0800

Rick Decker wrote:
> jstevh@msn.com wrote:
> > One of the most important points to make, which should end the
debate
> > and get someone to contact the US Government so that authorities
can be
> > informed is the closed form proof of surrogate factoring.
> >
> > By closed form I mean the proof that it is a complete solution
which
> > guarantees the factorization of a target number.
> >
> > However, after this point, the clock is really ticking, as anyone
can
> > use the information. You have your future in your hands. The
start
> > will be familiar. The closed form proof is short.
> >
> > Here we go.
> >
> > Take the two quadratics
> >
> > yx^2 + Ax - M^2 = 0
> >
> > and
> >
> > yz^2 + Az - j^2 = 0
> >
> > where A, j, and M are integers greater than 0 chosen, where M is
the
> > target to be factored, and you find that you can use T, where
> >
> > T = M^2 - j^2
> >
> > and substituting out y to solve for x and z gives
> >
> > 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, let A=1, and the first equation gives you rational z's where
those
> > are found by factoring TM^2, as simply
> >
> > Az = f_1 + f_2 + 2M^2
> >
> > where f_1 f_2 = TM^2.
> >
> > Now then there must be some integer z, which factors M, and for
that
> > integer z, you will get a rational x that is probably a fraction,
with
> > A=1.
> >
> > BUT, if x = x_num/x_denum, where x_num and x_denum are integers,
then
> >
> > let
> >
> > A = x_denum
> >
> > and you have
> >
> > sqrt((x_denum (x_num/x_denum) + 2j^2)^2 - 4Tj^2)
> >
> > which is just
> >
> > sqrt((x_num + 2j^2)^2 - 4Tj^2)
> >
> > as the A absorbed the x_denum.
> >
> > And notice that x_num is given by the integer factorization of
Tj^2, as
> > you just have
> >
> > x_num = 2j^2 - g_1 - g_2, where g_1 g_2 = Tj^2, and are integers.
>
> Okay, more or less, down to here.
> >
> > The A can always pull factors from the denominator in this way, so
the
> > full solution is in fact given by the factorization of Tj^2, which
must
> > give you an integer x_num, which will either be a factor of M or
can be
> > used to get a z which must factor M.
> >
> That's an interesting avenue, using either x or z, but I don't see
> why it must work.
>

Given any rational x that is a solution, if one exists, you can simply
let A equal the value of its denominator, proving some other x' exists
which IS an integer.

> > It's a closed solution.
> >
> > What does that mean practically?
> >
> > It means that someone tonite can take some large target M, find a
j,
> > and with that get a T, and then factor j and T, and use them to
then
> > get Tj^2, and using combinations of factors of Tj^2, guarantee
> > themselves a factorization of their target M.
>
> There's a good chance that for M in the range of the RSA challenges
> if someone started tonight, they would still be waiting for a
> factorization when the universe ended, using your method. That's
> not to say that your method is guaranteed to be impracticable,
> but you certainly haven't convinced anyone so far.

Well that's what you say, but where your proof?

I've actually looked at the equations showing how many combinations you
get by two for factoriztations.

If you have a number with only two prime factors, then you get 1
combination by two's, so if j were a number with only two prime factors
that's the number you'd have.

But, it probably is not and you have T to factor as well, and it tends
to have a lot of prime factors.

Now I've actually calculated, while you appear to be trying to convince
others to ignore my work without even bother to give mathematics, and
my calculations show that if you form a number by multiplying the first
one thousand primes together you get about 150 million combinations by
two's.

Now then Decker, how big of a number is formed by multiplying the first
1000 primes together?

I dare you to reply to this post and answer at least that one question.

> >
> > Now for some large number the number of combinations would be huge,
but
> > they'd be certain of success if they cycled through them all.
>
> Not necessarily. As you should know, when trying to factor M there
> are some values of j that give *no* useful factorizations. When that
> happens, all one can do is start anew with a new j.

Well then the theory must be wrong, as my analysis is that for any
integer j greater than 1 it should work.

If it doesn't, then I'm wrong, and I'd be very interested in seeing
why!!!

I have no interest in pushing wrong positions, but from what I've seen
it should work.

If I made a mistake in my analysis that mistake should be capable of
being shown.

> >
> > And that's a polynomial time problem, as the equation defining how
many
> > combinations of those factors there are is a polynomial one.
> >
> That's not what we mean by a poly-time solution. In fact, one might
> have to search for factors in a space that could be about as large as
M
> itself. If you're going to get a poly-time solution, you'll at least
> have to cut down the search to O((log M)^k), for some fixed k.
>

Why?

You aren't giving any information here, but just assertions without
explanation.

Why should I or anyone else just believe you?

> > Now I'm using these examples because they're easier to explain
with,
> > but the full surrogate factorization theory shows that only a
> > factorization of T is needed!
> >
> Even if that's so, it doesn't necessarily give you a good algorithm.
> Frankly, I'd love to see a proof that all one has to do is factor T.
>

It's not difficult. I've worked out the details on my Yahoo! site, but
before moving to the more complicated mathematics, I'd like to get past
the easy.

For those thinking I'm dodging, my answer is that certain posters
routinely try to move to the most complicated areas of my theories
where it's easy, well, to lie and confuse people about what's
mathematically true or not.

So I've learned to try and keep it simple.

> > So I'm at a primitive level with this discussion, as the full
theory
> > indicates that only a factorization of T is needed, which is what
I've
> > been trying to get to work.
> >
> > You can scoff at this if you want, but do the math. Trace it out.
> > Understand how A can absorb factors.
> >
> The value of A unimportant, as long as it's not zero. "Move on,
folks,
> there's nothing to see here."
>

Now you can see why. The poster is basically trying to deflect
interest in my work.

>
> Regards,
>
> Rick

Now Rick Decker has been at his game for years now, working to deflect
interest in my work, no matter what it is.

Now that's math society for real, not television or some movie, and I
have to get past that first.

The most important assertion Decker made in his post is that with

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)

derived in my original post that given a full factorization of Tj^2 to
get x, you cannot get a factorization of M.

If true then I'm wrong. It's math people. Either you're wrong, or
you're right.

Now I'm curious to know if I'm wrong, and then I'll be curious about
why.

That's called mathematical investigation.

Now Decker will play his games and work for whatever reasons motivate
such a person to deflect people from my research, but what I can do, is
do what I enjoy and continue that research despite people like him.

What you can do is pay attention to the facts.

James Harris

Relevant Pages

  • Re: Proof factoring solution is closed form
    ... >> informed is the closed form proof of surrogate factoring. ... >> guarantees the factorization of a target number. ... Now then Decker, how big of a number is formed by multiplying the first ... Now Rick Decker has been at his game for years now, ...
    (sci.math)
  • Re: Proof factoring solution is closed form
    ... >> factorization when the universe ended, ... > before moving to the more complicated mathematics, ... The poster is basically trying to deflect ... > Now Rick Decker has been at his game for years now, ...
    (sci.math)
  • Re: Proof factoring solution is closed form
    ... >> factorization when the universe ended, ... > before moving to the more complicated mathematics, ... The poster is basically trying to deflect ... > Now Rick Decker has been at his game for years now, ...
    (sci.crypt)
  • Re: Factoring problem, my assertion revisited
    ... [Rick Decker] ... > factorization has to satisfy a condition involving quadratic residues ... > I'm pretty sure I've posted a sample using James' method to factor 15. ... sense of what JSH might be saying. ...
    (sci.crypt)
  • Re: Factoring problem, my assertion revisited
    ... [Rick Decker] ... > factorization has to satisfy a condition involving quadratic residues ... > I'm pretty sure I've posted a sample using James' method to factor 15. ... sense of what JSH might be saying. ...
    (sci.math)
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