Re: Proof factoring solution is closed form

From: KeithK (me_at_nomail.com)
Date: 02/09/05 

Date: Wed, 9 Feb 2005 10:13:01 -0700

<jstevh@msn.com> wrote in message
news:1107951870.252176.91050@g14g2000cwa.googlegroups.com...
> 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.
>
Umm the number of combinations of n = 1000 things taken k=2 at a time is
n!/[k!*(n-k)!] = 1000!/[2!*(998)!] = 1000*999/2 = 499,500.

KeithK

> 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
>

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  • Re: Proof factoring solution is closed form
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