Re: Easy test of surrogate factoring

jstevh_at_msn.com
Date: 02/18/05 

Date: 18 Feb 2005 04:25:13 -0800

Paul Rubin wrote:
> Here's an even easier test of surrogate factoring: I gave you a list
> of fifty 20-digit composite numbers a couple weeks ago and you
haven't
> factored a single one. I think I know what test score to assign,
> based on that result.

No, it's not a fair test. I'll explain why, again, and I'll also
explain again, what the test is, and why my test is an easy one.

Ax= Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)

Az= Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)

shows that you have a set of rational Ax mapped to rational Az
solutions, where you have dependencies on the factorizations of TM^2
and Tj^2 to make the square roots rational.

Notice, you are guaranteed to have an integer Az that factors M, from
the first equation, which gives a rational Ax, and searches for
algorithms are basically about figuring out how to get all the possible
integer Az's.

However, you can do the time honored technique of working backwards to
see actual solutions, as is easily calculated if you *do* know the
prime factors of M, as then you can just pull out all the integer Az's
from the first equation.

It's a well-known technique to work backwards to figure out what's
going on.

My test is to work backwards.

If I'm wrong, then more knowledge doesn't help me, now does it?

I'd *hide* from the truth, but instead I want you to notice that
posters trying to convince you that I'm wrong are the ones who are
doing the running.

Now then, if sf is random, then the denominators of those Ax's found
that way will necessarily be random.

If it's not random, then some way exists to figure out how to get all
the integer Az's without knowing M.

Those are the two possibilities.

Now posters working hard to discount this research--apparently deciding
that it's Harris mathematics or JSH algebra--are avoiding the actual
issue from what I've seen in looking at posts in this thread.

The reason is simple, the mathematics doesn't work for them either way,
if their intent is to discount it.

That's what happens with major math results, human beings working hard
to dismiss them have to ignore facts.

Now my test is simple: calculate Ax, by taking an M with known factors,
and *look* at the prime factors of the denominator of Ax, and you will
find that they are the same prime factors as T has.

That proves that there are rules governing the value of the rationals
Ax's that results from the integer Az's.

And then, algorithms *can* be developed.

Now, if mathematicians were what they claimed then it wouldn't be
required that I actually produce a working program that can factor RSA
numbers, as that's an unreasonable request. It's an unresaonable
request as if I were at that point I wouldn't need to convince anyone,
as I could just demonstrate.

Worse, my ability or lack of ability to write such a program does not
disprove the mathematics!

And in this case that means someone else might, or may already have.

But mathematicians are NOT what they claim, and in this case they are
setting other people up to be responsible for their failures, if things
go badly.

Now I'll work at building that program to factor an RSA number, but
while time passes, it's you who may suffer the consequences, in a world
where hackers may now have the upper hand.

And mathematicians are de facto protecting them.

James Harris

Relevant Pages

  • Re: Easy test of surrogate factoring
    ... > Here's an even easier test of surrogate factoring: ... It's a well-known technique to work backwards to figure out what's ... The reason is simple, the mathematics doesn't work for them either way, ... That proves that there are rules governing the value of the rationals ...
    (sci.crypt)
  • Re: JSH: Nearly done
    ... > theory and method for factoring that I call surrogate factoring. ... > The mathematics though is surprisingly simple, ... but are in the field of rationals. ...
    (sci.math)
  • Re: JSH: Nearly done
    ... > theory and method for factoring that I call surrogate factoring. ... > The mathematics though is surprisingly simple, ... but are in the field of rationals. ...
    (sci.crypt)
  • Re: JSH: Nearly done
    ... > theory and method for factoring that I call surrogate factoring. ... > The mathematics though is surprisingly simple, ... I will take the liberty to describe what Harris is doing ... but are in the field of rationals. ...
    (sci.math)
  • Re: SF: Areas of confusion, infinity
    ... > notably the set of rationals. ... choosing any integer is the same as the probability of any other ... That's what make the factoring problem interesting, i.e., not ... > is naive with the surrogate factoring theorem. ...
    (sci.crypt)
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