Re: Easy test of surrogate factoring

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 02/18/05 

Date: Fri, 18 Feb 2005 08:25:07 -0600

On 18 Feb 2005 04:25:13 -0800, jstevh@msn.com wrote:

>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.

Wondering whether a new and earthshattering factoring method
can be used to factor numbers is not a fair test? Huh.

>I'll explain why,

Ok...

>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

Oops, I missed the explanation of why it's not a fair test.

************************

David C. Ullrich

Relevant Pages

  • Re: Easy test of surrogate factoring
    ... Wondering whether a new and earthshattering factoring method ... can be used to factor numbers is not a fair test? ... >That proves that there are rules governing the value of the rationals ... >But mathematicians are NOT what they claim, and in this case they are ...
    (sci.math)
  • Re: Factoring, SF, and transforms
    ... >> rationals using a transform that I call the surrogate factoring ... But math people have lied about the importance of my work, ... So part of my point is that "pure math" to mathematicians means, ...
    (sci.crypt)
  • 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)
  • Re: JSH: Contradictory behavior, issue of math fraud
    ... is not a hard problem after all, then how can mathematicians who not ... and that math journals routinely publish false papers!!! ... fraction of them were by the editor himself. ... implement your many variants of surrogate factoring. ...
    (sci.math)
  • JSH: Lets recrap
    ... Surrogate factoring preferentially yanks out small prime factors. ... but that was without modern computing technology, ... There is no way that I can see that mathematicians ignoring this ...
    (sci.math)
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