Re: SF: Two square, mystery celebration

jstevh_at_msn.com
Date: 04/23/05 

Date: 22 Apr 2005 15:27:07 -0700

Rick Decker wrote:
> Nora Baron wrote:
>
> > jst...@msn.com wrote:
> >
> >>Nora Baron wrote:
> >>
> >>>jst...@msn.com wrote:
> >>>
>
> <snip, though I hated to delete the accusation that Nora
> had maliciously tailored her example to further her lying agenda>
>
> >
> > My formula for the numerator of b_2 f_1 is
> >
> > b_2 f_1 = E / F, where
> >
> > E = (k_1 + k_2)*j^2 + k_1 (k_1 + 3k_2) + 2 (k_1)^2 k_2/j^2
> >
> > - (k_1 + k_2 + 2 j^2)*(k_1 + j^2) and
> >
> > F = 2 (k_1 + k_2 + 2 j^2 - M^2).
> >
> > So your question is really, how do I prove that this
> > formula actually works ?
> >
> > It follows from an identity.
> >
> > There are several identities that one might apply here.
> > One that *you* believe should work for the computation of
> > b_2 f_1 is:
> >
> > (m - n)^2 + 4*n*m = (m + n)^2.
> >
> > Let me call this the "+4" identity.
> >
> > Another related identity is:
> >
> > (m^2 - n^2)^2 + 4*m^2*n^2 = (m^2 + n^2)^2.
> >
> > This is the identity that is used to generate
> > Pythagorean triples. I will call this the
> > "Pythagorean" identity.
> >
> > Neither the "+4" identity nor the Pythagorean
> > identity apply to give the expressions that
> > I gave above for E and F. There is instead another
> > identity:
> >
> > Let U = c^2 (c + a)(2c + a + b)
> >
> > - 2(c^2 - ab)(c^2 + ac + bc + ab).
> >
> > Let V = 4 ab(c^2 - ab)(c^2 + ac + bc + ab)^2.
> >
> > Let W = (a + b)c^2 + a(a + 3b)c^2 + 2(a^2)bc.
> >
> > Then the identity is:
> >
> > U^2 + V = W^2.
> >
> > This holds for any positive numbers a, b, and c.
> >
> > I will call this the UVW identity.

So now you're making up stuff in the factoring arena.

You're just talking about freaking difference of squares and trying to
rename it.

Not very bright "Nora Baron" as there are too many people who care
about the subject.

There are no new identities beyond

b^2 - a^2 = n

which is the core identity called difference of squares.

Even you aren't so silly as to believe that the cryptology community is
going to accept that some sci.math poster just showed them up by
discovering new fundamental identities.
> >
> > You can check that this identity is true by
> > just expanding the left side and the right side and
> > comparing the terms.
> >
> > Now, how do you apply this identity to derive
> > the expressions I gave for E and F above?

And yes, readers of sci.crypt, such posts seem to work on sci.math with
a very odd audience.

So the "Nora Baron" poster is going to re-teach you difference of
squares.

> >
> > Let a = k_1, b = k_2, and
> >
> > c = M^2 = j^2 - (k_1 k_2)/j^2.
> >
> > Then you just crank it out.
> >
> > The "+4" identity that you thought would work to give
> > the result of your "theorem" does not work. The Pythagorean
> > identity also does not work. The UVW identity does work.
> >

It's freaking DIFFERENCE OF SQUARES.

> > It is probably a new identity. I don't see it as
> > particularly interesting by itself - not a publishable
> > result - but it might fit into a larger pattern of
> > identities, and it might be of interest if that is the case.

LOL. Oh my, laughing hard now, but it's so sad too.

ROFLOL.

So sci.crypt, one of the stars of the sci.math newsgroup is here to
tell you that the difference of squares is really the "UVW identity",
which that person say is "probably a new identity".

And yes, many times I've faced really bogus statements from the poster
"Nora Baron" which posters on sci.math would cheer!

I'd say, hey, that's nonsense. I even gave it a name: voodoo
mathematics.

The sci.math readership does not care.

They're like freaking weird as hell and rather odd people, but
dangerous as a bunch of them got together to shoot down a paper of mine
by gang emailing the editors of that now defunct journal.

So they are journal killers as well.

> >
> > In my view, this is the only nontrivial result to come
> > out of your "SFT". And it does not give any support for the
> > efficacy of your brand of surrogate factoring.
> >
> My hat's off to you, Nora. Cool work.
>
> <snip James' carping>

I left a lot in so readers can see what I have to deal with from
posters who seem to hate algebra.

It turns out I can easily prove them wrong, without going into nearly
so much...chatter.

The difference of squares is an ancient identity, and not some new
thing discovered by "Nora Baron" no matter how many sci.math'ers might
come along, like Rick Decker to proclaim "cool work".

What you see in

b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2

is a fundamantal block which prevents you from resolving the square
root without using some factorization of TM^2.

The triviality argument then boils down to the claim that the SFT gives
a factorization using factors

M and M, M^2 and 1,

and their negatives.

Notice that's an actual mathematical position as to where my work could
fail.

James Harris

Relevant Pages

  • Re: SF: Two square, mystery celebration
    ... >> That one is the easy square root. ... >> That poster is wrong. ... What you need to do "Nora Baron" is step out your calculations. ... You MUST use a factorization of M. ...
    (sci.crypt)
  • Re: JSH: Heart of dispute, number properties
    ... Third reply to a "Nora Baron" post. ... The "Nora Baron" poster will routinely simply delete out everything ... >> Mathematics has nothing to do with it. ... The sum of algebraic integers is an algebraic integer. ...
    (sci.math)
  • Re: JSH: Brainstorming over, for now
    ... [Nora Baron] ... I've just had a massive collapse in terms of getting surrogate ... That's from his "Big collapse on surrogate factoring" msg of March 10, ...
    (sci.math)
  • Re: SF: Back to theory
    ... Here's why I call "Nora Baron" a serial liar. ... For some years this poster has obsessively replied to my posts, ... So that's why I say the poster is a serial liar. ... Take your pick, lying ...
    (sci.math)
  • Re: SF: Back to theory
    ... Here's why I call "Nora Baron" a serial liar. ... For some years this poster has obsessively replied to my posts, ... So that's why I say the poster is a serial liar. ... Take your pick, lying ...
    (sci.crypt)
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