# Re: SF: Two square, mystery celebration

**From:** Proginoskes (*proginoskes_at_email.msn.com*)

**Date:** 04/23/05

**Next message:**David Bernier: "Re: SF: Abstracted SFT, issue of triviality"**Previous message:**mike: "Re: password derived key"**In reply to:**jstevh_at_msn.com: "Re: SF: Two square, mystery celebration"**Next in thread:**Rick Decker: "Re: SF: Two square, mystery celebration"**Reply:**Rick Decker: "Re: SF: Two square, mystery celebration"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ] [ attachment ]

Date: 22 Apr 2005 17:01:27 -0700

jst...@msn.com wrote:

*> 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.
*

Hey, your bulb's been dim ever since I started reading your posts.

So Nora Baron has this UVW identity, and the only thing you can do is

to attack her motives? Saying she's trying to take credit for the

difference-of-squares in a modified form? Constructing a case based on

something other than the mathematics?

Did you even BOTHER to check the mathematics? (Oh, that!) Any

fifth-string mathematician would go to the trouble. I don't think you

did, because one simple derivation proves that the math is wrong.

In Maple:

U := c^2* (c + a)*(2*c + a + b) - 2*(c^2 - a*b)*(c^2 + a*c + b*c +

a*b);

V := 4 *a*b*(c^2 - a*b)*(c^2 + a*c + b*c + a*b)^2;

W := (a + b)*c^2 + a*(a + 3*b)*c^2 + 2*(a^2)*b*c;

factor(U^2 + V - W^2);

(I factored it so that it would simplify the expression.) The response

is

c^3*(c-1)*(a+b)*(2*c*a^2+4*a^2*b+c^2*a+6*c*a*b+a*c+c^2*b+b*c)

which isn't even close to zero.

Granted, this takes a bit of work using P&P (pencil and paper), but it

still can be done.

Nora Baron made a mistake, and you goofed. You could have grabbed the

ball and run it in for a touchdown. But you fumbled it.

You responded like a crank when you could have responded like a

mathematician.

You could have done the math and found out that Nora Baron was wrong.

(Granted, she might be right 99% of the time, but if you wanted "proof"

that "she" was "lying", you couldn't hope for a better opportunity.)

Is it any wonder no one respects you?

*> 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.
*

Or "new" variants.

*> > >
*

*> > > 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.
*

If the readers of sci.math are complete incompentants, then why is it

that no one believes you've solved the factoring problem?

*> So the "Nora Baron" poster is going to re-teach you
*

*> difference of squares.
*

Just like you re-teach hyperbolas.

*> > >
*

*> > > 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!
*

How you turn one mistake into the plural "statements" is beyond me.

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

*> voodoo mathematics.
*

With the biggest subfield of Voodoo Mathematics being Surrogate

Factoring.

*> The sci.math readership does not care.
*

If I didn't care, I wouldn't be writing.

*> 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.
*

If you think a paper is so easy to kill, get rid of mine: "A New Proof

of the Independence Ratio of Triangle-Free Cubic Graphs" (with Robin

Thomas), Discrete Mathematics, Vol 233/1-3, pp. 233-237.

*> So they are journal killers as well.
*

How exactly did they kill off a journal? Do you have any evidence of

this?

*> > > 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.
*

*>>From human beings who make mistakes. I bet you that Nora Baron admits
*

to the mistake and doesn't sic the CIA and FBI on me, though, because I

have a higher opinion of "her" than you.

*> It turns out I can easily prove them wrong, without
*

*> going into nearly so much...chatter.
*

I proved the equality wrong. If you could have, why didn't you? Oh,

yes, you lost your key.

*> The difference of squares is an ancient identity, and
*

*> not some new thing discovered by "Nora Baron"
*

Now you're putting words into "her" mouth. She never said she had

discovered the difference of squares. She said she discovered an

identity which doesn't look like it could be derived from the

difference of squares, one which she hadn't seen before. That's why she

called it "new".

*> no matter how many sci.math'ers might come along,
*

*> like Rick Decker to proclaim "cool work".
*

Oh, no! Rick Decker (if that's his real name) will sic the FBI and CIA

on me as well! Oh, whatever shall I do?

*> 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.
*

"Resolving" is not the correct terminology. Do you mean "evaluating"

the square root? Proving that the answer is rational? (A sci.math'er,

maybe Decker or Baron, showed that the square root DOES turn out to be

rational, something that you never bothered to check. If you had ended

up with an equation like

b_2 f_1 = sqrt(5),

there would be no way to connect rational factors, since sqrt(5) is

irrational. And he didn't need to know how to factor M to do it; all he

needed was j and k_1.)

*> 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.
*

You've never actually said how to turn the SF Theorem into an algorithm

which produces the factorization. Once again, the sci.math people have

done your work for you, evaluating Ax, Az, and (-(Az - 2M^2)+/-

sqrt((Az - 2M^2)^2 - 4TM^2))/2. No one -- not even you -- has figured

out how to take the next step.

The SF Theorem says that once you've evaluated (-(Az - 2M^2)+/-

sqrt((Az - 2M^2)^2 - 4TM^2))/2 (let's call it r for short), then the

rational factors of M can be found from the rational factors of T, by

letting

b_2 = r / f_1.

The problem is that the statement

"The rational factors of M can be found from the rational factors of T,

by letting b_2 = r / f_1."

is true for ANY nonzero rational number r, regardless of how we get it.

You may as well choose r = 1 and avoid all the unnecessary

calculations.

*> Notice that's an actual mathematical position as to where
*

*> my work could fail.
*

Yeah, under "Voodoo Mathematics."

--- Christopher Heckman

P.S. If you think my criticism is harsh, you should remember that I was

one of your supporters for a long time. The last few weeks of

non-mathematical posts of yours has made me doubt that you have

something else up your sleeve.

In other words, you did yourself in.

**Next message:**David Bernier: "Re: SF: Abstracted SFT, issue of triviality"**Previous message:**mike: "Re: password derived key"**In reply to:**jstevh_at_msn.com: "Re: SF: Two square, mystery celebration"**Next in thread:**Rick Decker: "Re: SF: Two square, mystery celebration"**Reply:**Rick Decker: "Re: SF: Two square, mystery celebration"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ] [ attachment ]