# Re: Surrogate factoring, update

From: James Harris (jstevh_at_msn.com)
Date: 07/11/04

```Date: 11 Jul 2004 05:41:55 -0700

```

Marcel Martin <mm@ellipsa.no.sp.am.net> wrote in message news:<40F0DEB4.F55AEC35@ellipsa.no.sp.am.net>...
> James Harris a écrit :
> >
> > Jean-Luc Cooke <jlcooke@engsoc.org> wrote in message news:<ccp66s\$73c\$1@driftwood.ccs.carleton.ca>...

<deleted>

> > >
> > > Using quadratic recipcosity you can eliminate many possible values, but
> > > then end result is no net improvment in factorization problem.
> >
> > That's just one side of what I gave in my original post.
> > [...]
> > No, I didn't just come up with the standard congruence of squares.
> > Here are the equations pulled out:
> >
> > (jk - Tk + T)(jk + Tk + T) = T^4
> >
> > k = (-jT +/- T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2)

Notice that equation for k, and compare with the following one for j.

> >
> > and
> >
> > j = (-T +/- T sqrt(k^2 + T^2))/k
>
> Your method is essentially no more than a Fermat method.

That is true for the equation for j, but not true for the equation for
k, which is obvious to the eye.

The important difference is that as is typical with the Fermat method,
in the equation for j you see sqrt(k^2 + T^2), so k is defined by the
difference of factors of T^2.

For instance, trivially, let k = (T^2 - 1)/2, then

sqrt(T^4 - 2T^2 + 1 + 4T^2)/2 = T^2 + 1

where I used T^2 and 1, which are, of course, both factors of T^2.

However, that's just half of the equation set as you also have

k = (-jT +/- T^2 sqrt(j^2 - T^2 + 1))/(j^2 - T^2)

where you have sqrt(j^2 - T^2 + 1), and now it's the difference of
factors of

T^2 - 1

which is clearly different from before.

I know it can be a subtle point, especially if you look at

j = (-T +/- T sqrt(k^2 + T^2))/k

and seize on the familiar.

Basic research can be EXTRAORDINARILY difficult for some people,
especially if they see something that looks familiar to escape the
difficulty.

That's just human nature.

It's easier to just try and believe that something is not new,
especially if it's difficult to understand.

> Take both sides of (jk - Tk + T)(jk + Tk + T) = T^4 modulo T.
> It comes (jk)^2 = 0 mod T and, since, T has no square factors
> we can write jk = 0 mod T, i.e., jk = aT for some a.
>
> (aT - kT + T)(aT + kT + T) = T^4
> (a + 1 - k)(a + 1 + k)T^2 = T^4
> (a+1)^2 - k^2 = T^2
>
> Which is equivalent to something like n^2 = M^2 - E^2.

That's unnecessary work as k is already linked in a simple way with T
by my own work in my own replies. You're going in circles.

Consider now how I'll make use of your own equation.

It's the variable j, which is more interesting. Substituting with k =
aT/j into your own equation gives

(aT/j + 1)^2 - k^2 = T^2, so

(aT + j)^2 - j^2 k^2 = j^2 T^2, which is

(aT + j)^2 = j^2(k^2 + T^2)

which is a far more muddled picture.

I *deliberately* went looking for a factorization which mixed terms in
such a way that one of the variables could NOT be simply determined by
congruence of squares while one still is.

The mathematics here is not well-worked out, and the area is clearly
new, so I'm not surprised that several people have clung to what they

Basic research is difficult. I suggest for most you just sitting back
and waiting for experts in the field to comment is the best strategy.

> It leads nowhere, this is just a (uselessly complicated) Fermat's
> method.

I can understand your frustration. However, what I'm presenting isn't
textbook, so I don't have *easy* answers, and I can understand how
you'd like to seize upon the familiar.

It's human nature.

My suggestion to you is to sit back and wait for experts in the field,
who are willing to face the difficulties to emerge.

If mathematicians try to ignore this idea, however, others may not.

Essentially I found a way to link disparate factorization, specifially

factorizations of T and T^2 - 1 in the simplest form, and

factorization of sT and T^2 - s^2, in the generalized form,

where you can theoretically factor one to get the factorization of the
other.

If it is a practical idea, then if T = p_1 p_2, you could shift to
factoring some other number and pull out the factorization of T by
doing so, which would impact public key encryption.

I'm someone who just came up with an idea that I'm talking about, and
mathematicians may choose to ignore it, but if it is viable, then they
will be responsible for the consequences.

That is, people looking for someone to blame, if things go badly, will
have every right to go to mathematicians, and ask hard questions.

I already find it troubling that I have to emphasize that point and
deal with clumsy efforts to dismiss the idea, when the consequences IF
it's viable are so huge.

Make no mistake, if I lose money because someone comes off the
Internet exploiting an idea that some of you are trashing then I'd
feel it might right to sue you for damages, and I'm the one who came
up with this idea!!!

Here's where mathematicians prove their love of mathematics, and
especially of "pure math" which has here the potential of practical
application.

If it isn't practical, then it's just "pure math" and no one
(including me) need worry.

If it is practical and mathematicians ignore such a compelling idea,
then they not only prove their true disdain for mathematics versus
their claims, they also have the responsibility to cover damages.

That is, the world can come knocking on the doors of mathematicians if
things go badly since mathematicians are responsible here.

And in courts of law, liability could extend to personal property of
mathematicians IF mathematicians behave negligently.

Ignoring a viable idea is negligent.

James Harris

## Relevant Pages

• Re: Amateur takes on Wiless work
... >Mathematicians would check various elliptic curves and find they could ... Below is a copy of the paper "Advanced Polynomial Factorization" ... Factorization lemma, Ring of algebraic integers ... polynomial that are themselves polynomials. ...
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• Re: Surrogate factoring, update
... The important difference is that as is typical with the Fermat method, ... If mathematicians try to ignore this idea, however, others may not. ... Essentially I found a way to link disparate factorization, ... James Harris ...
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