Re: JSH: Why factoring solution must work

On Feb 8, 11:01 pm, JSH <jst...@xxxxxxxxx> wrote:
On Feb 8, 4:35 pm, Enrico <ungerne...@xxxxxxx> wrote:





On Feb 8, 10:21 am, JSH <jst...@xxxxxxxxx> wrote:

On Feb 8, 1:57 am, MichaelW <ms...@xxxxxxxxxx> wrote:

On Sat, 07 Feb 2009 13:30:27 -0800, JSH wrote:
Turns out you can prove through mathematical logic that given a solution
like:

(r(v) - c(v))(r(v) + c(v)) = D(s(v))^2

where r(v), c(v) and s(v) are non-zero integer functions of v, if D is a
composite, you MUST have a non-trivial factorization if

abs(r(v) - c(v)) < D and abs(r(v) + c(v)) < D.

So if you can FIND such function that smoothly traverse through all
possible solutions in integers, then you know that factoring D is just a
minima problem.

<snip>

James Harris

Hey the thread moved! Anyhow there has been a lot of progress and I am
ready to rebuild the factorisation code however I am getting stuck
following the modified algorithm.

It has been stated elsewhere in this thread that:

r +/- c = 8Dv^2 or 2(2v - (D - 1))^2

Yeah, I've seen that stated repeatedly by poster named Rick Decker.
I'm not confirming or denying it as I haven't checked!!!

However, the full method actually talks about using f_1*f_2 = D-1,
where I like f_2 = 2, f_2 = (D-1)/2 as I like easy and was doing a
demonstration factorization of 15.

I'm not sure it matters though, or I don't really care a lot at this
point.

Now once again sorry if I get this wrong so just let me know where my
assumptions are incorrect.

I am assuming we want the minimum for the function r(v). We have

(r(v)+c(v)) + (r(v)-c(v)) = 8Dv^2 + 2(D - 1 -2v)^2

That's a weird assumption since my original post notes that given

(r(v) + c(v))(r(v) - c(v)) = D(s(v))^2

where all are integer functions of v, that it must be true that if

abs(r(v) + c(v)) < D and abs(r(v) - c(v)) < D

then you MUST non-trivially factor D if D is a composite, which is a
mathematical absolute!

Mathematical absolute.  Total perfection.  Inviolate.  Unchangeable.

Absolute truth.

This simplifies to

r(v) = v^2 * (4D+4) + v * (4-4D) + D^2 - 2D + 1

So

r'(v) = v * (8D + 8) + (4-4D)

Now the quadratic equation for r(v) has a minimum when r'(v) = 0 or

v = (D-1)/(2D+2)

This is where I get stuck. r(v) is an integer function so obviously this
value of v is not correct (and I think it leads y^2 being negative)..
Obviously I have gone down the wrong path. Can you clarify? Remember I am

Yeah I did so above but I think I stated the actual conditions at the
top of my original post.

Maybe you were most honest when you noted you are a contractor who
would lose a lot of work (if I remember correctly not going back to
read that post) if RSA were to go away.

Might you be selectively thinking about this problem?  Ignoring
important information?  Hoping against hope that if you wish really,
really hard the mathematics will change?

Or maybe just hoping people will naively believe you if you keep
pushing doubts about a new factoring method possibly ending the RSA
era?

Factoring is not a hard problem.  All that happened is some people got
some things wrong for a few years.

Believe me, it HAS happened before.

Start thinking about how to get work in the post-RSA era.  I'm sure,
if you don't destroy your credibility with dumb postings, you will
find there are opportunities later.

Losing RSA is not the end of the world.  It's just more change.

James Harris- Hide quoted text -

- Show quoted text -

==================================================================

Given: D = 119

Find in integers:

R(v) ^ 2 - D*S(v) ^ 2 = C(v) ^ 2

There are an infinite number of families of solutions.
One of them is:

R(V) =  80 * V ^ 2 + 118 * V + 45

S(V) =  6 * V ^ 2 + 10 * V + 4

C(V) =  46 * V ^ 2 + 50 * V + 11

V=-4:   853 ^ 2 - 119 * 60 ^ 2 = 547 ^ 2
V=-3:   411 ^ 2 - 119 * 28 ^ 2 = 275 ^ 2
V=-2:   129 ^ 2 - 119 * 8 ^ 2 = 95 ^ 2
V=-1:   7 ^ 2 - 119 * 0 ^ 2 = 7 ^ 2
V=0:    45 ^ 2 - 119 * 4 ^ 2 = 11 ^ 2
V=1:   243 ^ 2 - 119 * 20 ^ 2 = 107 ^ 2
V=2:   601 ^ 2 - 119 * 48 ^ 2 = 295 ^ 2
V=3:   1119 ^ 2 - 119 * 88 ^ 2 = 575 ^ 2

How does the idea of solutions indicated by
the minima of polynomials tie in here?

If you find minima the equations have no choice but to trivially
factor D when it is a composite.

The minima has to exist, and finding it non-trivially factors D, when
D is a composite.

(It may not - I was trying to find S(v) as a
polynomial and found R(v) and C(v) also.)

I think there's a general formula for the
coefficients of R(v), S(v), and C(v), but
that will involve matrices and more time.

                                          Enrico

Short answer, I found that the factoring problem is a calculus problem
of finding minima.

AND showed how it connected to Pell's Equation.

Life's not as easy as proving something important because there are a
lot of social structures built up around people believing certain
things.

Mathematics is about absolute proof.

The proof will shred the disbelief, but it can take time.  It's like a
diamond cutting instrument.  It's only about time.

But some of you need to understand: you never really believed in
mathematical proof which is why I had to define it.

You believed in people.

James Harris- Hide quoted text -

- Show quoted text -

==========================================================
The thing that's bugging me is that with:

D = 119

R(V) = 80 * V ^ 2 + 118 * V + 45

S(V) = 6 * V ^ 2 + 10 * V + 4

C(V) = 46 * V ^ 2 + 50 * V + 11


R(v) ^ 2 - D*S(v) ^ 2 = C(v) ^ 2

Gives a factorization for ALL values of V.
I was expecting only 1 or 2 values of V to work.

I guess thats what I get for starting with the
solution and working backwards to the problem
statement. Meh.



Enrico
.

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