Re: JSH: Why factoring solution must work

On Feb 8, 12:21 pm, 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.

Wrong, for two reasons. First, finding v such that both

abs(r(v) + c(v)) < D and abs(r(v) - c(v)) < D will not in general

be possible. Second, it doesn't matter, since even if D were small
enough that the two conditions were simultaneously satisfied, it
wouldn't suffice to generate a nontrivial factorization of D.

As I've said before, it is a matter of simple algebra to show that

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

and

abs(r(v) - c(v)) = 8Dv^2

and neither of these will yield a nontrivial factorization of D
unless 2v+1 or v share a factor with D.

Here's an example. Let D = 77, f_2 = 2 and v = 1. Then, you have

x = 5784 / 5168 and y = 296 / 5168

Then

r + c = 5784 + 5168 = 10952 = 2(2 - 76)^2 = 2(74)^2

no factors of 77 here, obviously. We also have

r - c = 5784 - 5168 = 616 = 8(77)

here's where 77 lives. No information here, either.

I repeat, there's no need for spurious minimization,
since you _know_ what r + c and r - c are for
each value you pick for v.


Regards,

Rick
.

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