Re: JSH: One other option

On Sep 11, 5:19 pm, JSH <jst...@xxxxxxxxx> wrote:
On Sep 11, 2:03 am, "Jesse F. Hughes" <je...@xxxxxxxxxxxxx> wrote:

JSH <jst...@xxxxxxxxx> writes:
I'm running out of room for hope with this research and the expert
opinion of one of the world's leading math journals is against it.

So no reason not to just release what I have and hope.

Buck up, li'l camper. Next week, you'll have a new idea and there's
no reason this new idea won't work to solve the factoring problem.
Anyway, it *might* solve it, and should mathematicians want to bet their
careers, maybe even their lives, that it won't?

The subject of this thread is my other option which is just to toss
out what theory I have with the caveat that the paper giving that
theory--still not yet posted publicly by me--was rejected by the
Bulletin of the AMS.

It turns out Marcus pointed something out in another thread so I can
give a simpler definitive test of the latest theory which oddly takes
away any factoring of surrogates. Thanks Marcus!!!

Kind of odd really, but fascinating to contemplate as by the theory
the following algorithm should be applicable against an RSA sized
number factoring it in a maximum of 32 trials:

Note, still with x^2 = y^2 mod T, and k = 2x mod T.

So then, with x = floor(sqrt(T)), k = 2x,

Marcus noticed that x^2 needs to exceed T or it might not work.

Good work Marcus! So you need x = floor(sqrt(T)) + 1.

The reason why is that x>y is assumed. And Marcus found an example
where x^2<T, so there was no way that it could work.

The other equations are

n_min = floor((4(x+k-1) - 2k^2)/T)

and

n_max = floor(((x+k)^2 - 2k^2)/T) ,

you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min to
n_max, looking for a perfect square.

You have a limited number of possibles to check where I thought it was
32, but I might have got that wrong too, but it shouldn't be much
bigger.

I guess.


James Harris

.

Relevant Pages

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    (sci.crypt)
  • Re: Simple answer, surrogate factoring
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