Re: JSH: One other option

James Harris wrote:
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,

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

and

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

If I have understood you I believe that some simplification is
possible here. Substituting k=2x into the above equations we have

n_min = floor((12x - 4 - 8x^2)/T)
n_max = floor(x^2/T)

Note that if T is not a perfect square then (floor(sqrt(T)))^2 < T so
that n_max = 0. Meanwhile if T is very large then x^2 >> x and x^2 >>
4 so that n_min is approximately -8. Is this correct?

I'm afraid I don't see why you think that letting n range over these
values will guarantee that one finds a factor of T. Can you explain in
more detail?

.

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