So now with the full surrogate factoring theory, results are coming
fast and furious and I'll admit being very, very, very surprised that
an RSA number might be factored by p=3 and a fairly simple technique.

Those looking over the argument may recognize that there is only one
area where it's even maybe kind of looking like I didn't feel in the
blanks which is with how you find k.

But just try it. Factor a few numbers and you'll get that weird,
giddy out of this world feeling like maybe you stepped into the
Twilight Zone.

Reading over posters ranting and raving in reply to me is kind of
weird now. It's like there is something oddly wrong with them, but I
can't quite put my finger on it.

The challenges to factor an RSA public key though, seem to be
answerable now, and I'm mainly just absorbing the latest and the sense
of profound oddity of it all.

You can factor an RSA public key, if that key is 2 modulo 3, and if
with

z^2 = y^2 + public key

z is divisible by 3, and the math will just do it and kind of wink at
you as if it wasn't even hard.

And you get k by finding k such that abs(public key - 2k^2 ) is a
minimum and k is even, and you have two possibles k = 1 mod 3, or k =
-1 mod 3.

Then x = k/2, and z = 3x, and you get y and factor the public key.

Just like that.

Yup, I had reason to worry about factoring. Wacky. Factoring a
public key with p=3. Who would have thought the RSA system would
crash so profoundly?

Here's another example: T = (101)(103) = 10403, so trying k=-1 mod 3
(as I already know that k = 1 mod 3 won't work):

k = 68 is the maximal k, such that abs (10403 - 2k^2) is a minimum.

x = k/3 = 34. z = x + k = 102. 102^2 - 10403 = 1. y = 1, so

z-y = 101, z+y = 103

and that could just as easily been an RSA public key.

Still seems so weird though. So easy. All that work people did for
all those years and the answer is so easy.

James Harris