Re: JSH: SF: Finally, surrogate factoring

Rick Decker wrote:
Rick Decker wrote:



Tim Peters wrote:

[added "JSH:" to subject, spared sci.crypt and alt.math]

[jstevh@xxxxxxx]

...
And it has been posted in this thread that I DID get it wrong.

If you do it right, it shows a dependency on the factorization of T,
which is no good.

But what if you go the other way, isolating y on the right side,
instead of z?

So instead, you would complete the square twice isolating z on the left
and with the second you would get y inside the square on the right.

The reason for wondering is that if you solve for z using the four
linear equations, yup, it does solve in such a way that you can have a
difference of factors of T, but y does not.

Long-shot.

... and a miss.



I got

(42*z + 10*y - 3*f_2 + 19*f_1)^2 = (4*y + 3*f_2 - 5*f_1)^2 + 84*T

then, but didn't care enough to double-check it.

That's what I got, too. Mathematica agrees.

(Responding to my own post...)

So completing the square w.r.t y first yields

(2*y + 10*z + 5*f_1 - f_2)^2 = (4*z + 3*f_1 - f_2)^2 + 4*T

Completing the square w.r.t. z first yields

(42*z + 10*y - 3*f_2 + 19*f_1)^2 = (4*y + 3*f_2 - 5*f_1)^2 + 84*T


Hey, I was curious to see it, figured someone could put up a solution,
and I guess you're right, but, let's think about it a bit.

If you are correct, then

4y + 3f_2 - 5f_1 = h_1 - h_2

where h_1 h_2 = 21T = 21 g_1 g_2.

But I can also solve for y directly using the 4 linear equations,
expressing it directly as a solution of f_1, f_2, g_1, and g_2.

Given your solutions, there are more possible values for w, x, y and z
than there are possible using the 4 linear equations.

Now, I guess there is something I'm missing here.

After all, you say you're using Mathematica, and hey, why would you
lie?

After all, other people can check.

My reasoning can just be wrong here.

Tell me, is there something then that is forcing y to behave and follow
the solution that is forced upon it by the 4 linear equations, such
that somehow it is picking and choosing from 21 g_1 g_2 in such a way
as to only get one set of solutions?

Maybe there is and I can rescue something from this, you know, like
some extraordinary control that the algebra is using to pick only some
factorizations.


James Harris

.

Relevant Pages

  • Re: JSH: SF: Finally, surrogate factoring
    ... and with the second you would get y inside the square on the right. ... There is one explicit solution for y given by solving the 4 linear ... But from your original four linear equations we can derive ... surrogate factoring solution as the only way the algebra can avoid the ...
    (sci.math)
  • Re: JSH: SF: Finally, surrogate factoring
    ... and with the second you would get y inside the square on the right. ... There is one explicit solution for y given by solving the 4 linear ... Trouble is, 4 linear equations and 4 variables give ONE solution for y, ... surrogate factoring solution as the only way the algebra can avoid the ...
    (sci.math)