Re: JSH: SF: Finally, surrogate factoring
- From: Rick Decker <rdecker@xxxxxxxxxxxx>
- Date: Tue, 06 Jun 2006 10:09:51 -0400
Tim Peters wrote:
[added "JSH:" to subject, spared sci.crypt and alt.math]That's what I got, too. Mathematica agrees.
[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.
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.
Regards,
Rick
.
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