Re: JSH: SF: Finally, surrogate factoring



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

and rewriting both these as differences of squares yields the same
(useless) factorization of T:

T = (y + 3*z + f_1)(y + 7*z + 4*f_1 - f_2)

and it's not hard to verify that

g_1 = y + 3*z + f_1

g_2 = y + 7*z + 4*f_1 - f_2


Re-regards,

Rick

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