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Topic:
SF: Finally, surrogate factoring
Replies:
86
Last Post:
Jun 10, 2006 11:51 PM




Re: JSH: SF: Finally, surrogate factoring
Posted:
Jun 6, 2006 11:01 AM


Rick Decker wrote:
> > > Tim Peters wrote: > >> [added "JSH:" to subject, spared sci.crypt and alt.math] >> >> [jstevh@msn.com] >> >>> ... >>> 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. >>> >>> Longshot.
... 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 doublecheck 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
Reregards,
Rick



