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

 Messages: [ Previous | Next ]
 Rick Decker Posts: 1,356 Registered: 12/6/04
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,

>>
>>> 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

Date Subject Author
6/4/06 JAMES HARRIS
6/5/06 Doug Schwarz
6/5/06 Tim Peters
6/5/06 Doug Schwarz
6/5/06 Christopher J. Henrich
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