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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 10:09 AM

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.

>
>
> 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.

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
6/5/06 Gib Bogle
6/5/06 Proginoskes
6/5/06 William L. Bahn
6/5/06 JAMES HARRIS
6/5/06 Abstract Dissonance
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6/5/06 guenther.vonKnakspott@gmx.de
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6/5/06 Abstract Dissonance
6/5/06 Gordon Burditt
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6/5/06 Ed Weir \(ComCast\)
6/5/06 Abstract Dissonance
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6/5/06 gjedwards@gmail.com
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6/6/06 TC
6/6/06 Salami Man
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6/5/06 Rick Decker
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6/5/06 jshsucks@yahoo.com
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