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

 Messages: [ Previous | Next ]
 Tim Peters Posts: 2,034 Registered: 12/13/04
Re: JSH: SF: Finally, surrogate factoring
Posted: Jun 5, 2006 11:04 PM

[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.

Date Subject Author
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6/5/06 Doug Schwarz
6/5/06 Tim Peters
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6/5/06 Gib Bogle
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