Re: JSH: Nearly done

jstevh_at_msn.com
Date: 01/25/05 

Date: 25 Jan 2005 15:41:45 -0800

Sebastian Gottschalk wrote:
> jstevh@msn.com wrote:
> ^^^^^^^^^^^^^^
> Wow, you parents gave you a really bad name.
>
> > All of that is easy to prove, and not only did I prove it in a
paper, I
> > wrote a program that does it.
> >
> > See http://groups.yahoo.com/group/sufactor/
>
> Please have a look a close look at it.
> Essentially you've found a complicated way to rewrite the already
> well-known Quadratic Sieve. You just need to reconvert your formulas
to
> have single squareroots on one side, square them and inset some of
your
> substitions.
>

That's not possible. I see you probably are just trolling, but I'll
explain anyway, why that cannot be true.

Remember surrogate factoring involves at its most basic two
quadratics:

yx^2 + Ax - j^2 = T

and

yz^2 + Az - j^2 = 0

where the idea is to get all rationals, while A, j and T are given,
and currently I pick j as approximately floor(M/2), and A=8.

And T, of course, equals M^2 - j^2.

Now the requirement that z be an integer, tells you that

yz^2 + Az - j^2 = (a_1 z + b_1)(a_2 z + b_2)

where the a's and b's are rationals, and b_1 b_2 = - j^2.

And now using the a's and b's with the first quadratic

yx^2 + Ax - j^2 = (a_1 x + b_1)(a_2 x + b_2) = T

and letting f_1 f_2 = T, you get

a_1 x + b_1 = f_1, and

a_2 x + b_2 = f_2

and I'll just go ahead and give the solution for x, as that is

x = (b_1 f_2 + b_2 f_1 - 2b_1 b_2)/A.

Notice that x is defined by factors of j and T, not M.

If my method, as you naively or simply in a trolling way claim, were in
fact just congruence of squares, then x would be defined by the factors
of M.
But is not.

Dare you answer? Or is your trolling done?

James Harris