Re: JSH: Finally useful theory?

[added "JSH:" to subject, spared sci.crypt]

[jstevh@xxxxxxx]
I pondered why all those algebraic manipulations just kept giving me
equations that relied on my target and realized it was obvious,

You managed to do so without realizing that people have _told_ you why, over
& over & over again, for more than a year now? That's the first thing that
came to mind when you later accused someone else:

Why are you so stupid? I have to explain everything to you in such
detail.

Pot, meet the mother of all kettles.

so I thought about how to get something different, and had a realization.

Let

S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))

where S is the surrogate, and you have to multiply everything out and
complete the square, like usual.

But if my idea is a good one, you'll end up with this complicated
expression with k_1, k_2, k_3 and k_4 multipled times S, where you
would then pick that complicated piece to equal your target composite.

Huh? You expect x and y to vanish, or what?

Then you just pick any S, like S=15, and get squares for x and y using
it, and then use the sign ambiguity of the square roots.

Huh?

You see, one way the solution will factor your surrogate S.

The other way, changing the signs, it will factor your target, if this
idea works.

Huh?

Um, can someone solve out that equation for S to get the complicated
thing at the end?

I have the software, but can't make much sense of your words. After
multiplying out, isolating sqrt(x*y) on one side, and squaring to get rid of
the radical:

(s - k_3*k_1*x - k_4*k_2*y)^2 = (k_4*k_1 + k_3*k_2)^2*y*x

What do you want to do next? For example, multiply it all out again and
complete the square wrt x? If so, then we're at (with x only on the LHS,
inside a square -- but all the other symbols are in there too):

(2*k_3*k_1*s - 2*(k_3)^2*(k_1)^2*x + (k_4)^2*(k_1)^2*y +
(k_3)^2*(k_2)^2*y)^2
=
(k_4*k_1 + k_3*k_2)^2 *
y *
(4*k_3*k_1*s + (k_4)^2*(k_1)^2*y - 2*k_4*k_3*k_2*k_1*y +
(k_3)^2*(k_2)^2*y)

and this is starting ;-) to look like a fool's errand. If you then want to
expand the mess on the RHS and complete the square wrt y:

2 2 2 2 2 2 2
(2 k_3 k_1 s - 2 k_3 k_1 x + k_4 k_1 y + k_3 k_2 y) =

2
- (-(k_4 k_1 + k_3 k_2)

2 2 2 2 2
(2 k_3 k_1 s + k_4 k_1 y - 2 k_4 k_3 k_2 k_1 y + k_3 k_2 y)

2 2 4 2 3 3 2 4 2 2 2
+ 4 k_4 k_3 k_1 s + 8 k_4 k_3 k_2 k_1 s + 4 k_3 k_2 k_1 s )

2
/(k_4 k_1 - k_3 k_2)

Good luck with that :-)

...


.

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