Sunday, February 13, 2005


Surrogate factoring, room for error?

Well, considering the direct evidence that I may be wrong, it seems to me that the simplest explanation would be at the start:

yx^2 + Ax - M^2 = 0


yz^2 + Az - j^2 = 0

as that's an assertion of a truth.

Essentially I'm saying that there exists these two quadratics related in that given way, but if in fact no rationals exist such that all conditions can be met for a given natural M and j then that statement is false.

If it is true, then all else follows in the basic argument I've given.

So then, can that initial start be false?

Well I can solve out y to get

z^2 (M^2 - Ax) = x^2 (j^2 - Az)

and then I can solve for A to get

A = (z^2 M^2 - j^2 x^2)/xz(z-x)

so A is not as arbitrary as I imagined.

Hmmm…A again, where it's not clear that it's just settable, but then again, x and z are rationals, so more analysis is needed.

Switching to rationals for x and z, with z = a/b and x = c/d, I get

A = (a^2 d^2 M^2 - j^2 b^2 c^2)/ac(ad - bc)

so there must exist integers a, b, c and d such that A is an integer, or it's a fraction.

Yeah, but I can just choose an A, and then get rational x, y and z.

So, rationals is kind of a big set, there must exist a, b, c and d for any natural A chosen.

That means the start of my proof is valid, and

yx^2 + Ax - M^2 = 0


yz^2 + Az - j^2 = 0

must be true for some naturals M, j and A, with rational y, x and z.

Though I'm still convinced that if M is odd, j needs to be odd.

Mystery continues…why don't implementations work?

Where's the mistake?

Yup. And I had the reason correct at the beginning as if j is even when M is odd then both quadratics cannot necessarily be true, with rational nonzero x, y and z.

So that breaks things at the start of the argument, so nothing else follows.

A proof begins with a truth. The start of the proof is the set of quadratics for rational x, y and z.

If M is odd while j is even, then that start is not valid, as then no rational nonzero x, y and z exist.


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