Re: Surrogate factoring, room for error?
From: Rick Decker (rdecker_at_hamilton.edu)
Date: 02/14/05
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Date: Sun, 13 Feb 2005 23:01:39 -0500
Nora Baron wrote:
> jstevh@msn.com wrote:
>
<snip>
>
>>It has to be true then, as it begins with a truth and proceeds by
>>logical steps to a conclusion which then must be true.
>>
>
>
> Oh, so you're saying you want us to point out where you might
> have made a logical mistake! But you cannot be wrong, or else basic
> algebra is wrong! That's what you said!
>
> Of course, if you really want an answer, you might begin with the
> fact that you arrive at two equations, one in which x is defined
> in term of z and the other in which z is defined in terms of x.
> Thus very soon after you start you have a circular argument.
> You cannot prove z is rational until you have x, and you cannot
> prove x is rational because it is defined in terms of z, which
> is not known until you specify x ... Worse yet, you have to have
> both x and z defined and rational before you can find integer A such
> that Ax and Az are both integers. What you REALLY need is that
>
> (Ax - 2j^2)^2 + 4Tj^2
>
> is a perfect square. Given A, you can find rational x so that this
> is true, but again: your process defines A in terms of x and z.
> (If you have rational x and z, you can find A such that Ax and
> Az are integers ...) Again, you have a vicious circle. You have
> to start somehow with A or x or z, and then derive the others in
> terms of the first one.
>
> I think you have gotten yourself thoroughly confused on this.
>
> Nora B.
>
Yeah, it's a bit confusing, but that's generally where James starts.
Look back at his FLT construction and you'll see the same thing. By
analogy, I suspect that it will be several months before the interested
members of this ng manage to refine his constructions enough to
be generally acceptable.
Here he starts with 2 equations in 4 unknowns: M and j are given and you
have some freedom to pick y, A, x, and z. Now obviously (more or less)
x and z are related by those quadratics, so they can't be chosen
independently. In fact, Ax and Az are similarly related. In addition,
he's looking for integers for x and z so his most recent attempt
puts some constraints on A. There's an obvious solution for y in terms
of M^2, A and x (or z) so as near as I can tell you only have one
degree of freedom. That's good, since you want some wiggle room
but not too much. Unfortunately, you don't get quite enough wiggle
room, which is basically why James' method works (i.e., gives nontrivial
factors of M) for some j values, but not all. Whether this can be
fixed is unclear.
[Thinking about it, it occurs to me that there's really no reason
why James needs z to be rational, in light of the fact that all
he really needs is a rational (ideally integral) solution for x.
Of course relaxing the requirement on z screws you, since you
no longer have any requirements on x.]
Regards,
Rick
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