Re: I was right, surrogate factoring proof
jstevh_at_msn.com
Date: 02/13/05
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Date: 13 Feb 2005 10:15:50 -0800
Tim Peters wrote:
> [...]
>
> [JSH]
> >> In any event, the proof that the method has to work is easy.
There
> >> must exist integers Ax and Az, as shown above.
>
> [ošin]
> > And if we do not believe you this time, we are again liars, right?
>
> Of course -- that's always how it goes.
>
> James is certainly right that for any rationals x and z, there exists
an
> integer A such that Ax and Az are both integer. Alas, there are an
infinite
> number of such A, the set of all integer multiples of lcm(denom(x),
> denom(z)) (where lcm(a, b) is the least common multiple of a and b,
and
> assuming x and z are both in normalized form (gcd(numerator,
denominator) =
> 1)).
>
> Without a proof that an _appropriate_ A can be found efficiently, he
may
> well have a correct method (don't know -- didn't check), but of no
use. For
> example, here's a much simpler method of that kind:
>
Hmmm...an actual assertion in there against my proof?
Well, let me give just a bit of mathematics, and the theorem this time,
followed by proof, in line with the convention one poster keeps
bringing up.
Given a natural number M, a natural number j can be chosen such that
the factorization of M^2 - j^2 and j will factor M.
It can be shown that with non-zero integer A, and rationals x and z,
Ax = Az(-Az +/- sqrt((Az - 2M^2)^2 - 4TM^2))/(2j^2 - 2Az)
and
Az = Ax(-Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2M^2 - 2Ax)
where A is an integer such that Ax and Az are both integers.
Proof complete.
James Harris
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