Re: Computability
- From: The Dougster 22044 <DGoncz@xxxxxxxxxxxx>
- Date: Sun, 13 Jan 2008 20:02:36 -0800 (PST)
On Jan 13, 7:34 pm, The Dougster 22044 <DGo...@xxxxxxxxxxxx> wrote:
On Jan 10, 11:16 am, quasi <qu...@xxxxxxxx> wrote:Naw, that is wrong. I wrote it wrong, but we are mising the point.
You claimed to have checked it up to z = 256, but apparently you
missed the following counterexample:
(x,y,z) = (74,129,143) which has signature (6,3,3).
(x/y)^p == 143 - 105 == -38, not -1 which would be 142 mod 143
If (x/y)^p == -1 then (x/y)^2p == 1 and o(x/y,z) = 2p
But we lose information when we write o(x/y,z) = 2p. That doesn't
specify that
(x/y)^p == -1. (x/y)^p could be congruent to x for all we know! It
just doesn't say.
Now from (x/y)^p === -1 mod z I think there is a new characteistic
combination of x and y mod z that has an order of p or of 2p. Let's
see...
x^p + y^p = z^p
x^p + y^p == 0 mod z
x^p == -(y^p) mod z
x^p / -(y^p) == 1 mod z
1/(-1) * (x/y)^p == 1 mod z....
These are calculations in abstract algebra, I think. I'd better ask
for help here, but you see the point? There would be only the
inequality, without loss of generality, and the signature, but the
signature would have four parts, not just three. The part about
coprimality might be implied by the existence of the various
signatures.
So it would really boil down to just the signature.
Aside: I am looking at the conjoint and disjoint probabilities of the
inequality, the coprimality, and the signature, applying Bayesian
statistics. It's not going well because I only have had stats I, and
we just brushed by Bayes.
Doug
.
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