Re: Computability

On Sun, 13 Jan 2008 20:02:36 -0800 (PST), The Dougster 22044
<DGoncz@xxxxxxxxxxxx> wrote:

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.

Right.

Let me try to restate your conjecture ...

For greater simplicity, I won't bother with the order function.
Instead, I'll just use the divisibility relation.

Dougster's conjecture:

There do not exist positive integers x,y,z such that

(1) x < y < z < x+y

(2) x,y,z, are pairwise coprime

(3) z - y is not a multiple of x

(4) For some prime p > 2,

x | z^p - y^p
y | z^p - x^p
z | x^p + y^p

Remarks:

(1) For p = 3, your conjecture holds for z <= 1000.

(2) As you've previously noted, if a proof of your conjecture could be
had, that would yield an instant proof of FLT, however the known truth
of FLT does not appear to yield a proof of your conjecture.

(3) As far as trying to prove your conjecture, I would start with a
fixed prime, for example p = 3.

quasi

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