Re: FLTAA: Field structure p = 2, p>2
- From: "The Dougster" <DGoncz@xxxxxxx>
- Date: 14 Jan 2007 02:20:07 -0800
The Dougster wrote:
Chip Eastham wrote:
The Dougster wrote:
Chip Eastham wrote:
Hi, hagman:
Your example:
How about x=2, y=3, z=5, p=5:
Since we have z=y mod x, z=x mod y, x=-y mod z,
it follows that
(z/y,z/x,x/y)^p = (1,1,-1)^5 = (1,1,-1) in Mx x My x Mz.
is of the form x + y = z. In earlier threads the Dougster
restricts interest to 1 < x < y < z < x+y, as solutions
to x^p +y^p = z^p would have z < x+y when p > 1.
Doug was taking an abstract algebra course last term,
as he mentions, and AFAIK his interest in this arose
out his studies.
regards, chip
Hm. I must think more carefully about which restrictions to include
when restating the problem. Like exactly one of x,y,z even. I mean, is
that a restriction or a conclusion?
You ask that x,y,z be pairwise coprime, so that
at most one is even. Having xyz even appears
to be an additional restriction. I was never clear
if you'd found counterexamples with all x,y,z odd.
I agree you should come up with a concise
restatement of the problem with which to lead
off any new thread.
Writing a restatement would be and has been easy. Making it concise is
hard. Would anyone be interested in a wiki for this purpose?
OK!
http://fltma.pbwiki.com/
write for password.
Doug
.
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