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From: sonnens@mprvaxa
Newsgroups: net.math
Subject: Re: UCSB competition problem (solution)
Message-ID: <379@mprvaxa.UUCP>
Date: Thu, 17-Nov-83 18:26:48 EST
Article-I.D.: mprvaxa.379
Posted: Thu Nov 17 18:26:48 1983
Date-Received: Sun, 20-Nov-83 01:24:03 EST
Organization: Microtel Pacific Research, Burnaby BC
Lines: 22

Here's a solution to the following recently posed problem:

		 2              2
		x  = 1 + (y - z)

		 2              2
		y  = 2 + (x - z)

		 2              2
		z  = 3 + (x - y)


If b = sqrt(24), then (x,y,z) = (5/b,8/b,9/b) or -(5/b,8/b,9/b).

The trick is to write each equation as a difference of squares, which
can then be factored.  There are a lot of identical factors, which can
be cancelled out to reduce this to a system of three linear equations
(e.g., 2(x + y - z)(x - y + z) = (y + x - z)(y - x + z)).

Dan Sonnenschein
Microtel Pacific Research
...microsoft!uw-beaver!ubc-visi!mprvaxa!sonnens