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From: rjnoe@ihlts.UUCP
Newsgroups: net.math
Subject: Re: UCSB competition problem
Message-ID: <266@ihlts.UUCP>
Date: Tue, 15-Nov-83 15:11:52 EST
Article-I.D.: ihlts.266
Posted: Tue Nov 15 15:11:52 1983
Date-Received: Wed, 16-Nov-83 07:07:04 EST
References: <345@sun.UUCP>
Organization: AT&T Bell Labs, Naperville, Il
Lines: 31

Actually, only algebra is needed.  Rearrange to get:
		x^2 - (y - z)^2  =  1
		y^2 - (x - z)^2  =  2
		z^2 - (x - y)^2  =  3
and let
		A =  x + y - z
		B =  x - y + z
		C = -x + y + z
Then the system of equations is
		AB = 1
		CA = 2
		BC = 3
Which can be quickly solved to get A = sqrt(2/3), B = sqrt(3/2), and
C = sqrt(6).  Of course, the negatives of these work just as well.  Now
		2x = A + B = 5/sqrt(6)
		2y = A + C = 4*sqrt(2/3)
		2z = B + C = 3*sqrt(3/2)
therefore
		        5
		x = ---------
		    2*sqrt(6)

		y = 2*sqrt(2/3)


		z = (3/2) ^ (3/2)

As I've said, there is one other solution, taking the additive inverses of
all of these.
-- 
		Roger Noe		...ihnp4!ihlts!rjnoe