From - Thu Aug 27 15:14:19 1998
From: Graham Clemow <gmc@lambdal.demon.co.uk>
Newsgroups: sci.math
Subject: Re: an equation from a math competition
Date: Wed, 26 Aug 1998 16:23:37 +0100
Message-ID: <mOmdpCA5hC51Ewh9@lambdal.demon.co.uk>
References: <6rvei5$m3k$1@namib.north.de>

>Solve
>
>  (1/2)(x+y)(y+z)(z+x)+(x+y+z)^3 = 1-xyz
>
>  in the integers. (Austrian-Polish maths olympiad, 94?)
>


















Problems which are symmetrical in their variables can often
be simplified by transforming x,y,z to a,b,c where:
  a = x + y + z
  b = xy + yz + zx
  c = xyz

Using:
  (x+y)(y+z)(z+x) = (x + y + z)(xy + yz + zx) - xyz = ab - c
the problem equation:
  (1/2)(x+y)(y+z)(z+x)+(x+y+z)^3 = 1-xyz
becomes:
  (1/2)(ab-c) + a^3 = 1 - c
which reduces to:
  2(a^3) + ab + c = 2      (1)

For any number r:
  (r+x)(r+y)(r+z) = r^3 +a(r^2) + br + c
Substituting r = a gives:
  (a+x)(a+y)(a+z) = a^3 + a^3 + ab + c 
                  = 2     using (1) above  

Now if x, y and z are integer, then a = x+y+z is integer and
so the only possible triples of values (in any order) for
(a+x),(a+y),(a+z) are:
  1,    1,    2
 -1,   -1,    2
  1,   -1,   -2
 -1,    1,   -2

Now (a+x) + (a+y) + (a+z) = 3a + (x+y+z) = 4a and so the sum of each
triple must be divisible by 4 (since a is an integer) ruling out the
last two triples leaving the only solutions:
a=1, x=0,  y=0,  z=1  and permutations of x,y,z
a=0, x=-1, y=-1, z=2  and permutations of x,y,z

Hence the only solutions in (x,y,z) are:
(0,0,1), (0,1,0), (1,0,0), (-1,-1,2), (-1,2,-1), (2,-1,-1)

Regards
Graham Clemow