From - Thu Aug 27 15:14:19 1998 From: Graham Clemow Newsgroups: sci.math Subject: Re: an equation from a math competition Date: Wed, 26 Aug 1998 16:23:37 +0100 Message-ID: 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