Show that if m, n are positive integers such that (xm+n + ym+n + zm+n)/(m+n) = (xm + ym + zm)/m (xn + yn + zn)/n for all real x, y, z with sum 0, then {m, n} = {2, 3} or {2, 5}.
Solution
Put z = - x - y. If m and n are both odd, the lhs has a term in xm+n but the rhs does not. So at least one of m and n is even. Suppose both are even. Then comparing terms in xm+n we get 2/(m+n) = 4/mn. Put m = 2M, n = 2N, then MN = M+N. So M must divide N and N must divide M. Hence M = N = 2. So m = n = 4. But put x = y = 1, z = -2, the lhs is (1 + 1 + 256)/8 = 129/4 and the rhs is ( (1 + 1 + 16)/4)2 = 81/4. So one of m, n must be odd and the other even. Without loss of generality we may take m odd.
Comparing the terms in xm+n-1y, the coefficient on the lhs is 1. On the rhs it is 1 x 2/n. So we must have n = 2. Put x = y = 1, z = -2. We get lhs = (1 + 1 - 2m+2)/(m+2) and rhs = (1 + 1 + 4)/2 x (1 + 1 - 2m)/m. So (6 - m)2m-1= 2m+6. We cannot have m > 6 or the lhs is negative. Trying m = 1, 3, 5 we find that 3 and 5 are the only solutions.
Arguably, we still have to verify that m=3, n=2 and m=5, n=2 are solutions. That is just tedious algebra. [We do have equality for those values.]
© John Scholes
jscholes@kalva.demon.co.uk
21 Aug 2002