Let f(x, y, z) = x2 + y2 + z2 + xyz. a(x, y, z), b(x, y, z), b(x, y, z) are polynomials with real coefficients such that f(a(x, y, z), b(x, y, z), c(x, y, z)) = f(x, y, z). Prove or disprove that a(x, y, z), b(x, y, z), c(x, y, z) must be ±x, ±y, ±z in some order (with an even number of minus signs).
Solution
False. Another solution is x, -y, xy + z.
Fairly easy.
It is not immediately obvious whether this is true or false. The obvious line of attack is to try solutions which are almost x, y, z, but that cuts down the complexity. Specifically, we might try a(x, y, z) = x, b(x, y, z) = y. That requires c2 + xy c - z2 - xyz = 0, which is a quadratic in c. We know one solution is z, so the other is - xy - z (sum of roots is -xy, for example). It looks slightly more elegant if we switch two of the signs.
Of course, if this had not worked (the only solution turned out to be c(x, y, z) = z), then we would have achieved nothing, but we were lucky.
© John Scholes
jscholes@kalva.demon.co.uk
30 Sep 1999