IMO 1985 SL 12

Problem

3b.(GBR 4) A sequence of polynomials Pm(x, y, z), m = 0, 1, 2, . . ., in x, y, and z is defined by P0(x, y, z) = 1 and by Pm(x, y, z) = (x + z)(y + z)Pm−1(x, y, z + 1) −z2Pm−1(x, y, z) for m > 0. Prove that each Pm(x, y, z) is symmetric, in other words, is unaltered by any permutation of x, y, z.

Solution

We shall prove by induction on m that Pm(x, y, z) is symmetric and that (x + y)Pm(x, z, y + 1) −(x + z)Pm(x, y, z + 1) = (y −z)Pm(x, y, z) (1) holds for all x, y, z. This is trivial for m = 0. Assume now that it holds for m = n −1. Since obviously Pn(x, y, z) = Pn(y, x, z), the symmetry of Pn will follow if we prove that Pn(x, y, z) = Pn(x, z, y). Using (1) we have Pn(x, z, y) − Pn(x, y, z) = (y+z)[(x+y)Pn−1(x, z, y+1)−(x+z)Pn−1(x, y, z+1)]−(y2− z2)Pn−1(x, y, z) = (y + z)(y −z)Pn−1(x, y, z) −(y2 −z2)Pn−1(x, y, z) = 0. It remains to prove (1) for m = n. Using the already established symmetry we have (x + y)Pn(x, z, y + 1) −(x + z)Pn(x, y, z + 1) = (x + y)Pn(y + 1, z, x) −(x + z)Pn(z + 1, y, x) = (x + y)[(y + x + 1)(z + x)Pn−1(y + 1, z, x + 1) −x2Pn−1(y + 1, z, x)] −(x + z)[(z + x + 1)(y + x)Pn−1(z + 1, y, x + 1) −x2Pn−1(z + 1, y, x)] = (x + y)(x + z)(y −z)Pn−1(x + 1, y, z) −x2(y −z)Pn−1(x, y, z) = (y −z)Pn(z, y, x) = (y −z)Pn(x, y, z), as claimed.