IMO 1981 SL 6

Origin: CUB

Problem

Let P(z) and Q(z) be complex-variable polynomials, with degree not less than 1. Let Pk = {z \inC | P(z) = k}, Qk = {z \inC | Q(z) = k}. Let also P0 = Q0 and P1 = Q1. Prove that P(z) \equivQ(z).

Solution

Assume w.l.o.g. that n = degP \geqdegQ, and let P0 = {z1, z2, . . . , zk}, P1 = {zk+1, zk+2, . . . zk+m}. The polynomials P and Q match at k + m points z1, z2, . . . , zk+m; hence if we prove that k + m > n, the result will follow. By the assumption, P(x) = (x −z1)\alpha1 \cdot \cdot \cdot (x −zk)\alphak = (x −zk+1)\alphak+1 \cdot \cdot \cdot (x −zk+m)\alphak+m + 1 for some positive integers \alpha1, . . . , \alphak+m. Let us consider P ′(x). As we know, it is divisible by (x −zi)\alphai−1 for i = 1, 2, . . ., k + m; i.e., k+m

i=1 (x −zi)\alphai−1 | P ′(x). Therefore 2n −k −m = deg $k+m i=1 (x −zi)\alphai−1 \leqdegP ′ = n −1, i.e., k + m \geqn + 1, as we claimed.