IMO 1997 SL 10

Origin: CZE

Problem

Find all positive integers k for which the following statement is true: If F(x) is a polynomial with integer coefficients satisfying the condition 0 \leqF(c) \leqk for each c \in{0, 1, . . ., k + 1}, then F(0) = F(1) = \cdot \cdot \cdot = F(k + 1).

Solution

Suppose that k \geq4. Consider any polynomial F(x) with integer coeffi- cients such that 0 \leqF(x) \leqk for x = 0, 1, . . . , k+1. Since F(k+1)−F(0) is divisible by k + 1, we must have F(k + 1) = F(0). Hence F(x) −F(0) = x(x −k −1)Q(x) for some polynomial Q(x) with integer coefficients. In particular, F(x) − F(0) is divisible by x(k + 1 −x) > k + 1 for every x = 2, 3, . . . , k −1, so F(x) = F(0) must hold for any x = 2, 3, . . ., k −1. It follows that F(x) −F(0) = x(x −2)(x −3) \cdot \cdot \cdot (x −k + 1)(x −k −1)R(x)

for some polynomial R(x) with integer coefficients. Thus k \geq|F(1) − F(0)| = k(k −2)!|R(1)|, although k(k −2)! > k for k \geq4. In this case we have F(1) = F(0) and similarly F(k) = F(0). Hence, the statement is true for k \geq4. It is easy to find counterexamples for k \leq3. These are, for example, F(x) = ⎧ ⎨ ⎩ x(2 −x) for k = 1, x(3 −x) for k = 2, x(2 −x)2(4 −x) for k = 3.