IMO 1981 SL 13

Origin: ROM

Problem

Let P be a polynomial of degree n satisfying P(k) = n + 1 k −1 for k = 0, 1, . . ., n. Determine P(n + 1).

Solution

Lemma. For any polynomial P of degree at most n, n+1  i=0 (−1)i n + 1 i  P(i) = 0. (1) Proof. We shall use induction on n. For n = 0 it is trivial. Assume that it is true for n = k and suppose that P(x) is a polynomial of degree k + 1. Then P(x) −P(x + 1) clearly has degree at most k; hence (1) gives

0 = k+1  i=0 (−1)i k + 1 i  (P(i) −P(i + 1))

k+1  i=0 (−1)i k + 1 i  P(i) + k+2  i=1 (−1)i k + 1 i −1  P(i)

k+2  i=0 (−1)i k + 2 i  P(i). This completes the proof of the lemma. Now we apply the lemma to obtain the value of P(n + 1). Since P(i) =
n+1 i −1 for i = 0, 1, . . ., n, we have 0 = n+1  i=0 (−1)i n + 1 i  P(i) = (−1)n+1P(n + 1) + . 1, 2 | n; 0, 2 ∤n. It follows that P(n + 1) = . 1, 2 | n; 0, 2 ∤n.