n ≥ 4 Solution It is easy to find counter-examples for n=1,2,3. For example: x(2-x) has p(0) = 0, p(1) = 1, p(2) = 0 x(3-x) has p(0) = 0, p(1) = p(2) = 2, p(3) = 0 x(4-x)(x-2) 2 has p(0) = 0, p(1) = 3, p(2) = 0, p(3) = 3, p(4) = 0 Now suppose n ≥ 4 and 0 ≤ p(k) ≤ n for k = 0, 1, ... , n+1. For any polynomial p(n+1) - p(0) must be a multiple of n+1, so in this case p(n+1) must equal p(0). So p(x) - p(0) has factor x(n+1-x) which is > n for x = 2, 3, ... , n-1. Hence p(x) also = p(0) for x = 2, 3, ... , n-1. It remains to consider x = 1, x = n. But p(x) - p(0) must have factor (x-3)(n+1-x) (this is where we need n > 3) and so p(1) - p(0) must have factor 2n > n and hence be zero. p(x) - p(0) also has factor x(n-2-x) (again we need x > 3), so p(n) - p(0) has factor 2n and must be zero. Thanks to Kamil Duszenko 38th IMO shortlist 1997 © John Scholes jscholes@kalva.demon.co.uk 2 Dec2003 Last corrected/updated 2 Dec 2003
38th IMO 1997 shortlist Problem 10 Find all positive integers n such that if p(x) is a polynomial with integer coefficients such that 0 ≤ p(k) ≤ n for k = 0, 1, 2, ... , n+1 then p(0) = p(1) = ... = p(n+1).