43rd Swedish 2003 Problem 4 Find all real polynomials p(x) such that 1 + p(x) ≡ (p(x-1) + p(x+1) )/2.
x 2 + bx + c Solution Suppose p(x) has degree  2. So we may take p(x) = ax n + bx n-1 + cx n-2 + ... with a ≠ 0. Comparing coefficients of x n-2 we have c = (an(n-1) + 2c)/2, so a = 0. Contradiction. So p(x) must have degree 2. Suppose p(x) = ax 2 + bx + c. Then lhs = ax 2 + bx + (c+1) and rhs = (a(x 2 - 2x + 1) + b(x - 1) + c + a(x 2 + 2x + 1) + b(x + 1) + c)/2 = ax 2 + bx + c + a. So we have an identity provided a = 1. 43rd Swedish 2003 © John Scholes jscholes@kalva.demon.co.uk 11 January 2004 Last corrected/updated 11 Jan 04