32nd Putnam 1971 Problem A2 Find all possible polynomials f(x) such that f(0) = 0 and f(x 2 + 1) = f(x) 2 + 1. Solution
: f(x) = x. f(0) = 0. f(1) = f(0) 2 + 1 = 1. f(1 2 + 1) = f(1) 2 + 1 = 2. Similarly, by an easy induction we can get an arbitrarily large number of integers n for which f(n) = n. So if f has degree m, we can find at least m+1 integers on which it agrees with the polynomial p(x) = x. Hence it is identically equal to p. 32nd Putnam 1971 © John Scholes jscholes@kalva.demon.co.uk 7 Feb 2001