### 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**

Answer: 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