### 21st Putnam 1960

**Problem A5**

The real polynomial p(x) is such that for any real polynomial q(x), we have p(q(x)) = q(p(x)). Find p(x).

**Solution**

Take q(x) = x + k and set x = 0. Then we have p(k) = p(0) + k. This is true for all k, so p(x) must be x + c for some c. Now take q(x) = x^{2}. We get x^{2} + c = (x + c)^{2} = x^{2} + 2cx + c^{2}. Hence c = 0 and p(x) = x.

*Comment: it is even easier to take q(x) = k. Then p(k) = k for all k. In other words, p(x) = x.*

21st Putnam 1960

© John Scholes

jscholes@kalva.demon.co.uk

15 Feb 2002