32nd Putnam 1971

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Problem A2

Find all possible polynomials f(x) such that f(0) = 0 and f(x2 + 1) = f(x)2 + 1.

 

Solution

Answer: f(x) = x.

f(0) = 0. f(1) = f(0)2 + 1 = 1. f(12 + 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