I recall that there was a Putnam problem which went something like this:

Find all real functions satisfying

$$f(s^2+f(t)) = t+f(s)^2$$

for all $s,t \in \mathbb{R}$.

There was a cool trick to solving it that I wanted to remember. But I don't know which test it was from and google isn't much help for searching with equations.

Does anyone know which problem I am thinking of so I can look up that trick?

  • 1
    $\begingroup$ Are you saying this is the question and you're looking for the trick, or is it more like you're looking for the question's correct statement? $\endgroup$ – Patrick Da Silva Nov 10 '11 at 21:23
  • $\begingroup$ I was looking for both the question's correct statement, as well as the trick. I have encountered a similar problem which I wanted to try and apply the same trick to. It turns out my problem was much easier because I am allowed to assume differentiability, but I am still interested in this problem just for fun. $\endgroup$ – nullUser Nov 10 '11 at 22:26
  • $\begingroup$ Well I guess you got what you wanted. I actually found an answer too while playing with it but since there are plenty of solutions in the answers below I didn't bother. Thanks for posting it, it was fun =) $\endgroup$ – Patrick Da Silva Nov 10 '11 at 22:28

It was problem 2 in the International Mathematical Olympiad in 1992 (Russia).

You may read several solutions here.


1971 A2 was "Determine all polynomials $P(x)$ such that $P(x^2+1)=(P(x))^2+1$ and $P(0)=0$."


No idea. But I have a book called Putnam and Beyond by Gelca and Andreescu, and on page 185 they present a problem from a book called Functional Equations: A Problem Solving Approach by B. J. Venkatachala, from Prism Books PVT Ltd., 2002. I think the Ltd. means the publisher is British.

Almost, the publisher is (or was?) in India (Bangalore):



EDIT, December 3, 2011: The book is available, at least, from an online firm in India that is similar to Amazon.com




I cannot tell whether they ship outside India. But it does suggest that contacting the publisher by email is likely to work.


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