# 1992 IMO Problems/Problem 2

## Problem

Let $\mathbb{R}$ denote the set of all real numbers. Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that

$$f\left( x^{2}+f(y) \right)= y+(f(x))^{2} \hspace{0.5cm} \forall x,y \in \mathbb{R}$$

## Solution

[quote=probability1.01]Set x = 0 to get $f(f(y)) = y+f(0)^{2}$. We'll let $c = f(0)^{2}$, so $f(f(y)) = y+c$. Then

$f(a^{2}+f(f(b))) = f(b)+f(a)^{2}$ $f(a^{2}+b+c) = f(b)+f(a)^{2}$ $f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}$ $a^{2}+b+2c = b+(a+c)^{2}$ $2c = c^{2}+2ac$

Since this holds for all $a$, it follows that $c = 0$. Now we have $f(0) = 0 \implies f(f(y)) = y$. Note that f must be surjective since we may let y vary among all reals, and f must be injective since if $f(a) = f(b)$, then $a+f(x)^{2}= f(x^{2}+f(a)) = f(x^{2}+f(b)) = b+f(x)^{2}$. Finally, if $u > v$, then there is some $t$ s.t. $u = t^{2}+v$, and so $f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)$. Hence f is strictly increasing. It is now clear that since $f(f(y)) = y$, we must have $f(x) = x$ for all x.[/quote]