### 1992 IMO SL #6

Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that\[ f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}. \]

The answer is $\boxed{f(x)=x}$. It is easy to check that this works.

Let $P(x,y)$ denote the assertion. Using $P(0,y)$, we have $f(f(y))=y+f(0)^2$ which implies that $f$ is surjective. Also, if $f(x)=f(y)$, then $$x+f(0)^2=f(f(x))=f(f(y))=y+f(0)^2$$which implies $x=y$ or that $f$ is injective. Combining the two, we can say that $f$ is bijective. Now, consider $a \in \mathbb{R}$ such that $f(a)=0$. Using $P(a,f(y))$, we get $$f(a^2+f(f(y)))=f(y)+f(a)^2=f(y)$$ which implies that $y=f(f(y))+a^2$ since $f$ is injective. However, from before, we have $f(f(y))=y+f(0)^2$ which implies that $a^2 + f(0)^2=0$ and hence that $a=f(0)=0$ and $f(f(y))=y$. Using $P(x,0)$, we see that $f(x^2)=f(x)^2$ and furthermore $$f(x)^2=f(x^2)=f((-x)^2)=f(-x)^2$$ which implies that $f(-x)=-f(x)$ since $f$ is injective. Next, using $P(x,f(y))$, we obtain $$f(x^2+y)=f(x^2+f(f(y)))=f(y)+f(x)^2=f(y)+f(x^2).$$ Using this with the fact that $f(-x)=-f(x)$ yields $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Now note that for all $a \ge 0$, we have $$f(a)=f((\sqrt{a})^2)=f(\sqrt{a})^2 \ge 0.$$ Hence $f$ is bounded on an interval, from which we can say by the Cauchy functional equation that $f(x)=cx$ for some constant $c \in \mathbb{R}$. Substituting this into the equation yields that $f(x)=x$, as desired. $\square$

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