Baltic Way shortlist

Answer: the functions $f(x)=0$ and $f(x)=x$. A first observation is that $$f(1)=f(1^2)=f(1{)}^2,$$ so that $f(1)$ is either 0 or 1. Assume first that $f(1)=0$. For each positive integer $n$, we find $$f(n)=f(n\cdot 1^2)=nf(1{)}^2=0.$$ Given an arbitrary $x$, find $y$ so that $x^2+y^2$ becomes a positive integer $n$. Then $$f(x{)}^2+f(y{)}^2=f(x^2+y^2)=f(n)=0.$$ Consequently, $f(x)=0$ for all $x$. Now assume $f(0)=1$. We shall prove that $f(x)=x$ for all $x$. For each positive integer $n$, we find $$f(n)=f(n\cdot 1^2)=nf(1{)}^2=n.$$ For a non-negative rational number $\frac{p}{q}$, we find $$p^2=f(p^2)=f\left(q^2\cdot {\left(\frac{p}{q}\right)}^2\right)=q^{2}f{\left(\frac{p}{q}\right)}^2,$$ hence $f(x)=x$ also for rational numbers. Finally, let $x$ be an irrational number. Select a rational number $\frac{p}{q}>x$. Choosing $y$ so that $x^2+y^2=\frac{p^2}{q^2}$, we deduce $$\frac{p^2}{q^2}=f\left(\frac{p^2}{q^2}\right)=f(x^2+y^2)=f(x{)}^2+f(y{)}^2\ge f(x{)}^2,$$ hence $f(x)\le \frac{p}{q}$. Next, select a (positive) rational number $\frac{r}{s}<\sqrt{x}$, i.e. $\frac{r^2}{s^2}<x$. Choosing $z$ so that $\frac{r^2}{s^2}+z^2=x$, we deduce $$f(x)=f\left(\frac{r^2}{s^2}+z^2\right)=f{\left(\frac{r}{s}\right)}^2+f(z{)}^2=\frac{r^2}{s^2}+f(z{)}^2\ge \frac{r^2}{s^2},$$ hence $f(x)\ge \frac{r^2}{s^2}$. Together, these two bounds for $f(x)$ imply $f(x)=x$, and we are finished.