4. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$,

$$
f\left(x^{2}+f(y)\right)=(x-y)^{2} \cdot f(x+y) \text {. }
$$

The second round of category A takes place

## on Tuesday, January 16, 2001

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. These details will be communicated to the students before the competition begins.
4. Let $f$ be any of the sought functions. Notice that the left side of the given equation is an even function of the variable $x$. Therefore, when the number $x$ is replaced by its opposite number $-x$, the value of the right side of the equation does not change either:

$$
(x-y)^{2} \cdot f(x+y)=(-x-y)^{2} \cdot f(-x+y)
$$

or equivalently,

$$
(x-y)^{2} \cdot f(x+y)=(x+y)^{2} \cdot f(y-x) .
$$

Substitute here, for any $t \in \mathbb{R}$, the values $x=\frac{1}{2}(t-1)$ and $y=\frac{1}{2}(t+1)$, which are chosen so that $x+y=t$ and $y-x=1$. We get the relation $f(t)=t^{2} \cdot f(1)$ for every $t \in \mathbb{R}$. The function $f$ is thus necessarily of the form $f(x)=c x^{2}$, and the unknown coefficient $c$ can be determined by substituting into the equation from the problem statement (and thus also performing the verification): the equation $c\left(x^{2}+c y^{2}\right)^{2}=(x-y)^{2} \cdot c(x+y)^{2}$ is equivalent to the equation $c(c+1) y^{2}\left(2 x^{2}+(c-1) y^{2}\right)=0$, which is satisfied for any $x$ and $y$ if and only if $c=0$ or $c=-1$ (for example, substituting $x=y=1$ leads to the condition $c(c+1)^{2}=0$, so necessarily $\left.c \in\{0,-1\}\right)$.

The problem has exactly two solutions: the zero function $f_{1}(x)=0$ and the function $f_{2}(x)=-x^{2}$.