Belarus — Selection and Training Session

Answer : $a \in \{0\} \cup [1, \infty)$.
Let function $f: \mathbb{R}_{\ge 0} \to \mathbb{R}_{\ge 0}$ satisfy the equality
$$f(f(x) + f(y)) = y f(1 + y f(x)) \tag{1}$$
for all nonnegative real numbers $x$ and $y$. Set $f(0) = c$ and $f(1) = b$.
For $y = 0$ from equality (1) we get
$$f(f(x) + c) = 0. \tag{2}$$
for all $x \in [0, \infty)$.
For $x = f(x) + c$ from equality (1) we get
$$f(f(y)) = b y. \tag{3}$$
for all $y \in [0, \infty)$.
Suppose $b \neq 0$. From equality (3) it follows that function $f$ is injective. Hence from equality (2) it follows that $f(x) + c = f(0)$, a contradiction.

Therefore $b = 0$ and equality (3) is equivalent to
$$f(f(y)) = 0 \tag{4}$$

For $y = 0$ from equality (4) we get $f(c) = 0$ hence $f(0) = f(f(c)) = 0$. So $c = f(0) = 0$.
For $y = 1$ from equality (1) we get
$$f(1 + f(x)) = 0 \qquad (5)$$
for all $x \in [0, \infty)$.
Suppose that $f(a) = d \neq 0$ for some $a > 1$. Substituting $x = a$, $y = \frac{a-1}{d}$ to equality (1) we get
$$f(f(\frac{a-1}{d}) + d) = a - 1.$$
which leads to a contradiction with equality (5) for $x = f(\frac{a-1}{d}) + d$.
Hence for any $a \in \{0\} \cup [1, \infty)$ the equality $f(a) = 0$ is proved. The next example shows that there exists a function satisfying equality (1) such that $f(x) \neq 0$ for any $x \in (0, 1)$:
$$f(x) = \begin{cases} 0, & x = 0; \\ \frac{1}{x}, & x \in (0, 1); \\ 0, & x \geq a. \end{cases}$$