69th Belarusian Mathematical Olympiad

Answer: $f(x)=0$ for all $x\in \mathbb{R}$. Consider several preliminary substitutions to the problem condition $$f(f(x)+f(y))=(x+y)f(x+y).(1)$$ Substitution $x=y=0$ gives the equality $f(2f(0))=0$. Substitution $x=2f(0)$, $y=0$ gives $f(f(0))=0$. Finally, substitution $x=y=f(0)$ gives $f(0)=0$. Substitution $y=0$ to (1) implies that for all $x\in \mathbb{R}$ holds $f(f(x))=xf(x)$. Suppose that for some $a,b\in \mathbb{R}$ holds $f(a)=f(b)\ne 0$, then $af(a)=f(f(a))=f(f(b))=bf(b)$, whence $a=b$. Therefore $f$ can attend several times only zero. We will show that there exists $c\ne 0$ such that $f(c)=0$. Substitution $y=1-x$ to (1) implies $f(f(x)+f(1-x))=f(1)$. If $f(1)=0$, then $c=1$. Otherwise, $f(x)+f(1-x)=1$ for all $x\in \mathbb{R}$. Substitutions of $x=0$ and $x=1/2$ in the latter equation give $f(1)=1$ and $f(1/2)=1/2$. Finally for $x=1$ and $y=1/2$ the equality (1) gives $f(3/2)=3/2f(3/2)$, whence $f(3/2)=0$. Denote by $c$ an arbitrary nonzero number such that $f(c)=0$. Substitution $x=y=c$ to (1) implies $f(2c)=0$. Suppose that there exist a real number $t$ such that $f(t)\ne 0$. Putting to (1) successively $x=c$, $y=-c+t$ and $x=2c$, $y=-2c+t$, we obtain $$f(f(-c+t))=tf(t)=f(f(-2c+t)),\text{ where }tf(t)\ne 0.$$ Therefore, $f(-c+t)=f(-2c+t)\ne 0$, whence $-c+t=-2c+t$, which contradicts $c\ne 0$. Consequently such $t$ doesn't exist and $f(x)=0$ for all $x\in \mathbb{R}$.