66th Belarusian Mathematical Olympiad

Answer: $f(x)=a(x+2a)$, $g(x)=a(1-2a)(x+2a)$, $a\in \mathbb{R}$.

(Solution by A. Goloubitskaya, L. Manzhulina.) First, we claim that $f$ takes the value $0$. If $f(0)=0$, then there is nothing to prove. If $f(0)=b\ne 0$, then setting $x=0$ in the given equality $$f(x-2f(y))=xf(y)-yf(x)+g(x),(1)$$ we obtain $f(-2f(y))=-by+g(0)$. The right-hand side attains all real values since $b\ne 0$. So $f$ attains all real values including $0$. Let $c$ be such that $f(c)=0$. Set $y=c$ in (1): $f(x)=-cf(x)+g(x)$, whence $g(x)=(c+1)f(x)$. Therefore, (1) becomes $$f(x-2f(y))=xf(y)+(c+1-y)f(x).(2)$$ Let $a=f(c+1)$. Substituting $c+1$ for $y$ in (2), we obtain $f(x-2a)=ax$, whence $f(x)=ax+2a^2$. Since $f(c)=0$, we have $ac+2a^2=0$ and $$g(x)=(c+1)(ax+2a^2)=(-2a^2+a)x-4a^3+2a^2.$$ Easy verification shows that these functions $f$ and $g$ satisfy the given equality for all real $a$.