66th Belarusian Mathematical Olympiad

Answer: $a=0$ or $a=-1$.

By condition, there is $c$ such that $f(c)=a$. Substituting $c$ for $x$ in the equality $$f(f(x))=xf(x)-ax,(1)$$ we obtain $f(a)=f(f(c))=cf(c)-ac=ca-ac=0$.

Substituting $a$ for $x$ in (1), we obtain $$f(0)=f(f(a))=af(a)-a^2=-a^2(2)$$

Substituting $0$ for $x$ in (1), we obtain $f(-a^2)=f(f(0))=0\cdot f(0)-a\cdot 0=0$.

Substituting $-a^2$ for $x$ in (1), we obtain $$f(0)=f(f(-a^2))=-a^{2}f(-a^2)-a\cdot (-a^2)=a^3.(3)$$

From (2) and (3) it follows that $-a^2=a^3$, hence either $a=0$ or $a=-1$.

If $a=-1$, then it is easy to see that the function $$f(x)=\{\begin{array}{ll}-1& \text{for }x\ne -1,\\ 0& \text{for }x=-1\end{array}$$ satisfies the problem condition.

If $a=0$, then it is easy to see that the function $$f(x)=\{\begin{array}{ll}0& \text{for }x\ne 1,\\ 1& \text{for }x=1\end{array}$$ satisfies the problem condition.