Belorusija 2012

Answer: such function does not exist.

Suppose, contrary to our claim, that there exists a function satisfying the equality $f(f(x))=xf(x)+2x$ for all real $x$, and $f(\alpha )=-2$ for some $\alpha$. We have $f(-2)=f(f(\alpha ))=\alpha f(\alpha )+2\alpha =-2\alpha +2\alpha =0$. Then $f(0)=f(f(-2))=-2f(-2)-4=-4$. Further, $f(-4)=f(f(0))=0\cdot f(0)+2\cdot 0=0$. Now we have $f(0)=f(f(-4))=-4f(-4)-8=-4\cdot 0-8=-8$. But $f(0)=-4$, a contradiction.