62nd Belarusian Mathematical Olympiad

Answer: $a=0$.

Indeed, if $a=0$, then the function $f\equiv 0$ satisfies the condition.

Now let $a\ne 0$. Suppose that $f(\alpha )=0$ for some $\alpha$. We have $f(0)=f(f(\alpha ))=\alpha \cdot f(\alpha )+a=a$. Then $f(a)=f(f(0))=0\cdot f(0)+a=a$. Therefore, $a=f(a)=f(f(a))=a\cdot f(a)+a=a\cdot a+a=a^2+a$, i.e. $a=a^2+a$, and then $a=0$, a contradiction.