49th Mathematical Olympiad in Ukraine

It is evident that for every value of parameter $a$ there exists solution $x=y=z=0$, and so all we need is to find out when this solution is unique.

From the first equation we have that $y=-x-z$. Substitute $(-x-z)$ instead of $y$ into the second equation:

$-x^2-2xz-z^2+axz=0$

or

$x^2+xz(2-a)+z^2=0$.

We have obtained a quadratic equation with respect to $x$. Its discriminant $D=z^2(2-a{)}^2-4z^2=z^2(4a-a^2)$.

If $z=0$ then the system has zero solution, thus for all $z\ne 0$ our quadratic equation shouldn't have any solutions, that is $D=z^2(4a-a^2)<0$.

And the last inequality holds for $a\in (0,4)$.