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:

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

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 $z \neq 0$ our quadratic equation shouldn't have any solutions, that is $D = z^2(4a - a^2) < 0$.

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