## Task $3 / 73$

It is to be investigated for which real number $a$ the system of equations

$$
x^{2}+y^{2}=z \quad ; \quad x+y+z=a
$$

has exactly one real solution $(x ; y ; z)$. This solution is to be found.
In the given system, the variables $x$ and $y$ appear symmetrically. Therefore, for the required uniqueness of the solution, it is necessary that $x=y$ (since otherwise, if $x=x_{0}, y=y_{0}, z=z_{0}$ is a solution, then $x=y_{0}$, $y=x_{0}, z=z_{0}$ would also be a solution). Thus, the system takes the form

$$
2 x^{2}=z \quad ; \quad 2 x+z=a
$$

By eliminating $z$, we get $2 x^{2}+2 x-a=0$. For this equation to have only one real solution, it is necessary that the discriminant $D=\frac{2 a+1}{4}=0$. This immediately implies $a=-\frac{1}{2} ; x=-\frac{1}{2}$, and further $y=-\frac{1}{2}, z=\frac{1}{2}$.