A3 (3-1, Hungary) Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+z=a, \\
x^{2}+y^{2}+z^{2}=b^{2}, \\
x y=z^{2},
\end{array}\right.
$$

where $a, b$ are given real numbers.
And indicate, what conditions should $a, b$ satisfy so that the solutions $x, y, z$ are distinct positive numbers?
From (2), (3) it is easy to know
$$
(x+y+z)(x+y-z)=b^{2},
$$

When $a=0, b \neq 0$, from (1), (4) we know the original system of equations has no solution; when $a=b=0$, it is easy to know the system of equations has a unique solution: $x=y=z=0$; if $a \neq 0$, from (1), (4) we get
$$
x+y-z=\frac{b^{2}}{a} \text {. }
$$

Again, from (1), (3), (5), we get
$$
\begin{array}{c}
x+y=\frac{a^{2}+b^{2}}{2 a}, \\
z=\frac{a^{2}-b^{2}}{2 a}, \\
x y=\frac{\left(a^{2}-b^{2}\right)^{2}}{4 a^{2}} . \\
\text { When } \Delta=\frac{1}{4 a^{2}}\left[\left(a^{2}+b^{2}\right)^{2}-4\left(a^{2}-b^{2}\right)^{2}\right] \\
=\frac{1}{4 a^{2}}\left(3 a^{2}-b^{2}\right)\left(3 b^{2}-a^{2}\right) \geqslant 0 \text {, }
\end{array}
$$

the system of equations has real solutions
$$
\left\{\begin{array}{l}
x=\frac{1}{4 a}\left[a^{2}+b^{2} \pm \sqrt{\left(3 a^{2}-b^{2}\right)\left(3 b^{2}-a^{2}\right)}\right], \\
y=\frac{1}{4 a}\left[a^{2}+b^{2} \mp \sqrt{\left(3 a^{2}-b^{2}\right)\left(3 b^{2}-a^{2}\right)}\right], \\
z=\frac{1}{2 a}\left(a^{2}-b^{2}\right) .
\end{array}\right.
$$

To make all solutions positive, from (1) we must have $a>0, z>0$, which implies $a^{2}>b^{2}$. To make the solutions distinct real numbers, it is easy to know we must have $\Delta>0$, and from this and $3 a^{2}>b^{2}$ we get $3 b^{2}-a^{2}>0,|b|>0$.
All these inequalities can be simplified to
$$
0<\frac{a}{\sqrt{3}}<|b|<\sqrt{3}a
$$

Since $x, y$ are the roots of the quadratic equation $t^{2}-(x+y)t+xy=0$, we have $x, y>0$ and $x \neq y$ if and only if $\Delta>0$. From $x+y=\frac{a^{2}+b^{2}}{2a}>\frac{a^{2}-b^{2}}{2a}=z$ and $x y=\frac{\left(a^{2}-b^{2}\right)^{2}}{4 a^{2}}>\frac{\Delta}{2}$ (since $\left(a^{2}-b^{2}\right)^{2}>0$), we have $x>0, y>0$ and using $z=\sqrt{x y}$, we know that as long as $x \neq y, x, y, z$ are distinct positive numbers.

Note: From equation (1) we know the distance from the plane $x+y+z=a>0$ to the origin is $\frac{a}{\sqrt{3}}>0$; from equation (2) we know the sphere with center at the origin and radius $|b|$ is $x^{2}+y^{2}+z^{2}=b^{2}$. The plane and the sphere have common points if and only if $|b| \geqslant \frac{a}{\sqrt{3}}$, but when $|b|=\sqrt{3} a$, the plane and the sphere are tangent at one point, whose three coordinates are equal. Hence $0<\frac{a}{\sqrt{3}}<|b|<\sqrt{3}a$.

Since $x, y$ are distinct positive numbers, we have $x+y>2\sqrt{x y} \quad(x, y \text { are distinct positive numbers })$,

we get $\frac{1}{2}\left(a-\frac{a^{2}-b^{2}}{2 a}\right)>\frac{a^{2}-b^{2}}{2 a}$, which simplifies to $a^{2}>3b^{2}$. Therefore, we still have $0<a<\sqrt{3}|b|$.