Solve the following system of equations.

$$
\begin{aligned}
x+y+z & =a \\
x^{2}+y^{2}+z^{2} & =b^{2} \\
x y & =z^{2}
\end{aligned}
$$

where $a$ and $b$ are given numbers. What condition must the numbers $a$ and $b$ satisfy so that the numbers $x, y, z$ which are solutions to the system of equations are all positive and distinct?
From the square of the first equation to the second equation, the squares of the unknowns will cancel out:

$$
2[x y+(x+y) z]=a^{2}-b^{2}
$$

$x y$ and $x+y$ can be expressed in terms of $z$ based on (1) and (3), and thus we obtain a single-variable equation for $z$:

$$
2\left[z^{2}+(a-z) z\right]=a^{2}-b^{2}
$$

from which

$$
z=\frac{a^{2}-b^{2}}{2 a}
$$

provided, of course, that $a \neq 0$.

Now we know the product and sum of $x$ and $y$:

$$
x y=\left(\frac{a^{2}-b^{2}}{2 a}\right)^{2}, \quad x+y=a-z=\frac{a^{2}+b^{2}}{2 a}
$$

thus $x$ and $y$ are the roots of the following quadratic equation:

$$
u^{2}-\frac{a^{2}+b^{2}}{2 a} u+\left(\frac{a^{2}-b^{2}}{2 a}\right)^{2}=0
$$

The discriminant can be transformed as follows:

$$
D=\frac{1}{4 a^{2}}\left[\left(a^{2}+b^{2}\right)-\left(2 a^{2}-2 b^{2}\right)^{2}\right]=\frac{1}{4 a^{2}}\left(3 a^{2}-b^{2}\right)\left(3 b^{2}-a^{2}\right)
$$

thus

$$
u_{1,2}=\frac{a^{2}+b^{2} \pm \sqrt{\left(3 a^{2}-b^{2}\right)\left(3 b^{2}-a^{2}\right)}}{4 a}
$$

Accordingly, the system of equations has two solutions, in which the value of $z$ is common, and $x$ and $y$ are equal when swapped:

$$
\begin{array}{lll}
x_{1}=u_{1}, & y_{1}=u_{2}, & z_{1}=\frac{a^{2}-b^{2}}{2 a} \\
x_{2}=u_{2}, & y_{2}=u_{1}, & z_{2}=z_{1}
\end{array}
$$

For the positivity of the numbers $x, y, z$, according to (1), it is necessary that:

$$
a>0
$$

Thus $z$ is indeed positive if

$$
a^{2}-b^{2}>0, \quad \text { i.e., } \quad a>|b|
$$

With these, both expressions under (5) are positive, so $x, y$ are positive and distinct if and only if $D$ is positive. If $x, y$ are distinct, then according to (3), $z$ is not equal to either of them. According to (6), the first two-term expression of $D$ is positive, so the second one must also be positive:

$$
3 b^{2}>a^{2}
$$

The sought conditions can be summarized as follows:

$$
a>0, \quad b^{2}<a^{2}<3 b^{2}
$$

Géza Nagy (Debrecen, Reformed Church IV. Secondary School)