Solve the following system of equations:

$$
\begin{gathered}
x+y-z=2 a \\
x^{2}+y^{2}=z^{2} \\
m(x+y)=x y
\end{gathered}
$$

Under what conditions will the solutions of the system of equations be positive?
$$
\begin{gathered}
x+y-z=2 a \\
x^{2}+y^{2}=z^{2} \\
m(x+y)=x y
\end{gathered}
$$

Under what conditions will the roots of the system of equations be positive?

From 1) and 3), the following new equations can be derived:

$$
\begin{aligned}
& x+y=2 a+z \\
& x y=m(2 a+z)
\end{aligned}
$$

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

$$
u^{2}-(2 a+z) u+m(2 a+z)=0
$$

which gives

$$
u=\frac{1}{2}\left(2 a+z \pm \sqrt{(2 a+z)^{2}-4 m(2 a+z)}\right)
$$

However, equation 2) can also be written as:

$$
\begin{gathered}
(x+y)^{2}-2 x y-z^{2}=0 \\
(2 a+z)^{2}-2 m(2 a+z)-z^{2}=0 \\
(4 a-2 m) z+4 a^{2}-4 m a=0 \\
z=\frac{2 a(m-a)}{2 a-m}
\end{gathered}
$$

Substituting this value of $z$ into the expression for $u$, we get:

$$
\begin{gathered}
u=\frac{1}{2}\left(\frac{2 a^{2}}{2 a-m} \pm \sqrt{\frac{2 a^{2}}{2 a-m}\left(\frac{2 a^{2}}{2 a-m}-4 m\right)}\right) \\
u=\frac{a^{2} \pm 2 a^{2} \sqrt{a^{2}-4 a m+2 m^{2}}}{2 a-m}
\end{gathered}
$$

For $u$ to be real, it is necessary that:

$$
2 m^{2}-4 a m+a^{2} \geq 0
$$

or

$$
\begin{gathered}
2\left(\frac{m}{a}\right)^{2}-4\left(\frac{m}{a}\right)+1 \geq 0 \\
2\left(\frac{m}{a}-\frac{1}{2}(2+\sqrt{2})\right)\left(\frac{m}{a}-\frac{1}{2}(2-\sqrt{2})\right) \geq 0
\end{gathered}
$$

which holds if

$$
\frac{m}{a} \leq \frac{1}{2}(2-\sqrt{2}) \text { or } \frac{m}{a} \geq \frac{1}{2}(2+\sqrt{2})
$$

For both values of $u$ to be positive, it is necessary that their sum and product are simultaneously positive, i.e., from 6):

$$
\begin{aligned}
& 2 a+z>0 \text { and } m(2 a+z)>0 \\
& \frac{2 a^{2}}{2 a-m}>0 \text { and } \frac{2 a^{2} m}{2 a-m}>0
\end{aligned}
$$

which conditions are satisfied if

$$
2 a-m>0, \quad m>0, \quad a>0
$$

i.e.,

$$
0 < m < 2a, \quad a > 0
$$