## Task 4 - 100934

We are looking for all ordered triples of real numbers $(x, y, z)$ that are solutions to the system of equations:
(1) $x+y=2$
(2) $x y-z^{2}=1$
From the second equation, one obtains $x y=1+z^{2} \geq 1$ and thus

$$
0 \leq(x-y)^{2}=(x+y)^{2}-4 x y=2^{2}-4 x y \leq 4-4=0
$$

Therefore, equality must have held at every step in this chain of inequalities, so $x-y=0$ and $x y=1$, which means $x=y= \pm 1$, and from equation (2) also $z=0$ follows.

Thus, there can only be two solution triples $(x, y, z)$, namely $(-1,-1,0)$ and $(1,1,0)$. However, the verification confirms only the second, so $(1,1,0)$ is the only solution to the given system of equations.

## Second Solution:

Assume a triple $\left(x_{0}, y_{0}, z_{0}\right)$ is a solution to $\left({ }^{*}\right),\left({ }^{* *}\right)$. Then, by substituting $x_{0}=2-y_{0}$ into $(* *)$, we get $y_{0}^{2}-2 y_{0}+z_{0}^{2}+1=0$, thus

$$
\left(y_{0}-1\right)^{2}+z_{0}^{2}=0
$$

If $y_{0} \neq 1$ or $z_{0} \neq 0$, then $\left(y_{0}-1\right)^{2}>0$ or $z_{0}^2>0$, and since $\left(y_{0}-1\right)^{2} \geq 0$ and $z_{0}^{2} \geq 0$ always hold, in either case $\left(y_{0}-1\right)^{2}+z_{0}^{2}>0$, which contradicts (1). Therefore, from (1) it follows that $y_{0}=1$ and $z_{0}=0$ must hold.

From $\left(^{*}\right)$, it then follows that $x_{0}=1$. Thus, at most the triple $(1,1,0)$ can be a solution to the system of equations $\left(^{*}\right)$, $(* *)$.

Indeed, this is a solution, because for $x_{0}=1, y_{0}=1, z_{0}=0$,

$$
x_{0}+y_{0}=1+1=2 \quad \text { or } \quad x_{0} y_{0}-z_{0}^{2}=1-0=1
$$

## Adopted from $[2]$