## Task A-1.1.

Determine all triples $(x, y, z)$ of real numbers for which

$$
\begin{aligned}
& x^{2}-y=z^{2} \\
& y^{2}-z=x^{2} \\
& z^{2}-x=y^{2}
\end{aligned}
$$
## Solution.

If we sum all three equations, we conclude that $x+y+z=0$.

By substituting $z=-x-y$ into the first equation, we get $x^{2}-y=x^{2}+2 x y+y^{2}$, which simplifies to

$$
y(y+2 x+1)=0
$$

from which we get two cases: $y=0$ or $y+2 x+1=0$.

If $y=0$, then $z=-x$ and from the last equation, we see that $x(x-1)=0$, which means $x=0$ or $x=1$. This gives us the solutions $(0,0,0)$ and $(1,0,-1)$.

If $y=-2 x-1$, then $z=x+1$. Substituting into the second equation, we get $4 x^{2}+4 x+1-x-1=x^{2}$, which simplifies to $x(x+1)=0$. This gives us the solutions $(0,-1,1)$ and $(-1,1,0)$.

Finally, all solutions to the system are: $(0,0,0),(1,0,-1),(0,-1,1)$, and $(-1,1,0)$.