Find all real numbers \(x, y, z\) so that

\[
\begin{aligned}
x^{2} y + y^{2} z + z^{2} & = 0 \\
z^{3} + z^{2} y + z y^{3} + x^{2} y & = \frac{1}{4} \left( x^{4} + y^{4} \right)
\end{aligned}
\]
Answer: $x=y=z=0$.

$y=0 \Longrightarrow z^{2}=0 \Longrightarrow z=0 \Longrightarrow \frac{1}{4} x^{4}=0 \Longrightarrow x=0$. $x=y=z=0$ is a solution, so assume that $y \neq 0$. Then $z=0 \Longrightarrow x^{2} y=0 \Longrightarrow x=0 \Longrightarrow \frac{1}{4} y^{4}=0$, which is a contradiction. Hence $z \neq 0$. Now we solve the quadratic (first) equation w.r.t. $x, y$, and $z$.

$$
\begin{aligned}
& x= \pm \frac{\sqrt{-4 y^{3} z-4 y z^{2}}}{2 y} \\
& y=\frac{-x^{2} \pm \sqrt{x^{4}-4 z^{3}}}{2 z} \\
& z=\frac{-y^{2} \pm \sqrt{y^{4}-4 x^{2} y}}{2}
\end{aligned}
$$

The discriminants must be non-negative.

$$
\begin{gathered}
-4 y^{3} z-4 y z^{2} \geqslant 0 \\
x^{4}-4 z^{3} \geqslant 0 \\
y^{4}-4 x^{2} y \geqslant 0
\end{gathered}
$$

Adding the inequalities we get

$$
\begin{gathered}
y^{4}-4 x^{2} y+x^{4}-4 z^{3}-4 y^{3} z-4 y z^{2} \geqslant 0 \\
\frac{1}{4}\left(x^{4}+y^{4}\right) \geqslant z^{3}+z^{2} y+z y^{3}+x^{2} y
\end{gathered}
$$

But equation 2 says that $\frac{1}{4}\left(x^{4}+y^{4}\right)=z^{3}+z^{2} y+z y^{3}+x^{2} y$. This is only possible if all of the inequalities are in fact equalities, so we have

$$
\begin{gathered}
-4 y^{3} z-4 y z^{2}=0 \\
x^{4}-4 z^{3}=0 \\
y^{4}-4 x^{2} y=0
\end{gathered}
$$

This means that $x= \pm \frac{\sqrt{0}}{2 y}=0 \Longrightarrow y=\frac{-0^{2} \pm \sqrt{0}}{2 z}=0$, which is a contradiction. Hence the only solution is $x=y=z=0$.