Solve the following system of equations

$$
\begin{gathered}
\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)=455 \\
(x+y)=5
\end{gathered}
$$
$$
\begin{gathered}
\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)=455 \\
x+y=5
\end{gathered}
$$

Transform the first equation so that the left side is a function of $x+y$ and $xy$. It becomes:

$$
\left\{(x+y)^{2}-2 x y\right\}(x+y)\left\{(x+y)^{2}-3 x y\right\}=455
$$

After substituting the value of $x+y$:

$$
\begin{aligned}
& (25-2 x y)(25-3 x y)=91 \\
& \left.6 x^{2} y^{2}-125 x y\right)+534=0
\end{aligned}
$$

The roots of equation (3) are:

$$
\begin{gathered}
x y=\frac{125 \pm \sqrt{125^{2}-24 \cdot 534}}{12} \\
=\frac{125 \pm \sqrt{2809}}{12} \\
=\frac{125 \pm 53}{12}
\end{gathered}
$$

Thus,

$$
\begin{aligned}
x_{1} y_{1} & =\frac{89}{6} \\
x_{2} y_{2} & =6
\end{aligned}
$$

Therefore, $x_{1}$ and $y_{1}$ are the roots of

$$
6 z^{2}-30 z+89=0
$$

and $x_{2}$ and $y_{2}$ are the roots of

$$
z^{2}-5 z+6=0
$$

Thus,

$$
\begin{gathered}
x_{1}=\frac{30 \pm \sqrt{900-24 \cdot 89}}{12} \\
x_{1}=\frac{30 \pm \sqrt{-1236}}{12} \\
x_{1}=\frac{15 \pm i \sqrt{309}}{6} \\
y_{1}=\frac{15 \mp i \sqrt{309}}{6} \\
x_{2}=\frac{5 \pm \sqrt{25-24}}{2} \\
x_{2}=\frac{5 \pm 1}{2} \\
x_{2}=3 \text { or } 2 \\
y_{2}=2 \text { or } 3 .
\end{gathered}
$$

(Mihály Seidner, 8th grade student, Losoncz)

The problem was also solved by: Ignácz Bergstein, 8th grade student, Nyíregyháza; József Greiner, 8th grade student, Pécs; Tivadar Heymann, 7th grade student, Győr; József Kaczvinszky, 8th grade student, Budapest, 4th district; Sándor Kugel, 8th grade student, Losoncz; Sándor Pollák, 7th grade student, Győr; István Sztrapkovits, 8th grade student, S.-A.-Ujhely; Gergely Jorga, 8th grade student, Arad.