Solve the following system of equations:

$$
\begin{aligned}
& \sqrt[5]{x}+\sqrt[5]{y}=7 \\
& x+y=3157
\end{aligned}
$$
$$
\begin{aligned}
& \sqrt[5]{x}+\sqrt[5]{y}=7 \\
& x+y=3157
\end{aligned}
$$

Let's substitute $x$ and $y$ with $u^{5}$ and $v^{5}$. The equations then become:

$$
\begin{gathered}
u+v=7 \\
u^{5}+v^{5}=3157
\end{gathered}
$$

If we raise the third equation to the fifth power and subtract the fourth equation from it, we get the following equations:

$$
\begin{gathered}
5 u^{4} v+10 u^{3} v^{2}+10 u^{2} v^{3}+5 u v^{4}=13650 \\
u v\left(u^{3}+2 u^{2} v+2 u v^{2}+v^{3}\right)=2730 \\
u v(u+v)\left(u^{2}+u v+v^{2}\right)=2730 \\
u v\left(u^{2}+u v+v^{2}\right)=390 \\
u v\left[(u+v)^{2}-u v\right]=390 \\
u v(49-u v)=390 \\
(u v)^{2}-49(u v)+390=0
\end{gathered}
$$

From this,

$$
\begin{gathered}
u v=\frac{49 \pm \sqrt{49^{2}-4 \cdot 390}}{2} \\
u v=\frac{49 \pm 29}{2} \\
u_{1} v_{1}=39 \\
u_{2} v_{2}=10
\end{gathered}
$$

Thus, $u$ and $v$ are the roots of the following quadratic equations:

$$
\begin{aligned}
& z^{2}-7 z+39=0 \\
& z^{2}-7 z+10=0
\end{aligned}
$$

That is,

$$
\begin{aligned}
& z=\frac{7 \pm \sqrt{49-4 \cdot 39}}{2} \\
& z^{\prime}=\frac{7 \pm \sqrt{49-4 \cdot 10}}{2}
\end{aligned}
$$

Thus,

$$
\begin{gathered}
u_{1}=\frac{1}{2}(7+\sqrt{-107}) \\
v_{1}=\frac{1}{2}(7-\sqrt{-107}) \\
u_{2}=5 \\
v_{2}=2
\end{gathered}
$$

Therefore,

$$
\begin{gathered}
x_{1}=\frac{1}{32}(7+\sqrt{-107})^{5} \\
y_{1}=\frac{1}{32}(7-\sqrt{-107})^{5} \\
x_{2}=3125 \\
y_{2}=32
\end{gathered}
$$

Finally, they can also be written as:

$$
\begin{aligned}
& x_{1}=1578.5-905.5 \sqrt{-107} \\
& y_{1}=1578.5+905.5 \sqrt{-107}
\end{aligned}
$$

Schönner Odilo and Seidner Mihály Losoncz; Sztrapkovits István, S.-A.-Ujhely.