Solve the following equation:

$$
\sqrt[5]{37-x}+\sqrt[5]{x-4}=3
$$
Let

then

putting into (1):

$$
33-y^{5}=(3-y)^{5}
$$

or rearranging the equation

$$
y^{4}-6 y^{3}+18 y^{2}-27 y+14=0
$$

or

$$
(y-1)(y-2)\left(y^{2}-4 y+7\right)=0
$$

from which

$$
y_{1}=1, \quad y_{2}=2, \quad y_{3}=\frac{3+i \sqrt{19}}{2}, \quad y_{4}=\frac{3-i \sqrt{19}}{2}
$$

and thus from (2)

$$
\begin{aligned}
x_{1}=5, \quad x_{2} & =36, \quad x_{3}=\left(\frac{3+i \sqrt{19}}{2}\right)^{5}+4=\frac{41-59 i \sqrt{19}}{2} \\
x_{4} & =\left(\frac{3-i \sqrt{19}}{2}\right)^{5}+4=\frac{41+59 i \sqrt{19}}{2}
\end{aligned}
$$

(Béla Koffler, Budapest.)