Consider the real numbers $x$ for which $\operatorname{tg} x + \operatorname{ctg} x$ is a positive integer. Determine among them those for which $\operatorname{tg}^{3} x + \operatorname{ctg}^{3} x$ is a prime number.
Solution. Let $a=\operatorname{tg} x+\operatorname{ctg} x$. According to the conditions of the problem, the sum is positive, and $a$ is a number and the sum of its reciprocal, so it is known that $a \geq 2$, where equality holds only if $\operatorname{tg} x=\operatorname{ctg} x=1$. We also know that

$$
\begin{aligned}
\operatorname{tg}^{3} x+\operatorname{ctg}^{3} x & =(\operatorname{tg} x+\operatorname{ctg} x)\left(\operatorname{tg}^{2} x-\operatorname{tg} x \operatorname{ctg} x+\operatorname{ctg}^{2} x\right)= \\
& =(\operatorname{tg} x+\operatorname{ctg} x)\left[(\operatorname{tg} x+\operatorname{ctg} x)^{2}-3 \operatorname{tg} x \operatorname{ctg} x\right]=a\left(a^{2}-3\right)
\end{aligned}
$$

If $a \geq 2$ is a positive integer, then $a^{2}-3$ is also a positive integer, so the product can only be a prime number if $a$ is a prime number and $a^{2}-3=1$.

$$
a^{2}-3=(\operatorname{tg} x+\operatorname{ctg} x)^{2}-3=1, \quad \operatorname{tg} x+\operatorname{ctg} x=2
$$

This only holds if $\operatorname{tg} x=\operatorname{ctg} x=1$. Therefore, the solutions to the equation are: $x=\frac{\pi}{4}+k \cdot \pi$, where $k \in \mathbb{Z}$.