4A. In the set of integers, solve the equation

$$
\operatorname{arctg} \frac{1}{x}+\operatorname{arctg} \frac{1}{y}=\operatorname{arctg} \frac{1}{10}
$$
Solution. If we use that $\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}$ and $\operatorname{tg}(\operatorname{arctg} \alpha)=\alpha$, for every $\alpha \in(-\infty,+\infty), \operatorname{tg} \alpha \operatorname{tg} \beta \neq 1$, we get that the given equation is equivalent in turn to the equations

$$
\operatorname{tg}\left(\operatorname{arctg} \frac{1}{1}+\operatorname{arctg} \frac{1}{y}\right)=\operatorname{tg}\left(\operatorname{arctg} \frac{1}{10}\right)
$$

$$
\begin{aligned}
& \frac{\frac{1}{x}+\frac{1}{y}}{1-\frac{1}{x} \cdot \frac{1}{y}}=\frac{1}{10} \\
& (x-10)(y-10)=101
\end{aligned}
$$

Since 101 is a prime number, from the last equation we get the systems

$$
\begin{array}{ll}
\left\{\begin{array}{l}
x-10=1 \\
y-10=101
\end{array}\right. & \left\{\begin{array}{l}
x-10=-1 \\
y-10=-101
\end{array}\right. \\
\left\{\begin{array}{l}
x-10=101 \\
y-10=1
\end{array}\right. & \left\{\begin{array}{l}
x-10=-101 \\
y-10=-1
\end{array}\right.
\end{array}
$$

whose solutions are the ordered pairs $(11,111),(9,-91),(111,11),(-91,9)$, respectively. Finally, the solution is any ordered pair $(x, y) \in\{(11,111),(9,-91),(111,11),(-91,9)\}$.