Determine all integer numbers $x$ and $y$ such that

$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{19}
$$
The original equation assumes $x \neq 0$ and $y \neq 0$, so we can consider the equivalent equation

$$
x y=19(x+y), \quad \text { with } \quad x \neq 0, y \neq 0
$$

Since we are looking for integer solutions and 19 is a prime number, this equality implies that $x$ or $y$ must be divisible by 19. As the equation is symmetric with respect to the variables $x$ and $y$, we can assume that $x$ is divisible by 19, that is, $x=19 k$, for some integer $k \neq 0$. Thus, the original equation is also equivalent to

$$
k y=19 k+y, \quad \text { with } \quad k \neq 0, y \neq 0
$$

From this equality, we obtain that $19 k+y$ is divisible by $k$. Since $19 k$ is already divisible by $k$, we conclude that $y$ is divisible by $k$ (prove this), that is, $y=k m$, for some integer $m \neq 0$. It follows that

$$
k k m=k y=19 k+y=19 k+k m=(19+m) k
$$

that is, $k m=19+m$, which is equal to $19=(k-1) m$. Thus, we establish that the integers $m$ and $k-1$ are divisors of the prime number 19. Since $k \neq 0$, it follows that $k-1 \neq-1$, leaving us with only three options, as follows.

- If $m=19$ and $k-1=1$, then $x=38$ and $y=38$;
- If $m=1$ and $k-1=19$, then $x=380$ and $y=20$;
- If $m=-1$ and $k-1=-19$, then $x=-342$ and $y=18$.

Thus, by symmetry, we obtain the only five pairs of integers $(x, y)$ that are solutions to the given equation:

$$
(38,38),(380,20),(-342,18),(20,380) \text { and }(18,-342)
$$