Find all positive integers $x$ and $y$ for which

$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{4}
$$

(A. Bohiniková)

Hint. Can both unknowns be greater than, for example, 14 at the same time?
For the same unknown, the solution is apparently $x=y=8$.

Different unknowns cannot both be simultaneously smaller, or larger than 8 (otherwise, the left side of the equation would be larger, or smaller than $\frac{1}{4}$). One unknown must therefore be smaller than 8 and the other larger than 8. Given the symmetry of the equation, it suffices to consider the case where $x<8$ and $y>8$, so $x$ can only take values from 1 to 7.

If we express $y$ in terms of $x$ from the equation, we get

$$
y=\frac{4 x}{x-4}
$$

From this, it is clear that $y$ is positive if and only if $x>4$. Thus, $x$ can only take values from 5 to 7. For these three possibilities, it suffices to check if $y$ is an integer:

- for $x=5$, $y=20$,
- for $x=6$, $y=12$,
- for $x=7$, $y=\frac{28}{3}$.

In total, we see that all pairs $(x, y)$ that are solutions to the problem are $(5,20)$, $(6,12)$, $(8,8)$, $(12,6)$, and $(20,5)$.

Another solution. If we express $y$ in terms of $x$ from the equation, we get (1). This expression can be further modified to "whole part plus remainder":

$$
y=\frac{4 x}{x-4}=\frac{4(x-4)+16}{x-4}=4+\frac{16}{x-4}.
$$

From this, it is clear that $y$ is a positive integer if and only if $x-4$ is a positive integer divisor of the number 16, and there are exactly five such divisors:

- for $x-4=1$, $x=5$ and $y=20$,
- for $x-4=2$, $x=6$ and $y=12$,
- for $x-4=4$, $x=8$ and $y=8$,
- for $x-4=8$, $x=12$ and $y=6$,
- for $x-4=16$, $x=20$ and $y=5$.

This list includes all solutions to the problem.

Note. When modifying the given equation to the form (1), one can encounter an equivalent equation

$$
x y-4 x-4 y=0 \text{. }
$$

The left side matches the three terms in the expansion of the expression $(x-4)(y-4)$, except for 16. Adding 16 to both sides of equation (3) gives

$$
(x-4)(y-4)=16 \text{. }
$$

All solutions can thus be found using all possible factorizations of the number 16 into the product of two positive integers. These ideas are just a non-fractional interpretation of the modification (2) and the subsequent procedure.