Let $p$ and $q$ be different prime numbers. In how many ways can the fraction $1 / p q$ be decomposed into the sum of the reciprocals of 2 different natural numbers?
The two natural numbers - let's denote them by $a$ and $b$ - each have a reciprocal that is smaller than $1 / pq$, so each of them is greater than $pq$. Let then $a = pq + c$, $b = pq + d$, where $c, d > 0$, are integers, and for now, let's assume that $c < pq$, i.e., $c$ is a divisor of $p^2 q^2$ that is smaller than $pq$, and $d$ is its co-divisor. The divisors of $p^2 q^2$ can thus be arranged as follows:

$$
\begin{array}{lll}
1, & p, & p^2 \\
q, & pq, & p^2 q \\
q^2, & pq^2, & p^2 q^2
\end{array}
$$

Their number is 9, excluding $pq$, 4 pairs are given for $c, d$, and thus for $a, b$.

Therefore, $1 / pq$ can be decomposed into the sum of the reciprocals of two different natural numbers in 8 ways, if we also consider decompositions different if they only differ in the order of the terms. If, however, we do not consider different those decompositions that can be obtained from each other by swapping the order of the terms, then the number of possible decompositions is 4.