Let $p$ and $q$ be prime numbers. How many positive integer pairs $(x, y)$ are there such that

$$
\frac{p}{x}+\frac{q}{y}=1 ?
$$
Multiplying the equation by $xy$ and rearranging,

\[
\begin{gathered}
p y + q x = x y \\
p q = p q + x y - p y - q x = (x - p)(y - q)
\end{gathered}
\]

Since $p$ and $q$ are prime numbers, the divisors of $pq$ can only be $\pm 1, \pm p, \pm q, \pm pq$. The possible cases can be summarized in the following table:

|  | $x - p$ | $y - q$ | $x$ | $y$ |
| :---: | :---: | :---: | :---: | :---: |
| 1. | $p$ | $q$ | $2 p$ | $2 q$ |
| 2. | $q$ | $p$ | $p + q$ | $p + q$ |
| 3. | $pq$ | 1 | $pq + p$ | $q + 1$ |
| 4. | 1 | $pq$ | $p + 1$ | $pq + q$ |
| 5. | $-p$ | $-q$ | 0 | 0 |
| 6. | $-q$ | $-p$ | $p - q$ | $q - p$ |
| 7. | $-pq$ | -1 | $p - pq$ | $q - 1$ |
| 8. | -1 | $-pq$ | $p - 1$ | $q - pq$ |

In the 5th pair, $x = y = 0$, which is excluded from the domain of the original equation. In the 6th pair, at least one of $x$ or $y$ is not positive, but we are looking for positive solutions. In the 7th pair, $x = p - pq = p(1 - q) < 0$, since $q$ is a prime number and thus at least 2, while $p$ is positive; similarly, the 8th case can also be excluded, as $q - pq < 0$.

We still need to check how many different pairs the first 4 cases, where positivity is clearly satisfied, provide. Let $x_i$ and $y_i$ denote the solution corresponding to the $i$-th case. If $(x_i, y_i) = (x_2, y_2)$, then $2p = p + q = 2q$, which implies $p = q$. If $(x_i, y_i) = (x_3, y_3)$, then $q + p = 2p$, which implies $p(q - 1) = 0$, contradicting the fact that $p$ and $q$ are primes. If $(x_i, y_i) = (x_4, y_4)$, then $2p = p + 1$, which is only possible if $p = 1$, which is not a prime.

If $(x_2, y_2) = (x_3, y_3)$, then $p + q = pq + p$, which implies $q(1 - p) = 0$, which is also impossible; if $(x_2, y_2) = (x_4, y_4)$, then $p + q = p + 1$, which implies $q = 1$, which is not a prime; and finally, if $(x_3, y_3) = (x_4, y_4)$, then $pq + p = p + 1$, which implies $pq = 1$, which is also impossible, since $p$ and $q$ are primes.

In summary: if $p \neq q$, then each pair is different, and there are 4 solutions; if $p = q$, then the first two pairs are the same, so there are 3 solutions.

Based on the work of Nyul Gábor (Debrecen, Fazekas M. Gymnasium, 2nd year).