Let $p$ be a fixed prime. Find all pairs $(x,y)$ of positive numbers satisfying $p(x-y) = xy$.
1. Start with the given equation:
   \[
   p(x - y) = xy
   \]
2. Expand and rearrange the equation:
   \[
   px - py = xy
   \]
   \[
   xy - px + py = 0
   \]
3. Factor the equation:
   \[
   xy - px + py = 0
   \]
   \[
   xy - px + py = 0 \implies xy - px + py = 0
   \]
   \[
   xy - px + py = 0 \implies (x + p)(y - p) = -p^2
   \]
4. Since \(x + p\) and \(y - p\) are factors of \(-p^2\), and both \(x\) and \(y\) are positive, we need to consider the positive divisors of \(p^2\). The positive divisors of \(p^2\) are \(1, p, p^2\).

5. Since \(x + p\) must be positive and greater than \(p\), and \(y - p\) must be positive and less than \(p\), we can conclude:
   \[
   x + p = p^2 \quad \text{and} \quad y - p = 1
   \]
6. Solving for \(x\) and \(y\):
   \[
   x + p = p^2 \implies x = p^2 - p
   \]
   \[
   y - p = 1 \implies y = p + 1
   \]

The final answer is \(\boxed{(x, y) = (p^2 - p, p + 1)}\)