4. (CZS 3) Let $n$ be a positive integer and let $p$ be a prime number, $p>3$. Find at least $3(n+1)$ [easier version: $2(n+1)]$ sequences of positive integers $x, y, z$ satisfying 
$$ 
x y z=p^{n}(x+y+z) 
$$ 
that do not differ only by permutation.
4. Let \( x = p^{\alpha} x', y = p^{\beta} y', z = p^{\gamma} z' \) with \( p \nmid x' y' z' \) and \( \alpha \geq \beta \geq \gamma \). From the given equation it follows that \( p^{n}(x + y) = z(x y - p^{n}) \) and consequently \( z' \mid x + y \). Since also \( p^{\gamma} \mid x + y \), we have \( z \mid x + y \), i.e., \( x + y = q z \). The given equation together with the last condition gives us 
\[ x y = p^{n}(q + 1) \quad \text{and} \quad x + y = q z. \]
Conversely, every solution of (1) gives a solution of the given equation. For \( q = 1 \) and \( q = 2 \) we obtain the following classes of \( n + 1 \) solutions each:
\[ \begin{array}{ll}
q = 1: (x, y, z) = \left(2 p^{i}, p^{n-i}, 2 p^{i} + p^{n-i}\right) & \text{for } i = 0, 1, 2, \ldots, n \\
q = 2: (x, y, z) = \left(3 p^{j}, p^{n-j}, \frac{3 p^{j} + p^{n-j}}{2}\right) & \text{for } j = 0, 1, 2, \ldots, n
\end{array} \]
For \( n = 2k \) these two classes have a common solution \( \left(2 p^{k}, p^{k}, 3 p^{k}\right) \); otherwise, all these solutions are distinct. One further solution is given by \( (x, y, z) = \left(1, \frac{p^{n}(p^{n} + 3)}{2}, p^{2} + 2\right) \), not included in the above classes for \( p > 3 \). Thus we have found \( 2(n + 1) \) solutions. Another type of solution is obtained if we put \( q = p^{k} + p^{n-k} \). This yields the solutions
\[ (x, y, z) = \left(p^{k}, p^{n} + p^{n-k} + p^{2n-2k}, p^{n-k} + 1\right) \quad \text{for } k = 0, 1, \ldots, n \]
For \( k < n \) these are indeed new solutions. So far, we have found \( 3(n + 1) - 1 \) or \( 3(n + 1) \) solutions. One more solution is given by \( (x, y, z) = \left(p, p^{n} + p^{n-1}, p^{n-1} + p^{n-2} + 1\right) \).