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 $x=p^{\sigma} x^{\prime}, y=p^{\beta} y^{\prime}, z=p^{\gamma} z^{\prime}$ with $p \nmid x^{\prime} y^{\prime} z^{\prime}$ and $\alpha \geq \beta \geq \gamma$. From the given equation it follows that $p^{n}(x+y)=z\left(x y-p^{n}\right)$ and consequently $z^{\prime} \mid x+y$. 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
$$
x y=p^{n}(q+1) \text { and } \quad x+y=q z .
$$

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 $n=2 k$ 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 $(x, y, z)=\left(1, p^{n}\left(p^{n}+3\right) / 2, p^{2}+2\right)$, 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
$$
(x, y, z)=\left(p^{k}, p^{n}+p^{n-k}+p^{2 n-2 k}, p^{n-k}+1\right) \quad \text { for } k=0,1, \ldots, n .
$$

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 $(x, y, z)=\left(p, p^{n}+\right.$ $\left.p^{n-1}, p^{n-1}+p^{n-2}+1\right)$.