$12 \cdot 148 p$ is a prime number, find all positive integer solutions $(x, y, z, p)$ of the equation $x^{p}+y^{p}=p^{z}$.
(Bulgarian Mathematical Olympiad, 1994)
[Solution] If $(x, y, z, p)$ is a set of positive integer solutions to the equation
$$
x^{p}+y^{p}=p^{z}
$$

and since $p$ is a prime number, then
$$
(x, y)=p^{k}.
$$

Here $k$ is a non-negative integer. From (2) we have
$$
x=p^{k} x_{0}, \quad y=p^{k} y_{0}.
$$

Here $\left(x_{0}, y_{0}\right)=1, x_{0} \in N, y_{0} \in N$. Substituting (3) into (1) gives $x_{0}^{p}+y_{0}^{p}=p^{z-k p}$. Since $x_{0}, y_{0}$ are positive integers, then $p^{z-k p} \geqslant 2$,
thus $z_{0}=z-k p$ is a positive integer.
This gives $x_{0}^{p}+y_{0}^{p}=p^{z_{0}}$.
Equation (5) has the same form as equation (1), but in equation (5), $x_{0}$ and $y_{0}$ are coprime. At this point, from equation (5), $x_{0}$ and $y_{0}$ are not multiples of $p$.
When $p=2$, from (5) we know that $x_{0}$ and $y_{0}$ are both odd, and equation (5) becomes
$$
x_{0}^{2}+y_{0}^{2}=2^{z_{0}}
$$

Since
$$
x_{0}^{2}+y_{0}^{2} \equiv 2(\bmod 4),
$$

then
$$
z_{0}=1.
$$

Substituting $z_{0}=1$ into (6) gives $x_{0}=1, y_{0}=1$.
Thus, from (3) we get $x=2^{k}, y=2^{k}$, and substituting into equation (1) gives
$$
\begin{array}{l}
2^{z}=x^{2}+y^{2}=2^{2 k}+2^{2 k}=2^{2 k+1}, \\
z=2 k+1
\end{array}
$$

Thus, the solution is $(x, y, z, p)=\left(2^{k}, 2^{k}, 2 k+1,2\right)$.
When $p$ is an odd prime, since $x_{0}+y_{0} \mid x_{0}^{p}+y_{0}^{p}$, and from (5) we know that $x_{0}+y_{0}$ is a factor of $p^{z_{0}}$, i.e., there exists a positive integer $t_{0}$ such that
$$
x_{0}+y_{0}=p^{t_{0}}.
$$

Since $p \geqslant 3$, then $x_{0}+y_{0} \geqslant 3, x_{0}^{p}+y_{0}^{p}>x_{0}+y_{0} \geqslant 3$, considering equation (5), then $1 \leqslant t_{0} \leqslant z_{0}-1$,

Thus, $x_{0}^{p}+\left(p^{t}-x_{0}\right)^{p}=p^{z_{0}}$, expanding the second term on the left side, and noting that $p$ is odd, the equation becomes
$$
p^{p t_{0}}-C_{p}^{1} p^{(p-1) t_{0}} x_{0}+\cdots-C_{p}^{p-2} p^{2 t_{0}} x_{0}^{p-2}+C_{p}^{1} p^{t_{0}} x_{0}^{p-1}=p^{z_{0}},
$$

Since $p$ is an odd prime, $C_{p}^{j}(j=1,2, \cdots, p-2)$ is a multiple of $p$. The left side of the above equation, except for the last term $C_{p}^{1} p^{t} x_{0}^{p-1}=p_{0}^{t_{0}+1} x_{0}^{p-1}$, all other terms are multiples of $p^{2 t_{0}+1}$, thus $p^{2} 0$ must be a multiple of $p^{t_{0}+1}$, and it must not be a multiple of $p^{t_{0}}+2$, so $z_{0}=t_{0}+1$.
At this point, from (7) we have $x_{0}+y_{0}=p^{z_{0}-1}$.
From (5) we also have $x_{0}^{p}+y_{0}^{p}=p\left(x_{0}+y_{0}\right)$.
On the other hand, it is easy to prove $u^{n-1}>n(u \geqslant 2, u \in N)$,
thus we also have $x_{0}^{p}+y_{0}^{p}=x_{0} \cdot x_{0}^{p-1}+y_{0} \cdot y_{0}^{p-1}>p\left(x_{0}+y_{0}\right)$.
(8) and (9) are contradictory. Therefore, at least one of $x_{0}$ and $y_{0}$ must be 1. Without loss of generality, let $x_{0}=1$.
From $y_{0}=p^{z_{0}-1}-1$,
and (8), $y_{0}^{p}-p y_{0}=p-1$.
Thus, $y_{0}$ is a factor of $p-1$, from (1) we know
$z_{0}-1=1$. That is, $z_{0}=2$. Hence, $y_{0}=p-1$.
Substituting into (11) gives $(p-1)^{p}=p(p-1)+p-1=p^{2}-1$.
Since $p$ is an odd prime, we get $(p-1)^{p-1}=p+1$.
When $p>3$, it is easy to prove $(p-1)^{p-1}>p+1$.
Therefore, $p=3$. $y_{0}=p-1=2$,
then $x=3^{k}, y=2 \cdot 3^{k}, z=2+3 k$.
By symmetry, we also have $x=2 \cdot 3^{k}, y=3^{k}, z=2+3 k$.
Thus, we have another set of solutions $(x, y, z, p)=\left(3^{k}, 2 \cdot 3^{k}, 2+3 k, 3\right)$,
$$
(x, y, z, p)=\left(2 \cdot 3^{k}, 3^{k}, 2+3 k, 3\right).
$$

Therefore, all solutions to this problem are $\left(2^{k}, 2^{k}, 2 k+1,2\right),\left(3^{k}, 2 \cdot 3^{k}, 2+3 k, 3\right)$, $\left(2 \cdot 3^{k}, 3^{k}, 2+3 k, 3\right)$.