Second Pump Find all real numbers $p$ such that the cubic equation
$$
5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p
$$

has three roots that are all natural numbers.
The original problem is equivalent to
$$
5 x^{2}-5 p x+66 p-1=0
$$

having both roots as natural numbers.
Method 1: Let $u, v$ be the roots of equation (2), then
$$
\left\{\begin{array}{l}
u+v=p, \\
u v=\frac{1}{5}(66 p-1) .
\end{array}\right.
$$

Eliminating the parameter $p$ yields
$$
\begin{array}{l}
5 u v=66 u+66 v-1, \\
u=\frac{66 v-1}{5 v-66}=\frac{66}{5}+\frac{4351}{5(5 v-66)}, \\
5 u=66+\frac{19 \times 229}{5 v-66} .
\end{array}
$$

Clearly, $5 v-66>0$, and $5 v-66$ is a positive divisor of $19 \times 229$ that is congruent to 4 modulo 5, i.e.,
$5 v-66=19$ or 229.
$$
\left\{\begin{array} { l } 
{ u = 5 9 ; } \\
{ v = 1 7 ; }
\end{array} \text { or } \left\{\begin{array}{l}
u=17, \\
v=59 .
\end{array}\right.\right.
$$

Therefore, $p=u+v=76$.
Method 2: Clearly, $p$ should be a natural number, and the discriminant
$$
\Delta=25 p^{2}-20(66 p-1)=(5 p-132)^{2}-4351 \times 4
$$

is a perfect square. Let $A=5 p-132, \Delta=B^{2}(B \geqslant 0)$, then
$$
A^{2}-B^{2}=4 \times 19 \times 229 \text {. }
$$
$A, B$ have the same parity, and they can only both be even (if $A, B$ are both odd, then $8\left|\left(A^{2}-1\right), 8\right|\left(B^{2}-1\right)$; i.e., $8 \mid\left(A^{2}-B^{2}\right)$, which is a contradiction).
$$
\begin{array}{l}
\text { Let } A=2 A_{1}, B=2 B_{1} \text {, then } \\
A_{1}^{2}-B_{1}^{2}=19 \times 229 .
\end{array}
$$

From $A_{1}-B_{1} \leqslant A_{1}+B_{1}$, we get
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ A _ { 1 } - B _ { 1 } = - 4 3 5 1 , } \\
{ A _ { 1 } + B _ { 1 } = - 1 ; }
\end{array} \left\{\begin{array}{l}
A_{1}-B_{1}=-229, \\
A_{1}+B_{1}=-19 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ A _ { 1 } - B _ { 1 } = 1 , } \\
{ A _ { 1 } + B _ { 1 } = 4 3 5 1 ; }
\end{array} \left\{\begin{array}{l}
A_{1}-B_{1}=19, \\
A_{1}+\overline{8}=229 .
\end{array}\right.\right.
\end{array}
$$

Thus, $5 p-132==2, A_{1}=--432,-21 \hat{2}, 4352,240$.
The positive integer $\phi$ can only be 75.