21. (1995 National High School Mathematics Competition) 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 two roots that are natural numbers.
21. Solution 1 Since $5-5(p+1)+(77 p-1)+1=66 p$, then $x=1$ is a natural number solution to the original cubic equation.

By synthetic division, the original cubic equation is reduced to a quadratic equation
$5 x^{2}-5 p x+66 p-1=0$.
This problem is converted to: find all real numbers $p$ such that equation (1) has two natural number solutions.
Let $u, v(u \leqslant v)$ be the two natural number solutions of equation (1). By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
v+u=p, \\
v u=\frac{1}{5}(66 p-1) .
\end{array}\right.
$$

Eliminating $p$ from (2) and (3) yields
$5 u v=66(u+v)-1$.
From (4), it is known that $u, v$ cannot be divisible by $2,3,11$.
From (4), we get $v=\frac{66 u-1}{5 u-66}$.
Since $u, v$ are natural numbers, then
$u>\frac{66}{5}$, so $u \geqslant 14$.
Also, $2 \times u, 3 \times u$, so $u \geqslant 17$.
Furthermore, by $v \geqslant u$ we get $\frac{66 u-1}{5 u-66} \geqslant u$, which means $5 u^{2}-132 u+1 \leqslant 0$.
Thus, $u \leqslant \frac{66+\sqrt{66^{2}-5}}{5}q$ leading to $\left\{\begin{array}{l}m_{0}-q_{0}=1,19,229,4351, \\ m_{0}+q_{0}=4351,229,19,1 .\end{array}\right.$
Solving this, we get $m_{0}= \pm 2176, \pm 124$.
Therefore, $5 p-132=2 m_{0}= \pm 4352, \pm 248$.
From (2) in Solution 1, we know that $p$ is a natural number, and from (3), we know that $u, v$ are odd numbers, and from (2) again, we know that $p$ is even.
Thus, we can solve to get $p=76$.