2. 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.
Solve:
$$
5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1-66 p=(x-1)\left[5 x^{2}-5 p x+(66 p-1)\right] \text {, }
$$

Let the two roots $a<b$ of the equation $5 x^{2}-5 p x+(66 p-1)=0$ both be natural numbers, then
$$
\begin{array}{l}
\left\{\left.\begin{array}{l}
a+b=p, \\
a b=\frac{66 p-1}{5}
\end{array} \Rightarrow 5 \right| 66 p-1 \Rightarrow p=5 k+1(k \in \mathbf{N})\right. \text {. } \\
\text { Thus }\left\{\begin{array}{l}
a+b=5 k+1, \\
a b=66 k+13
\end{array} \Rightarrow 25 a b-66 \times 5(a+b)=-5\right. \\
\Rightarrow(5 a-66)(5 b-66)=66^{2}-5=4351=19 \times 229 \text {. } \\
\text { Therefore }\left\{\begin{array} { l } 
{ 5 a - 6 6 = 1 9 , } \\
{ 5 b - 6 6 = 2 2 9 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=17, \\
b=59
\end{array} \Rightarrow a+b=76\right.\right. \text {. } \\
\end{array}
$$

Therefore, there exists a unique $p=76$, such that the three roots of the cubic equation $5 x^{3}-5(p+1) x^{2}+(71 p-1) x+1=66 p$ are all natural numbers.