36. Find the prime number $p$, given that the number $13 p+1$ is a perfect cube.
36. If $13 p+1=n^{3}$, then $13 p=(n-1)\left(n^{2}+n+1\right)$. Since on the left side of the last equation we have the product of two prime numbers, there are only three possible cases:

$$
\left\{\begin{array} { l } 
{ n - 1 = 1 } \\
{ n ^ { 2 } + n + 1 = 1 3 p ; }
\end{array} \quad \left\{\begin{array} { l } 
{ n - 1 = \mathrm { p } } \\
{ n ^ { 2 } + n + 1 = 1 3 ; }
\end{array} \quad \left\{\begin{array}{l}
n-1=13 \\
n^{2}+n+1=p
\end{array}\right.\right.\right.
$$

The first system has no solutions in natural numbers. From the second and third systems, we find respectively: $n=3$, $p=2$ and $n=14, p=211$. Indeed, $13 \cdot 2+1=3^{3}$ and $13 \cdot 211+$ $+1=14^{3}$