Task 1. Determine a prime number $p$, such that the number $2 p+1$ is a perfect cube of some natural number.
Solution. From the condition $2 p+1=n^{3}$, it follows that $n$ is an odd number. Furthermore,

$$
2 p=n^{3}-1=(n-1)\left(n^{2}+n+1\right)
$$

Since $n$ is an odd number, let $n=2 k+1, k \in \mathbb{N}$. Then

$$
\begin{aligned}
& 2 p=2 k\left(4 k^{2}+6 k+3\right) \\
& p=k\left(4 k^{2}+6 k+3\right)
\end{aligned}
$$

The number $p$ is prime, so from the last equation it follows that $k=1$, and

$$
p=13,2 p+1=27=3^{3}
$$