## Task $2 / 80$

All prime numbers $p$ are to be determined that satisfy the equation $2 p+1=m^{3}$ with $m \in N$.
Since $2 p+1$ is an odd number, if there exists an $m \in \mathbb{N}$ such that $2 p+1=m^{3}$, then $m$ must also be odd: $m=2 k+1$ with $k \in \mathbb{N}$. This leads to

$$
2 p+1=(2 k+1)^{3}=8 k^{3}+12 k^{2}+6 k+1 \rightarrow p=k\left(4 k^{2}+6 k+3\right)
$$

Given that $p$ is a prime number by assumption, it follows that either $k=1$ and $4 k^{2}+6 k+3=p$ or $k=p$ and $4 k^{2}+6 k+3=1$. However, the latter (quadratic) equation in $k$ has no solution in the natural numbers. Therefore, $k=1$, $p=4 k^{2}+6 k+3=13$, and thus $m=3$.

Therefore, there is exactly one prime number $p$ with the required property: $p=13$.