## Task 22/81

We are looking for all natural numbers $n$ that satisfy the equation $p_{2}^{3}-2=p_{1} \cdot n$; here, $p_{1} ; p_{2}$ is a prime twin pair.
Given that $p_{1} ; p_{2}$ is a pair of twin primes, it follows that $p_{2}=p_{1}+2$ and thus

$$
p_{2}^{3}-2=\left(p_{1}+2\right)^{3}-2=p_{1}^{3}+6 p_{1}^{2}+12 p_{1}+8-2=p_{1}\left(p_{1}^{2}+6 p_{1}+12+\frac{6}{p_{1}}\right)=p_{1} \cdot n
$$

Since $n \in \mathbb{N}$, it follows that $\frac{6}{p_{1}} \in \mathbb{N}$, hence $p_{1}=2$ or $p_{1}=3$. The first possibility is excluded; otherwise, $p_{2}=p_{1}+2=4$ would contradict the primality of $p_{2}$. On the other hand, $p_{1}=3$ yields the twin prime $p_{2}=5$, and we have

$$
n=3^{2}+6 \cdot 3+12+\frac{6}{3}=41
$$

The verification confirms the correctness. Therefore, there exists exactly one natural number $n$ that satisfies the given equation (which is also a prime number).