## Task 14/85

Determine all prime twin pairs $\left(p_{1} ; p_{2}\right)$ of the form $p_{1}=2 p-1, p_{2}=2 p+1$, where $p$ is also a prime number!
Of the three consecutive natural numbers $2 p-1, 2 p$, and $2 p+1$, one is certainly divisible by 3.

1. Let this be $p_{1}=2 p-1$. Due to the prime property of $p_{1}$, then

$$
p_{1}=2 p-1=3 \quad ; \quad p=2 \quad ; \quad p_{2}=2 p+1=5
$$

Thus, the pair $\left(p_{11} ; p_{21}\right)=(3 ; 5)$ is found as an element of the solution set.

2. Let this be $2 p$. Due to the prime property of $p$, then

$$
p=3 \quad ; \quad p_{1}=2 p-1=5 \quad ; \quad p_{2}=2 p+1=7
$$

and the pair $\left(p_{12} ; p_{22}\right)=(5 ; 7)$ is found as an element of the solution set.

3. Let this be $p=2 p+1$. Due to the prime property of $p_{2}$, then $p_{2}=2 p+3$ and $p=1$, which contradicts the required prime property of $p$. Therefore, this possibility is ruled out. Since the case distinction is complete, there can be no further elements of the solution set.