## Task 15/87

Determine all prime numbers $p_{1}$ for which

$$
p_{2}=p_{1}+2^{n} ; \quad p_{3}=p_{1}+2^{n+1} ; \quad p_{4}=n p_{1}-3
$$

are also prime numbers. What values can $n$ take?
If $p_{1}=2$, then $p_{2}$ and $p_{3}$ would be even numbers and, because $p_{3}>p_{2}>p_{1}$, they would not be prime numbers. Therefore, $p 1 \geq 3$.

Since $p_{2}=p_{1}+2^{n}, p_{3}=p_{1}+2^{n+1}=p_{1}+2 \cdot 2^{n}=p_{2}+2^{n}$, the prime numbers $p_{1} ; p_{2} ; p_{3}$ form an arithmetic sequence of the first order with the difference $d=2^{n} \neq 3 k(k \in N)$. Thus, exactly one of these prime numbers is divisible by 3 and, due to the prime number property, it must be equal to 3. Because $p_{3}>p_{2}>p_{1}$, this can only be $p_{1}$: $p_{1}=3$.

It follows that $p_{4}=n p_{1}-3=n \cdot 3-3=3(n \mid 1)$. From the prime number property of $p_{4}$, it immediately follows that $p_{4}=3, n-1=1$, thus $n=2, p_{2}=7, p_{3}=11$. This provides a solution. Due to the inevitability of the derivation, this is also the only one.