## Task 30/77

We are looking for an ordered quadruple of four prime numbers $p_{i}$ with $i=1 ; 2 ; 3 ; 4$, that satisfy the following conditions:

1. It is $p_{1}<p_{2}<p_{3}<p_{4}$.
2. $p_{3}$ and $p_{4}$ form a prime twin pair.
3. The sum of the two smaller prime numbers is equal to the arithmetic mean of the two larger ones and equal to twice the square of a fifth prime number.
4. The product $p_{1} \cdot p 2$ is maximal.

Form the triples $\left(p_{1} ; p_{2} ; 100 p_{3}+p_{4}\right)$ and $\left(p_{1} ; p_{2} ; 100 p_{3}+p_{1} p_{2}\right)$!
From 2., considering 1.: $p_{3}=p_{4}+2$. From 3. follows $\frac{p_{3}+p_{4}}{2}>2 p_{5}^{2}$, thus

$$
p_{3}+p_{4}=2 p_{4}+2=2\left(p_{4}+1\right)=4 p_{5}^{2} \rightarrow p_{4}+1=2 p_{5}^{2}
$$

Given $p_{4}5 \cdot 13$ (condition 4.), $p_{1}=7, p_{2}=11$ (considering 1.). The prime quadruplet is thus $(7 ; 11 ; 19 ; 17)$, and the sought triples are (7.11.1917) and (7.11.1977).