## Task 1/64

In the sequence of natural numbers, there are two groups of four consecutive prime numbers each, symmetrically arranged around a pair of twin primes. The distance between the smallest and the largest of these prime numbers is 70, and their product is equal to 3959.

Which are these eight prime numbers and the twin prime pair?
We call the center of the symmetric grouping $x$. Then the smallest of these primes is equal to $x-35$ and the largest is equal to $x+35$, and it follows that

$$
(x-35)(x+35)=3959 \rightarrow x^{2}-1225=3959 \rightarrow x_{1}=72 ; x_{2}=-72
$$

Thus, 71 and 73 form the prime twin. The smallest of the sought primes is $72-35=37$, and the largest is $72+35=107$. The two groups of four consecutive primes are $37 ; 41 ; 43 ; 47$ and $97 ; 101 ; 103 ; 107$. Evaluating the $x_{2}$ value leads to the corresponding negative numbers.