## Task 15/70

Find all prime numbers that can be represented both as the sum and as the difference of two prime numbers.
Let $p$ be one of the sought prime numbers. If $p$ can be expressed as the sum of two prime numbers, it is certainly true that $p>2$. Consequently, $p$ is odd. If further $p$ can be represented as the difference of two prime numbers, i.e., $p=p_{1}-q_{1}$ where $p_{1}$ and $q_{1}$ are also prime numbers, then $p_{1}>p$, and thus $p_{1}$ is odd.

It immediately follows that $q_{1}$ is even, otherwise $p$ would be even (contradicting the above statement). Therefore, $q_{1}=2$, as there are no other even prime numbers.

Similarly, from $p=p_{2}+q_{2}$, it follows that exactly one of the two prime numbers $p_{2}$ and $q_{2}$ must be even and thus equal to 2; without loss of generality (due to the commutativity of addition), let this be $q_{2}$. Then we have

$$
p_{2}+2=p=p_{1}-2
$$

The sought prime numbers are thus the middle numbers of prime triplets.

According to problem 10/69 (issue 4/1969), there is exactly one prime triplet, namely 3;5;7. Therefore, $p=5$ is the only prime number that can be expressed both as the sum and as the difference of two prime numbers: $3+2=5=7-2$.