1. find all triples $(p, q, r)$ of prime numbers such that the three differences

$$
|p-q|, \quad|q-r|, \quad|r-p|
$$

are all prime numbers.

## Solution:
Due to the symmetry of the problem, we can assume $p2$, then $r-p$ is greater than 2 and even, i.e. not prime, contradiction. If $r>q+2$, then it follows by the same argument that $r-q$ cannot be prime. So $p=2$ and $r=q+2$ apply. According to the condition, $q-p=q-2$ is also prime. Of the three numbers $q-2, q, q+2$, exactly one is divisible by 3 and is also prime, so $q-2=3$ and

$$
(p, q, r)=(2,5,7)
$$

## Remarks and scoring scheme:

Reducing the problem to the question of when $q-2, q, q+2$ can all be prime numbers was worth $3 \mathrm{P}$. There were partial points for the two important observations $p=2$ and $r=q+2$. Whoever could prove one of the two received 2P, for both together there were 3P. Anyone who could not prove either of the two, but showed that exactly one of the differences $|p-q|,|q-r|,|r-p|$ is equal to 2, received 1P.

Whoever showed that of three prime numbers of the form $q-2, q, q+2$ one must be divisible by 3 received 5P, whoever also noticed that this must therefore be equal to 3 received $6 \mathrm{P}$.