amparvardi wrote: Find all prime numbers $p, q$ and $r$ such that $p>q>r$ and the numbers $p-q, p-r$ and $q-r$ are also prime. Assume that $r> 2$ then all the three prime numbers are odd, so $p-r$ and $q-r$ are both even prime numbers, so $p-r=q-r=2$ which is absurde since $p>q$. therefore $r=2$, then the problem becomes about to find prime numbers $p>q$ for which, $p-q$ ; $p-2$ and $c-2$ are all primes. since $p>q>2$ , $p,q$ are both odd, and so $p-q=2$, hence, numbers $p$ , $p-2$ and $p-4$ should be primes. since $p>q>2$ we have $p>3$ hene we have two case: if $p\equiv 2[3]$, so $p-2=3$ but this gives $p-4=1$ which isn't a prime. if $p\equiv 1[3]$, so $p-4=3$, this gives $p-2=5$ and $p=7$ those numbers are all primes, so, the only solution for the problem is $(p,q,r)=(7,5,2)$.