X OBM

The smallest prime is $2$, so it cannot be written as the sum of two primes. All other primes $p$ are odd, so if $p$ can be written as the sum of two primes and the difference of two primes, then in each case one of the primes must be $2$. Thus we have $p=q-2=r+2$ for some primes $q,r$. If $r\equiv 1(\mathrm{mod}3)$, then $p$ is a multiple of $3$, so $p$ must be $3$. But then $r=1$, which is not a prime. So there are no solutions with $r\equiv 1(\mathrm{mod}3)$. If $r\equiv 2(\mathrm{mod}3)$, then $q$ is a multiple of $3$, so $q$ must be $3$, but then $p=1$, which is not a prime. So there are no solutions with $r\equiv 2(\mathrm{mod}3)$. Hence $r$ must be a multiple of $3$. But $r$ is prime, so $r$ must be $3$. Hence $q=7$ and $p=5$.