Brazilian Mathematical Olympiad

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 $r \equiv 1 \pmod{3}$, 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 $r \equiv 1 \pmod{3}$. If $r \equiv 2 \pmod{3}$, 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 $r \equiv 2 \pmod{3}$. Hence $r$ must be a multiple of $3$. But $r$ is prime, so $r$ must be $3$. Hence $q = 7$ and $p = 5$.