Estonian Math Competitions

Let $p$ be a cute prime. The numbers $pq-2$, $pq$ and $pq+2$ give all the possible remainders modulo $3$, so one of them must be divisible by $3$. If $3\mid pq-2$, then $pq-2=3$ and $pq=5$, which is impossible, as $p$, $q$ are primes. Similarly if $3\mid pq+2$, then $pq+2=3$ and $pq=1$, which is also impossible. So $3\mid pq$, which means that either $p=3$ or $q=3$. Given that in the case $p=3$ we can have $q=3$, we have shown that $p$ is cute iff both $3p-2$ and $3p+2$ are primes.

Therefore $p$ is wonderful if $p$, $p+2$, $3p-2$, $3p+2$, $3p+4$ and $3p+8$ are all primes. We know that the numbers $3p-2$, $3p$, $3p+2$, $3p+4$ and $3p+6$ give all of the possible remainders modulo $5$, so one of those numbers is divisible by $5$. If it's $3p-2$, $3p+2$ or $3p+4$, then due to primality it must be equal to $5$, but then $p$ is not prime. If it's $3p$ or $3p+6$, then either $5\mid p$ or $5\mid p+2$ respectively. Thus $p=3$ or $p=5$. We can verify that both work.