Given a subset of prime numbers say $A$. It is given that for $p,q\in A$ we also must have $(pq+4)\in A$ . Show that $A=\emptyset$.
My work so far: It is obvious that $2,3\notin A$ . because all the primes $\equiv \pm 1(\bmod 6)$ If $3\in A$ and another odd prime $p\in A$ then we have $(pq+4)\equiv 1(\bmod 6)$
My claim all the primes in $A$ must be $\equiv-1(\bmod 6)$ If not then say their exist two primes $p',q'\equiv 1(\bmod 6)$ but $p'q'+4\equiv -1(\bmod 6)$ then it is a prime in the form $6k-1$.
If $p'\equiv-1(\bmod 6)$ and $q'\equiv 1(\bmod 6)$ then we can generate another prime $r=(p'q'+4)\equiv 3(\bmod 6)$ contradiction. So all the primes $p,q$ must be in the form $6k-1$, and so does $pq+4$ . But i am stuck now..