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..

  • $\begingroup$ Your chain of reasoning is backwards! I think I can follow it, but only if I start at the end. $\endgroup$ – TonyK Apr 4 '14 at 7:41
  • 1
    $\begingroup$ What is $\phi$? $\endgroup$ – Jack M Apr 4 '14 at 8:08
  • $\begingroup$ @Jack: empty set. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 4 '14 at 8:15

If $A\neq\emptyset$, and $p\in A$ then, $p\equiv 1\bmod{3}$ or $p\equiv 2\bmod{3}$ (because you know $p\neq3$). Moreover, $p^2+4\in A$ and also $p(p^2+4)+4\in A$.

If $p\equiv 1\bmod{3}$ then $p^2+4\equiv 1+1\bmod{3}\equiv 2\bmod{3}$, and $p(p^2+4)+4\equiv 2+1\equiv 0\bmod3$: $A$ has a non prime number!

If $p\equiv 2\bmod{3}$, a similar reasonig gives the same conclusion.


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