Set of prime integers Let $S$ be a set of primes such that $a,b\in S$ ($a$ and $b$ need not be distinct) implies $ab+4\in S$. Show that $S$ must be empty. (Hint use modulo 7)
I don't have an idea how to use the hint, any further hint will be appreacited

Suppose $p \in S$.
Suppose $p\equiv 1 \pmod 7$ and $p\in S$.  Then $p_2=p^2 +4 \equiv 5 \pmod 7$ is prime and in $S$.  so $p_3 =p_1p_2 + 4\equiv 9\equiv 2\pmod 7$.  So $p_4 = p_3p_2 +4\equiv 2*5+4 \equiv 0 \pmod 7$ is in $S$.  But $7|p_4$ and $p_4$ is prime so $p_4 = p_3p_2 + 4 =7$ so $p_3p_2=3$ but $3$ is primes so that's impossible as $p_3p_2$ are prime so neither are equal to $1$.
So $S$ contains no primes $\pmod 1 \pmod 7$.
suppose $p\equiv 2\pmod 7$.  Then $p_2 = p^2 +4 \equiv 8 \equiv 1 \pmod 7$ which we proved is impossible..
Suppose $p \equiv 3\pmod 7$ then $p_2 = p^2 + 4\equiv 6\pmod 7$.  Then $p_3=pp_2 + 4\equiv 18\equiv 4\pmod 7$.  And $pp_3 + 4\equiv 16 \equiv 2\pmod 7$ which was impossible.
Suppose $p\equiv 4 \pmod 7$. Then $p_2= p^2+4 \equiv 6$ And $p_3=pp_2 + 4 \equiv 28\equiv 0$ so $7|p_3$ so $p_3$ (which is prime) is equal to $7$ and $pp_2 \equiv 3 \pmod 7$ which was impossible two paragraphs up and is still impossible.
Suppose $p\equiv 5 \pmod 7$ then $p_2=p^2 + 4\equiv 29 \equiv 1$.  That's out.
Suppose $p \equiv 6 \pmod 7$ then $p_2 = p^2 + 4 \equiv 5\pmod 7$.  Nope.
So that only leaves $p\equiv 0 \pmod 7$ and as $p$ prime then $7$ is the only element in $S$.  But $7^2 + 4 =53\in S$ which is a contradiction.
so $S$ is empty.
