$G(B) = \{k \in \Bbb{Z} : \#(A+k) \cap B = \infty \text{ whenever } \#(A \cap B) = \infty\, A \subset \Bbb{Z}\}$ is a group. Question...

Question

Let $B \subset \Bbb{Z}$ be any infinite subset. Define $G(B) = \{k \in \Bbb{Z} : \#(A+k) \cap B = \infty \text{ whenever } \#(A \cap B) = \infty\, A \subset \Bbb{Z}\}$. It forms an additive subgroup of $\Bbb{Z}$. Let $P^2 = \{pq \in \Bbb{Z}: p,q \text{ prime}\}$. How do I prove that $1 \in G(P^2)$, which amounts to saying $\#(A + 1) \cap P^2 = \infty$ whenever $\# A\cap P^2 = \infty$?

Answer 1

Let $p$ be prime and $A=pP$, in which case clearly $A\cap P$ is infinite. Then $A+p\cap P^2$ is finite because $pq+p=p(q+1)$ is in $P^2$ is only possible with $q=2$. We conclude $p\notin G(P^2)$. As $G(P^2)$ is additive we conclude $1\notin G(B)$.

(Inspired by Igor Rivin's answer): Let $n\in \Bbb N$ be composite. Then there are infinitely many primes $\equiv 1\pmod n$ Let $A$ be the set of products of two such primes, an infinite subset of $P^2$. Then $A+n-1$ contains only multiples of $n$ and at most one of these multiples is the product of two primes. We conclude that $G(B)$ contains no positive composite integers. As $a\in G(B)$ implies that the composite $6|a|$ is $\in G(B)$, we conclude $$ G(P^2)=\{0\}.$$

Answer 2

This is false. Let $A$ be the set of products $p q,$ where $p, q$ are prime, and $p\equiv 1 \mod 4,$ $q\equiv 3 \mod 4.$ Then, every element in $A+1$ is divisible by $4,$ and so is not in $P^2.$