Number of elements in shifted powersets of positive integers Let us split the set of positive integers $(\mathbb{Z}^+)$ into the following power sets:   
$$
\begin{aligned}
A &= \{1, n\times(6m+1),\ldots\} \\
B &= \{n\times(6k-1),\ldots\}
\end{aligned}
$$
Where we choose $m$ and $k$ so that $(6m+1)$ and $(6k-1)$ cover all prime numbers, and $n$ is every positive integer from 1 to infinity.
Therefore $\mathbb{Z}^+ = A U B$, because every prime and composite number is included in either $A$ or $B$, or in some cases, both in $A$ and $B$. 1 is included in $A$.
Let us define $A'$ as adding $x$ to each element in $A$, and define $B'$ as adding $y$ to each element in $B$, where $x$ and $y$ are different positive integers.
Am I right to assume that if we find at least 1 common element in $A'$ and $B'$, then if we take $\mathbb{Z}' = A' + B'$, then $\mathbb{Z}-\mathbb{Z}'$ will contain infinitely many elements?
I assume that it is true if we speak about power sets. 
Are you aware of any proof on this? 
If not, what do you think about my conjecture?
 A: Bela, this is interesting. First as is, it is not the case that $A\cup B=\mathbb{Z}^+$. In fact, even if this were true, then $A'+B'$ would consist only of positive integers, and any negative integer is not in $A'+B'$. So, I'll modify your sets such that $A\cup  B=\mathbb{Z}$.
Note that, $A\cup B=\neq \mathbb{Z}^+$. For instance, to which set, numbers of form, $3^k$ or $2^k$ belong to? Nevertheless, suppose we modify this, and add $\{3n:n\in\mathbb{Z}^+\}$ to $A$ and $\{2n:n\in\mathbb{Z}^+\}$ to $B$, I believe $A\cup B=\mathbb{Z}^+$. Let us also allow $n\in\mathbb{Z}$, so that $A\cup B=\mathbb{Z}$.
Construct $A'=A+x$ and $B'=B+y$, where $x$ and $y$ are fixed (and I suppose known). Now, I claim that, for every $n\in\mathbb{Z}$, there exists $a'\in A'$ and $b'\in B'$ such that $a'+b'=n-x-y$. To see this, let $n-x-y\equiv c\pmod{p}$. Fix a  prime $p\equiv 1\pmod{6}$, and observe that, the number $p(c+1)\equiv c+1\pmod{6}$, and is in $A$. Now $n-x-y-p(c+1)\equiv -1\pmod{6}$. It is very well known that such a number must have a prime divisor $q$, so that $q\equiv -1\pmod{6}$. For this divisor, you see that, $(n-x-y-p(c+1))/q \triangleq T\in\mathbb{Z}$, and $Tq\in B$. Hence, $Tq+p(c+1)+x+y=n$, with $Tq+y\in B'$ and $p(c+1)\in A'$.
