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?
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 $p(c+1)\in A'$.