Let $O = $ the set of odd primes in $\Bbb{N}$. And $M = \Delta O = \{ x-y: x,y \in O\}$. Then we can take either set $X$ modulo $n$ for any $n \geq 2$: $\overline{X} = \{ x + (n) : x \in X\}$.
Then we have that $\overline{M} = \overline{\Delta O} = \Delta\overline{O}$ i.e. moduloing a set commutes with differencing.
There is a major open problem which is that $M = (2)$ the ideal generated by $2$ in $\Bbb{Z}$. That is to say for every integer $z \in \Bbb{Z}$ there exists an odd prime solution to $p - q = 2z$.
My question is: if it can be proven that $\overline{M} = (2)$ in every ring $\Bbb{Z}/2^n$ then does it follow that $M = (2)$ in $\Bbb{Z}$? That is to say, given a proof of that lemma, could the open problem be closed?
Can we make use of the fact that $f_n : \Bbb{Z}/(2^{n}) \twoheadrightarrow \Bbb{Z}/(2^{n-1})$ (ring homomorphism) for all $n \geq 2$?
That would give us an inverse system in the category of rings and the $2$-adic integer ring $\widehat{\Bbb{Z}_2} = \lim\limits_{\leftarrow} \Bbb{Z}/(2^i)$, the categorical limit. For which we know that there is an injection of rings:
$$ g : \Bbb{Z} \hookrightarrow \widehat{\Bbb{Z}_2}. $$
Let $(2)_{n}$ be the ideal $2 \Bbb{Z}/2^{n}$. Then given $f_n : \Bbb{Z}/2^n \twoheadrightarrow \Bbb{Z}/2^{n-1}$ for all $n \geq 2$ we have:
$$ f_n((2)_n) = 2 f_n(\Bbb{Z}/2^n) = 2\Bbb{Z}/2^{n-1} = (2)_{n-1} $$
So we have a commutative sequence of commutative squares in the category of rings:
$$ \require{AMScd} \begin{CD} @>{...}>> (2)_n @>{g_n}>> (2)_{n-1} @>{...}>> \\ ...@V{i}VV @V{j}VV ... \\ @>{...}>> \Bbb{Z}/2^n @>{f_n}>> \Bbb{Z}/2^{n-1} @>{...}>> \\ \end{CD} $$
where $i,j$ are inclusions and $g_n$ is the restriction of $f_n$ to $(2)_n$ is a surjective homomorphism of rings (every ideal is also a subring).
I worked out some of the math in a related question:
Limit of ideals modulo $p^n$ isomorphic to an ideal of the ring of $p$-adic integers?
Identify $\mathbb{Z}$ with its image in $\mathbb{Z}_p$. Then if an ideal $J \subset \mathbb{Z}_p$ satisfies $J \cap \mathbb{Z} = p\mathbb{Z}$, you must have $J \supset (p\mathbb{Z})\mathbb{Z}_p$ (the extended ideal). But $\mathbb{Z}_p$ is a DVR with max ideal $(p\mathbb{Z})\mathbb{Z}_p$, whence $J = (p\mathbb{Z})\mathbb{Z}_p$.
Furthermore, if an ideal $M \subset \mathbb{Z}_p$ is such that $\pi_n(M) = (p+\mathbb{Z}/p^n\mathbb{Z})$ for all $n \geq 1$ in $\mathbb{N}$, then $M$ contains the limit of the inverse system defined by those ideals (which is a sub-inverse system to that defining $\mathbb{Z}_p$) which is exactly the maximal ideal $(p\mathbb{Z})\mathbb{Z}_p$.
Edit: As per Gerry's comment, this isn't enough to close out the conjecture; you'd have to also know that $M$ is actually an ideal.