Problem: Let $S \subseteq \ZZ$ be such that:
\begin{enumerate}
  \ii[(a)] there exist $a,b \in S$ with
  $\gcd(a,b) = \gcd(a-2, b-2) = 1$;
  \ii[(b)] if $x$ and $y$ are elements of $S$ (possibly equal),
  then $x^2-y$ also belongs to $S$.
\end{enumerate}
Prove that $S = \ZZ$.

Solution: Call an integer $d > 0$ \emph{shifty} if
$S = S+d$ (meaning $S$ is invariant under shifting by $d$).

First, note that if $u, v \in S$, then for any $x \in S$,
\[ v^2 - (u^2-x) = (v^2-u^2) + x \in S. \]
Since we can easily check that $|S| > 1$
and $S \neq \{n, -n\}$ we conclude there exists a shifty integer.

We claim $1$ is shifty, which implies the problem.
Assume for contradiction that $1$ is not shifty.
Then for GCD reasons the set of shifty integers must be $d \ZZ$ for some $d \ge 2$.

\begin{claim*}
  We have $S \subseteq \left\{ x : x^2 \equiv m \pmod d \right\}$ for some fixed $m$.
\end{claim*}
\begin{proof}
  Otherwise if we take any $p,q \in S$ with distinct squares modulo $d$,
  then $q^2-p^2 \not\equiv 0 \pmod d$ is shifty, which is impossible.
\end{proof}

Now take $a,b \in S$ as in (a).
In that case we need to have
\[ a^2 \equiv b^2 \equiv (a^2-a)^2 \equiv (b^2-b)^2 \pmod d. \]
Passing to a prime $p \mid d$, we have the following:
\begin{itemize}
  \ii Since $a^2 \equiv (a^2-a)^2 \pmod p$ or equivalently $a^3(a-2) \equiv 0 \pmod p$,
  either $a \equiv 0 \pmod p$ or $a \equiv 2 \pmod p$.
  \ii Similarly, either $b \equiv 0 \pmod p$ or $b \equiv 2 \pmod p$.
  \ii Since $a^2 \equiv b^2 \pmod p$, or $a \equiv \pm b \pmod p$, we find either
  $a \equiv b \equiv 0 \pmod p$ or $a \equiv b \equiv 2 \pmod p$ (even if $p=2$).
\end{itemize}
This is a contradiction.

\begin{remark*}
  The condition (a) cannot be dropped, since otherwise we may take
  $S = \left\{ 2 \pmod p \right\}$ or $S = \left\{ 0 \pmod p \right\}$, say.
\end{remark*}