Problem: Let $s_1, s_2, s_3, \dots$ be an infinite,
nonconstant sequence of rational numbers,
meaning it is not the case that $s_1 = s_2 = s_3 = \dots$.
Suppose that $t_1, t_2, t_3, \dots$ is also an infinite,
nonconstant sequence of rational numbers
with the property that $(s_i - s_j)(t_i - t_j)$
is an integer for all $i$ and $j$.
Prove that there exists a rational number $r$
such that $(s_i - s_j)r$ and $(t_i - t_j)/r$
are integers for all $i$ and $j$.

\end{enumerate}

Solution: First we eliminate the silly edge case:
\begin{claim*}
  For some $i$ and $j$, we have
  $(s_i - s_j)(t_i - t_j) \neq 0$.
\end{claim*}
\begin{proof}
  Assume not.
  WLOG $s_1 \neq s_2$, then $t_1 = t_2$.
  Now select $i$ such that $t_i \neq t_1 = t_2$.
  Then either $t_i - s_1 \neq 0$
  or $t_i - s_2 \neq 0$, contradiction.
\end{proof}

So, WLOG (by permutation) that $n = (s_1-s_2)(t_1-t_2) \neq 0$.
By shifting and scaling appropriately,
we may assume
\[ s_1 = t_1 = 0, \quad s_2 = 1, \quad t_2 = n. \]
Thus we deduce \[ s_i t_i \in \ZZ, \quad s_i t_j + s_j t_i \in \ZZ
  \qquad \forall i,j. \]

\begin{claim*}
  For any index $i$,
  $t_i \in \ZZ$, $s_i \in \frac 1n \ZZ$.
\end{claim*}
\begin{proof}
  We have $s_i t_i \in \ZZ$
  and $t_i + n s_i \in \ZZ$ by problem condition.
  By looking at $\nu_p$ of this,
  we conclude $ns_i, t_i \in \ZZ$
  (since if either as negative $p$-adic value,
  so does the other, and then $s_i t_i \notin \ZZ$).
\end{proof}

Last claim:
\begin{claim*}
  If $d = \gcd t_\bullet$, then $ds_i \in \ZZ$ for all $i$.
\end{claim*}
\begin{proof}
  Consider a prime $p \mid n$, and let $e = \nu_p(t_j)$.
  We will show $\nu_p(s_i) \ge -e$ for any $i$.

  This is already true for $i = j$,
  so assume $i \neq j$.
  Assume for contradiction $\nu_p(s_i) < -e$.
  Then $\nu_p(t_i) > e = \nu_p(t_k)$.
  Since $\nu_p(s_k) \ge -e$ we deduce
  $\nu_p(s_i t_k) < \nu_p(s_k t_i)$;
  so $\nu_p(s_i t_k) \ge 0$ and $\nu_p(s_i) \ge -e$ as desired.
\end{proof}