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\begin{theorem*}[1]
  If $2, 3 \nmid q$, then
  \[
    \frac{\ord(q)}{(\log q)^2} \to \infty
  .\]
\end{theorem*}

\begin{theorem*}[2]
  For all $\eps > 0$, there are only finitely many
  pairs of multiplicatively independent $a, b \in
  \mathcal{S}$ such that
  \[
    \gcd(a - 1, b - 1) > \max(a, b)^\eps
  .\]
\end{theorem*}

\begin{proof}[Proof of Theorem 1 using Theorem 2]
  Let
  \[
    \Lambda = \{(n, k) \in \Zbb^2 : 2^n \cdot 2^k
    \equiv \pmod{q}\}
  .\]
  This is a subgroup of $\Zbb^2$, and $|\Zbb^2 /
  \Lambda\ = \ord(q)$ The volume $\Rbb^2 /
  \Lambda$ is $\ord(q)$.

  Our aim is to find $(n_1, k_1), (n_2, k_2) \in
  \Lambda \cap \Zbb_{\ge 0}^2$ linearly
  idnependent and $n_1, k_1, n_2, k_2 \le C
  \ord(q) / \log q$, where $C$ is absolute.

  If we can do this, then: $q \mid \gcd(2^{n_1}
  2^{k_1}  - 2^{n_2} 3^{k_2} - 1)$. By Theorem
  (2), since $2^{n_1} 3^{k_1} - 1$ and $2^{n_2}
  3^{k_2} - 1$ are multiplicatively independent,
  we would get
  \begin{align*}
    q
    &< \max(2^{n_1} 3^{k_1}, 2^{n_2} 3^{k_2})^\eps
    \\
    &< \exp(C \ord(q) / \log q)^\eps
  \end{align*}
  Taking $\log$:
  \begin{align*}
    \log q
    &< C \cdot \eps \ord(q) / \log q \\
    \ord(q)
    &> C^{-1} \cdot \eps^{-1} \cdot (\log q)^2
  \end{align*}
  provided $q$ is sufficiently large in terms of
  $\eps$.

  Now to the proof of the above stated aim: Let
  $(\tilde{n}_1, \tilde{k}_1), (\tilde{n}_2,
  \tilde{k}_3 \in \Lambda$ that generate $\Lambda$
  and such that their angle is as close to
  $\frac{\pi}{2}$ as possible.

  Then this angle is between $\frac{\pi}{3}$ and
  $\frac{2\pi}{3}$:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/d84f7ea40554416b.png}
  \end{center}
  The area of the parallelogram spanned by
  $(\tilde{n}_1, \tilde{k}_1)$ and $(\tilde{n}_2,
  \tilde{k}_2)$ is at least
  \[
    \frac{2}{\sqrt{3}} \|(\tilde{n}_1,
    \tilde{k}_1)\|_2 \|(\tilde{n}_2,
    \tilde{k}_2)\|_2 \le \ord(q)
  .\]
  Minkowski's second theorem in the geometry of
  numbers.

  We know that $q \mid 2^{|\tilde{n}_1|} \cdot
  3^{|\tilde{k}_1|} - 1$ or $q \mid
  2^{|\tilde{n}_1|} - 3^{|\tilde{k}_1|}$. Then:
  either $|\tilde{n}_1|$ or $|\tilde{k}_1|$ has to
  be $\ge \half \log_3(q)$. In particular:
  $\|(\tilde{n}_1, \tilde{k}_1)\|_2 \ge c \log q$
  (for some absolute constant $c$).

  Then $\|(\tilde{n}_1, \tilde{k}_1)\|_2,
  \|(\tilde{n}_2, \tilde{k}_2)\|_2 \le c
  \frac{\ord(q)}{\log q}$.
\end{proof}

\begin{proposition}
  Let $L \in \Qbb[X_1, \ldots, X_n]$ be a linear
  form. Then there exists $C = C(L)$ such that any
  solution $x_1, \ldots, x_n \in \mathcal{S}$ of
  $L(x_1, \ldots, x_n) = 0$ satisfies
  \[
    |x_i - x_i|_\infty |x_i - x_j|_2 |x_i - x_j|_3
    < C
    \tag{$*$}
  \]
  for some $i \neq j \in \{1, \ldots, n\}$.
\end{proposition}

\begin{remark*}
  For $x \in \Zbb$ such that $x = 2^n 3^k y$ with
  $n, k \in \Zbb_{\ge 0}$, $2, 3 \nmid y$, then
  \[
    |x|_\infty |x|_2 |x|_3 = |y|
  .\]
  Note that $(*)$ is invariant under
  multiplication by elements of $\mathcal{S}$.
\end{remark*}

\begin{theorem*}
  Let $V$ be a vector space of dimension $n$ over
  $\ol{\Qbb}$. Let $S \subset M_\Qbb$ be finite
  with $\infty \in S$. For each $v \in S$, let
  $\Lambda_1^{(v)}, \ldots, \Lambda_n^{(v)}$ be a
  basis of $V^*$. Furthermore, let
  $\Lambda_1^{(0)}, \ldots, \Lambda_n^{(0)}$ be
  another basis. Fix an extension of each
  $|\bullet|_v$ from $\Qbb$ to $\ol{\Qbb}$.

  Then for all $\eps > 0$, there are finitely many
  $\varphi_1, \ldots, \varphi_n \in V^*$ such that
  all solutions $x \in V$ of
  \[
    \prod_{v \in S} \prod_{j = 1}^{n}
    |\Lambda_j^{(v)}(x)|_v \le
    H(\Lambda_1^{(0)}(x), \ldots,
    \Lambda_n^{(0)}(x))^{-\eps}
  \]
  with $\Lambda_1^{(0)}(x), \ldots,
  \Lambda_n^{(0)}(x) \in \Zbb$ satisfy
  $\varphi_i(x) = 0$ for some $i = 1, \ldots, n$.
\end{theorem*}

\begin{proof}[Proof of Proposition]
  By induction on $n$. Suppose $n = 2$. As we
  observed the conclusion, is invariant under
  dividing $x_1, x_2$ by the same element of
  $\mathcal{S}$. Now $\gcd(x_1, x_2) \in
  \mathcal{S}$. So it is enough to prove for
  solutions with $\gcd(x_1, x_2) = 1$.

  Let $L(X_1, X_2) = aX_1 + bX_2$. Then $ax_1 +
  bx_2 = 0$ with $\gcd(x_1, x_2) = 1$ implies $x_1
  \mid b$ and $x_2 \mid a$.

  So there are finitely many possibilities for
  $x_1, x_2$ in terms of $L$. Pick $C$ that works
  for all.

  (to be continued).
\end{proof}