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``generalised $S$-unit equations''.

Let $K$ be a number field: $\mathcal{O}_K = \{x
\in K : |x|_v \le 1 \text{ for all $v \in M_{K,
f}$}\}$. Let $S \in M_K$ be a finite set
containing $M_{K, \infty}$: $\mathcal{O}_{K, s} =
\{x \in K : |x|_v \le 1 \text{ for all $v \notin
S$}\}$ (``$S$-integers''). $\mathcal{O}_{K,
s}^\times$ units in $\mathcal{O}_{K, s}$
(``$S$-units'').

Unit eqution $x + y = 1$ with $x, y$ units.

\begin{proof}[Proof (continued)]
  Induction on $d$. $d = 2$ was checked before.

  Suppose $d > 2$, and the claims hold for $d -
  1$. We make some simplifying assumptions to be
  specified later. We apply the subspace theorem
  on $\Qbb^{d - 1} = V$. The reference basis is
  $\Lambda_j^{(0)} = X_j$, $j = 1, \ldots, d - 1$.
  As a first approximation, we try
  $\Lambda_j^{(v)} = X_j$ for all $j, v$.
  Let $S = \{\infty, 2, 3\}$.
  Let $x = (x_1, \ldots, x_d)$ be a solution of
  $L(x_1, \ldots, x_d) = 0$. Then
  \[
    \prod_{v \in S} \prod_{j = 1}^{d - 1}
    |\Lambda_j^{(v)}(x)|_v = 1
  .\]
  We can replace $\Lambda_1^{(w)}$ by
  \[
    \frac{a_1}{a_n} X_1 + \cdots + \frac{a_{d -
    1}}{a_d} X_{d - 1}
  ,\]
  where $L = a_1 X_1 + \cdots + a_d X_d$. Then we
  replace $|x_1|_w$ by $|x_n|_w$. We do this for
  some choice $w$.

  Now back to the simplifying assumptions:
  We assume that $|x|_\infty$ is maximal for $j =
  n$. Then $|x_n|_2 |x_n|_3 = |x_n|_\infty^{-1}$.
  So let $w \in \{2, 3\}$ such that $|x_n|_w \le
  |x_n|_\infty^{-\half}$. We may also assume
  $|x_1|_w = 1$. For this, we may need to divide
  $x$ by the common divisor, and rearrange the
  indices.

  For these augmented $\Lambda_j^{(v)}$'s, we get
  \[
    \prod_{v \in S} \prod_{j = 1}^{d - 1}
    |\Lambda_j^{(v)}(x)|_v \le |x_n|_\infty^{-\half}
    \le H(\Lambda_1^{(0)}(x), \ldots, \Lambda_{d -
    1}^{(0)}(x))^{-\half}
  .\]
  So the subspace theorem applies with $\eps =
  \half$. So $x_1, \ldots, x_{d - 1}$ satisfies
  one of finitely many linear equations. Apply the
  induction hypothesis for each of them.
\end{proof}

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

\begin{proof}
  Fix some $\eps > 0$. Let $a, b \in \mathcal{S}$
  multiplicatively independent and such that
  \[
    d = \gcd(a - 1, b - 1) > \max(a, b)^\eps
  .\]
  Our goal is to show $d < C$ for some $C =
  C(\eps)$. Note: $2, 3 \nmid d$, because
  otherwise $2 \nmid a, b$ or $3 \nmid a, b$. Then
  $a$ and $b$ would be a power of the same prime.
  Not possible due to multiplicative independence.

  Fix some $n \in \Zbb_{>0}$ sufficiently large
  depending on $\eps$. We apply the subspace
  theorem on $V = \Qbb^{n^2} / \{(x, \ldots, x) :
  x \in \Qbb\}$.

  We will evaluate our functionals at the point
  $e / d = (e_1 / d, \ldots, e_{n^2} / d)$ where $e_1,
  \ldots, e_{n^2}$ is an enumeration of $a^k b^l$
  for $k = 0, \ldots, n - 1$, $l = 0, \ldots, n -
  1$ such that $e_1 = 1$, $e_{n^2} = a^{n - 1}
  b^{n - 1}$.

  Note: $\frac{e_i}{d} - \frac{e_j}{d} \in \Zbb$.
  This is because $e_i \equiv 1 \pmod d$. Also:
  $\left| \frac{e_i}{d} - \frac{e_j}{d} \right|_v
  \le \min \left( \left| \frac{e_i}{d} \right|_v,
  \left| \frac{e_j}{d} \right|_v \right)$ for all
  $v \in S = \{\infty, 2, 3\}$.
  The coordinates on $\Qbb^{n^2}$ will be denoted
  by $Y_1, \ldots, Y_{n^2}$. All our linear forms
  on $V$ will be of the form $Y_i - Y_j$ for some
  $i \neq j$. This is indeed well defined on $V$.
  Reference basis $\Lambda_j^{(0)} = Y_j -
  Y_{n^2}$.
  \[
    H\left(\Lambda_1^{(0)} \left( \frac{e}{d}
    \right), \ldots, \Lambda_{n^2 - 1}^{(0)}
    \left( \frac{e}{d} \right) \right)
    \le a^n b^n
  .\]
  For $v = \infty$:
  \begin{align*}
    \Lambda_j^{(\infty)}
    &= Y_{j + 1} - Y_1 \\
    |\Lambda_j^{(\infty)}(e / d)|
    &= |e_{j + 1} / d|_{\infty} \\
    \Lambda_j^{(v)}
    &= Y_{j} - Y_{n^2} \\
    |\Lambda_j^{(v)}(e / d)|_v
    &= |e_j|_v \\
    \prod_{j = 1}^{n^2 - 1}
    |\Lambda_j^{(\infty)}(e / d)|_\infty
    &\le \left( \prod_{j = 1}^{n^2} |e_j /
    d|_\infty \right) \cdot d \\
    \prod_{j = 1}^{n^2 - 1} |\Lambda_j^{(v)}(e /
    d)|_v
    &\le \left( \prod_{j = 1}^{n^2} |e_j / d|_v
    \right) / |a^{n - 1} b^{n - 1}|_v \\
    \prod_{v \in S} \prod_{j = 1}^{n^2 - 1}
    |\Lambda_j(e / d)|
    &\le d \cdot (a^{n - 1} b^{n - 1}) \cdot
    d^{-n^2} \\
    |e_i / d|_\infty
    |e_j / d|_2 |e_j / d|_3
    &= \frac{1}{d}
  \end{align*}