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\begin{lemma*}
  Let $f, P \in \Zbb[X, Y]$ without common factors
  in $\Zbb[X, Y]$. Then the system of equations
  $f(X, Y) = P(X, Y) = 0$ has only finitely many
  solutions.
\end{lemma*}

\begin{proof}
  $\Zbb[X, Y] \cong \Zbb[X][Y]$ (poynomials in $Y$
  with coefficients in $\Zbb[X]$). $f, P$ have no
  common factors in $\Zbb[X][Y]$. Then Gauss's
  lemma gives us that they have no common factors
  in $\Qbb(X)[Y]$. This is because $\Zbb[X]$ is a
  UFD and $\Qbb(X)$ is its quotient field.

  Since $\Qbb(X)[Y]$ is a Euclidean domain, there
  exists $F, G \in \Qbb(X)$ such that
  \[
    F \cdot P + G \cdot f = 1
  .\]
  Multiply by the common denominator $D$ of $F,
  G$, and we get
  \[
    \tilde{F} \cdot P + \tilde{G} \cdot f = D(X)
  \]
  for some $\tilde{F}, \tilde{G} \in \Zbb[X]$.
  Hence the common solutions of $f = P = 0$ has
  finitely many $X$-coordinates. Then swap $X$ and
  $Y$.
\end{proof}

\begin{theorem*}
  Let $F \in \Zbb[X, Y]$ homogeneous of degree
  $d$, without repeated factors. Let $G \in
  \Zbb[X, Y]$ of degree $< d$. Assume $F - G$ is
  irreducible in $\Zbb[X, Y]$. Then there are at
  most finitely many solutions of $F(X, Y) = G(X,
  Y)$ with $X, Y \in \Zbb$.
\end{theorem*}

$F(X, Y) = (X - \alpha_1 Y) \cdots (X - \alpha_d
Y)$. $\Gamma = \Gamma_0 \cup \Gamma_1 \cup \cdots
\cup \Gamma_d$. $P = F - G$, $I = P \cdot
\ol{\Qbb}[X, Y]$. $V = \ol{\Qbb}[X, Y]^{(D)} / I
\cap \ol{\Qbb}[X, Y]^{(D)}$. $\ord_{p_j}(f) =
\sup(t \in \Zbb : f(X, Y) \cdot Y^t \text{ bounded
on $\Gamma_j$})$. $n = \dim V$.

$\forall j \exists l_1, \ldots, l_n \in V$ a basis
such that $\ord_{p_j}(l_i) \ge -D + i - 1$. $n =
\dim V > dD - d(d - 1)$.

\textbf{Subspace Theorem:} Let $V$ be a vector
space of dimension $n$ over $\ol{\Qbb}$. Let $l_1,
\ldots, l_n$, $l_1^{(0)}, \ldots, l_n^{(0)} \in V$
be two bases. $\forall \eps > 0$
there exists $f_1, \ldots, f_m \in V_{\neq 0}$
such that $\forall \varphi \in V^*$ that satisfies
\[
  \prod_{i = 1}^{n} |\varphi(l_j)|
  \le H(\ub{\varphi(l_1^{(0)})}_{\in \Zbb}, \ldots,
  \ub{\varphi(l_n^{(0)})}_{\in \Zbb})^{-\eps}
\]
then $\varphi(f_j) = 0$ for some $j = 1, \ldots,
m$.

\begin{proof}[Proof of Schinzel's Theorem]
  We show that $\Zbb^2 \cap \Gamma_j$ is finite
  for any $j = 1, \ldots, d$. Let $l_1, \ldots,
  l_n \in V$ be a basis with $\ord(l_i) \ge -D + i
  - 1$. Then
  \begin{align*}
    \prod_{i = 1}^{n} |l_i(X, Y)|
    &\le C \cdot Y^{\sum -\ord_{p_j}(l_i)}
    &&\text{on $\Gamma_j$} \\
    &\le C \cdot Y^{-D - 1}
  \end{align*}
  if $n \ge 2D + 2$. We set $D$ to be large enough
  so that this holds.

  Recall the reference basis $l_j^{(0)}$ are
  suitable monomials of degree $\le D$, so
  \[
    |l_i^{(0)} (X, Y)| < CY^D
  .\]
  Then for $x, y \in \Zbb^2 \cap \Gamma_j$, we
  have:
  \[
    H(l_i^{(0)}(x, y), \ldots, l_n^{(0)}(x, y))
    \le C \cdot |Y|^D
  .\]
  Hence
  \[
    \prod_{i = 1}^{n} |l_i(x, y)|
    < H(l_1^{(0)}(x, y), \ldots)^{-1}
  \]
  provided $y$ is still large.

  By the subspace theorem, $f_i(x, y) = 0$ for
  some $i = 1, \ldots, m$.

  To apply the lemma, we need $f_i \in \Zbb[X,
  Y]$. This can be assumed: indeed, multiplying
  $f_i$ by an element of $\ol{\Qbb}$, we can make
  the leading coefficient to be in $\Zbb$, and all
  other coefficients will be algebraic integers.
  Then replace $f_i$ by the sum of its Galois
  conjugates.
\end{proof}

\begin{theorem*}
  For $q \in \Zbb_{> 0}$ with $\gcd(q, 6) = 1$, we
  write $\ord(q)$ for the order of the
  multiplicative group generated by $2, 3$ in
  $\Zbb / q\Zbb$.

  Then:
  \[
    \lim_{q \to \infty} \frac{\ord(q)}{(\log q)^2}
    = \infty
  .\]
\end{theorem*}

\begin{remark*}
  $2^n 3^m$ for $n < \half \log_2 q$, $m < \half
  \log_3 q$. Hence
  \[
    \ord(q) \ge \left( \half \log_2 q \right)
    \left( \half \log_3 q \right)
  .\]
\end{remark*}

\begin{theorem*}[Corvaja, Zannier; Hern\'andez,
  Luca]
  Write $\mathcal{S} = \{2^n 3^m : n, m \in
  \Zbb_{\ge 0}\}$. Then for all $\eps > 0$, there
  are only finitely many pairs of multiplicatively
  independent $a, b \in \mathcal{S}$ such that
  \[
    \gcd(a - 1, b - 1) \ge \max(a, b)^\eps
  .\]
\end{theorem*}

$a, b$ are multiplicatively independent if there
does not exist $n, m \in \Zbb$ such that $a^n =
b^m$.

\textbf{Fact:} there exist infinitely many $n$
such that
\[
  \gcd(2^n - 1, 3^n - 1) \ge 3^{n^{c / \log \log n}}
.\]
