Problem: Suppose $a_1 < a_2 < \dots < a_{2024}$ is an arithmetic sequence of positive integers,
and $b_1 < b_2 < \dots < b_{2024}$ is a geometric sequence of positive integers.
Find the maximum possible number of integers that could appear in both sequences,
over all possible choices of the two sequences.

Solution: \paragraph{Answer.} $11$ terms.
\paragraph{Construction.} Let $a_i = i$ and $b_i = 2^{i-1}$.
\paragraph{Bound.}
We show a $\nu_p$-based approach communicated by Derek Liu,
which seems to be the shortest one.
At first, we completely ignore the geometric sequence $b_i$
and focus only on the arithmetic sequence.

\begin{claim*}
  Let $p$ be any prime, and consider the sequence
  \[ \nu_p(a_1), \; \nu_p(a_2), \; \dots, \; \nu_p(a_{2024}). \]
  Set $C \coloneq \left\lfloor \log_p(2023) \right\rfloor$.
  Then there are at most $C+2$ different values in this sequence.
\end{claim*}
\begin{proof}
  By scaling, assume the $a_i$ do not have a common factor,
  so that $a_i = a + di$ where $\gcd(a,d) = 1$.
  \begin{itemize}
    \ii If $p \mid d$, then $p \nmid a$ and $\nu_p(a_i)$ is constant.

    \ii Otherwise, assume $p \nmid d$.
    We will in fact prove that every term in the sequence
    is contained in $\{0, 1, \dots, C\}$ with at most one exception.

    Define $M \coloneq \max_i \nu_p(a_i)$.
    If $M \le C$, there is nothing to prove.
    Otherwise, fix some index $m$ such that $\nu_p(a_m) = M$.
    We know $\nu_p(i-m) \le C$ since $|i-m| \le 2023$.
    But now for any other index $i \neq m$;
    \begin{align*}
      \nu_p(d(i-m)) &= \nu_p(i-m) \le C < M = \nu_p(a_m) \\
      \implies \nu_p(a_i) &= \nu_p(a_m + d(i-m)) = \nu_p(i-m) \le C.
    \end{align*}
    so $\nu_p(a_m)$ is the unique exceptional term of the sequence exceeding $C$.
    \qedhere
  \end{itemize}
\end{proof}

\begin{remark*}
  The bound in the claim is best possible by taking
  $a_i = p^M  + (i-1)$ for any $M > C$.
  Then indeed, the sequence $\nu_p(a_i)$ takes on values in $\{0,1,\dots,C\}$ for $i>1$
  while $\nu_p(a_1) = M$.
\end{remark*}

Back to the original problem with $(b_i)$.
Consider the common ratio $r \in \QQ$ of the geometric sequence $(b_i)$.
If $p$ is any prime with $\nu_p(r) \neq 0$,
then every term of $(b_i)$ has a different $\nu_p$.
So there are two cases to think about.
\begin{itemize}
  \ii Suppose any $p \ge 3$ has $\nu_p(r) \neq 0$.
  Then there are at most
  \[ 2 + \log_p(2023) = 2 + \log_3(2023) \approx 8.929 < 11 \]
  overlapping terms, as needed.

  \ii Otherwise, suppose $r$ is a power of $2$ (in particular, $r \ge 2$ is an integer).
  We already have an upper bound of $12$; we need to improve it to $11$.

  As in the proof before, we may assume WLOG by scaling down by a power of $2$
  that the common difference of $a_i$ is odd.
  (This may cause some $b_i$ to become rational non-integers, but that's OK.
  However, unless $\nu_2(a_i)$ is constant, the $a_i$ will still be integers.)

  Then in order for the bound of $12$ to be achieved,
  the sequence $\nu_2(a_i)$ must be contained in $\{0,1,\dots,10,M\}$ for some $M \ge 11$.
  In particular, we only need to work with $r = 2$.

  Denote by $b$ the unique odd-integer term in the geometric sequence,
  which must appear among $(a_i)$.
  Then $2b$ appears too, so the common difference of $a_i$ is at most $b$.

  But if $(a_i)$ is an arithmetic progression of integers that contains $b$
  and has common difference at most $b$,
  then no term of the sequence can ever exceed $b + 2023 \cdot b = 2024b$.
  Hence $2^M b$ cannot appear for any $M \ge 11$. This completes the proof.
\end{itemize}

\begin{remark*}
  There are several other approaches to the problem, but most take some time to execute.
  The primary issue is that the common difference of the $a_i$'s could
  share prime factors with the common ratio in the $b_i$'s,
  which means that merely trying to write out a lot of modular arithmetic equations
  leads to a lot of potential technical traps that are not pleasant to defuse.

  One unusual thing is that many solutions end up proving a bound of $12$
  (in the case the common ratio is $2$) and then having to adjust it to $11$ later.
\end{remark*}