Problem: Suppose we have distinct positive integers $a$, $b$, $c$, $d$
and an odd prime $p$ not dividing any of them,
and an integer $M$ such that if one considers the infinite sequence
\begin{align*}
  ca &- db \\
  ca^2 &- db^2 \\
  ca^3 &- db^3 \\
  ca^4 &- db^4 \\
  &\vdots
\end{align*}
and looks at the highest power of $p$ that divides each of them,
these powers are not all zero, and are all at most $M$.
Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$)
such that whenever $p$ divides an element of this sequence,
the maximum power of $p$ that divides that element is exactly $p^T$.

\end{enumerate}

Solution: By orders, the indices of terms divisible by $p$
is an arithmetic subsequence of $\NN$:
say they are $\kappa$, $\kappa + \lambda$, $\kappa + 2\lambda$, \dots,
where $\lambda$ is the order of $a/b$.
That means we want
\[ \nu_p \left( c a^{\kappa + n\lambda} - d b^{\kappa + n \lambda} \right)
  = \nu_p \left(
    \left( \frac{a^\lambda}{b^\lambda} \right)^n
    - \frac{d a^\kappa}{c b^{\kappa}} \right)
\]
to be constant.
Thus, we have reduced the problem to the following proposition:
\begin{proposition*}
  Let $p$ be an odd prime.
  Let $x, y \in \QQ_{>0}$ such that $x \equiv y \equiv 1 \pmod p$.
  If the sequence $\nu_p\left( x^n - y \right)$ of positive integers
  is nonconstant, then it is unbounded.
\end{proposition*}

For this it would be sufficient to prove the following claim.
\begin{claim*}
  Let $p$ be an odd prime.
  Let $x, y \in \QQ_{>0}$ such that $x \equiv y \equiv 1 \pmod p$.
  Suppose $m$ and $n$ are positive integers such that
  \[ d = \nu_p(x^n-y) < \nu_p(x^m-y) = e. \]
  Then there exists $\ell$ such that $\nu_p(x^\ell - y) \ge e + 1$.
\end{claim*}
\begin{proof}
First, note that
$\nu_p(x^m - x^n) = \nu_p\left( (x^m-y) - (x^n-y) \right) = d$
and so by exponent lifting we can find \emph{some} $k$ such that
\[ \nu_p(x^k-1) = e \]
namely $k = p^{e-d}|m-n|$.
(In fact, one could also choose more carefully
$k = p^{e-d} \cdot \gcd(m-n, p^\infty)$,
so that $k$ is a power of $p$.)

Suppose we set $x^k = p^e u + 1$ and $x^m = p^e v + y$
where $u,v \in \QQ$ aren't divisible by $p$.
Now for any integer $1 \le r \le p-1$ we consider
\begin{align*}
  x^{kr+m}-y &= (p^e u + 1)^r \cdot (p^e v + y) - y \\
  &= p^e \left( v + yu \cdot r \right) + p^{2e} \left( \dots \right).
\end{align*}
By selecting $r$ with $r \equiv -v/u \pmod p$,
we ensure $p^{e+1} \mid x^{kr+m} - y$,
hence $\ell = kr+m$ is as desired.
\end{proof}

\begin{remark*}
  One way to motivate the proof of the claim is as follows.
  Suppose we are given $\nu_p(x^m-y) = e$,
  and we wish to find $\ell$ such that $\nu_p(x^\ell - y) > e$.
  Then, it is necessary (albeit insufficient) that
  \[ x^{\ell-m} \equiv 1 \pmod{p^e} \text{ but }
    x^{\ell-m} \not\equiv 1 \pmod{p^{e+1}}. \]
  In particular, we need $\nu_p(x^{\ell-m}-1) = e$ exactly.
  So the $k$ in the claim must exist if we are going to succeed.

  On the other hand, if $k$ is \emph{some} integer for which
  $\nu_p(x^k-1) = e$, then by choosing $\ell-m$ to be some
  multiple of $k$ with no extra factors of $p$,
  we hope that we can get $\nu_p(x^\ell-y) = e+1$.
  That's why we write $\ell = kr+m$ and see what happens when we expand.
\end{remark*}