Problem: Let $a$ and $b$ be positive integers.
Show that if $4ab - 1$ divides $(4a^{2} - 1)^{2}$, then $a = b$.

Solution: As usual,
\[ 4ab-1 \mid (4a^2-1)^2 \iff 4ab-1 \mid (4ab \cdot a-b)^2
  \iff 4ab-1 \mid (a-b)^2. \]
Then we use a typical Vieta jumping argument.
Define \[ k = \frac{(a-b)^2}{4ab-1}. \]
Note that $k = 0 \iff a = b$.
So we will prove that $k > 0$ leads to a contradiction.

Indeed, suppose $(a, b)$ is a minimal solution with $a > b$
(we have $a \neq b$ since $k \neq 0$).
By Vieta jumping, $(b, \frac{b^2+k}{a})$ is also such a solution.
But now
\begin{align*}
  \frac{b^2+k}{a} \ge a &\implies k \ge a^2 - b^2 \\
  &\implies \frac{(a-b)^2}{4ab-1} \ge a^2-b^2 \\
  &\implies a-b \ge (4ab-1)(a+b)
\end{align*}
which is absurd for $a,b \in \ZZ_{>0}$.
(In the last step we divided by $a-b > 0$.)