Why is $ \frac{5}{64}((161+72\sqrt{5})^{-n}+(161+72\sqrt{5})^{n}-2)$ always a perfect square? I'm working on a puzzle, and the solution requires me somehow establishing that
$$ f(n):=\frac{5}{64}\Big(\big(161+72\sqrt{5}\big)^{-n}+\big(161+72\sqrt{5}\big)^{n}-2\Big)$$
is a perfect square for $n\in \mathbb{Z}_{\geq 0}$.
I've done a lot of simplification to get to this point, and am stuck here. I can provide the context of the puzzle if necessary, but it's pretty far removed from what I have here. The goal is basically to show that a formula generates solutions to a given equation.
Any tips on how to proceed?
Here's the first few values:
$$\begin{array}{|c|c|}
\hline
n&\text{value}\\ \hline
0&0\\ \hline
1& 5^2 \\ \hline
 2&90^2 \\ \hline
 3& 1615^2\\ \hline
 4& 28980^2\\ \hline
\end{array}$$

Note that $161+72\sqrt{5} = (161-72\sqrt{5})^{-1}$ and $161+72\sqrt{5}=(9+4\sqrt{5})^2,$ and therefore
\begin{align*}
\big(161+72\sqrt{5}\big)^{-n}+\big(161+72\sqrt{5}\big)^{n}-2 &= \big( (161+72\sqrt{5})^{n/2} - (161-72\sqrt{5})^{n/2} \big)^2 \\
&= \big( (9+4\sqrt{5})^n - (9-4\sqrt{5})^n \big)^2.
\end{align*}
It therefore suffices to prove that $(9+4\sqrt{5})^n - (9-4\sqrt{5})^n$ is always $\sqrt5$ times an integer that is a multiple of $8$, which can be done using the binomial expansions of $(9\pm 4\sqrt{5})^n$.
