Let $n \in \mathbb{N}$, show that $$f(n) = \frac{(2+\sqrt{3})^{1+2n}+(2\sqrt{3})^{1+2n}+2}{6}$$ is a square.
For $n=1$ to $5$ we have ($n, \ \sqrt{f(n)}, \ f(n)$):
$\begin{array}{lll}
1. & 3. & 9. \\
2. & 11. & 121. \\
3. & 41. & 1681. \\
4. & 153. & 23409. \\
5. & 571. & 326041.
\end{array}$
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