<s>[INST] How to write $[(2+\sqrt{3})^n + (2-\sqrt{3})^n + (2+\sqrt{3})^{n+1} + (2-\sqrt{3})^{n+1}]/6$ to the form $a^2 + 2 b^2$ ($a, b \in \mathbb{N}$). We know that $(2+\sqrt{3})^n + (2-\sqrt{3})^n$ is an integer (See here).
However, we want to write the formula
\begin{align}
&\frac{3+\sqrt{3}}{6} (2+\sqrt{3})^n + \frac{3-\sqrt{3}}{6} (2-\sqrt{3})^n\\
&=\frac{1}{6} \left[(2+\sqrt{3})^{n+1} + (2-\sqrt{3})^{n+1} + (2+\sqrt{3})^n + (2-\sqrt{3})^n\right]
\end{align}
to the form
$$a^2 + 2\,b^2,\ (a, b \in \mathbb{N}).$$
How?
 [/INST] Call your formula $f(n)$. Since $2\pm\sqrt{3}$ are the roots of $x^2-4x+1$, it's easy to show that $f(n)$ is  determined by $f(0)=1$, $f(1)=3$ and the recursive relation $f(n+1)=4f(n)-f(n-1)$.
After looking at @J.W.Tanner comment, I found out that we actually have
\begin{align*}
f(2n)&=f(n)^2+2\left(\sum_{0\leq k\leq n-1}f(k)\right)^2\\
f(2n+1)&=f(n)^2+2\left(\sum_{0\leq k\leq n}f(k)\right)^2
\end{align*}
Now, I don't really know how difficult is to prove this by induction, but it is an interesting pattern.
</s>