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 $a^2 + 2 b^2$ ($a, b \in \mathbb{N}$).

Question

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?

Answer 1

Let $f(n)= \dfrac{3+\sqrt{3}}{6} (2+\sqrt{3})^n + \dfrac{3-\sqrt{3}}{6} (2-\sqrt{3})^n$ and $g(n)=\dfrac{\sqrt3} 6(2+\sqrt3)^n-\dfrac{\sqrt3}6(2-\sqrt3)^n$.

Can you show $f(n)^2+2\times g(n)^2=f(2n)$ and $f(n)^2+2\times g(n+1)^2=f(2n+1)?$

Answer 2

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.