Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$ 
Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$ where $n$ is a positive integer.

We have \begin{align*}\lfloor(1+\sqrt{3})^{2n-1}\rfloor &= (\sqrt{3}+1)^{2n-1}-(\sqrt{3}-1)^{2n-1}\\&=\dfrac{(4+2\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(4-2\sqrt{3})^n}{1-\sqrt{3}}\\&=2^n\left(\dfrac{(2+\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(2-\sqrt{3})^n}{1-\sqrt{3}}\right).\end{align*} Thus we must show that the second term is an integer. How can we continue?

Set $x_n=\frac{(2+\sqrt{3})^n}{1+\sqrt{3}}+\frac{(2-\sqrt{3})^n}{1-\sqrt{3}}$. Show that $x_1, x_2$ are integer and $x_{n+2}=4x_{n+1}-x_{n}$ holds, thus by induction $x_n$ is an integer for all $n \in \mathbb{N}$.
