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?

Hint: $1 \pm \sqrt{3}$ are the roots of $t^2-2t-2=0\,$, so $\;a_{n} = (1+\sqrt{3})^{n}+(1-\sqrt{3})^{n}$ is the solution to the recurrence $a_{n+2}=2a_{n+1}+2a_{n}\,$ with $a_0 = a_1 = 2$.
