Let $n$ be a nonnegative integer. Show that $\lfloor (2+\sqrt{3})^n \rfloor $ is odd and that $2^{n+1}$ divides $\lfloor (1+\sqrt{3})^{2n} \rfloor+1 $.
My attempt:
$ u_{n}=(2+\sqrt{3})^n+(2-\sqrt{3})^n=\sum_{k=0}^n{n \choose k}2^{n-k}(3^{k/2}+(-1)^k3^{k/2})\in\mathbb{2N} $
$ 0\leq (2-\sqrt{3})^n \leq1$
$ (2+\sqrt{3})^n\leq u_{n}\leq 1+(2+\sqrt{3})^n $
$ (2+\sqrt{3})^n-1\leq u_{n}-1\leq (2+\sqrt{3})^n $
$ \lfloor (2+\sqrt{3})^n \rfloor=u_{n}-1\in\mathbb{2N}+1 $