$\lfloor (2+\sqrt{3})^n \rfloor $ is odd

Question

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:

$$ u_{n}=(2+\sqrt{3})^n+(2-\sqrt{3})^n=\sum_{k=0}^n{n \choose k}2^{n-k}(3^{k/2}+(-1)^k3^{k/2})\in\mathbb{2N} $$

$$ 0\leq (2-\sqrt{3})^n \leq1$$

$$ (2+\sqrt{3})^n\leq u_{n}\leq 1+(2+\sqrt{3})^n $$

$$ (2+\sqrt{3})^n-1\leq u_{n}-1\leq (2+\sqrt{3})^n $$

$$ \lfloor (2+\sqrt{3})^n \rfloor=u_{n}-1\in\mathbb{2N}+1 $$

Answer 1

Hint for the first part: Consider $u_n = (2+\sqrt{3})^{n} + (2-\sqrt{3})^{n}$. Prove that $u_n$ is always an even integer and that $u_n = \lceil (2+\sqrt{3})^n \rceil$. Use that $(2-\sqrt{3})^{n}\to 0$.

(This has now been incorporated into the edited question.)

Hint for the second part: Consider $v_n = (1+\sqrt{3})^{n} + (1-\sqrt{3})^{n}$. Find a second-order recursion for $v_n$ based on the quadratic equation that defines $1\pm\sqrt{3}$.

Answer 2

You can use recurrences such as $$f(n)=4f(n-1)-f(n-2)+2$$ or $$f(n)=5f(n-1)-5f(n-2)+f(n-3)$$ starting at $f(0)=1, f(1)=3$.

Then show the various results by induction.