$\lfloor (2+\sqrt{3})^n \rfloor $ is odd 
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:
$$ u_{n}=(2+\sqrt{3})^n+(2-\sqrt{3})^n=\sum_{k=0}^n{n \choose k}2^{n-k}(3^{k/2}+(-1)^k3^{k/2})\in\mathbb{2N} $$
$$ 0\leq (2-\sqrt{3})^n \leq1$$
$$ (2+\sqrt{3})^n\leq u_{n}\leq 1+(2+\sqrt{3})^n $$
$$ (2+\sqrt{3})^n-1\leq u_{n}-1\leq (2+\sqrt{3})^n $$
$$ \lfloor (2+\sqrt{3})^n \rfloor=u_{n}-1\in\mathbb{2N}+1 $$
 A: 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. 
A: For the first claim:
Here is a more general discussion.
As is often the case,
none of this is original,
and I did work it out
on my own for fun.
To make one root of
$f(x)
=x^2-2ax+b
=0$
with
$a, b > 0$
in $(0, 1)$.
The roots are
$x_{\pm}
=\dfrac{2a\pm\sqrt{4a^2-4b}}{2}
=a\pm\sqrt{a^2-b}
$.
$\begin{array}\\
x_{-}
&=a-\sqrt{a^2-b}\\
&=(a-\sqrt{a^2-b})\dfrac{a+\sqrt{a^2-b}}{a+\sqrt{a^2-b}}\\
&=\dfrac{a^2-(a^2-b)}{a+\sqrt{a^2-b}}\\
&=\dfrac{b}{a+\sqrt{a^2-b}}\\
\end{array}
$
So we want
$0 < b < a+\sqrt{a^2-b}
$
so that
$ b < a^2$
and
$(b-a)^2 < a^2-b
$
or
$b^2-2ab+a^2 < a^2-b
$
or
$b^2+b < 2ab
$
or
$b< 2a-1$.
If $a=2$ then
$b < 3$
so, if $b$ is an integer,
$b \in \{1, 2\}
$
so
$a^2-b
\in \{3, 2\}
$
so
$\{x_-, x_+\}
=2\pm\sqrt{3},
2\pm\sqrt{2}
$.
Let $c = a^2-b$.
$\begin{array}\\
x_+^n+x_-^n
&=(a+\sqrt{c})^n+(a-\sqrt{c})^n\\
&=\sum_{k=0}^n \binom{n}{k}a^kc^{(n-k)/2}+\sum_{k=0}^n \binom{n}{k}a^k(-1)^{n-k}c^{(n-k)/2}\\
&=\sum_{k=0}^n \binom{n}{k}a^kc^{(n-k)/2}(1+(-1)^{n-k})\\
&=\sum_{k=0}^n \binom{n}{k}a^{n-k}c^{k/2}(1+(-1)^{k})\\
&=\sum_{k=0}^{\lfloor n/2 \rfloor}2\binom{n}{2k}a^{n-2k}c^{k}\\
&=2\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}a^{n-2k}c^{k}\\
\end{array}
$
Since
$0 < x_-^n < 1$.
$\lfloor x_+^n-1 \rfloor
$
is odd and
$\lfloor x_+^n\rfloor
=2\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}a^{n-2k}c^{k}-1
$
so
$\begin{array}\\
x_+^n-\lfloor x_+^n \rfloor
&=x_+^n-2\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}a^{n-2k}c^{k}+1\\
&=1-x_-^n\\
&\to_- 1\\
\end{array}
$
If $a=2, b=1$
then
$x_+^n-\lfloor x_+^n \rfloor
=1-(2-\sqrt{3})^n
=1-\dfrac{1}{(2+\sqrt{3})^n}
$.
A: 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 $1\pm\sqrt{3}$.
A: It is not hard to prove that$$(\forall n\in\mathbb N):\left(2+\sqrt3\right)^n+\left(2-\sqrt3\right)^n\in2\mathbb N.$$This, together with the fact that $2-\sqrt3\in(0,1),$ is enough to prove that $\left\lfloor\left(2+\sqrt3\right)^n\right\rfloor$ is an odd integer.
