Divisibility problem involving the floor function I am trying to prove the following:

$$2^{m+1}\mid\lfloor(1+\sqrt{3})^{2m+1} \rfloor$$ but $2^{m+2}\nmid\lfloor(1+\sqrt{3})^{2m+1} \rfloor$ for every natural number $m$.

I tried to use induction on $m$ but am not sure how to use the induction hypothesis. (The base case is easy to verify). Is there some other way to approach this problem?
Thanks.
 A: Consider the recurrence $x_0=2, x_1=20$ and $x_n=8x_{n-1}-4x_{n-2}$ for $n \geq 2$. It has characteristic polynomial $x^2-8x+4$, with roots $4 \pm 2\sqrt{3}=(1 \pm \sqrt{3})^2$, so $x_n=a(1+\sqrt{3})^{2n}+b(1-\sqrt{3})^{2n}$ for some contants $a, b$.
Using the values for $x_0, x_1$, we get: 
$$2=a+b$$
$$20=a(1+\sqrt{3})^2+b(1-\sqrt{3})^2=4(a+b)+2\sqrt{3}(a-b)=8+2\sqrt{3}(a-b)$$
$$a-b=\frac{12}{2\sqrt{3}}=2\sqrt{3}$$
$$a=\frac{2+2\sqrt{3}}{2}=1+\sqrt{3}, b=\frac{2-2\sqrt{3}}{2}=1-\sqrt{3}$$
Therefore $x_n=(1+\sqrt{3})^{2n+1}+(1-\sqrt{3})^{2n+1}$. Note that $x_n$ is an integer, and $-1<(1-\sqrt{3})^{2n+1}<0$, so $\lfloor(1+\sqrt{3})^{2n+1}\rfloor=\lfloor x_n-(1-\sqrt{3})^{2n+1} \rfloor=x_n$.
The problem is thus equivalent to proving that $2^{n+1} \| x_n$. We proceed by induction.
When $n=0$, clearly $2^1 \|2=x_0$. 
When $n=1$, clearly $2^2 \|20=x_1$.
Suppose that the statement holds for $0 \leq n \leq k, k \geq 1$. Then $2^{k} \|x_{k-1}$ and $2^{k+1}\|x_k$. Therefore $x_{k+1}=8x_k-4x_{k-1} \equiv 2^{k+2} \pmod{2^{k+3}}$, so $2^{k+2} \|x_{k+1}$.
We are thus done by induction.
A: A different approach in the language of elementary theory of algebraic numbers. 
I abbreviate $\alpha=1+\sqrt3$ and $\overline{\alpha}=1-\sqrt3=-\alpha(2-\sqrt3)$
As observed by Gerry Myerson and Ivan Loh we have i) $-1<\overline{\alpha}<0$, and ii) $\alpha^{2m+1}+\overline{\alpha}^{2m+1}\in\mathbb{Z}$, so
$$x_m:=\lfloor(1+\sqrt{3})^{2m+1} \rfloor=\alpha^{2m+1}+\overline{\alpha}^{2m+1}=
\alpha^{2m+1}\left(1-(2-\sqrt3)^{2m+1}\right).\qquad(*)$$
I shall work in the ring $R=\mathbb{Z}[\sqrt3]=\{a+b\sqrt3\mid a,b\in\mathbb{Z}\}$. From algebraic number theory I use the facts that $R$ is a PID and that the prime $2$ ramifies in $R$ as $(2)=(\alpha)^2$. In other words $\alpha^2$ and $2$ are associates, as
$$
\alpha^2=4+2\sqrt3=2(2+\sqrt3)
$$
and $2+\sqrt3$ is a unit of $R$, because $(2+\sqrt3)(2-\sqrt3)=1$.
Let us consider the ideal $I=2R=(\alpha)^2\subset R$. As $I=2R$, it has four cosets in $R$: a useful set of representatives being $0,1,\sqrt3,1+\sqrt3$. We see that $2-\sqrt3\equiv\sqrt3\pmod{I}.$ As $\sqrt3^2=3\equiv1\pmod I$ we get that
$$
(2-\sqrt3)^{2m+1}\equiv\sqrt3\pmod I
$$
for all integers $m$. Plugging this into our equation $(*)$ we see that
$$
x_m=\alpha^{2m+1}q_m,
$$
where $q_m$ is an element of $R$ such that
$$
q_m\equiv 1+\sqrt3 \pmod I.
$$
So in the ring $R$ we have $\alpha\mid q_m$, but $\alpha^2\nmid q_m$.
Therefore $\alpha^{2m+2}$ is the highest power of $\alpha$ dividing $x_m$. As $\alpha^2$ and $2$ are associates, and we have unique factorization, this tells us that $2^{m+1}$ is the highest power of two dividing $x_m$.
