Show that $\lfloor(2+\sqrt{3})^n\rfloor$ is an odd integer whenever n is a nonnegative integer I know its not right to post a question if I don't have any proofs of my own because most of you will think that I used it as a homework generator. But, can you help me generate proof in this problem? I really don't have any idea how or where to start. Yes, I can illustrate the problem but I can't find ways to start my proof.
Show that $\lfloor(2+\sqrt{3})^n\rfloor$ is an odd integer whenever n is a nonnegative integer
 A: \begin{align}
\left ( 2+\sqrt{3} \right )^{n}+\left ( 2-\sqrt{3} \right )^{n} 
&=\sum_{i=0}^n\binom{n}{i}2^{n-i}\sqrt{3}^i+\sum_{i=0}^n\binom{n}{i}2^{n-i}(-1)^i\sqrt{3}^i
\\
&=2\sum_{i=0}^{n/2}\binom{n}{2i}2^{n-2i}\sqrt{3}^{2i}
\\
&=2\sum_{i=0}^{n/2}\binom{n}{2i}2^{n-2i}3^{i}
\end{align}
So $(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}$ is even. Since $0<2-\sqrt{3}<1$, $0<(2-\sqrt{3})^{n}<1$. Thus $\lfloor(2+\sqrt{3})^{n}\rfloor$ is odd. 
A: Another way to solve this question is to notice that the sequence given by $a_n = (2 + \sqrt{3})^n + (2-\sqrt{3})^n$ satisfies the recursive relations $a_{n+2} = 4a_{n+1} - a_n$. Therefore as $4$ is even we have that $a_{n+2}$ and $a_n$ have the same parity. So as $a_2 = 14$ is even, so are all even indexed terms of the sequence. Similarly $a_1 = 4$ is even and so does all odd indexed terms.
And now as $\lfloor (2 + \sqrt{3})^n \rfloor = a_n - 1$, as User1006 explained, we have the wanted result.
A: The sequence
$$
a_n=\left(2+\sqrt3\right)^n+\left(2-\sqrt3\right)^n
$$
satisfies the recursion $a_n=4a_{n-1}-a_{n-2}$. Furthermore, since $a_0=2$ and $a_1=4$, all the terms of the sequence are even. Thus, for $n\ge0$,
$$
\left(2+\sqrt3\right)^n=a_n-\overbrace{\left(2-\sqrt3\right)^n}^{\in(0,1]}
$$
Therefore,
$$
\left\lfloor\left(2+\sqrt3\right)^n\right\rfloor=\overbrace{\ \ a_n-1\ \ }^\text{odd}
$$
