Prove that $\lfloor{(2+\sqrt{3})^n}\rfloor$ is an odd number? How can we prove that $\lfloor{(2+\sqrt{3})^n}\rfloor$ is an odd number? My teacher did something like this:
$$\lfloor{(2+\sqrt{3})^n}\rfloor=(2+\sqrt{3})^n+(2-\sqrt{3})^n-1$$
And she said a few more words, but I understood nothing. Could you please tell me what she meant by that?
 A: Consider $a_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n$. The second degree monic polynomial having $2+\sqrt{3}$ and $2-\sqrt{3}$ as roots is $X^2-4X+1$. This means that $a_n$ satisfies the recursion
$$
a_0=2, a_1=4,\quad a_{n+2}=4a_{n+1}-a_n
$$
Since the two initial terms are even, we get that $a_n$ is even for every $n$.
Moreover, $0<2-\sqrt{3}<1$, so also $0<(2-\sqrt{3})^n<1$. Therefore,
$$
\lfloor(2+\sqrt{3})^n\rfloor=a_n-1
$$
is odd.
A: If you expand both $(2+\sqrt{3})^n$ and $(2-\sqrt{3})^n$ using the binomial formula, all the terms in the binomial expansions involving $\sqrt{3}$ cancel, so that the right-hand side is in fact an integer.
Note that all terms of each expansion, except for the last term, have even coefficients. The last term is $3^{n/2}+3^{n/2}$ (an even number) if $n$ is even, and is zero if $n$ is odd. Thus the sum of the two powers is even so the left-hand side is odd.
A: $$(2+\sqrt{3})^n=\binom{n}{0}2^{n}\sqrt{3}^{0}+ \binom{n}{1}2^{n-1}\sqrt{3}^{1}+\binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+\binom{n}{n}2^{0}\sqrt{3}^{n}\\
(2-\sqrt{3})^n=\binom{n}{0}2^{n}\sqrt{3}^{0}- \binom{n}{1}2^{n-1}\sqrt{3}^{1}+\binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+(-1)^n\binom{n}{n}2^{0}\sqrt{3}^{n}$$ note that for sum of them 
$$ (2+\sqrt{3})^n+(2-\sqrt{3})^n=2\left ( \binom{n}{0}2^{n}\sqrt{3}^{0}+ \binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+\binom{n}{n}2^{0}\sqrt{3}^{n} \right )=2Q\\if \space n=even$$
$$ (2+\sqrt{3})^n+(2-\sqrt{3})^n=2\left ( \binom{n}{0}2^{n}\sqrt{3}^{0}+ \binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+\binom{n-1}{n}2^{1}\sqrt{3}^{n-1} \right )=2Q\\if \space n=odd$$
both sum is integer ,and now
  $$ \left \lfloor (2+\sqrt{3})^n \right \rfloor=\left \lfloor 2Q-(2-\sqrt{3})^n \right \rfloor$$ note that 
$$0<(2-\sqrt{3})^n<1\\2Q-1<2Q-(2-\sqrt{3})^n<2Q\\$$ so 
$$ \left \lfloor (2+\sqrt{3})^n \right \rfloor=\left \lfloor 2Q-(2-\sqrt{3})^n \right \rfloor=2Q-1$$
