<s>[INST] 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?
 [/INST] $$(2+\sqrt{3})^n=\binom{n}{0}2^{n}\sqrt{3}^{0}+ \binom{n}{1}2^{n-1}\sqrt{3}^{1}+\binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+\binom{n}{n}2^{0}\sqrt{3}^{n}\\
(2-\sqrt{3})^n=\binom{n}{0}2^{n}\sqrt{3}^{0}- \binom{n}{1}2^{n-1}\sqrt{3}^{1}+\binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+(-1)^n\binom{n}{n}2^{0}\sqrt{3}^{n}$$ 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 
$$ \left \lfloor (2+\sqrt{3})^n \right \rfloor=\left \lfloor 2Q-(2-\sqrt{3})^n \right \rfloor=2Q-1$$
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