9
$\begingroup$

How can I show that $\lfloor (2+\sqrt{3})^n \rfloor $ is odd and that $2^{n+1}$ divides $\lfloor (1+\sqrt{3})^{2n} \rfloor+1 $ ?

$$ u_{n}=(2+\sqrt{3})^n+(2-\sqrt{3})^n=\sum_{k=0}^n{n \choose k}2^{n-k}(3^{k/2}+(-1)^k3^{k/2})\in\mathbb{2N} $$

$$ 0\leq (2-\sqrt{3})^n \leq1$$

$$ (2+\sqrt{3})^n\leq u_{n}\leq 1+(2+\sqrt{3})^n $$

$$ (2+\sqrt{3})^n-1\leq u_{n}-1\leq (2+\sqrt{3})^n $$

$$ \lfloor (2+\sqrt{3})^n \rfloor=u_{n}-1\in\mathbb{2N}+1 $$

  • 0
    Related to http://math.stackexchange.com/questions/48508/how-are-the-integral-parts-of-9-4-sqrt5n-and-9-4-sqrt5n-relate2011-12-23

2 Answers 2