# $\lfloor (2+\sqrt{3})^n \rfloor$ is odd

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

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

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

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

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

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

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Related to math.stackexchange.com/questions/48508/… –  lhf Dec 23 '11 at 11:30

Hint for the first part: Consider $u_n = (2+\sqrt{3})^{n} + (2-\sqrt{3})^{n}$. Prove that $u_n$ is always an even integer and that $u_n = \lceil (2+\sqrt{3})^n \rceil$. Use that $(2-\sqrt{3})^{n}\to 0$.

(This has now been incorporated into the edited question.)

Hint for the second part: Consider $v_n = (1+\sqrt{3})^{n} + (1-\sqrt{3})^{n}$. Find a second-order recursion for $v_n$ based on the quadratic equation that defines $1\pm\sqrt{3}$.

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You don't really need $(2-\sqrt 3)^n\to 0$, just that $0<(2-\sqrt 3)^n<1$, right? And you've confused $u_n$ and $u_2n$ in the second part. –  Thomas Andrews Dec 22 '11 at 15:18
@Thomas, right, just that $0<(2-\sqrt 3)^n<1$. I've fixed the typo, thanks. –  lhf Dec 22 '11 at 15:23
Could someone detail the second part? –  Chon Dec 22 '11 at 17:20
@PlaneChon-Ju, actually, I don't think that the second part is true. The numbers in the sequence are 2,4,14,52,194,724,2702,10084,37634,140452,524174,1956244,7300802,27246964 and they are either 2 or 4 times an odd number. –  lhf Dec 22 '11 at 18:40
Sorry, the quantity is actually $\lfloor (1+\sqrt{3})^{2n} \rfloor+1$ –  Chon Dec 22 '11 at 21:19

You can use recurrences such as $$f(n)=4f(n-1)-f(n-2)+2$$ or $$f(n)=5f(n-1)-5f(n-2)+f(n-3)$$ starting at $f(0)=1, f(1)=3$.

Then show the various results by induction.

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