# Divisibility of $\lfloor (1+\sqrt{3})^{2m+1}\rfloor$ by $2^k$

I want to prove that the largest exponent $k$ of $2^k$ such that $2^k|\lfloor (1+\sqrt{3})^{2m+1}\rfloor$ is $m+1$ if $m\geq 1$. My instinct tells me that I should regard $(1+\sqrt{3})^{2m+1}$ as $2^{2m+1}(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2})^{2m+1}$ because $\left(\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2=1$. But I don't know how to go on from this point. Could you give me any hint?

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## 1 Answer

One way to prove it is to use

$$\left\lfloor (1+\sqrt{3})^{2m+1}\right\rfloor = (1+\sqrt{3})^{2m+1} + (1-\sqrt{3})^{2m+1}.$$

Then look at the sequence

$$a_n = (1+\sqrt{3})^n + (1-\sqrt{3})^n,$$

and prove the result by induction, using the linear recurrence of the sequence $(a_n)$.

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That method is really clever! –  Aran Komatsuzaki Jun 5 '14 at 11:30