Show that $<b_n\sqrt{3}>=a_n$ ($<x>$: rounding function)

Question

I would appreciate if somebody could help me with the following problem

Q: Let $(2+\sqrt{3})^n=a_n+b_n\sqrt{3}$ $(a_n,b_n,n\in\mathbb{N})$ .

Show that $<b_n\sqrt{3}>=a_n$ ($<x>$: rounding function)

Answer 1

Consider instead the sequence $(2+\sqrt{3})^n$ with the same restrictions. Note that since $$ (2-\sqrt{3})^n=2^n-\binom{n}{1}2^{n-1}\sqrt{3}+\cdots +(-1)^k\binom{n}{k}2^{n-k}\sqrt{3}^k+\cdots $$ The negated terms are exactly those in the form $m\sqrt{3}$ for some integer $m$. But as $$ (2+\sqrt{3})^n=2^n+\binom{n}{1}2^{n-1}\sqrt{3}+\cdots +\binom{n}{k}2^{n-k}\sqrt{3}^k+\cdots $$ we see that $(2-\sqrt{3})^n=a_n-b_n\sqrt{3}$. Note that $(2-\sqrt{3})^n<0.5$ for all positive integers $n$, so $a_n-b_n\sqrt{3}<0.5$ always, which implies $<b_n\sqrt{3}>=a_n$ as desired.