<s>[INST] Prove that $\lceil(\sqrt{3}+1)^{2n}\rceil$ is divisible by $2^{n+1}$. Let $n$ be a positive integer. Prove that $\lceil(\sqrt{3}+1)^{2n}\rceil$ is divisible by $2^{n+1}$.
I tried rewriting $\lceil(\sqrt{3}+1)^{2n}\rceil$ as $m*2^{n+1}$ for some m, but couldn't get anywhere.
 [/INST] Let $(1+\sqrt{3})^{2n}=a_n+b_n\sqrt{3}$. Then $(1-\sqrt{3})^{2n}=a_n-b_n\sqrt{3}$.
Thus $(1+\sqrt{3})^{2n}+(1-\sqrt{3})^{2n}$ is the integer $2a_n$. Since $|1-\sqrt{3}|\lt 1$, we have 
$$2a_n=\left\lceil(\sqrt{3}+1)^{2n}\right\rceil.$$
Note that 
$$(1+\sqrt{3})^{2n+2}=(1+\sqrt{3})^{2n}(4+2\sqrt{3})=(a_n+b_n\sqrt{3})(4+2\sqrt{3}).$$ Thus 
$$a_{n+1}=4a_n+6b_n \quad\text{and}\quad b_{n+1}=2a_n+4b_n.$$
Thus the largest power of $2$ that divides $2a_k$ increases by at least $1$ when we increment $k$. But $2a_0=2$. This completes the proof.
Remark: Whenever $a+b\sqrt{c}$ has a problem, where $a$ and $b$ and $c$ are integers, and $c$ is not a perfect square, its conjugate $a-b\sqrt{c}$ is likely to be helpful. 
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