<s>[INST] Showing that $\lceil (\sqrt{3} + 1)^{2n} \rceil$ is divisible by $2^{n+1}$. I have a question which has fluxommed me and my pals for the past few days. Any help or solution is welcome

Show using Binomial theorem that the integer just after $(3^{1/2} + 1)^{2n}$ is divisble by $2^{n+1}$.

here n belongs to natural numbers (positive integers).
 [/INST] This is similar to Henry's answer, but things are simpler if we use $4+2\sqrt{3}=(1+\sqrt{3})^2$ and $4-2\sqrt{3}=(1-\sqrt{3})^2$ instead. These satisfy $x^2-8x+4=0$.
The sequence defined by
$$
a_0=2\text{, }a_1=8\text{, and }a_n=8a_{n-1}-4a_{n-2}\tag{1}
$$
has the solution
$$
\begin{align}
a_n
&=(4+2\sqrt{3})^n+(4-2\sqrt{3})^n\\
&=(1+\sqrt{3})^{2n}+(1-\sqrt{3})^{2n}\\
&=\left\lceil(\sqrt{3}+1)^{2n}\right\rceil\text{ for }n>0\tag{2}
\end{align}
$$
Let $b_n=a_n2^{-n-1}$. Then, $(1)$ becomes
$$
b_0=1\text{, }b_1=2\text{, and }b_n=4b_{n-1}-b_{n-2}\tag{3}
$$
Recursion $(3)$ insures that $b_n\in\mathbb{Z}$ for all $n\ge0$. Thus, for $n>0$, $(2)$ yields
$$
\begin{align}
\left\lceil(\sqrt{3}+1)^{2n}\right\rceil
&=a_n\\
&=2^{n+1}b_n\\
&\in2^{n+1}\mathbb{Z}\tag{4}
\end{align}
$$
</s>