Let $n$ be a non-negative integer. Show that $2^{n+1}$ divides the value of $\left\lfloor{(1+\sqrt{3})^{2n+1}}\right\rfloor$ where function $\lfloor{x}\rfloor$ returns the largest integer not exceeding the give real number $x$.
First $$0 < \sqrt{3} - 1 < 1\implies 0 < \left(\sqrt{3}-1\right)^{2n+1} < 1$$
Next, by binomial expansion, it can be shown that $$\left(\sqrt{3}+1\right)^{2n+1} - \left(\sqrt{3}-1\right)^{2n+1}$$
is an integer because all the odd powers of $\sqrt{3}$ are cancelled. It follows that $$\left\lfloor{\left(1+\sqrt{3}\right)^{2n+1}}\right\rfloor = \left(\sqrt{3}+1\right)^{2n+1}- \left(\sqrt{3}-1\right)^{2n+1}$$
Thus, it is sufficient to show $\left(\sqrt{3}+1\right)^{2n+1} - \left(\sqrt{3}-1\right)^{2n+1}$ is a multiple of $2^{n+1}$. $$\begin{align*} &\left(\sqrt{3}+1\right)^{2n+1} - \left(\sqrt{3}-1\right)^{2n+1}\\&=\ \left(\sqrt{3}+1\right)^{2n}\left(\sqrt{3}+1\right) - \left(\sqrt{3}-1\right)^{2n}\left(\sqrt{3}-1\right)\\&=\ \underbrace{\left(\sqrt{3}+1\right)^{2n} + \left(\sqrt{3}-1\right)^{2n}}_{A}+\underbrace{\sqrt{3}\left(\left(\sqrt{3}+1\right)^{2n}-\left(\sqrt{3}-1\right)^{2n}\right)}_{B} \end{align*}$$
By the conclusion of # 2693, $A$ is a multiple of $2^{n+1}$. So it is sufficient to show that part $B$ is a multiple of $2^{n+1}$ too. $$\begin{align*} &\sqrt{3}\left(\left(\sqrt{3}+1\right)^{2n}-\left(\sqrt{3}-1\right)^{2n}\right) \\&=\ \sqrt{3}\left(\left(4+2\sqrt{3}\right)^n - \left(4-2\sqrt{3}\right)^n\right)\\&=\ \sqrt{3}\cdot 2^n\left(\left(2+\sqrt{3}\right)^n-\left(2-\sqrt{3}\right)^n\right)\\&=\ 2^n\cdot\sqrt{3}\cdot 2\cdot\left(\binom{n}{1}\sqrt{3} + \binom{n}{3}\left(\sqrt{3}\right)^3+\cdots\right)\\&=\ 2^{n+1}\left(\binom{n}{1}\cdot 3 + \binom{n}{3}\cdot 3^2 + \cdots\right)\end{align*}$$
Therefore, $B$ is a multiple of $2^{n+1}$. It follows that $(A+B)$ is a multiple of $2^{n+1}$ which means the claim holds.