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.

Hint: The value is equal to $( \sqrt{3} + 1)^{2n} + (\sqrt{3}-1 )^{2n}$
Hint: Expanding the square, we get $ (4 + 2\sqrt{3})^n + (4-2\sqrt{3})^n$
This term is clearly a multiple of $2^n$, which we can factor out.
Hint: $(2 + \sqrt{3})^n + (2-\sqrt{3})^n$ is even, by the Binomial Theorem expansion.
