Show that $ \left\lceil( \sqrt3 +1)^{2n}\right\rceil$ where $n \in \mathbb{N}$ is divisible by $2^{n+1}$.
I wrote the binomial expansion of $ ( \sqrt3 +1)^{2n}$ and $( \sqrt3 -1)^{2n}$ and then added them to confirm that the next integer is even.
Afterwards I applied $AM \ge GM$ on the two terms to get $ ( \sqrt3 +1)^{2n} + ( \sqrt3 -1)^{2n} \ge (2^{n+1})$.
Now I'm unable to figure out the next step. Any help would be appreciated. :)