Finding the Ceiling function of $(\sqrt{3}+ 1)^{2n}$

Question

Ceiling function of $(\sqrt{3}+ 1)^{2n}$ is $(\sqrt{3} + 1)^{2n} + (\sqrt{3} - 1)^{2n}$.

While solving a problem that states $2^{n+1}$ divides ceiling function of $(\sqrt{3}+ 1)^{2n}$. I went through the solution and there was a line stated that, Ceiling function of $(\sqrt{3}+ 1)^{2n}$ is $(\sqrt{3} + 1)^{2n} + (\sqrt{3} - 1)^{2n}$. I couldn't get the proof and why did it say so.

I tried by putting values and it satisfied but couldn't get where did it came from.

Answer 1

Denote $a_n = (\sqrt{3} + 1)^{2n}$, and $b_n = a_n + (\sqrt{3} - 1)^{2n}$. The fact that $b_n$ is the ceiling of $a_n$ follows from two observations:

1. $a_n < b_n < a_n + 1$, which is a simple estimate.
2. $b_n$ is an integer (this can be seen using the binomial formula and grouping terms with the same power of $\sqrt{3}$).

Answer 2

Since $1=\sqrt{1}<\sqrt{3}<\sqrt{4}=2$ we can infer that $0<\sqrt{3}-1<1$, it follows that for every $n\in\mathbb{N}$ we have the inequality $0<\left(\sqrt{3}-1\right)^{2n}<1$. So, it is suffice to show that

$$ \left(\sqrt{3}+1\right)^{2n}+\left(\sqrt{3}-1\right)^{2n}\in\mathbb{N} $$

Now, from the binomial formula it follows that

$$ \left(\sqrt{3}+1\right)^{2n}+\left(\sqrt{3}-1\right)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}(\sqrt{3})^k(1+(-1)^{2n-k})=2\sum_{k=0}^n\binom{2n}{2k}3^k\in\mathbb{N} $$