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.
 A: Let $(1+\sqrt{3})^{2n}=a_n+b_n\sqrt{3}$. Then $(1-\sqrt{3})^{2n}=a_n-b_n\sqrt{3}$.
Thus $(1+\sqrt{3})^{2n}+(1-\sqrt{3})^{2n}$ is the integer $2a_n$. Since $|1-\sqrt{3}|\lt 1$, we have 
$$2a_n=\left\lceil(\sqrt{3}+1)^{2n}\right\rceil.$$
Note that 
$$(1+\sqrt{3})^{2n+2}=(1+\sqrt{3})^{2n}(4+2\sqrt{3})=(a_n+b_n\sqrt{3})(4+2\sqrt{3}).$$ Thus 
$$a_{n+1}=4a_n+6b_n \quad\text{and}\quad b_{n+1}=2a_n+4b_n.$$
Thus the largest power of $2$ that divides $2a_k$ increases by at least $1$ when we increment $k$. But $2a_0=2$. This completes the proof.
Remark: Whenever $a+b\sqrt{c}$ has a problem, where $a$ and $b$ and $c$ are integers, and $c$ is not a perfect square, its conjugate $a-b\sqrt{c}$ is likely to be helpful. 
A: 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.
A: A succession of hints:


*

*Show that $a_n=(\sqrt{3}+1)^{n}+(1-\sqrt{3})^{n}$ is an integer

*Show that $\lceil(\sqrt{3}+1)^{2n}\rceil = (\sqrt{3}+1)^{2n}+(1-\sqrt{3})^{2n} (=a_{2n})$ (note that the latter term is positive since it's being raised to an even power, and that $|1-\sqrt{3}|\lt 1$).

*Show that the sequence $a_n=(\sqrt{3}+1)^{n}+(1-\sqrt{3})^{n}$ satisfies the recurrence relation $a_{n+2} = 2a_{n+1}+2a_n$ with suitable initial conditions.  (This is actually easier to do in the other direction - look up the theory of linear recurrence relations for more details.  It's related to the Binet formula for the Fibonacci numbers.)

*Show by induction using the recurrence relation that the successive $a_n$ are divisible by higher and higher powers of 2.

