TITLE: Prove that $\lceil(\sqrt{3}+1)^{2n}\rceil$ is divisible by $2^{n+1}$.
QUESTION [9 upvotes]: 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.

REPLY [8 votes]: 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.

REPLY [7 votes]: 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.

REPLY [5 votes]: 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.