<s>[INST] Showing that $\lceil (\sqrt{3} + 1)^{2n} \rceil$ is divisible by $2^{n+1}$. I have a question which has fluxommed me and my pals for the past few days. Any help or solution is welcome

Show using Binomial theorem that the integer just after $(3^{1/2} + 1)^{2n}$ is divisble by $2^{n+1}$.

here n belongs to natural numbers (positive integers).
 [/INST] This is a classic, the standard proof is to consider
$$ (\sqrt{3} + 1)^{2n} = x_n + y_n \sqrt{3}$$
for integer $x_n$, $y_n$, (which you can see using Binomial theorem) and show that
$$ (\sqrt{3} - 1)^{2n} = x_n - y_n \sqrt{3}$$
again using Binomial theorem.
Now use the fact that $\sqrt{3} - 1 \lt 1$ and that
$$(\sqrt{3} + 1)^{2n} + (\sqrt{3} - 1)^{2n} = 2x_n$$
Thus the integer you are looking for is $2x_n$.
Get a recurrence for $x_n$ and $y_n$ and use induction.
To get the recurrence:
we have that $(\sqrt{3} + 1)^2 = 4 + 2\sqrt{3}$ and so
$$ x_{n+1} + y_{n+1} = (x_n + y_n \sqrt{3})(4 + 2 \sqrt{3}) = (4x_n + 6y_n) + (2x_n + 4y_n) \sqrt{3}$$
Thus
$$ x_{n+1} = 4x_n + 6y_n$$
$$ y_{n+1} = 2x_n + 4y_n$$
Write $$x_{n+2} = 4x_{n+1} + 6y_{n+1}$$ and eliminate $y_n$ and $y_{n+1}$
(I believe it comes to $x_{n+2} = 8x_{n+1} - 4 x_n$)
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