Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$ 
Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$ where $n$ is a positive integer.

We have \begin{align*}\lfloor(1+\sqrt{3})^{2n-1}\rfloor &= (\sqrt{3}+1)^{2n-1}-(\sqrt{3}-1)^{2n-1}\\&=\dfrac{(4+2\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(4-2\sqrt{3})^n}{1-\sqrt{3}}\\&=2^n\left(\dfrac{(2+\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(2-\sqrt{3})^n}{1-\sqrt{3}}\right).\end{align*} Thus we must show that the second term is an integer. How can we continue?

$$(\sqrt{3}+1)^{2n-1}-(\sqrt{3}-1)^{2n-1}$$
Take n = 2
$$(\sqrt{3}+1)^{3}-(\sqrt{3}-1)^{3}$$
Look at, binomial expansion of the powers
First term,
$$ 3\sqrt(3) + 3*3 + 3\sqrt(3) + 1 $$
Minus second term,
$$ 3\sqrt(3) - 3*3 + 3\sqrt(3) - 1 $$
Gives
$$ 2*3*3 + 2 $$
So you get only every second term. Generalize to higher powers and get a general formula. See that it divides by $2^n$
