Prove the integer part of $(1+\sqrt{3})^{2006}$ is odd

Question

Prove that the integer part (flooring ) of $(1+\sqrt{3})^{2006}$ is odd. Here's what I did: $(1+\sqrt{3})^{2006}=(4+2\sqrt{3})^{1003}={1003\choose 0}4^{1003}+{1003\choose 2}4^{1001}.(2\sqrt{3})^2+\cdots+{1003\choose 1002}4.(2\sqrt{3})^{1002}) + ({1003\choose1}.4^{1002}.2\sqrt{3}+{1003\choose 3}4^{1000}.(2\sqrt{3})^3+\cdots+{1003\choose 1003}(2\sqrt{3})^{1003})$. It is clear that the expression in the left parentheses is even. So now the task is to prove the floor of the expression in the other parentheses is odd. I have no idea how to proceed with this.

Answer 1

Hint:

Observe that $$(\sqrt3+1)^{2n}+(\sqrt3-1)^{2n}$$ is an even integer

Now $\sqrt3-1=\dfrac{3-1}{\sqrt3+1}<1$

$\implies0<(\sqrt3-1)^{2n}<1$