$\lfloor (2+\sqrt{3})^n \rfloor $ is odd

Hint for the first part: Consider $u_n = (2+\sqrt{3})^{n} + (2-\sqrt{3})^{n}$. Prove that $u_n$ is always an even integer and that $u_n = \lceil (2+\sqrt{3})^n \rceil$. Use that $(2-\sqrt{3})^{n}\to 0$.

(This has now been incorporated into the edited question.)

Hint for the second part: Consider $v_n = (1+\sqrt{3})^{n} + (1-\sqrt{3})^{n}$. Find a second-order recursion for $v_n$ based on the quadratic equation that defines $1\pm\sqrt{3}$.

It is not hard to prove that$$(\forall n\in\mathbb N):\left(2+\sqrt3\right)^n+\left(2-\sqrt3\right)^n\in2\mathbb N.$$This, together with the fact that $2-\sqrt3\in(0,1),$ is enough to prove that $\left\lfloor\left(2+\sqrt3\right)^n\right\rfloor$ is an odd integer.

You can use recurrences such as $$f(n)=4f(n-1)-f(n-2)+2$$ or $$f(n)=5f(n-1)-5f(n-2)+f(n-3)$$ starting at $f(0)=1, f(1)=3$.

Then show the various results by induction.