TITLE: Recurrence relation satisfied by $\lfloor(1+\sqrt{3})^n\rfloor$
QUESTION [7 upvotes]: This is a follow up to a question I had asked earlier about a linear recurrence relationship satsified by $\lfloor(1+\sqrt{5})^n\rfloor$. I messed up there, and I actually meant to ask about $L(n)=\lfloor(1+\sqrt{3})^n\rfloor$.
Following Douglas' suggestion I have determined that the values (at least the first 1000) satisfy the following recurrence: 
$L(2n+5)=8L(2n+3)-4L(2n+1)$
The question is how do I prove something like this. I can prove the recurrence for the values inside the floor function, but floor function in general does not commute with addition and multiplication. 
Explicitly, it's easy to show
$(1+\sqrt{3})^{2n+5}=8(1+\sqrt{3})^{2n+3}-4(1+\sqrt{3})^{2n+1}$
but I am not sure how to prove the recurrence from here.

REPLY [10 votes]: Prove the same relation for $1-\sqrt 3$, and check that the resulting powers are all smaller than 1, all negative, and when added to the same powers of $1+\sqrt 3$ you end up with an integer: For all $k=1,2,\dots$, $$(1+\sqrt 3)^k+(1-\sqrt 3)^k $$ is an integer. 
There are several ways of checking this fact. For example, by induction. Or by using the Binomial theorem.