TITLE: Alternate proof that for every natural number $n,\ \left\lfloor\left(\frac{7+\sqrt{37}}{2}\right)^n\right\rfloor$ is divisible by $3$
QUESTION [15 upvotes]: Original Problem:

Prove that for every natural number $n$,$$\left\lfloor\left(\frac{7+\sqrt{37}}{2}\right)^n\right\rfloor$$ is divisible by $3$.

I found the problem in the book Winning Solutions. I do have a solution at hand which I was able to construct duly from the given hints: At first, take $a_n=\left(\frac{7+\sqrt{37}}{2}\right)^n+\left(\frac{7-\sqrt{37}}{2}\right)^n$ and then prove that $a_{n+2}=7a_{n+1}-3a_n$ which in turn, can be used to prove that $3\ |\ a_n-1$.The proof is complete once we can prove that $a_n-1=\left\lfloor\left(\frac{7+\sqrt{37}}{2}\right)^n\right\rfloor$.
Questions:

Why would anyone think of coming up with such an ingenious recurrence? I am simply at a loss to understand the motivation behind the whole series of hints. They led me to a solution but I doubt I would have come up with it by myself.
Can anyone please show me a more natural proof?

Thanks!

REPLY [3 votes]: Here is an answer that uses some algebra; I don't know if it suits the OP's background, but it may help someone.
Let $\alpha = (7 + \sqrt{37})/2,$ and consider the ring $R := \mathbb Z[\alpha] = \mathbb Z[(1+\sqrt{37})/2]$.
Let $x \mapsto \overline{x}$ denote the automorphism of this ring given by $\sqrt{37} \mapsto -\sqrt{37}.$  If $x \in R$ then $x + \overline{x}, x\overline{x}  \in \mathbb Z$.  
Note that $\alpha\overline{\alpha} = 3$, and so $(\alpha) \subset R$ is an ideal
such that $R/(\alpha) = \mathbb F_3$.  Furthermore, since $\alpha \equiv 0 \bmod \alpha$ (obviously), and since $\alpha + \overline{\alpha} = 7 \equiv 1 \bmod 3$, we see that the image of $\overline{\alpha}$ in $\mathbb F_3$ is $1$.
Thus the image of the integer $\alpha^n + \overline{\alpha}^n$ in $\mathbb F_3$ is equal to $0 + 1^n = 1,$ and so 
$$\alpha^n + \overline{\alpha}^n \equiv 1 \bmod 3.$$
(In the other answers this is checked using a recursion, or more elementary algebra.)
After this the solution is the same as in all the other answer; one has to
note that $0 < \overline{\alpha}^n < 1$, so that the floor of $\alpha^n$ is
$0 \bmod 3$.