Using induction to prove $3$ divides $\left \lfloor\left(\frac {7+\sqrt {37}}{2}\right)^n \right\rfloor$ How can I use induction to prove that $$\left \lfloor\left(\cfrac {7+\sqrt {37}}{2}\right)^n \right\rfloor$$ is divisible by $3$ for every natural number $n$?

Consider the recurrence given by
$$x_{n+1} = 7x_n - 3 x_{n-1}$$ where $x_0 = 2$, $x_1  = 7$.
Note that $$x_{n+1} \equiv (7x_n - 3 x_{n-1}) \pmod{3} \equiv (x_n + 3(2x_n - x_{n-1})) \pmod{3} \equiv x_n \pmod{3}$$
Since $x_1 \equiv 1 \pmod{3}$, we have that $x_n \equiv 1 \pmod{n}$. Ths solution to this recurrence is given by
$$x_n = \left( \dfrac{7+\sqrt{37}}{2} \right)^n + \left( \dfrac{7-\sqrt{37}}{2}\right)^n \equiv 1 \pmod{3}$$
Further, $0 < \dfrac{7-\sqrt{37}}{2} < 1$. This means $$3M < \left( \dfrac{7+\sqrt{37}}{2} \right)^n < 3M+1$$ where $M \in \mathbb{Z}$. Hence, we have that
$$3 \text{ divides }\left \lfloor \left (\dfrac{7+\sqrt{37}}{2}\right)^n \right \rfloor$$
