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 $n$?

Hint: $$a_n = \left(\frac{7+\sqrt{37}}{2}\right)^n + \left(\frac{7-\sqrt{37}}{2}\right)^n$$
is always an integer, and there is a recurrence relation for $a_n$.  Since $0<\frac{7-\sqrt{37}}2<1$, you want to show that $a_n-1$ is divisble by $3$. Prove it by working out the recurrence relationship and use induction.
This result can be generalized. For integers $n>0$ and $d>0$:
$$\left\lfloor\left(\frac{2d+1+\sqrt{4d^2+1}}2\right)^n\right\rfloor$$
is divisible by $d$.
