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 $n$?

Since
$$
\left(x-\frac{7+\sqrt{37}}2\right)\left(x-\frac{7-\sqrt{37}}2\right)=\left(x-\frac72\right)^2-\frac{37}4=x^2-7x+3\;,
$$
the recurrence relation
$$x_n=7x_{n-1}-3x_{n-2}$$
has solutions
$$
x_n=c_1\left(\frac{7+\sqrt{37}}2\right)^n+c_2\left(\frac{7-\sqrt{37}}2\right)^n\;.
$$
With $c_1=c_2=1$ we have $x_0=2$ and $x_1=7$. Since $0\lt\frac{7-\sqrt{37}}2\lt1$ and $x_n$ is an integer,
$$
\left\lfloor\left(\frac{7+\sqrt{37}}2\right)^n\right\rfloor=x_n-1\;.
$$
Now you just need to use the recurrence relation to show that the property that $x_1$ has remainder $1$ mod $3$ is inherited by the remaining terms.
