Find when $\frac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}$ is an integer 
Find all natural numbers $n$ such that $$\dfrac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}$$ is an integer.

For $n = 1$ we have $\dfrac{18}{3} = 6$.
For $n = 2$ we have $\dfrac{126}{18} = 7$.
For $n = 3$ we have $\dfrac{972}{126} = \dfrac{54}{7}$
For $n = 4$ we have $\dfrac{7938}{972} = \dfrac{49}{6}$.
I don't seem to find any other integers for the fraction for $n > 2$. How can we find all of the $n$ such that it is an integer?
 A: Hint
$$\dfrac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}=3 \dfrac{1+2^n+3^n}{1+2^{n-1}+3^{n-1}}$$
Hint 2: If $n \geq 2$ then 
$$2 < \dfrac{1+2^n+3^n}{1+2^{n-1}+3^{n-1}}<3$$
So you need to solve
$$\dfrac{1+2^n+3^n}{1+2^{n-1}+3^{n-1}}=\frac{7}{3}  \mbox{ and }\\
\dfrac{1+2^n+3^n}{1+2^{n-1}+3^{n-1}}=\frac{8}{3}$$
A: $\begin{array}\\
\dfrac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}
&=\dfrac{3^n+6^n+9^n}{3^{n-1}+6^{n-1}+9^{n-1}}-9+9\\
&=\dfrac{3^n+6^n+9^n-9(3^{n-1}+6^{n-1}+9^{n-1})}{3^{n-1}+6^{n-1}+9^{n-1}}+9\\
&=\dfrac{3^n+6^n+9^n-9\cdot 3^{n-1}-9\cdot 6^{n-1}-9^{n}}{3^{n-1}+6^{n-1}+9^{n-1}}+9\\
&=\dfrac{3^n+6^n-9\cdot 3^{n-1}-9\cdot 6^{n-1}}{3^{n-1}+6^{n-1}+9^{n-1}}+9\\
&=\dfrac{3^n+6^n- 3^{n+1}-\frac23  6^{n}}{3^{n-1}+6^{n-1}+9^{n-1}}+9\\
&=\dfrac{\frac13 6^n-2\cdot 3^{n}}{3^{n-1}+6^{n-1}+9^{n-1}}+9\\
&=\dfrac{3\cdot (2/3)^{n-1}-9\cdot (1/3)^{n-1}}{(1/3)^{n-1}+(2/3)^{n-1}+1}+9\\
\end{array}
$
For
$n \ge 3$,
$0 \lt 3\cdot (2/3)^{n-1}-9\cdot (1/3)^{n-1}
\lt 1$,
so this is never an integer
for $n \ge 2$.
A: Hint: It looks like considering the top and bottom $\mod 4$ narrows it down quite a bit.
