$$Problem$$

Given $ (a_n)_{n\in N}$ the sequence defined by:

$$a_1=15,\ a_2=18,\ a_{n+2}=6a_{n+1}-7a_n^4,\ \forall n\in \mathbb{N}$$

Prove by induction that $\forall n \in N$

a) $3^n|a_n$

b) $3^{n+1}\not| \ a_n $

I don't know how to face this problem properly, here's what I did. Is it >ok? Is there something I'm missing or a proper way to solve it?

$$My \ solution$$ a)

First I define $P(n)$ like this: $$P(n): 3^n|a_n$$

Then I test the base cases $$P(1): 3|15, \ true \\ P(2):3^2|18, \ true$$

So I assume that $P(n)$ (already defined), and $$P(n+1):3^{n+1}|a_{n+1}$$ are true and try to assure that they imply $$P(n+2):3^{n+2}|a_{n+2}$$ So now, I try to prove by induction that the last statement is true, so I write it in other way (with the definition of $a_{n+2}$) in order to operate, like this: $$6a_{n+1}-7a_n^4=3^{n+2}q\ \ ,q\in \mathbb{Z}$$

then $$\frac{6a_{n+1}}{3^{n+2}}-\frac{7a_n^4}{3^{n+2}}=q$$

and by hypothesis I know that $\frac{a_{k+1}}{3^{k+1}}\ and \frac{a_k}{3^k}$ are integers, I'll call them $s$ and $t$ respectively:

$$\frac{6s}{3}-\frac{7a_n^3t}{3^2}=q\ (with\ s,t,q \in \mathbb{Z})$$ $$2s-7\dfrac{a_n}{3}\dfrac{a_n}{3}a_nt=q$$ and again we know that if $3^n | a_n$ then $3| a_n$ because $3|3^n$, so $\frac{a_n}{3}\in \mathbb{Z}$.

Finally defining $r$ as $\frac{a_n}{3}$ I get the equation $$q=2s-7r^2a_nt$$ which I'm sure has solutions $\in \mathbb{Z}$ since all terms are integers.

Is it enough to prove that $P(n+2)$ is true?

b)

I don't know how to start this one.