We can see that $a_n=\frac{(3x+7)^{F_n}-1}{3}$ where $F_n$ is the Fibonacci sequence ($F_0=0$, $F_1=1$, $F_2=1$, and $F_{n+1}=F_n+F_{n-1}$). We can see this through induction: it is true for $n=0$ and $n=1$; we can check its truth with $n=2$ also. The inductive step, then, is to plug them in to the equation and equate each side. We get: \begin{align*}&a_{n-1}+3a_{n-1}a_{n-2}+a_{n-2}\\ &=\frac{(3x+7)^{F_{n-1}}-1}{3}+3\left(\frac{(3x+7)^{F_{n-1}}-1}{3}\right)\left(\frac{(3x+7)^{F_{n-2}}-1}{3}\right)+\frac{(3x+7)^{F_{n-2}}-1}{3}\\ &=\frac{(3x+7)^{F_{n-1}}-1+(3x+7)^{F_{n-1}}(3x+7)^{F_{n-2}}-(3x+7)^{F_{n-1}}-(3x+7)^{F_{n-2}}+1+(3x+7)^{F_{n-2}}-1}{3}\\ &=\frac{(3x+7)^{F_{n-1}+F_{n-2}}-1}{3}\\ &=\frac{(3x+7)^{F_n}-1}{3}=a_n\end{align*} and it is proven. Therefore we can answer the two questions. (a) If $k$ divides $m$, then $nk=m$ for some $n$. Then we compare $a_k$ and $a_m$: $a_k=\frac{(3x+7)^{F_k}-1}{3}$ and $a_m=\frac{(3x+7)^{F_{nk}}-1}{3}$. It's well known that $a^m-1$ divides $a^n-1$ iff $m|n$, so $a_k$ divides $a_m$ iff $F_k$ divides $F_{nk}$, which is another well known fact about Fibonacci numbers. Therefore, part (a) is true. (b) We expand the polynomial: $a_n=\frac{(3x)^{F_n}+7\cdot F_n\cdot(3x)^{F_n-1}+\ldots-1}{3}$ $=3^{F_n-1}x^{F_n}+7\cdot F_n\cdot3^{F_n-2}x^{F_n-1}+\ldots-\frac{1}{3}$ We can check that $n=1$ and $n=2$ satisfy it, and $n=3$ does not. If $n>3$ then the sum of the roots is the negative coefficient of $x^{F_n-2}$ divided by the leading coefficient. This equals $-\frac{7\cdot F_n\cdot 3^{F_n-2}}{3^{F_n-1}}=-\frac{7F_n}{3}$ so $F_n$ must be divisible by 3. This happens when $n=4k$ for some $k$. Therefore, the answer is $n=1, 2, 4k$ for all $k\geq1$.

a)Define $b_n=3a_n$. Then the equation becomes $b_n=b_{n-1}+b_{n-1}b_{n-2}+b_{n-2}$. Now,add 1 to both sides and we get $b_{n}+1=(b_{n-1}+1)(b_{n-2}+1)$. Let $b_{n}+1=c_{n}$. Then $c_{n}=c_{n-1}c_{n-2}$. By using logarithms we get a Fibonacci-type recurrence, thus $c_n=(3x+7)^{F_{n}}$ and $a_n=\frac{(3x+7)^{F_n}-1}{3}$. It is known that if $m$ divides $n$ then $F_m$ divides $F_n$,and $x^m-1$ divides $x^n-1$. Thus, if $m$ divides $n$ then $a_m$ divides $a_n$. b) For $n=1$,$n=2$ it is true. For $n>2$ the roots of the polynomial are $r_k= \frac{\omega ^{k}-7}{3}$ where $\omega$ is a $F_n$th primitive root of unity. Summing them up we get $S=-7F_n/3$ (the roots of unity sum up to zero) which is an integer only if $n=4k$.