Rational Points on Fibonacci-like Sequence of Polynomials Let $\{a_n\}$ be a sequence of polynomials in $\mathbb{Q}[x,y]$ with $a_0=0,a_1=1$, and
$$a_n=xa_{n-1}+ya_{n-2}$$
The first few look like
$$a_3:y+x^2$$
$$a_4:2xy+x^3=x(2y+x^2)$$
$$a_5:y^2+3x^2y+x^4$$
$$a_6:3xy^2+4x^3y+x^5=x(y+x^2)(3y+x^2)$$
Note that $a_6=0$ has many obvious rational points, for example $(3,-9)$. 
I conjecture: For $n>6$, $a_n=0$ has no nontrivial rational points. How do/would I prove this?

First of all, you can do the same sort of trick to get a closed form for this as we use for any other linear recurrence.
The roots of $U^2-xU -y=0$ are $u_1,u_2=\frac{x\pm\sqrt{x^2+4y}}{2}$.
And $a_n = bu_1^n + cu_2^n$ for some $b,c\in R=\mathbb Q(x,y)[\sqrt{x^2+4y}]$.
Solving for $b$ and $c$, we get $b=-c$ and $b(u_1-u_2)=b\sqrt{x^2+4y}=1$. 
So $$a_n = \frac{1}{\sqrt{x^2+4y}}\left(\left(\frac{x+\sqrt{x^2+4y}}{2}\right)^n - \left(\frac{x-\sqrt{x^2+4y}}{2}\right)^n\right)$$
Since $u_1u_2=-y$, $u_1^n=u_2^n$ for specific rational $x,y$ means that $u_1^{2n}=(-y)^n$, so $-\frac{u_1^2}{y}$ must be an $n$th root of $1$. Now:
$$-\frac{u_1^2}{y} = \frac{-x^2-2y - x\sqrt{x^2+4y}}{2y}$$ which, if $x,y$ are rational, is quadratic over $\mathbb Q$. The only primitive roots of unity that are at most quadratic over $\mathbb Q$ are the square roots, fourth roots, the cube roots, and the sixth roots.
Note that this also shows that if $(x_0,y_0)$ is a rational point of $a_n$, then it is a rational point of $a_{kn}$ for any $k$. This actually extends to any point, because the argument that $-\frac{u_1^2}{y}$ must be a root of unity holds true whenever $u_1^n=u_2^n$ and $y\neq 0$.
It's worth adding in the separate case when $x^2+4y=0$, which is the case when $u_1=u_2=\frac{x}{2}$ and then the closed formula is of the form:
$$a_n = (bn+c)\left(\frac{x}{2}\right)^n$$
So $c=0$ and $bx/2 = 1$ so $b=\frac{2}{x}$ and the formula in this case is:
$$a_n = n\left(\frac{x}{2}\right)^{n-1}$$ 
This is only zero when $x=0$ and hence $y=0$.
One way to see that $a_n$ is a factor of $a_{nk}$ is to note that:
$$a_{nk}=\frac{1}{\sqrt{x^2+4y}}(u_1^{nk}-u_2^{nk}) = \frac{1}{\sqrt{x^2+4y}}(u_1^n-u_2^n)\left(...\right)$$
And just show the elided section must be a polynomial in $x,y.$ It's not hard - it is a combination of expressions of the form $u_1^k+u_2^k$, which causes all the square roots to cancel out.
What happens in the cases given? That is, when $x^2=-y,-2y,-3y$?
Then you get:
$$\begin{align}
(x,y)=(1,-1)&\implies -\frac{u_1^2}{y} = \frac{1-\sqrt{3}}{-2}=\frac{-1+\sqrt{3}}{2}\\
(x,y)=(2,-2)&\implies-\frac{u_1^2}{y}=\frac{0-2\sqrt{-4}}{-4}=i\\
(x,y)=(3,-3)&\implies-\frac{u_1^2}{y}=\frac{-3-3\sqrt{-3}}{-6}=\frac{1+\sqrt{3}}{2}
\end{align}$$
Also, for $a_2$, we get any $(0,y)$ is a root, and $-\frac{u_1^2}{y}=-1$.
So you have all the quadratic and linear roots of unity represented, excepting $1$ itself.
