How can we show that all the solutions are given by $X_m(a), Y_m(a)$?

Question

Let $F$ be an integral domain with characteristic $2$. Let $a\in F[t]$ and $a \notin F$. Let $\alpha (a)$ be a root of the equation $x^2+ax+1=0$. We define two sequences $X_m(a), Y_m(a) \in F[t], m \in \mathbb{Z}$ by
$$X_m(a)+\alpha (a)Y_m(a)=(\alpha (a))^m=(a+\alpha (a))^{-m} \tag 1$$

Lemma.

Let $F$ be an integral domain with characteristic $p=2$. Let $a \in F[t], a \notin F$. For all $m, n \in \mathbb{Z}$ we have :

1. $X_m(a)$ (resp. $Y_m(a)$) is equal to the polynomial obtained by substituting $a$ for $t$ in $X_m(t)$ (resp. $Y_m(t)$).
   The degree of the polynomial $X_m(t)$ is $m-2$, if $m \geq 2$.
   The degree of the polynomial $Y_m(t)$ is $m-1$, if $m \geq 2$.
   $X_{-m}=X_m(a)+aY_m(a)$
   $Y_{-m}(a)=Y_m(a)$

2. All solutions $X, Y \in F[t]$ of the equation $$X^2+aXY+Y^2=1\tag 2$$ are given by $X_m(a), Y_m(a)$, with $m \in \mathbb{Z}$.

I want to prove this lemma but I am facing some difficulties at $2$.

First I showed that $(X_m(a), Y_m(a))$ is a solution of the equation $(2)$. But how could we show that all the solutions of the equation $(2)$ are given by $X_m(a), Y_m(a)$.

I have no idea how to do that... Could you give me some hints?

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EDIT:

At the paper that I am looking I found the corresponding lemma for the case that the characteristic is not $2$ and its proof:

PROOF.

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I want to try to do the same for the case $\text{char}=2$ but first I have to clarify some points at the proof above.

Why do we consider the field $K=R(t)(a)$?

Is $S$ the set of points at which the functions of $K$ are not defined?

Answer 1

Suppose $(X,Y)$ is a solution where $X,Y \neq 0$.

Then, by looking at the degrees in the equation, we either have $\deg X = \deg a + \deg Y$ or $\deg Y = \deg a + \deg X$.
Suppose we are in the first case and let $X' = X + aY$. $(X',Y)$ is again a solution of the equation, and looking at the first coefficients of $X,Y,a$, we see that the coefficient of $T^{\deg X}$ in $X'$ vanishes (here we use that $F$ is an integral domain), so $\deg X' < \deg X$, and so we get a "smaller" solution.

We can repeat this process until we reach a solution where $X=0$ or $Y=0$. In that case it is easy to see that the only solutions are $(0,1)$ and $(1,0)$.

So every solution can be obtained by starting from one of those two and applying the operations $(X,Y) \mapsto (X,Y+aX)$ and $(X,Y) \mapsto (X+aY,Y)$, and those are exactly the $(X_m(a), Y_m(a))$ for $m \in \Bbb Z$