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mathmari
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Hey! 
I am reading the proof of the following lemma but I got stuck at some points... Let $R$ be any integral domain of characteristic zero.
We consider the Pell equation $$X^2-(T^2-1)Y^2=1 \tag 1$$ over $R[T]$.
Let $U$ be an element in the algebraic closure of $R[T]$ satisfying $$U^2=T^2-1 \tag 2$$
Define two sequences $X_n, Y_n, n=0, 1, 2, \dots$, of polynomials in $\mathbb{Z}[T]$, by setting $$X_n+UY_n=(T+U)^n\tag 3$$
Lemma.
The solutions of $(1)$ in $R[T]$ are given precisely by $$X=\pm X_n, Y=\pm Y_n, n=0, 1, 2, \dots$$
Proof.
$(1)$ is equivalent too $$(X-UY)(X+UY)=1 \tag 4$$
From $(3)$ and $(2)$ follows $$X_n-UY_n=(T-U)^n=(T+U)^{-n}$$
Hence the $X_n, Y_n$ are solutions of $(1)$.
Conversely, suppose $X$ and $Y$ in $R[T]$ satisfy $(1)$.
Let us parametrise the curve $(2)$ by $$T=\frac{t^2+1}{t^2-1}\ \ ,\ \ U=\frac{2t}{t^2-1}$$
The rational functions $X+UY$ and $X-UY$ in $t$ have poles only at $t=\pm 1$.
Moreover $(4)$ implies they have zeroes only at $t=\pm 1$.
Hence $$X+UY=c\left ( \frac{t+1}{t-1}\right )^m=c(T+U)^m, c \in R, m \in \mathbb{Z}$$
Thus also $X-UY=c(T-U)^m$.
But substistuting this in $(4)$ gives $c^2=1$, which proves the lemma by $(3)$.
Could you explain to me the following part of the proof?
Moreover $(4)$ implies they have zeroes only at $t=\pm 1$.
Hence $$X+UY=c\left ( \frac{t+1}{t-1}\right )^m=c(T+U)^m, c \in R, m\in \mathbb{Z}$$
Thus also $X-UY=c(T-U)^m$.
But substistuting this in $(4)$ gives $c^2=1$, which proves the lemma by $(3)$.
Why do we have from the relation $(4)$ that the functions $X+UY$ and $X-UY$ have zeroes only at $t=\pm 1$ ?
How do we get to the equation $X+UY=c\left ( \frac{t+1}{t-1}\right )^m=c(T+U)^m$ ? How does $c$ appear? 11
I am reading the proof of the following lemma but I got stuck at some points... Let $R$ be any integral domain of characteristic zero.
We consider the Pell equation $$X^2-(T^2-1)Y^2=1 \tag 1$$ over $R[T]$.
Let $U$ be an element in the algebraic closure of $R[T]$ satisfying $$U^2=T^2-1 \tag 2$$
Define two sequences $X_n, Y_n, n=0, 1, 2, \dots$, of polynomials in $\mathbb{Z}[T]$, by setting $$X_n+UY_n=(T+U)^n\tag 3$$
Lemma.
The solutions of $(1)$ in $R[T]$ are given precisely by $$X=\pm X_n, Y=\pm Y_n, n=0, 1, 2, \dots$$
Proof.
$(1)$ is equivalent too $$(X-UY)(X+UY)=1 \tag 4$$
From $(3)$ and $(2)$ follows $$X_n-UY_n=(T-U)^n=(T+U)^{-n}$$
Hence the $X_n, Y_n$ are solutions of $(1)$.
Conversely, suppose $X$ and $Y$ in $R[T]$ satisfy $(1)$.
Let us parametrise the curve $(2)$ by $$T=\frac{t^2+1}{t^2-1}\ \ ,\ \ U=\frac{2t}{t^2-1}$$
The rational functions $X+UY$ and $X-UY$ in $t$ have poles only at $t=\pm 1$.
Moreover $(4)$ implies they have zeroes only at $t=\pm 1$.
Hence $$X+UY=c\left ( \frac{t+1}{t-1}\right )^m=c(T+U)^m, c \in R, m \in \mathbb{Z}$$
Thus also $X-UY=c(T-U)^m$.
But substistuting this in $(4)$ gives $c^2=1$, which proves the lemma by $(3)$.
Could you explain to me the following part of the proof?
Moreover $(4)$ implies they have zeroes only at $t=\pm 1$.
Hence $$X+UY=c\left ( \frac{t+1}{t-1}\right )^m=c(T+U)^m, c \in R, m\in \mathbb{Z}$$
Thus also $X-UY=c(T-U)^m$.
But substistuting this in $(4)$ gives $c^2=1$, which proves the lemma by $(3)$.
Why do we have from the relation $(4)$ that the functions $X+UY$ and $X-UY$ have zeroes only at $t=\pm 1$ ?
How do we get to the equation $X+UY=c\left ( \frac{t+1}{t-1}\right )^m=c(T+U)^m$ ? How does $c$ appear? 11