MHB Why does $Y_n$ have that specific form?

[mathmari]

Hey!

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 1.

The solutions of $(1)$ in $R[T]$arwe given precisely y $$X=\pm X_n , Y=\pm Y_n , n=0, 1, 2, \dots$$ We write $V \sim W$ to denote that the polynomials $V$ and $W$ in $R [T]$ take the same value at $T = 1$. Notice that the relation $Z \sim 0$ is diophantine over $R [T]$ with coefficients in $\mathbb{Z}[T]$, indeed $$Z \sim 0 \leftrightarrow \exists X \in R[T]: Z=(T-1)X$$

Lemma 2.

We have $Y_n \sim n$, for $n=0, 1, 2, \dots $.

Proof.

From $(3)$ and $(2)$ follows $$Y_n=\sum_{i=1, i \text{ odd }}^{n}\binom{n}{i}(T^2-1)^{(i-1)/2}T^{n-i}$$ Substitute now $T=1$.
Could you explain to me the proof of the Lemma $2$?

I haven't understood it...

Why is $Y_n$ of that form?

 

[Euge]

Hi mathmari,

The form you see for $Y_n$ results from applying the binomial theorem to the right-hand side of $(3)$, then using $(2)$ to express $Y_n$ all in terms of $T$. Indeed, by the binomial theorem,

$$X_n + UY_n = (T + U)^n = \sum_{k = 0}^n \binom{n}{k}T^{n-k}U^k.$$

The terms of the sum for which $k$ is even is $X_n$. Consider all terms of the sum above for which $k$ is odd. For all such $k$, $(k-1)/2$ is a positive integer and $U^k =UU^{k-1} = U(U^2)^{(k-1)/2} = U(T^2 - 1)^{(k-1)/2}$. Hence

$$\sum_{0\le k \le n,\, k\, \text{odd}} \binom{n}{k}T^{n-k}U^k = U\sum_{1\le k\le n,\, k\, \text{odd}} \binom{n}{k}(T^2 - 1)^{(k-1)/2}T^{n-k} = UY_n,$$

and so

$$Y_n = \sum_{1\le k\le n,\, k\, \text{odd}} \binom{n}{k}(T^2 - 1)^{(k-1)/2}T^{n-k}.$$

Now

$$Y_n = n + \sum_{3\le k \le n,\, k\, \text{odd}} \binom{n}{k}(T^2 - 1)^{(k-1)/2}T^{n-k} \qquad (*)$$

and when $T = 1$, all terms in the sum on the right-hand side of $(*)$ are zero. Therefore, $T_n \sim n$.

 

[mathmari]

The terms of the sum for which $k$ is even is $X_n$.

Why are the terms of the sum for which $k$ is even, $X_n$ ?

Does this mean that the terms of the sum for which $k$ is odd, is $UY_n$ ? Why?

$$Y_n = n + \sum_{3\le k \le n,\, k\, \text{odd}} \binom{n}{k}(T^2 - 1)^{(k-1)/2}T^{n-k} \qquad (*)$$

Why does this hold?

 

[Euge]

Why are the terms of the sum for which $k$ is even, $X_n$ ?

Does this mean that the terms of the sum for which $k$ is odd, is $UY_n$ ? Why?

Perhaps I should put it this way. We break up the sum $\sum\limits_{k = 0}^n \binom{n}{k}T^{n-k}U^k$ into even and odd parts. Let $A_n$ be the sum of the even parts and $B_n$ the sum of the odd parts. As I've shown, $B_n = UC_n$, where $C_n$ is the element of $\Bbb Z[T]$ given by $C_n = \sum\limits_{1 \le k \le n,\, k\, \text{odd}} \binom{n}{k}(T^2 - 1)^{(k-1)/2}T^k$. Now we have expressed $(T + U)^n = A_n + UC_n$, where $A_n, B_n \in \Bbb Z[T]$ have been determined. Thus $X_n = A_n$ and $Y_n = C_n$. I hope this makes more sense.As for your last question, the $n$ appearing from the right-hand side of $(*)$ comes from the $k = 1$ term of $Y_n$. That's all there is to it.

Edit: By the way, $A_n \in \Bbb Z[T]$ since we can express

$$A_n = \sum_{0\le k\le n,\, k\, \text{even}} \binom{n}{k}T^{n-k}(T^2-1)^{k/2}.$$

 

[mathmari]

Perhaps I should put it this way. We break up the sum $\sum\limits_{k = 0}^n \binom{n}{k}T^{n-k}U^k$ into even and odd parts. Let $A_n$ be the sum of the even parts and $B_n$ the sum of the odd parts. As I've shown, $B_n = UC_n$, where $C_n$ is the element of $\Bbb Z[T]$ given by $C_n = \sum\limits_{1 \le k \le n,\, k\, \text{odd}} \binom{n}{k}(T^2 - 1)^{(k-1)/2}T^k$. Now we have expressed $(T + U)^n = A_n + UC_n$, where $A_n, B_n \in \Bbb Z[T]$ have been determined. Thus $X_n = A_n$ and $Y_n = C_n$. I hope this makes more sense.As for your last question, the $n$ appearing from the right-hand side of $(*)$ comes from the $k = 1$ term of $Y_n$. That's all there is to it.

Edit: By the way, $A_n \in \Bbb Z[T]$ since we can express

$$A_n = \sum_{0\le k\le n,\, k\, \text{even}} \binom{n}{k}T^{n-k}(T^2-1)^{k/2}.$$

I got it! Thanks for the explanation! (Sun)