We denote by $\mathrm{GL}_2(\mathbb{Z})$ the set of invertible elements of $\mathcal{M}_2(\mathbb{Z})$. We denote $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $\tau = \operatorname{Tr} A$, $\delta = \det A$. Let $A$ be an element of $\mathrm{GL}_2(\mathbb{Z})$ for which there exist $\sigma \in \mathbb{Z}$ and $\nu \in \{-1,1\}$ satisfying: i. $E_{n-1}(\sigma, \nu)$ divides $b$, $c$ and $a - d$. ii. $\tau = D_n(\sigma, \nu)$ and $\delta = \nu^n$. For simplicity, we denote $p = E_{n-1}(\sigma, \nu)$. We then define a matrix $B = \begin{pmatrix} r & s \\ t & u \end{pmatrix}$ with $$r = \frac{1}{2}\left(\sigma + \frac{a-d}{p}\right) \quad s = \frac{b}{p} \quad t = \frac{c}{p} \quad u = \frac{1}{2}\left(\sigma - \frac{a-d}{p}\right)$$ a) By introducing a complex root of the polynomial $X^2 - \sigma X + \nu$ and using the relation $D_n(x + a/x, a) = x^n + a^n/x^n$, show that $$\tau^2 - 4\delta = p^2(\sigma^2 - 4\nu) \quad \text{then} \quad ru - st = \nu$$ Deduce that $B$ belongs to $\mathrm{GL}_2(\mathbb{Z})$. b) Show that $A = B^n$.
We denote by $\mathrm{GL}_2(\mathbb{Z})$ the set of invertible elements of $\mathcal{M}_2(\mathbb{Z})$. We denote $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $\tau = \operatorname{Tr} A$, $\delta = \det A$.
Let $A$ be an element of $\mathrm{GL}_2(\mathbb{Z})$ for which there exist $\sigma \in \mathbb{Z}$ and $\nu \in \{-1,1\}$ satisfying:
i. $E_{n-1}(\sigma, \nu)$ divides $b$, $c$ and $a - d$.
ii. $\tau = D_n(\sigma, \nu)$ and $\delta = \nu^n$.
For simplicity, we denote $p = E_{n-1}(\sigma, \nu)$. We then define a matrix $B = \begin{pmatrix} r & s \\ t & u \end{pmatrix}$ with
$$r = \frac{1}{2}\left(\sigma + \frac{a-d}{p}\right) \quad s = \frac{b}{p} \quad t = \frac{c}{p} \quad u = \frac{1}{2}\left(\sigma - \frac{a-d}{p}\right)$$
a) By introducing a complex root of the polynomial $X^2 - \sigma X + \nu$ and using the relation $D_n(x + a/x, a) = x^n + a^n/x^n$, show that
$$\tau^2 - 4\delta = p^2(\sigma^2 - 4\nu) \quad \text{then} \quad ru - st = \nu$$
Deduce that $B$ belongs to $\mathrm{GL}_2(\mathbb{Z})$.
b) Show that $A = B^n$.