We denote by $\mathrm{GL}_2(\mathbb{Z})$ the set of invertible elements of $\mathcal{M}_2(\mathbb{Z})$. For $B \in \mathrm{GL}_2(\mathbb{Z})$ with $\sigma = \operatorname{Tr} B$ and $\nu = \det B$, we have $B^n = E_{n-1}(\sigma,\nu) \cdot B - \nu E_{n-2}(\sigma,\nu) \cdot I_2$ and $\operatorname{Tr}(B^n) = D_n(\sigma,\nu)$. We denote $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $\tau = \operatorname{Tr} A$, $\delta = \det A$. Deduce that if $A$ is an $n$-th power ($n \geqslant 2$) in $\mathrm{GL}_2(\mathbb{Z})$, then there exist $\sigma \in \mathbb{Z}$ and $\nu \in \{-1,1\}$ such that i. $E_{n-1}(\sigma, \nu)$ divides $b$, $c$ and $a - d$. One will briefly justify that $E_{n-1}(\sigma, \nu)$ is indeed an integer. ii. $\tau = D_n(\sigma, \nu)$ and $\delta = \nu^n$.
We denote by $\mathrm{GL}_2(\mathbb{Z})$ the set of invertible elements of $\mathcal{M}_2(\mathbb{Z})$. For $B \in \mathrm{GL}_2(\mathbb{Z})$ with $\sigma = \operatorname{Tr} B$ and $\nu = \det B$, we have $B^n = E_{n-1}(\sigma,\nu) \cdot B - \nu E_{n-2}(\sigma,\nu) \cdot I_2$ and $\operatorname{Tr}(B^n) = D_n(\sigma,\nu)$.
We denote $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $\tau = \operatorname{Tr} A$, $\delta = \det A$.
Deduce that if $A$ is an $n$-th power ($n \geqslant 2$) in $\mathrm{GL}_2(\mathbb{Z})$, then there exist $\sigma \in \mathbb{Z}$ and $\nu \in \{-1,1\}$ such that
i. $E_{n-1}(\sigma, \nu)$ divides $b$, $c$ and $a - d$. One will briefly justify that $E_{n-1}(\sigma, \nu)$ is indeed an integer.
ii. $\tau = D_n(\sigma, \nu)$ and $\delta = \nu^n$.