If $n\in\textbf{N}$, what are the matrices $A\in M_{n}(\textbf{Z}/n\textbf{Z})$ such that $A^3=I_{n}$ ?
An exercise I found and have no clue how to do. Even if I assume $n$ is prime and we are talking about matrices whose coefficients belong to a field, I don't really see how to proceed. Help?
edit: both the $n$ in the size of the matrices and the $n$ in $\textbf{Z}/n\textbf{Z}$ are the same by the way
We assume that $n$ is a prime. Let $J_n$ be the nilpotent Jordan block of dimension $n$. $x^3-1=(x-1)(x^2+x+1)$. Then $A$ is semi-simple except if:
Case 1. $n=3$. Since $(x^3-1)=(x-1)^3$, $A=I+N$ where $N=0$ or $N\sim diag(J_2,0)$ or $N\sim J_3$.
Case 2.1. $n\not= 3$ and $x^2+x+1$ has no roots in $\mathbb{F}_{n}$, that is $-3$ is not a square. Then $A$ is diagonalizable over $\mathbb{F}_{n^2}$ and $x^2+x+1$ is irreducible over $\mathbb{F}_{n}$. Thus $\chi_A(x)$ has the form $(x-1)^{n-2p}(x^2+x+1)^p$, that implies $A=I_n$ or $A\sim diag(I_{n-2p},U_1,\cdots U_p)$ where $U_i=\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$ (the companion matrix of $x^2+x+1$).
Case 2.2. $n\not= 3$ and $x^2+x+1$ has $2$ roots $\alpha,\beta\in\mathbb{F}_{n}$ (or $-3$ is a square, as when $n=7$). As Jim wrote above, $A$ is diagonalizable over $\mathbb{F}_n$ and its eigenvalues belong to the set $\{1,\alpha,\beta\}$.