I want to find $A \in M_{n}(\mathbb{Q})$ not equal to identity matrix such that $A^{n+1} = I_n$.
From my previous post, $A \in M_{n\times n}(\mathbb{Q})$ with $A\neq I_n$, $A^p=I_n$ then $p\leq n+1$.
Followings are my trials:
Note that the matrix equation gives $t^{n+1} - 1$, but since $t^{n+1} -1 = (t-1) (1+t+t^2+ \cdots + t^{n-1})$. Note that minimal polynomial divides this and since $m_A(t) \neq t-1$, it divides $(1+t+\cdots + t^{n-1})$. But this can be reducible ...
We have $p(t)=t^{n+1}-1=(t-1) (1+t+t^2+\dots+t^n)$.
Consider the companion matrix of $\chi(t)=1+t+t^2+\ldots+t^n$. It is $$A_n=\begin{bmatrix} 0 & 0 & \dots & 0 & -1 \\ 1 & 0 & \dots & 0 & -1 \\ 0 & 1 & \dots & 0 & -1\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & -1 \end{bmatrix}.$$ It means its characteristic polynomial is $\chi$. By Cayley-Hamilton's theorem, we get $p(A_n) =(A_n-I_n) \cdot \chi(A_n)=(A_n-I_n)\cdot O_n=O_n$, e. g. $$\begin{bmatrix}0&-1\\1&-1\end{bmatrix}^3=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$
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