$A\in M_n(\mathbb{Q})$ satisfy $A^5=I$, and 1 isn't an eigenvalue. Please help me find answer for the following task:
Prove, that $A\in M_n(\mathbb{Q})$ satisfy $A^5=I$, and 1 isn't an eigenvalue. Show, that $4 \mid n$
 A: If $P(x)=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$ then $P(A)=0$. 

Prove that:


*

*$A-I$ is invertible,

*If $Q(x)=x^4+x^3+x^2+x+1$ then $Q(A)=0$,

*$Q(x)$ is irreducible over $\mathbb Q$,

*If $\chi_A(x)$ is the characteristic polynomial of $A$ then $\chi_A(x)=Q^k(x)$ for some $k\in\mathbb N$.


Now compare the degrees of $\chi_A(x)$ and $Q(x)$.
A: Because $P(A)=0$ with $P(X)=X^5-1$, $A$ is diagonalizable over $\mathbb{C}$. So we can write $\mathbb{C}^n = E_1 \oplus E_2 \oplus E_3 \oplus E_4$, where $A$ is homothetic of coefficient $\lambda_k= e^{i2k\pi/5}$ on $E_k$. 
Because $Q$ is real, $E_3= \overline{E_2}$ and $E_4=\overline{E_1}$, hence $\dim(E_1)=\dim(E_4)$ and $\dim(E_3)=\dim(E_2)$.
Moreover, $\displaystyle \sum\limits_{i=1}^4 \lambda_i \dim(E_i)=\text{tr}(Q) \in \mathbb{Q}$. Because $\displaystyle \cos \left( \frac{\pi}{5} \right)= \frac{1+\sqrt{5}}{4}$, it is equivalent to $\dim(E_1)=\dim(E_2)$.
Therefore, $n=\dim(\mathbb{C}^n)= \dim(E_1)+\dim(E_2)+ \dim(E_3)+\dim(E_4)= 4 \dim(E_1)$; hence $4 \mid n$.
