TITLE: How do you prove that there does not exist a $3 \times 3$ matrix over $\Bbb{Q}$ such as $A^8=I$ and $A^4 \ne I$?
QUESTION [6 upvotes]: I am kind of stuck with the following problem:
Prove that there does not exist a $3 \times 3$ matrix over $\Bbb{Q}$ such as $A^8=I$ and $A^4 \ne I$. 
I already tried some things and got that if $A^8=I$, then $(A^4)^2=I$.
And I guess a similar reasoning would work for this one.   I bet that I am missing something obvious but I can't find what is it :(

REPLY [8 votes]: Hint: The minimal polynomial of A divides $(x^8−1)=(x^4−1)(x^4+1)$, and $(x^4+1)$ is irreducible over $\mathbb{Q}$