TITLE: Why does $A^8=I_3$ implies $A^4=I_3$?
QUESTION [5 upvotes]: Let $A\in M_3(\mathbb{Q})$ such that $A^8=I_3$. Prove that $A^4=I_3$

(We denote by $m_X$ the minimal polynomial of matrix X and by $p_X$ its characteristic polynomial and by $gr \space X$ its degree)
The solution is the following:
$1)$ $m_a$ divides $P\in\mathbb{Q}[X]$, $$P(x)=x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$$
$2)$ Since $A^4\neq I_3$, then $p_A$ would have at least one common root with $x^4+1$
$3)$ Since $x^4+1$ is irreducible over $\mathbb{Q}[X]$ we have that $x^4+1|m_A$
$4)$ $A\in M_3(\mathbb{Q})$ so $gr\space m_A\leq3$ and $gr\space x^4+1=4$ , contradiction
I understood only points $1)$ and $4)$. Can you please help me to understand $2)$ and $3)$? Are they even right?

REPLY [2 votes]: I think this version of the same argument is clearer and does involve contradiction:

$m_A$ divides $x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$.
the degree of $m_A$ is at most $3$ because $A\in M_3(\mathbb{Q})$.
Therefore, $m_A$ divides $(x-1)(x+1)(x^2+1)=x^4-1$.
Write $x^4-1=m_A(x) q(X)$.
Then, $A^4-I=m_A(A) q(A)=0$.