Show that there do not exist 3 $\times$ 3 matrices $A$ over $\mathbb{Q}$ such that $A^8 = I $and $A^4 \neq I.$. Show that there do not exist 3 $\times$ 3 matrices $A$ over $\mathbb{Q}$ such that $A^8 = I $and $A^4 \neq I.$.
I am aware that the minimal polynomial of $A$ divides $(x^8−1)=(x^4−1)(x^4+1)$.If the minimal polynomial divides $x^4+1$ then it will have roots outside $\mathbb{Q}$.The roots of the minimal polynomial are also roots of Characteristic polynomial of A , thus Characteristic polynomial of $A$ has roots outside $\mathbb{Q}$. I am unable to progress from this point onwards.
I would really appreciate some help.
Thanks !

Hint: Prove that $x^4+1$ is irreducible in $\Bbb{Q}[x]$. What factors of $x^4+1$ can thus occur as factors of the minimal polynomial of a $3\times3$ matrices with entries from $\Bbb{Q}$?
