A matrix of order 8 over $\mathbb{F}_3$ What is an example of an invertible matrix of size 2x2 with coefficients in $\mathbb{F}_3$ that has exact order 8? 
I have found by computation that the condition that the 8th power of a matrix 
$\begin{bmatrix}a & b\\c & d\end{bmatrix}$ is the identity is
$$
b c (a + d)^2 (a^2 + 2 b c + d^2)^2 + ((a^2 + b c)^2 + 
   b c (a + d)^2)^2=1, \qquad b (a + d) (a^2 + 2 b c + d^2) (a^4 + 4 a^2 b c + 2 b^2 c^2 + 
   4 a b c d + 4 b c d^2 + d^4)=0, \qquad c (a + d) (a^2 + 2 b c + d^2) (a^4 + 4 a^2 b c + 2 b^2 c^2 + 
   4 a b c d + 4 b c d^2 + d^4)=0, \qquad b c (a + d)^2 (a^2 + 2 b c + 
    d^2)^2 + (b c (a + d)^2 + (b c + d^2)^2)^2=1
$$
and the condition for invertibility is $ad\neq bc$. If the 4th power is not the identity, then no power that is not a multiple of 8 is not the identity (because we could cancel out to either get that the first power is the identity or that the second power is the identity, both lead to contradiction). That is another cumbersome condition to write out.
I hope somebody can suggest a nicer way.

If $A$ satisfies $A^8 = I$, then all of the eigenvalues $\lambda$ of $A$ satisfy $\lambda^8 = 1$. Rearranging and factoring gives
$$0 = \lambda^8 - 1 = (\lambda^4 + 1)(\lambda^4 - 1) .$$ Since we want $A^4 \neq I$ (otherwise the order of $A$ divides $4$), at least one eigenvalue of $A$ must be a solution of $\lambda^4 + 1$.

But $$0 = \lambda^4 + 1 = (\lambda^2 - \lambda - 1) (\lambda^2 + \lambda - 1) ,$$ and both of the quadratic factors are irreducible, so the eigenvalues of $A$ must be the roots of $\lambda^2 \pm \lambda - 1$ for some choice of $\pm$, and hence that quantity is the characteristic polynomial of $A$. Any monic polynomial is the characteristic polynomial of its companion matrix, giving the solutions $$\pmatrix{0 & 1 \\ 1 & \mp 1} .$$ In fact, the uniqueness of the Jordan normal form implies that all matrices of order $8$ in $GL(2, \Bbb F_3)$ are conjugates of these.

