Question: What's the easiest and simplest way to show that there isn't $A \in M\_2(\mathbb R)$ such that $A^{2004}=\begin{pmatrix}
-1 &0 \\ 
 0&-2 
\end{pmatrix}$?

Thank you.

Answer: 
Suppose $A^{2004}=\begin{pmatrix}
-1 &0 \\ 
 0&-2 
\end{pmatrix}.$

Then $B=A^{1002}$ is a real 2x2 matrix such that
$B^{2}=\begin{pmatrix}
-1 &0 \\ 
 0&-2 
\end{pmatrix}.$
We show that this is not possible. 

Let $\quad B=\begin{pmatrix}
a &b \\ 
 c&d 
\end{pmatrix}.$

Then
$B^{2}=\begin{pmatrix}
a^{2}+bc &ab+bd \\ 
 ac+cd&d^{2}+bc 
\end{pmatrix}=\begin{pmatrix}
-1 &0 \\ 
 0&-2 
\end{pmatrix}\quad \quad \quad (1)$

Considering the (1,2) entry in equation (1) yields 
$$ab+bd=0.$$If $b \neq 0 $ then we must have
$a=-d$ 
but then the diagonal entries of $B^{2}$ would be equal contradicting (1). If $b=0$ then by considering the (1,1) entry in equation (1) we see that $a^{2}=-1$, which is not possible for $a$ real.
We assume that such a $A$ exists.
We have $A^2-\mathrm{Tr}(A)A +(\det A)I\_2= 0$ hence $A^{2004}$ can be written as $\alpha I\_2+\beta A$ where $\alpha$ and $\beta$ are real, since $\mathrm{Tr}(A)$ and $\det A$ are real (and $\beta\neq 0$ because $A^{2004}\neq \alpha I\_2$). Since $\beta A =A^{2004}-\alpha I\_2$, $A$ is diagonal with real entries, namely $A= \begin{pmatrix}d\_1&0\\0&d\_2\end{pmatrix}$ but since $d\_1$ is real we can have $d\_1^{2004}=-1$. 
We can take $2p$ where $p\in\mathbb N$ instead of $2004$.