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: 
If $\lambda$ and $\mu$ are the eigenvalues of $A$ then $\lambda^{2004}=-1$ and $\mu^{2004}=-2$. This implies that $\lambda$ and $\mu$ are complex, non-conjugate eigenvalues of $A$. But for any real matrix all non-real eigenvalues come in conjugated pairs.
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 $2004$.