Solving a simple matrix polynomial Does there exist a $2\times 2$ Matrix $A$ such that
$A-A^2=\begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$ ?

This is a very unusual question. Let me try to give a conceptual proof that it has complex but no real solutions. First working over the complex numbers note that $\begin{pmatrix}3&1\\1&4\end{pmatrix}$ is diagonalizable as shown in NotNotLogical's answer (the eigenvalues are real). So conjugate the equation by $P$ and $P^{-1}$ so that it becomes,
$$B-B^2=\begin{pmatrix}\lambda &0\\0&\mu\end{pmatrix}$$ 
where $B=PAP^{-1}$ and I write $\lambda$, $\mu$ for the eigenvalues. 
Now one can find diagonal $B$ by solving two quadratic equations over $\mathbb{C}$,
$$x-x^2=\lambda$$ and 
$$y-y^2=\mu$$ Thus $A$ exists.
In any case $A$ must have a complex eigenvalue, say $\alpha$ it follows then that $\alpha-\alpha^2$ is an eigenvalue of  $\begin{pmatrix}3&1\\1&4\end{pmatrix}$ from which it follows that $\alpha$ is not real. Therefore over $\mathbb{C}$ there are two distinct eigenvalues and thus $A$ is diagonalizable over $\mathbb{C}$. Now if $B$ is the diagonalization of $A$ over the complex numbers then $B-B^2$ is diagonal so by the uniqueness of diagonalization $B$ must satisfy the equation above. And this implies that the given matrix  $\begin{pmatrix}3&1\\1&4\end{pmatrix}$ can be diagonalized by a complex matrix, but we also know that it is diagonalized by a real matrix, by uniqueness we have that $A$ is diagonalized by a real matrix and we know this is false, so there is no real solution.
Well its little messy, but maybe it can be streamlined.
