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

Let $A = \begin{bmatrix} a & b\\ c & d\end{bmatrix}$, then $A^2 = \begin{bmatrix} a & b\\ c & d\end{bmatrix}\cdot \begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd\\ ac + cd & bc + d^2\end{bmatrix}$. Thus:
$A - A^2 = \begin{bmatrix} a-a^2-bc & b-ab-bd\\ c-ac-cd & d-bc-d^2\end{bmatrix} = \begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$. Thus:
$a-a^2-bc = 3$
$d-bc-d^2 = 4$
$b-ab-bd = 1$
$c-ac-cd = 1$.
Subtract the last $2$ equations to each other:
$b-c - a(b-c) - d(b-c) = 0 \to (b-c)(1-a-d) = 0$.
And subtract the first $2$ equations to obtain:
$a-d -(a^2-d^2) = -1 \to (a-d)(1-a-d) = -1$. Thus: $b-c = 0$, and $b = c$. Let's rewrite the system:
$a-a^2-b^2 = 3$
$d-b^2-d^2 = 4$
$b-ab-bd = 1$. Using Wolframalpha:
$a \approx 0.5 - 0.479i$
$b \approx 1.588i$
$d \approx 0.5 + 1.109i$.
Thus all roots are imaginary numbers, which means there is no matrix $A$ with real entries for the equation.
