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

Completing the square, we find that the given equation is equivalent to
$$ (A-\tfrac12 I)^2 = \tfrac14 I - \left[\begin{matrix} 3 & 1 \\ 1 & 4 \end{matrix}\right] = -\frac14 \left[\begin{matrix} 11 & 4 \\ 4 & 15 \end{matrix}\right] \tag{$\ast$} $$
If you wanted $A$ to be real, we can now show it can't be done by setting
$$ A-\tfrac12 I = \left[\begin{matrix} a & b \\ c & d\end{matrix}\right] $$
which yields
$$ \left[\begin{matrix} a^2+bc & b(a+d) \\ c(a+d) & bc+d^2 \end{matrix}\right] = \left[\begin{matrix} -11/4 & -1 \\ -1 & -15/4 \end{matrix}\right] $$
Since $b(a+d)$ and $c(a+d)$ are equal (and not zero), we get $b=c$; but then the diagonal entries are $a^2+b^2$ and $b^2+d^2$, which are nonnegative, contrary to what we see.
If you don't mind $A$ having complex entries, you can note that the RHS of ($\ast$) is symmetric, so it's diagonalizable; thus it has a (complex) square root, and adding $\frac12 I$ to that square root will give $A$.
