Considering $A \in M_2(C) $ and its minimal polynomial $\pi_A$, I have to find a relation between something with $\pi_A$ and the fact that A admits a square root (which means $\exists X \in M_2(C), X^2=A$).
What seems the more logical to me would be that $\pi_A$ splits with simple roots.
It works in one way: if it splits with simple roots, then A is diagonalizable so it is similar to a diagonal matrix. Any diagonal matrix admits a square root and I have proved that if a matrix M admits a square root, then any matrix similar to M admits one. Thus A admits a square root.
In the other way, let's assume A admits a square root B. We are in C so $\pi_A $ splits but I have to show its roots are simple. If its degree is one, its fine. If it is 2, it is equal to the characteristic polynomial of A : $X^2-tr(A)X+det(A) $. So if the roots are not simple, $\Delta=tr(A)^2-4det(A)=0$ thus $tr(A)=\pm 2 \sqrt{det(A)} $ thus $tr (X^2)=\pm 2det(X) $ but I dont see any contradiction... Am I following a false lead?
Thanks.
If $m_A(x) = (x - \lambda)^2$ has a double root, then $A$ is similar to
$$ A' = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}. $$
If $\lambda \neq 0$, then $A'$ has a square root
$$ B' = \begin{pmatrix} \sqrt{\lambda} & \frac{1}{2\sqrt{\lambda}} \\ 0 & \sqrt{\lambda} \end{pmatrix} $$
where $\sqrt{\lambda}$ is some choice of a complex square root of $\lambda$ and.
If $\lambda = 0$, then one can readily verify that $A'$ doesn't have a square root. Thus, the condition for $A$ to have a square root is that $m_A(x) \neq x^2$.