This question must have been asked but I did not find any complete proof on the site.
Let $A \in End(\mathbb R^n)$ be any endormorphism. Prove that the set of solutions of equation $B^2=A$ is finite or uncountable.
I think the proof is quite long and complicated so any link to a paper or else is welcome.
Sketch of a proof:
It suffices to show that this is true for $End(\Bbb C^n)$.
First, we note that if $A$ has $2$-dimensional eigenspace associated with any eigenvalue, then $A$ has uncountably many square roots. In particular, it suffices to note that $$ \pmatrix{\lambda & t\\&-\lambda}^2 = \pmatrix{\lambda^2\\&\lambda^2} $$ for any $t \in \Bbb C$. Thus, it suffices to study non-derogatory matrices. In particular, it suffices to study their Jordan forms, which have one block per eigenvalue.
First, show that any Jordan block has finitely many square roots (in particular, the $\lambda = 0$ block has no square roots if its size is at least $2$). Then, one must consider the case of direct sums of distinct Jordan blocks.
Note in particular that anything that commutes with a non-derogatory matrix must be a polynomial of that matrix (cf. Horn and Johnson). Thus, a square root of a non-derogatory matrix in the form described above must also be a (conformal) direct sum.