Sizes of square matrices such that $\mathrm{A}^2=-\mathrm I$ does not exist

Question

This question Prove that there is no 5 × 5 matrix A such that $A^2 = −I$ Raises the questions:

For what values of n $n \times n$ matrix $A$ there is no $\mathrm {A}$ exists s.t. $\mathrm {A}^2=-\mathrm {I}$

or more generally

For what values of $\mathrm {B},p$, $n \times n$ matrix there is no $\mathrm {A}$ exists s.t. $A^p=\mathrm {B}$

Answer 1

Whenever $n$ is odd. For example, $det(A)^2=det(A^2)=(-1)^n$, so whenever $n$ is odd, you need a square root of $-1$.

Answer 2

Since for A real $$\det(A^2)=[\det(A)]^2\ge0$$ such $A$ doesn't exist for all $n$ odd and $B$ such that $\det(B)<0$.