Given $S$ a real symmetric matrix. The original question is to find an if and only if condition for there to exist an anti-symmetric matrix $A$ such that $S=A^2.$ My guess is that the condition should be $S$ having all non-positive eigencalues.
One direction is trivial: if such an $A$ exists, then $S=-A^TA$ and we know $A^TA$ is positive-definite and thus have non-negative eigenvalues. So $S$ has non-positive eigenvalues.
Then I am kind stuck at the other direction. Given such an $S$, we can diagonalize $-S$ with unitary $U$:$$-S=US'U^T$$ with $S'$ have positive diagonal elements. Then we can take square root of $S'$ i.e. $-S=U\sqrt{S'}\sqrt{S'}U^T.$ Then put $A=\sqrt{S'}U^T$ we have $$S=-A^TA.$$ But I cannot proceed to show $-A^TA=A^2$ i.e. $A$ is anti-symmetric. So I am wondering how to proceed from there or is my initial iff condition correct? Any hint is appreciated!