Square root of a specific 3x3 matrix

Assume that there exists a real number $t$ and a real matrix $B$ such that $A(t)=B^2$.

Note that $-1$ is an eigenvalue of $A(t)$, hence $A(t)+I=(B-\mathrm{i}I)(B+\mathrm{i}I)$ is singular. This implies that $B-\mathrm{i}I$ or $B+\mathrm{i}I$ is singular. Since $B$ is real valued, this means that both $B-\mathrm{i}I$ and $B+\mathrm{i}I$ are singular. Likewise, $1$ is an eigenvalue of $A(t)$, hence $A(t)-I=(B-I)(B+I)$ is singular. This implies that $B-I$ or $B+I$ is singular. Hence the eigenvalues of $B$ are $\{\mathrm{i},-\mathrm{i},1\}$ or $\{\mathrm{i},-\mathrm{i},-1\}$.

In both cases, $B$ has three distinct eigenvalues hence $B$ is diagonalizable on $\mathbb{C}$. This implies that the eigenvalues of $A(t)$ are $-1$ (twice) and $1$ (once) and that $A(t)$ is diagonalizable as well. Hence $t=-1$. We now look at the matrix $A(-1)$.

One can check that $A(-1)$ is diagonalizable hence $A(-1)$ is similar to a diagonal matrix with diagonal $(1,-1,-1)$. Both $I_1$ (the $1\times1$ matrix with coefficient $1$) and $-I_2$ (the $2\times2$ diagonal matrix with diagonal coefficients $-1$) have square roots: take $I_1$ for $I_1$ and the rotation matrix $\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$ for $-I_2$. Hence $A(-1)$ is a square.

Finally $A(t)$ is a square if and only if $t=-1$.