We denote by $\mathbb{H}$ the real quaternion division ring, that is, $\mathbb{H}=\{a+bi+cj+dk\mid a,b,c,d\in\mathbb{R}\}$ where the following multiplication conditions are imposed: $i^2=j^2=k^2=-1,ij=k,ji=-k,jk=i,kj=-i,ki=j,ik=-j$, and every $a\in\mathbb{R}$ commutes with $i,j,k.$ Note that $\det A$ is understood the Dieudonné determinant for $A\in M_n(\mathbb{H}).$
Let $A\in M_n(\mathbb{H})$ and $\det A=1.$ We have shown that if there exist $B,C\in M_n(\mathbb{H})$ such that $A=BCB^{-1}C^{-1},$ then there exists $D\in M_n(\mathbb{H})$ such that $\det D=1$ and $A=D^2.$
I want to prove this converse. But I don't still do it, and I don't know if this converse is true. Thank for all your support.