<s>[INST] Proving $(A+B)^{-1} = A^{-1} + B^{-1}$ when there exists $J$ so that $J^2 = -I$ I'm having trouble proving this biconditional statement:

Prove that there exist $A, B \in \mathcal{M}_{n\times n}(\mathbb{R}^n)$ such that $(A+B)^{-1} = A^{-1} + B^{-1}$ if and only if there exists $J \in \mathcal{M}_{n \times n}(\mathbb{R}^n)$ such that $J^2 = -I_n$.

Here, $I_n$ is the $n \times n$ identity matrix. So far, none of my ideas have panned out and I don't feel like I have enough information to solve this problem. Any hints would be appreciated!
 [/INST] $(\Longrightarrow)$
Since
$(A+B)(A^{-1}+B^{-1})= I + AB^{-1}+BA^{-1} + I=(A+B)(A+B)^{-1}=I$
$AB^{-1}+BA^{-1}=AB^{-1}+(AB^{-1})^{-1}=-I$ and $ (AB^{-1})^2+I=-AB^{-1}$
and if we let $X=AB^{-1}$, then we have $X^2+X+I=O$.
We can do a change of variables $X=Y+C$
$(Y+C)^2+(Y+C)+I=Y^2 +(2C+I)Y+C^2+C+I$
and by letting $C=-\frac{1}{2}I$, then we get $Y^2+0Y+\frac{3}{4}I=Y^2+\frac{3}{4}I=O$
now $\left(\frac{2}{\sqrt{3}}Y\right)^2+I=Z^2+I=0$ and $Z=\frac{2}{\sqrt{3}}\left(X+\frac{1}{2}I\right)$ satisfies the properties of $J$.

$(\Longleftarrow)$
Given $J$ s.t. $J^2=-I$, then let $Z=-\frac{1}{2}I+\frac{\sqrt{3}}{2}J$
then $\left(-\frac{1}{2}I+\frac{\sqrt{3}}{2}J\right)\left(-\frac{1}{2}I-\frac{\sqrt{3}}{2}J\right)= \frac{1}{4}I^2-\frac{3}{4}J^2= \frac{1}{4}I+\frac{3}{4}I=I$ and $Z$ is invertible.
$Z$ also satisfies $Z^2 + Z +I =O$
$\left(-\frac{1}{2}I+\frac{\sqrt{3}}{2}J\right)^2 + \left(-\frac{1}{2}I+\frac{\sqrt{3}}{2}J\right)+ I=\frac{1}{4}I^2-\frac{\sqrt{3}}{2}J+\frac{3}{4}J^2-\frac{1}{2}I+\frac{\sqrt{3}}{2}J+I$
$=\frac{1}{4}I-\frac{3}{4}I-\frac{1}{2}I+I=O$
Given an invertible matrix $B$, let $ZB=A$, and then $Z=AB^{-1}$ where $A$ is invertible because $Z$ and $B$ are inverible.
$(A^{-1}+B^{-1})=(A+B)^{-1}$ because $Z+Z^{-1}=AB^{-1}+BA^{-1} = -\frac{1}{2}I + \frac{\sqrt{3}}{2}J -\frac{1}{2}I -\frac{\sqrt{3}}{2}J =-I$
thus $I+AB^{-1}+BA^{-1}+I=(A+B)(A^{-1}+B^{-1})=I$.
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