Non existence of skew-symmetric matrices with the difference of their square is diagonal of some type By using Maple I can show that there is no  skew-symmetric matrices $J_1$ and $J_2$ of order 4 satisfying
$$J_1^2-J_2^2=\left(\begin{array}{cccc}-\frac12Trace(J_1^2)&0&0&0\\
     0&\frac12Trace(J_2^2)&0&0\\0&0&-1 &0\\0&0&0&-1\end{array} \right).$$
I guess that there is a more elegant proof of this fact using some properties of skew-symmetric matrices I don't know.
 A: If the matrices are real, we may argue as follows. Since $J_1^2$ is the square of a skew-symmetric matrix, its nonzero eigenvalues must form pairs of repeated non-positive eigenvalues. Therefore, the spectrum must be in the form of $\{-a,-a,-b,-b\}$ for some $a\ge b\ge0$. Similarly, the spectrum of $J_2^2$ is $\{-c,-c,-d,-d\}$ for some $c\ge d\ge0$. So, the equation in question can be rewritten as
$$
J_1^2=J_2^2+\operatorname{diag}\left(a+b,\,-c-d,\,-1,-1\right).\tag{1}
$$
Denote the trailing principal $3\times3$ submatrices of $J_1^2$ and $J_2^2$ by $A_3$ and $C_3$ respectively. Then $(1)$ implies that
$$
A_3=C_3+\operatorname{diag}\left(-c-d,\,-1,-1\right)\preceq C_3-mI_3,\tag{2}
$$
where $-m=\max\{-c-d,\,-1\}$. Also, by the interlacing inequality for bordered matrices (cf. Horn and Johnson, Matrix Analysis, theorem 4.3.8 on p.185 in the 1st edition or theorem 4.3.17 on p.242 in the 2nd edition), we have $\lambda_\min(A_3)=-a$, $\lambda_\max(A_3)=-b$, $\lambda_\min(C_3)=-c$ and $\lambda_\max(C_3)=-d$. Therefore $(2)$ gives
$$
\begin{cases}
-a\le-c-m,\\
-b\le-d-m.
\end{cases}\tag{3}
$$
Take traces on both sides of $(1)$, we get $-2(a+b)=-2(c+d)+(a+b)-(c+d)-2$, i.e.
$$
a+b=c+d+\frac23.\tag{4}
$$
So, by adding up the two inequalities in $(3)$, we obtain $-m\ge-\frac13$. It follows from the definition of $m$ and from $(4)$ that
\begin{align}
m=c+d&\le\frac13, \tag{5}\\
a\le a+b=c+d+\frac23&\le1. \tag{6}
\end{align}
We now claim that
$$
d+1>a.\tag{7}
$$
Suppose the contrary that $d+1\le a$. From $(6)$, we also have $a\le1$. Since $a$ and $d$ are nonnegative, this is possible only if $a=1$ and $d=0$. But then by $(6)$ again, we must have $b=0$ and $c=\frac13$, which is a contradiction to $(3)$ (which implies that $b\ge d+m=d+(c+d)\ge c$). Thus $(7)$ must hold.
Next, denote by $A_2$ and $C_2$ the trailing principal $2\times2$ submatrices of $A_3$ and $C_3$. By the interlacing inequality,
\begin{align}
\left[\lambda_\min(A_2),\lambda_\max(A_2)\right]
\subseteq\left[\lambda_\min(A_3),\lambda_\max(A_3)\right]&=\{-a,-b\},\tag{8}\\
\left[\lambda_\min(C_2),\lambda_\max(C_2)\right]
\subseteq\left[\lambda_\min(C_3),\lambda_\max(C_3)\right]&=\{-c,-d\}.
\end{align}
In turn,
$$
\left[\lambda_\min(C_2-I_2),\lambda_\max(C_2-I_2)\right]
=\left[\lambda_\min(C_2)-1,\lambda_\max(C_2)-1\right]
\subseteq[-c-1,-d-1].\tag{9}
$$
We now arrive at a contradiction: by $(1)$, $A_2=C_2-I_2$. Therefore $\left[\lambda_\min(A_2),\lambda_\max(A_2)\right]$ must coincide with $\left[\lambda_\min(C_2-I_2),\lambda_\max(C_2-I)\right]$. Yet, $(7),(8)$ and $(9)$ show that this is impossible, because the two intervals are disjoint. So, we conclude the equation in question is not solvable.
