Square of squares

There is a famous unsolved problem in mathematics, called the square of squares, mainly popularised by numberphile. The goal is the create a 3x3 square meeting the following two criteria:

- 9 unique square numbers
- Adding to the same value along all rows, columns and diagonals.

Today, we are not going to solve this problem, but a slight adaptation of it. In particular, we are going to apply the following two changes:

- Numbers can instead of being a perfect square, also be the negative of a perfect square.
- The centre of the square has to be zero.

Our squares then looks as follows: $$\begin{matrix} b^2&-d^2&c^2\\ a^2&0 & -a^2\\ -c^2&d^2&-b^2 \end{matrix} $$ I will now prove that such a square cannot exist.
The above configuration leads to 2 equations: $$a^2 + b^2 = c^2$$ $$b^2 + c^2 = d^2$$ as $a^2 + 2b^2 = d^2$, $a$ and $d$ have the same parity. Define: $x := \frac{d-a}{2}$ and $y := \frac{d+a}{2}$ then is $$a^2 = (x-y)^2$$ $$b^2 = 2xy$$ $$c^2 = x^2 + y^2$$ $$d^2 = (x+y)^2$$ If we look at the second equation, $b^2$ is even, so $b^2$ is a multiple of 4. therefor $x$ or $y$ has to be even. As all formulas are symmetric over switching $x$ and $y$, one can assume without loss of generality that $x$ is even. Now if $b' := b/2$ and $x' := x/2$ then $$b'^2 = x'y$$ If a product is a perfect square, each factor can be written as a common factor times a square, so $x' = q\xi^2$ and $y = q\gamma^2$. Plugging this in in the third equation gives: $$c^2 = q^2(4\xi^4 + \gamma^4)$$ So $4\xi^4 + \gamma^4$ is a perfect square. Let's call it $n^2$. $$n^2 = 4\xi^4 + \gamma^4$$ Now assume we have the smallest non-zero solution for $n,\xi,\gamma$ If $n$ and $\gamma$ are even, then $n' := n/2$ and $\gamma' := \gamma/2$. $$n'^2 = \xi^4 + 4\gamma'^4$$ So our solution was not the smallest, and this leads to a contradiction. Now if we assume $n$ and $\gamma$ are odd. $$4\xi^4 = (\gamma^2-n)(\gamma^2+n)$$ as $$\gamma^2 \equiv 1 (mod 4)$$ and $n$ is odd, one of the factors on the right side is $2 (mod 4)$ and the other $0 (mod 4)$. SO for one factor the power of prime factor $2$ is $1$, and for the other it is even. So the power of prime factor 2 for the full right side is odd. As the left side is square, the power of prime factor $2$ is even. So as both sides of the equation have a different prime decomposition, this leads to a contradiction. Therefore, there is no nontrivial solution. This being said, thare is obviouslty a trivial solution, where all nine sqaures are zero.