Solving the system $a^2-c^2=x^2-z^2$, $ab=xy$, $ac=xz$, $bc=yz$ I've stumbled upon these equations, and am struggling to find a manual way to solve this in $\Re$:

$$\begin{align}
a^2-c^2&=x^2-z^2 \\
ab&=xy \\
ac&=xz \\
bc&=yz\end{align}$$

I've used Wolfram Alpha to compute this and I found that this is only possible if:
$$a=x, \quad y=b, \quad z=c$$
Is there an easy manual way to compute this?
I've tried moving around variables with no luck.

There is in fact another solution: $a=-x$, $b=-y$, $c=-z$.
Multiply the bottom three equations together:
$$(abc)^2=(xyz)^2$$
$$abc=\pm xyz$$
This leads to $c=\pm z$, $b=\pm y$, $a=\pm x$. The first equation is automatically satisfied after these relations. To see that all $\pm$ assignments must be the same, try letting (say) $a=x$ but $b=-y$ and see that it leads to a contradiction.
