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.
 A: 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.
A: Let uppercase letters denote the logarithm of the corresponding lowercase parameter, in absolute value. The last three equations read
$$\begin{cases}X+Y&=A+B,\\Y+Z&=B+C,\\Z+X&=C+A,\end{cases}$$
which is an easy system, giving the single solution $X=A,Y=B,Z=C$. Hence $(x,y,z)=(\pm a,\pm b,\pm c)$.
Now to discuss the signs, it suffices to remark that $x=a,y=b,z=c$ is a solution regardless the signs of $a,b,c$, and if we try to change the sign of one unknown, we need to change the other two. Finally
$$(x,y,z)=(a,b,c)\lor (x,y,z)=-(a,b,c).$$
In all cases, the first equation is implicitly fulfilled.
A: Ok, So I found another way to do this,
Let,
$A = a + ci$
$B= x + zi$
Now,
$A^2 = a^2 - c^2 + aci$ 
$B^2 = x^2 - z^2 + xzi$ 
$A^2 = B^2$
$(a + ci)^2 = (x + zi)^2$
$a +ci = \pm(x + zi)$
So, ($a = x$ and $c=z$) or ($a = -x$ and $c=-z$)
Substituting into $ab=xy$ gives
$(a,b,c) = \pm(x,y,z)$
A: This is an homogeneous system so making $y = \lambda x$ and $z = \mu x$ we have
$$
a^2-c^2=x^2(1-\mu^2)\\
ab = x^2\lambda\\
ac = x^2\mu\\
bc = x^2\lambda\mu
$$
and making now $A = \frac ax,B = \frac bx, C = \frac cx$
$$
A^2-C^2=1-\mu^2\\
AB = \lambda\\
AC = \mu\\
BC = \lambda\mu
$$
so we conclude easily
$$
A^2B^2C^2=\lambda^2\mu^2\\
A^2-C^2= 1-A^2C^2\Rightarrow A^2=1\Rightarrow a^2=x^2\Rightarrow b^2= y^2\Rightarrow c^2= z^2
$$
