An identity attributed to Catalan is:
$A^3 = B^2 + C^2 + D^2$
where $ A = x^2 + y^2 + z^2$
$B = x(3z^2 - x^2 - y^2)$
$C = y(3z^2 - x^2 - y^2)$
$D = z(z^2 - 3x^2 - 3y^2) $ .
This can be used to show (one of several ways) that the set of integers expressible as the sum of three squares is closed under taking powers (despite the fact that it is not closed under taking products). $\;$ The curious look of this identity prompts the following questions about its origin, nature and symmetry.
Questions: 1) It is possible to invent a plausible series of steps whereby one might derive such an identity but does anyone know historically how Catalan actually came up with it?
2) What does it mean? $\,$ Can it be interpreted as a kind of geometric transformation or perhaps some result from invariant theory? $\,$ In this regard, it might be helpful to know if it's just an isolated result or whether it can be seen in a natural way as belonging to a family.
3) It is obvious that the left hand side is invariant under all permutations of the variables x,y,z (in addition to sign changes like x $\mapsto$ -x). $\,$ However, the right side is only obviously symmetric under interchanging x and y .$\,$ Is there some way to tell that the right hand side is also symmetric while swapping y and z (without multiplying everything out and without invoking the identity itself)?
Thanks
This Link should take you to the correct page (II,268) in Dickson
$x$ and $y$ do not really appear separately. If you write $r^2 = x^2 + y^2$ it becomes
$$ \left( z^2 + r^2 \right)^3 = \left( z^2 - 3 r^2 \right)^2 \; z^2 + \left( 3z^2 - r^2 \right)^2 \; r^2 $$
In turn, with $i^2 = -1,$ compare $$ (z+ir)^3 = z(z^2 - 3 r^2) + i r (3 z^2 - r^2) $$