Let’s say I have \begin{align} \tag{1} x_1^2 = y_1^2+y_2^2+y_3^2+y_4^2, \end{align} where all the numbers are integers, and $x_1$ is odd. Evidently, $x_1^2$ is not of the form $4^k(8m+7)$, and so [by well-known classical results] it can be written as the sum of three squares.
Is there an established transformation/parameterization to turn $x_1^2=y_1^2+y_2^2+y_3^2+y_4^2$ into $x_1^2=z_1^2+z_2^2+z_3^2$, where the $z_i$ are integer (or rational) functions of the $y_i$?
I know Bradley et al. give formulas for (1), e.g. \begin{align} x_1 &= (uz+vy+wx)^2+m^2(m^2+x^2+y^2+z^2+u^2+v^2+w^2), \\ y_1 &= (uz+vy+wx)^2-m^2(m^2+x^2+y^2+z^2-u^2-v^2-w^2), \\ y_2 &= 2m(um^2+uz^2+xvm-ywm+xwz+yvz), \\ y_3 &= 2m(vm^2+vy^2+zwm-xum+zuy+xwy), \\ y_4 &= 2m(wm^2+wx^2+yum-zvm+yvx+zux), \end{align} and Catalan gave the complete solution for the [2.1.3] equation, so I’ll do the grunt work if necessary… I was just hoping this particular wheel was already invented.
Comment: We may try this idea; suppose we have two Pythagorean triples:
$a^2+b^2=c^2$
$d^2+e^2=f^2$
Multiplying both sides we get:
$(ad)^2+(ae)^2+(bd)^2+(be)^2=(cf)^2=(af)^2+(bf)^2=(af)^2+(be)^2+(bd)^2$
But we do not want identical terms on both sides, so we have to find numbers like $s$ and $t$ such that:
$A=(be)^2+(bd)^2=s^2+t^2$
It leads to this question : are there numbers(not necessarily perfect square) like $A$ which can be written as the sum of two squares in two different forms? We can try this analytically; suppose our two triples are of the following forms:
$(4i)^2+(4i^2-1)^2=(4i^2+1)^2$
$(2i+1)^2+[2i(i+1)]^2=[2i(i+1)+1]^2$
After multiplying both sides we finally get:
$A=[8i^3(i+1))]^2+[2(i^2+i)(4i^2-1)]^2=s^2+t^2$
Now we have a Diophantine equation with three unknown to solve.