Combining sums and/or differences of squares I'd like to combine a sum of as many squares as possible into a sum of as few squares as possible.  The signs of the squares doesn't matter.
For example, the Brahmagupta-Fibonacci Identity combines a sum of four squares into a sum of two squares.  Thus the compression ratio is 2:1.  Using this identity, we can, with careful calculations, apply it recursively to get an even better compression ratio.  In other words, take two sums of four squares and combine them together to get one sum of four squares, and then combine this to get one sum of two squares.  Thus we have a ratio of 4:1.
THE QUESTION'
Are there are any other formulas that take a sum of as many squares as possible and combine them into as few squares as possible?

The "compression ratio" can be made as high as you want.

I. Euler-Aida Ammei

$$(-x_0^2+x_1^2+x_2^2+\dots+x_n^2)^2+(2x_0x_1)^2+(2x_0x_2)^2+\dots+(2x_0x_n)^2 = (x_0^2+x_1^2+x_2^2+\dots+x_n^2)^2$$
Thus, there can be arbitrarily many squares on the LHS.

II. Fauquembergue $n$-Squares Identity

Using the Brahmagupta-Fibonacci,
$$(a^2+b^2)(c^2+d^2) = u_1^2+u_2^2 = (ac+bd)^2+(ad-bc)^2$$
then,
$$(a^2+b^2-c^2-d^2+x)^2 + (2u_1)^2 + (2u_2)^2 + 4x(c^2+d^2) = (a^2+b^2+c^2+d^2+x)^2$$
where $x$ is arbitrary and can be chosen as any sum of $n$ squares. 

III. Using the Euler-Four Square

From,
$$(a^2+b^2+c^2+d^2)(e^2+f^2+g^2+h^2) = v_1^2+v_2^2+v_3^2+v_4^2$$
then,
$$(a^2+b^2+c^2+d^2-e^2-f^2-g^2-h^2+x)^2 + (2v_1)^2 + (2v_2)^2 + (2v_3)^2 + (2v_4)^2 + 4x(e^2+f^2+g^2+h^2) = (a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+x)^2$$
where $x$ can again be any sum of squares. (I trust you know the expressions for the $v_i$.) Of course, a similar $n$-square identity can be made using the Degen-Graves Eight-Square. 
For more on sums of $n$ squares, more details at my site.  
