On products of ternary quadratic forms $\prod_{i=1}^3 (ax_i^2+by_i^2+cz_i^2) = ax_0^2+by_0^2+cz_0^2$

The equation,

$$ (ax_1^2+by_1^2)(ax_2^2+by_2^2) = ax_0^2+by_0^2\tag1$$

has the well-known solution when $a=b=1$,

$$ (x_1^2+y_1^2)(x_2^2+y_2^2) = (x_1 y_2 + x_2 y_1)^2 + (x_1 x_2 - y_1 y_2)^2$$

Hence the product of the sum of two squares is itself the sum of two squares. If we use one more factor, then it has an identity for general $a,b$, namely,

$$ (ax_1^2+by_1^2)(ax_2^2+by_2^2)(ax_3^2+by_3^2) = ax_0^2+by_0^2\tag2$$

where,

$$x_0 =a \color{blue}{x_1 x_2 x_3} + b \big(-\color{blue}{x_1} y_2 y_3 + \color{blue}{x_2} y_1 y_3 + \color{blue}{x_3} y_1 y_2 \big)$$

$$y_0 =a \big(-x_2 x_3 \color{brown}{y_1} + x_1 x_3 \color{brown}{y_2} + x_1 x_2 \color{brown}{y_3}\big) + b \color{brown}{y_1 y_2 y_3}$$

High-lighted this way, one can immediately see the pattern.

Q: Is there a similar identity for ternary quadratic forms,

$$ (ax_1^2+by_1^2+cz_1^2)(ax_2^2+by_2^2+cz_2^2)(ax_3^2+by_3^2+cz_3^2) = ax_0^2+by_0^2+cz_0^2\tag3$$

such that $x_0, y_0, z_0$ are integer functions in terms of the other $x_i, y_i, z_i$ just like for $(2)$?

Solution 1:

No, there cannot be a completely general identity. There cannot even be such an identity for $(a,b,c)=(1,1,1)$, since $$ (1+0+0)(1+1+1)(0+1+4) = 15 $$ and by Legendre's three-square theorem $15$ cannot be written as the sum of three squares.

Since we can add factors of $(1+0+0)$ without changing the product, it is clear that additional factors cannot resolve the problem in this case.