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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 $(2)$?

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  • $\begingroup$ Well, first you have to solve the equation. $$(a_1x_1^2+b_1y_1^2+c_1z_1^2)(a_2x_2^2+b_2y_2^2+c_2z_2^2)=ax^2+by^2+cz^2$$ Then you can move on to a more complicated formula. $\endgroup$
    – individ
    Dec 11, 2015 at 14:22
  • $\begingroup$ @individ: The problem with that approach is that there might be no solution. For example, the equation $(1)$, namely, $$(ax_1^2+by_1^2)(ax_2^2+by_2^2) = ax_0^2+by_0^2$$ has no known solution for general $a,b$, but becomes solvable when a third factor is used like in $(2)$. $\endgroup$
    – Tito Piezas III
    Dec 11, 2015 at 14:37
  • $\begingroup$ Isn't it a known theorem that there are such formulas for 2,4,8, and 16 terms, but no others? I can't remember where I saw that… $\endgroup$
    – Kieren MacMillan
    Aug 6, 2016 at 11:54
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    $\begingroup$ @KierenMacMillan:That is for $n$-nary quadratic forms with unit coefficients for $=2,4,8$ (The case $n=16$ and higher already involves rational functions.) This post asks for more general coefficients. $\endgroup$
    – Tito Piezas III
    Aug 6, 2016 at 15:36
  • $\begingroup$ In (2) and what follows, are there $\pm$ options like the two different options in (1) and what follows? $\endgroup$
    – Kieren MacMillan
    Oct 16, 2016 at 0:20

1 Answer 1

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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.

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  • $\begingroup$ Hm, that is rather deflating. I guess the next case would be, $$\prod_{i=1}^n (ax_i^2+by_i^2+cz_i^2+dt_i^2) = ax_0^2+by_0^2+cz_0^2+dt_0^2$$ for either $n=4$ or $5$. $\endgroup$
    – Tito Piezas III
    Dec 16, 2015 at 5:03
  • $\begingroup$ @TitoPiezasIII, works for $n=4$ because we get quaternion multiplication with $w^2 + a x^2 + b y^2 + ab z^2.$ Pretty sure $w^2 + ac x^2 + bc y^2 + ab z^2$ also works. In both cases it is a multiplication of pairs of coefficient quadruples, $x_0 \, 1 + i x_1 \sqrt a + j x_2 \sqrt b + k x_3 \sqrt {ab}$ $\endgroup$
    – Will Jagy
    Dec 17, 2015 at 18:04

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