Resolving a rational system of equations with too many unkowns

Question

A little bit of context.

While working a larger proof (the proof is quite related to this question I asked), I stumbled upon the following problem.

The question.

Can we find $x_1,\ldots,x_{24}\in\mathbb Q$ such that for all $a,b,c,d\in\mathbb Q$:

$$ (x_1a+x_2b+x_3c+x_4d)(x_5a+x_6b+x_7c+x_8d)-(x_{9}a+x_{10}b+x_{11}c+x_{12}d)(x_{13}a+x_{14}b+x_{15}c+x_{16}d)-(x_{17}a+x_{18}b+x_{19}c+x_{20}d)(x_{21}a+x_{22}b+x_{23}c+x_{24}d)$$ $$=7a^2-b^2-c^2-d^2\quad ?$$

What I tried.

I have tried to do a systematic resolution of the underlying system, but this is too massive to compute.

I have tried arbitrarily fixing some of the $x_i$ in order to simplify the system, but I always end up in one of the following cases:

• no solution,

• irrational solutions,

• complex solutions.

None of this cases is acceptable.

Any leads would be greatly appreciated.

Answer 1

From Legendre $3$-square theorem the right hand side can never vanish unless all terms are zero.

However, no mather what the $x_i$ are, it is possible with three linear conditions on $(a,b,c,d)$ to force the left hand side to vanish, meaning the left hand side has infinitely many rational zeroes.

Answer 2

There may be a solution, though one may not have searched hard enough. For example, if your expression with the squares was,

$$D = 5a^2+b^2-c^2-d^2$$

then I found,

$$\left(a+\frac b4+\frac c4+\frac d2\right)\big(4a-b+c-2d\big)-(-a + b + c + d)(a + b - c + d) -\left(-\frac {9b}4+\frac {9c}4-\frac{3d}2\right)\left(b+c+\frac{2d}3\right) = 5a^2+b^2-c^2-d^2$$

In fact,

$$D = k\,a^2+b^2-c^2-d^2$$

where $k=\frac{4n^2+1}{n}$ and other families are solvable, though I haven't come across one where $k=7$.