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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.

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  • $\begingroup$ You can just remove all the terms $x_{17}$ through $x_{24}$. They are redundant with the preceding eight. Replace $x_9$ with $x_9+x_{17}$ and so on. $\endgroup$ – Ross Millikan Jan 21 '19 at 15:18
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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.

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  • $\begingroup$ Thanks for the answer. Can we find explicitly these linear conditions on $(a,b,c,d,)$? $\endgroup$ – E. Joseph Jan 22 '19 at 10:46
  • $\begingroup$ Vanish one factor per term in your left hand side. $\endgroup$ – Mindlack Jan 22 '19 at 10:47
  • $\begingroup$ This answer may not be completely correct. The argument based on the 3-square does not seem to distinguish between the forms $$7a^2-b^2-c^2-d^2\\ 5a^2+b^2-c^2-d^2$$ and the latter one has the requisite identity given in my answer. $\endgroup$ – Tito Piezas III Jun 2 '19 at 15:46
  • $\begingroup$ @Tito Piezas III: my proof is in two steps. First, I show that the right hand side has no nontrivial zero (for which I used Legendre’s theorem). Second, I show that no matter what rational numbers the $x_i$ are, there is always a line $L \subset \mathbb{Q}^4$ such that the left hand side vanishes for every $(a,b,c,d) \in L$. This second step is absolutely independent of the quadratic form in the right hand side. On the other hand, $5a^2+b^2-c^2-d^2=0$ has non-trivial rational solutions (multiples of $(1,2,3,0)$), thus the first step doesn’t work in that case. $\endgroup$ – Mindlack Jun 4 '19 at 8:39
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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 $k=7$.

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  • $\begingroup$ Thanks for the answer, though it is not quite the same in my opinion : you have changed the sign in front of $b^2$ ($-\to +$), and I was interested in $7a^2-b^2-c^2-d^2$ since $7a^2-b^2-c^2-d^2=0$ has no rational solution ; but $5a^2+b^2-c^2-d^2=0$ does... $\endgroup$ – E. Joseph Jun 3 '19 at 11:20
  • $\begingroup$ @E.Joseph: I don't think that if $y_1 a^2+y_2 b^2+y_3 c^2+ y_4 d^2 = 0$ has no rational solutions, then it is a guarantee there are no 24 non-trivial rational $x_i$. If I have time, I may investigate this further. $\endgroup$ – Tito Piezas III Jun 3 '19 at 12:33
  • $\begingroup$ If you find a form $y_1a^2+y_2b^2+y_3c^3+y_4d^4$ with no rational solution but such that there is a $24$ non-trivial rational solution, I would be really interested to hear about it! $\endgroup$ – E. Joseph Jun 3 '19 at 12:35
  • $\begingroup$ @E.Joseph The form $D=ka^2+b^2-c^2-d^2$ with $k = (4n^2+1)/n$ has the 24 rational $x_i$. One need only check if there are $k$ such that $D=0$ is not solvable. $\endgroup$ – Tito Piezas III Jun 3 '19 at 12:40
  • $\begingroup$ Yes exactly, I will think about this problem $\endgroup$ – E. Joseph Jun 3 '19 at 12:42

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