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Let $k$ be any field (maybe need to be algebraically closed), and $n$ be a positive integer, $k[x_1,\ldots,x_n]$ be the polynomial ring. Then consider the equation $$\sum_{i=1}^n l_i(x)x_i^2=q(x)(x_1+\ldots+ x_n)$$ where $l_i(x)\in k[x_1,\ldots,x_n]$ is a homogeneous polynomial of degree $1$, and $q(x)\in k[x_1,\ldots,x_n]$ is a homogeneous polynomial of degree $2$. We want to find the solution $(l_i,q)$ for the equation above. Easy to see $$l_i=x_1+\ldots+ x_n$$ $$q=x_1^2+\ldots+ x_n^2$$ is a trivial solution (and any $\mathbb C$ multiple, i.e. $c_i l_i$ for $c_i\in k$). My question is

is there any other non-trivial solution?


In general I am concerned about the solution to $$\sum_{i=1}^n l_i(x)x_i^2=q(x)(\text{a general linear equation})$$ but let's see the special case first. I have no idea how to do this.


Edit

Robert's answer shows there are other solutions for $n<5$. Now I want to know the case $n=5$.

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  • $\begingroup$ For $n=2$, take $l_i(x)=x_i$ and $q=x_1^2-x_1x_2+x_2^2$ as an example. $\endgroup$ – Mohan Oct 14 '18 at 22:20
  • $\begingroup$ @Mohan Thanks for your example. But maybe I should require $n>2$ since this is too special. $\endgroup$ – User X Oct 14 '18 at 22:27
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Algebraically closed is not an issue: this involves solving a linear system of equations for the coefficients of the $l_i$ and $q$, with integer coefficients.

You will find other nontrivial solutions. For example, for each $i$ you could have $l_i = x_1 + \ldots + x_n$, all other $l_j = 0$, $q(x) = x_i^2$.

In the case $n=3$ you could also take $l_1(x) = x_2 - x_3$, $l_2(x) = x_1$, $l_3(x) = -x_1$, $q(x) = x_1 x_2 - x_1 x_3$ (and similarly for permutations of the variables).

In the case $n=4$ you could take $$ \eqalign{l_{{1}}(x)&=x_{{2}}-x_{{4}}\cr l_{{2}}(x)&=x_{{1}}-x_{{3}}\cr l_{{3}}(x)&=-x_{{2}}+x_{{4}}\cr l_{{4}}(x)&=-x_{{1}}+x_{{3}}\cr q(x)&=x_{{1}}x_{{2}}-x_{{1}}x_{{4}}-x_{{2}}x_{{3}}+x_{{3}}x_{{4}}\cr}$$ (and again permutations of this).

I suspect this type of solution doesn't exist for $n\ge 5$.

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  • $\begingroup$ Thanks for your answer! But in fact I am mostly interested in the case $n=5$. I guess there is no solution for $n=5$, and thought it is a general fact for all $n$ (now you shows it is not). So, is there a way to show no solution for $n=5$? $\endgroup$ – User X Oct 15 '18 at 6:35
  • $\begingroup$ And note that the first solution you provided $l_i = x_1 + \ldots + x_n$, all other $l_j = 0$, $q(x) = x_i^2$, should be regarded as an trivial solution (the multiple of $l_i$, coefficients vary for different $i$, sorry for did not make it clear) $\endgroup$ – User X Oct 15 '18 at 6:51
  • $\begingroup$ Ok I have proved there is no trivial solution for $n\geq 5$. $\endgroup$ – User X Oct 15 '18 at 19:05
  • $\begingroup$ You mean no nontrivial solution? $\endgroup$ – Robert Israel Oct 16 '18 at 3:04
  • $\begingroup$ Yeah, it is a typo :) The proof is just by assume a solution and compare coefficients on both sides. $\endgroup$ – User X Oct 16 '18 at 21:18

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