Solution to a special type of polynomial equation

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 $$n=5$$.

• For $n=2$, take $l_i(x)=x_i$ and $q=x_1^2-x_1x_2+x_2^2$ as an example. – Mohan Oct 14 '18 at 22:20
• @Mohan Thanks for your example. But maybe I should require $n>2$ since this is too special. – User X Oct 14 '18 at 22:27

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 $$n\ge 5$$.

• 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 $n=5$? – User X Oct 15 '18 at 6:35
• 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) – User X Oct 15 '18 at 6:51
• Ok I have proved there is no trivial solution for $n\geq 5$. – User X Oct 15 '18 at 19:05
• You mean no nontrivial solution? – Robert Israel Oct 16 '18 at 3:04
• Yeah, it is a typo :) The proof is just by assume a solution and compare coefficients on both sides. – User X Oct 16 '18 at 21:18