Recently I've surprised myself when I thought of one very basic problem generalization that I never thought of before, but still wasn't able to make any progress.
So, we all know that using radicals we can solve a univariate polynomial up to degree 4. We are given one polynomial $P$ of degree at most $4$, one variable, $x$, and are asked to solve the next system (only one equation) for $x$: $$P(x)=0.$$
Next, we all know that we can solve an arbitrarily large system of linear equations. We are given $n^2$ coefficients $a_{ij}$, $n$ coefficients $b_i$ and $n$ variables $x_1,x_2,\dots,x_n$ and are asked to solve the next system: $$\begin{matrix} a_{11}x_1+a_{12}x_2+\dots+a_{1n}x_n+b_1=0,\\ a_{21}x_1+a_{22}x_2+\dots+a_{2n}x_n+b_2=0,\\ \dots\\ a_{n1}x_1+a_{n2}x_2+\dots+a_{nn}x_n+b_n=0.\\ \end{matrix}$$ Of course, the usual compact way to write this system is using objects of linear algebra such as matrices and vectors and it is indeed a way to solve it. But what if, otherwise, we chose to compactify the system using polynomials instead. Then, we are given $n$ polynomials $P_1,P_2,\dots,P_n$ in $n$ variables, each of the total degree $1$, we are given $n$ variables $x_1,x_2,\dots,x_n$ and are asked to solve the system $$\begin{matrix} P_1(x_1,x_2,\dots,x_n)=0,\\ P_2(x_1,x_2,\dots,x_n)=0,\\ \dots\\ P_n(x_1,x_2,\dots,x_n)=0. \end{matrix}$$ This is just a way to write it, we would still solve it after firstly translating into language of linear algebra. Here is the question: The first problem incorporates $1$ polynomial and variable and extends the degree all up to $5$. The other problem incorporates arbitrary (but same) number of polynomials and variables but fixes the degree to $1$. My question is is there a way to solve a system which has both metrics higher than $1$? Like, we are given $n$ polynomials $P_1,P_2,\dots,P_n$ in $n$ variables, each of the total degree at most $k\leq 4$, we are given $n$ variables $x_1,x_2,\dots,x_n$ and are asked to solve the previous system.
For what pairs $(n,k)$ is this system solvable?
In case $k=2$ and $n\geq 2$, I only found that I should expect up to $2^n$ solutions, but I haven't found any algorithm nor formulas. The main problem in my attempt is that I don't know how to deal with coefficients since their number of $n\cdot\big(\binom{n}{2}+n+1\big)$ of them is very not friendly.
Any hint or algorithm is welcome.