I want to solve this system of equations. I took these equations as a result of a problem I was trying to solve. I write equations in the simplest possible form.
Let $A$ and $B$ be Real or Complex coefficients,where $A\neq 0$, $B\neq 0$ and $x,y,z,u,v$ be variables, where $x\neq 0,y\neq 0,z\neq 0,u\neq 0,v\neq 0.$
I want to solve this system of equations:
$$\begin{cases}3z^2uB+3x+2uzA+xz^2A=0 &(1)\\
3y+2vzA+u^2A+2xzuA+yz^2A+3vz^2B+3x^2+3zu^2B=0 &(2)\\
3y^2+v^2A+2xuvA+2yzvA+3v^2zB+yu^2A+3vu^2B+3x^2y=0 &(3)\\
3xy^2+xv^2A+2yuvA+3v^2uB=0 &(4)\end{cases} $$
$\Huge{\text{My attempts:}}$
$3y^2+2vzyA+u^2yA+2xzuyA+y^2z^2A+3vz^2yB+3x^2y+3zu^2yB-3y^2-v^2A-2xuvA-2yzvA+3v^2zB-yu^2A-3vu^2B-3x^2y=0\Longrightarrow (yz-v)(2xuA+3vzB+yzA+vA+3u^2B)=0$
Let, $yz=v$, then applying $v=yz$ we get from $(1)$ and $(4)$
$yz^2A+zvA=0 \Longrightarrow zy+v=0 \Longrightarrow 2v=0 \Longrightarrow v=0$ which is contradiction. So, $yz\neq v$
We get, $2xuA+3vzB+yzA+vA+3u^2B=0.$
There exist a solution for $v=0, x=0, z=0$ and $3y+u^2A=0$ which gives infinitely many solutions. $x=y=z=u=v=0$ is trival solution.
I can not continue from $yz\neq v$.
I am looking for a solution for non-zero variables $x,y,z,u,v$ where $A\neq0, B\neq 0.$ The solution does not have to be in the real numbers set.
For special case $A=B=1$ we have,
$$\begin{cases}3z^2u+3x+2uz+xz^2=0 &(1)\\
3y+2vz+u^2+2xzu+yz^2+3vz^2+3x^2+3zu^2=0 &(2)\\
3y^2+v^2+2xuv+2yzv+3v^2z+yu^2+3vu^2+3x^2y=0 &(3)\\
3xy^2+xv^2+2yuv+3v^2u=0 &(4)\end{cases} $$
Is it possible to solve this system of equations for non-zero variables?
Best Answer
Comment. For some values of $A$ and $B$ it is trivial to find solutions. For example, for $A=2$ and $B=-3$ we can take $x=2,y=1,z=1,u=2,v=1$. For $A=B=1$ one can take $$ x= 1,\; y= 1, \; z = - \frac{1}{3}, u=\frac{28}{3}, v=-\frac{1}{3}. $$ In this case we have infinitely many solutions for $A=B=1$, e.g., $$ y=x,\; v=xz,\; u=-\frac{xz^2+3x}{3z^2+2z}, \, z^3+z^2+1=0 $$ for arbitrary $x\neq 0$.