# Why does the following nonlinear system have 21 solutions? y + x y 2 </msup>

Why does the following nonlinear system have 21 solutions?
$y+x{y}^{2}-{x}^{3}+2x{z}^{4}=0\phantom{\rule{0ex}{0ex}}-x-{y}^{3}-3{x}^{2}y+3y{z}^{4}=0\phantom{\rule{0ex}{0ex}}-\frac{5}{2}{y}^{2}{z}^{3}-2{x}^{2}{z}^{3}-\frac{{z}^{7}}{2}=0$
there is one real and twenty complex solutions: $\left(0,0,0\right)$
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Gornil2
Mark the solution $\left(0,0,0\right)$ and remove ${z}^{3}$ from the last equation. The reduced systems for $z=0$ has two cubic equations, so at most $3\cdot 3=9$ solutions.Then z only occurs as ${z}^{4}$. Multiply the first equation by $x$, the second by $y$ and express all equations in $a={x}^{2}$, $b={y}^{2}$, $c=xy$ and $d={z}^{4}$, adding the equation $ab-{c}^{2}=0$. Then the system reads as
$\begin{array}{rl}c+{c}^{2}-{a}^{2}+2ad& =0\\ -a-bc-3ac+3cd& =0\\ -5b-4a-d& =0\\ ab-{c}^{2}& =0\end{array}$
This system has the Bezout bound $8$ for the number of solutions for $\left(a,b,c,d\right)$, each of these will have at most $8$ solutions for $\left(x,y,z\right)$. This reduces the estimate to $64+9+1=74$ solutions from the $5\cdot 5\cdot 7=175$ or $5\cdot 5\cdot 4+3\cdot 3=109$ for the Bezout bound of the (reduced) original system.