Invalid subtraction when solving system of equations? I'm trying to solve these two equations:
$$\begin{cases}
1-4x(x^2+y^2)=0 \\
1-4y(x^2+y^2)=0
\end{cases}$$
and I tried to do it by subtracting the first equation from the second, yielding $(4x-4y)(x^2+y^2)=0$. Clearly this is satisfied when $x=y$, which gives $(x,y)=(\frac{1}{2},\frac{1}{2})$, or when $x^2+y^2=0$, which gives $(x,y)=(0,0)$.
But when I plug them back in, it's obvious that $(x,y)=(0,0)$ is not a solution. Where did I go wrong?


$$\begin{cases}
1-4x(x^2+y^2)=0 \\
1-4y(x^2+y^2)=0
\end{cases}$$

Let $S$ the set of solutions.
You can summarise the logic steps:


*

*If $(x,y)$ is a solution:
$$4x(x^2+y^2)=4y(x^2+y^2)
\implies x^2+y^2=0 \text{ or }x=y
\\
\implies (x,y)=(0,0) \text{ or }
\\
x=y=\frac{1}{4(x^2+y^2)}
=\frac{1}{8x^2}\implies x=y=\frac 1{8^{1/2}}=\frac 12
$$
You have proved $S\subset\{(1/2,1/2),(0,0)\}$.

*If $x=(1/2,1/2)$ the equation is true. If $x=(0,0)$ the equation is false.
You have proved $S=\{(1/2,1/2)\}$.

