<s>[INST] Replacing in equation introduces more solutions Let's say I have an equation $y=2-x^2-y^2$.
now, since I know that $y$ is exactly the same as $2-x^2-y^2$ I can create the following, equation by replacing $y$ with $2-x^2-y^2$.
$y=2-x^2-(2-x^2-y^2)^2$
doing this replacement introduces new solutions such as $(-1, 0)$. Replacements in other various equations have similar results, although some do not change the equation at all! 
What mechanic introduces these new solutions, and what are they?
Edit: one such example of an equation where no solutions are introduced via replacement is $y=x^2+y^2$. That will give $y=x^2+(x^2+y^2)^2$ which upon graphing is the same graph as the original, $y=x^2+y^2$.
Here is an image of the iteration of this replacement on the same function, just for fun.
 [/INST] We know that every $(x,y)$ such that 
$$y=2-x^2-y^2\quad\text{i.e.}\quad x^2+y^2+y-2=0$$
satisfies
$$y=2-x^2-(2-x^2-y^2)^2\quad \text{i.e.}\quad (2-x^2-y^2)^2+x^2+y-2=0\tag1$$
This implies that $(2-x^2-y^2)^2+x^2+y-2$ is divisible by $x^2+y^2+y-2$ :
$$(2-x^2-y^2)^2+x^2+y-2=(x^2+y^2+y-2)(x^2+y^2-y-1)$$
Hence, 
$$(1)\iff x^2+y^2+y-2=0\quad\text{or}\quad x^2+y^2-y-1=0.$$

Similarly, for the second example, we know that $(x^2+y^2)^2+x^2-y$ is divisible by $x^2+y^2-y$ :
$$(x^2+y^2)^2+x^2-y=(x^2+y^2-y)(x^2+y^2+y+1)$$
Note here that there are no $(x,y)\in\mathbb R$ such that
$$x^2+y^2+y+1=0\iff x^2+\left(y+\frac 12\right)^2=-\frac 34$$
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