TITLE: Replacing in equation introduces more solutions
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: Suppose first that $f$ is a real-valued function of one variable. The equation
$$
x = f(x)
\tag{1}
$$
acts as a condition, selecting values of $x$ for which (1) is true. Substituting (1) into itself gives a new condition,
$$
x = f\bigl(f(x)\bigr) = f^{[2]}(x),
\tag{2}
$$
and so forth.
Certainly every solution of (1) is a solution of (2). If the function $f$ is not injective (one-to-one), however, (2) can have solutions that are not solutions of (1).
For example, if $f(x) = 4x(1 - x)$, then $f$ maps $[0, 1]$ onto $[0, 1]$. Each point except $x = \frac{1}{2}$ has two preimages, and $f$ maps each interval $[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$ bijectively to $[0, 1]$. It follows that $f \circ f$ maps each half-interval onto $[0, 1]$, and each point of $[0, 1]$ has two preimages in each half-interval, so $f^{[2]} = f \circ f$ has four fixed points, etc. (Diagram below.) In this example, the $n$-fold composition of $f$ with itself, $f^{[n]}$, has $2^{n}$ fixed points, i.e., the equation
$$
x = f^{[n]}(x) = (\underbrace{f \circ \dots \circ f}_{n \text{times}})(x)
\tag{n}
$$
has $2^{n}$ solutions, even though (n) is obtained from (1) by successively substituting (1) into itself.

Your situation is analogous: Starting from
$$
y = 2 - x^{2} - y^{2} = f(x, y),
\tag{1a}
$$
you substitute (1a) into itself, obtaining
$$
y = 2 - x^{2} - f(x, y)^{2} = f\bigl(x, f(x, y)\bigr),
\tag{2a}
$$
and so forth.
In your example, $f$ has qualitatively similar behavior along the $y$-axis to the "logistic map" $f(x) = 4x(1 - x)$, and the solution sets
$$
y = f\Bigl(\dots f\bigl(x, f(x, y)\bigr)\dots\Bigr)
$$
become increasingly complicated with successive iteration.
This type of phenomenon (e.g., the precise location/shape of the solutions) is generally complicated (chaotic in the technical sense). Wikipedia pages of possible interest include:

The logistic map
The tent map
The horseshoe map
Dynamical systems