How to find all solutions of $x^2-y^2=x+y$? For this equation $$x^2-y^2=x+y$$ I know there are 2 functional solutions, $y=x-1$, and $y=-x$.
The second solution we can see immediately because both sides will always be $0$.
The first solution is also easy to get:$$(x-y)(x+y)=x+y\ \ ///\div(x+y)\ \ \ [y\neq-x]\\x-y=1\\y=x-1$$
But how do I solve this in a formal way? So that I'll find all solutions, and show that there aren't other solutions?

$x^2 - y^2 = (x-y)(x+y) = x+y$
If $x+y=0$ then all $\{(x, -x)|x \in \mathbb R\}$ are solutions.
If $x+y\ne 0$ then $(x-y) = 1$ and $y = x -1$ so all $\{(x, x-1)|x \in \mathbb R\}$ are solutions.
So solutions are $\{(x, -x)|x \in \mathbb R\}\cup\{(x, x-1)|x \in \mathbb R\}$.
....
or $x^2 - y^2 = x+y$
$x^2 - y^2 -(x+y) = 0$
$(x+y)[(x-y) - 1] = 0$
so either $x+y = 0$ or $x-y-1 = 0$ so....
