Here's the context of this problem.
Solve:
$x^2=y^3-3y^2+2y$
$y^2=x^3-3x^2+2x$
We subtract the second equation from the first and obtain $$(x-y)(x^2+y^2+xy-2x-2y+2)=0$$ The first factor yields the solutions $0,2\pm\sqrt2$ and the second one can be rewritten as $$(x+y)^2+(x-2)^2+(y-2)^2=4$$ I have checked Wolfram Alpha and it tells me that there are no more real solution to the above system.This means that the above equation has no solution.Now I have to prove it.
Obviously,none of the expressions $(x+y)^2,(x-2)^2,(y-2)^2$ can be $4$ otherwise the LHS would be larger than $4$. Hence,we have the inequalities $$|x+y|<2$$$$|x-2|<2$$$$|y-2|<2$$ which(unless I am mistaken) yield $$0<x<2$$ $$0<y<2$$ I cannot proceed any further.Perhaps there is some clever way to write the whole expression solely using squares but I cannot see it immediately.
I would appreciate it if the method used to solve this is elementary.Some help will be appreciated.