rogue wrote: Solve equation $\sqrt{1+2x-xy}+\sqrt{1+2y-xy}=2.$ So $\sqrt{1+2x-xy}=1+t$ and $\sqrt{1+2y-xy}=1-t$ for some $t\in[-1,1]$ $\iff$ $2x-xy=t^2+2t$ and $2y-xy=t^2-2t$ Which very easily gives $\boxed{(x,y)\in\{(t,-t),(2+t,2-t)\}\text{ }\forall t\in[-1,+1]}$