## Algebra 2 (1st Edition)

$(1,0),(0,1)$
When squaring $(1-y^2)$ he got an incorrect result; the correct result is: $1-2y^2+y^4$, then our equation is: $1-2y^2+y^4+y^2-2+2y^2-2y=-1\\y^4+y^2-2y=0\\y(y^3+y-2)=0\\y(y-1)(y^2+y+2)=0$ The third term has a positive discriminant, so it cannot be $0$. Thus, $y=0$ or $y=1$. Thus the solutions are: $(1,0),(0,1)$