Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 32 / 1 / iv

The inverse on this Weierstrass equation of an elliptic curve is $(x,y)\mapsto (x,-y-1)$. A line $y=\lambda x+\nu$ through the points being added has third intersection $R$ with $x(R)={\lambda }^2-x_1-x_2$, and the sum is $-R$.

At $P=(0,0)$, implicit differentiation gives $\lambda =(3x^2-1)/(2y+1)=-1$. The tangent is $y=-x$. Substitution into the equation gives $x^2=x^3$, so its third intersection is $(1,-1)$, whose inverse is $(1,0)$. Next the line through $(0,0)$ and $(1,0)$ is $y=0$; its remaining intersection is $(-1,0)$, whose inverse is $(-1,-1)$. Consequently