Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 28 / 2 / ii / Solution

For this Weierstrass equation of an elliptic curve, the inverse is $(x,y)\mapsto (x,13-y)$. At a point with $2y-13\ne 0$, implicit differentiation gives the tangent slopeIf the tangent line is $y=mx+b$, substitution into the equation shows that the three intersection abscissae sum to $m^2-1$. Two of them equal the tangency abscissa $x$, so the third is $x_3=m^2-1-2x$. Reflecting the third intersection by the inverse formula gives the elliptic curve group law from Riemann-Roch doubling formulaFor $P$, the slope is $13/(-13)=-1$, giving $x(2P)=0$ and $y(2P)=13$. For $Q$, the slope is $(12-4+13)/(6-13)=-3$, giving $x(2Q)=9-1+4=12$ and $y(2Q)=13-3+(-3)(-2-12)=52$. HenceIn particular $P$ has order exactly three: it is nonidentity and $2P=-P$.