Exercise 5.7.2 (The inflection points of an elliptic curve are aligned three by three)

Proof. In the context of elliptic curves, an inflection point of the curve is a point $A$ such that the tangent at this point has intersection multiplicity $3$ with the curve (see definition p.254). Therefore the tangent does not contain any other point on the curve than $A$, so that $\mathit{AA}=A$. In particular, since $O$ is an inflection point, $\mathit{OO}=O$.

Put $C=\mathit{OA}$. Then $\mathit{AC}=O$, and $A+A=O(\mathit{AA})=\mathit{OA}=C$, thus $3A=A+C=O(\mathit{AC})=\mathit{OO}=O$. This shows that $3A=O$.

Conversely, if $3A=O$, then $A+A=-A$. Put $B=-A$. By construction of the point $B$, $\mathit{AB}=\mathit{OO}$ (see figure (b) p.265). Since $O$ is an inflection point, $\mathit{OO}=O$, so $\mathit{AB}=O$, and $B=\mathit{OA}$.

Moreover $B=A+A=O(\mathit{AA})$, thus $\mathit{OA}=O(\mathit{AA})$, so $O+A=O(\mathit{OA})=O(O(\mathit{AA}))=O+\mathit{AA}$. Therefore $A=\mathit{AA}$. This shows that the tangent at point $A$ has a unique intersection point with the curve, with multiplicity $3$.

So $3A=O$ if and only if $A$ is an inflection point.

Suppose that $A$, $B$ are inflection points. Then $3A=3B=O$, thus $3(A+B)=3A+3B=O$, and $3(-A-B)=-3(A+B)=O$. By the first part (and Theorem 5.23), the point $\mathit{AB}=-(A+B)$ is an inflection point. □