The addition law and part (a) give the approximate parallelogram identity
$$
h(P+Q)+h(P-Q)=2h(P)+2h(Q)+O(1).
$$
Replace $P,Q$ by $2^nP,2^nQ$, divide by $4^n$, and let $n\to\infty$. The error disappears, leaving the exact <parallelogram law>
$$
\widehat h(P+Q)+\widehat h(P-Q)=2\widehat h(P)+2\widehat h(Q).
$$
Together with $\widehat h(2P)=4\widehat h(P)$, induction on $|m|$ yields
$$
\boxed{\widehat h(mP)=m^2\widehat h(P)}
$$
for every integer $m$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 125 / 5 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-125/5/b/ii/solution