 Solution

ID: past-exam-of-the-analytic-number-theory-and-elliptic-curves-course-of-the-university-of-oxford/2008/b/8/a/i/solution

Fix

P

and put

e_{m} = h_{x} (m P) - m^{2} h_{x} (P)

for

m \geq 0

. Then

e_{0} = e_{1} = 0

. Apply the given approximate parallelogram identity with its two arguments

m P

and

P

. The quadratic terms cancel exactly, leaving

∣ e_{m + 1} - 2 e_{m} + e_{m - 1} ∣ \leq C (m \geq 1) .

d_{m} = e_{m} - e_{m - 1}

. Starting from

d_{1} = 0

, the inequality gives

∣ d_{m} ∣ \leq (m - 1) C

. Summing from one to

m

therefore yields

∣ h_{x} (m P) - m^{2} h_{x} (P) ∣ \leq \frac{m ( m - 1 )}{2} C .

Thus

C_{m} = m (m - 1) C /2

works independently of

P

, including

m = 1

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