Question

Let E be an elliptic curve involving the equation x³ + ax + b = y² over the finite field Fp. Suppose you have the additional information that x³ + ax + b is never zero for any X ∈ F_p. Show that E must have an odd number of points. (Hint: don’t forget O)

Answer 1

Let E : y 2 = x 3 +ax+b be an elliptic curve over Z/(n).

Let P be a point on the E all of whose coordinates are in (Z/(n)) and let X = (x0, y0) be the natural lift of P to Z. Consider the family of curves over Z F(E, X) = {E(α, β)||α|, |β| ≤ 3n 3 , E passes X and E reduces to E modulo n} where E(α, β) is defined by the equation y 2 = x 3 + αx + β. Then for sufficiently large n, a random curve in the family has rank 1 with probability greater than some constant. In light of the general heuristic assumption [1, 2, 21], it would be actually reasonable to expect that a random curve in F(E, X) has rank 1 with probability around 1/2, and rank 2 with probability around 1/2. How do we sample a random curve in the family? To do this, we choose a random integer i < n and set α = a+in, then set β = y 2 0 −x 3 0 −αx0. It is easy to see that E passes through X. x3+ax+b is never showing zero for any X element Fp. so E always have a odd number.