Point Order on Elliptic Curve

Question

i got stuck doing the exercise 4.5 d) in the Book Elliptic Curves: Number Theory and Cryptography by L. Washington and would be grateful for hints.

Let $ p \equiv 1 \text{ (mod 4)}$ be prime and let $E$ be given by $y^2 = x^3 -kx$, where $k \not\equiv 0 \text{ (mod $p$)}$. Let $k$ be a square but not a fourth power mod $p$. Show that exactly one of the curves $y^2 = x^3 -x$ and $y^2 = x^3 -kx$ has a point of order $4$ defined over $\mathbf{F_p}$.

I tried to find a point such that $2P = (\pm\sqrt{k},0)$, but that did not work out and solving the division polynomial $\psi_4$ also didnt work.

Answer 1

• $k=d^2$ then $y^2=x^3-kx$ is isomorphic to $E_d:y^2=d(x^3-x)$.

• $E_1:y^2=x^3-x$.

• Since $d$ is not a square, for all $a\in \Bbb{F}_p$, either $(a,\pm \sqrt{a^3-a})\in E_1(\Bbb{F}_p)$ or $(a,\pm \sqrt{d(a^3-a)})\in E_d(\Bbb{F}_p)$

• $\#E_1(\Bbb{F}_p)+\#E_d(\Bbb{F}_p)= 2(p-3)+8\equiv 4\bmod 8$

• Both $E_1(\Bbb{F}_p)$ and $E_d(\Bbb{F}_p)$ contain the 2-torsion (the points coming from $a=\infty$ or $a^3-a=0$)

• Thus, exactly one of $\# E_1(\Bbb{F}_p)$,$\# E_d(\Bbb{F}_p)$ is $\equiv 0\bmod 8$, the other is $\equiv 4\bmod 8$.

• Exactly one of $ E_1(\Bbb{F}_p)$,$E_d(\Bbb{F}_p)$ contains some point of order $4$.