Torsion points on elliptic curve

Question

Please help or hints me to solve this question:

Suppose that $E: y^2+y= x^3+x$ be an elliptic curve over $\Bbb F_2$, We know that $E$ is supersingular. Show that $E[5] ‎‎\subseteq‎‎ E(\Bbb F_{16})$.

Answer 1

Extended hints:

A point $P$ is of order $5$ iff $[4]P=-P$. For this to happen it is necessary that the points $[4]P$ and $[P]$ share the $x$-coordinate, but don't share the $y$-coordinate (if $[4]P=P$, then $P$ will be 3-torsion). This suggests the following approach:

• Figure out the point doubling formula for this curve.
• Iterate the above to figure out the point quadrupling formula.

Then show that this leads to a system of equations with all the solutions in $\Bbb{F}_{16}$.

Answer 2

My (never totally reliable) computation says that for this curve, $[2](\xi,\eta)=(\xi^4+1,\eta^4+\xi^4+1)$, and that therefore $[4](\xi,\eta)=(\xi^{16},\eta^{16}+1)$. Since $[-1](\xi,\eta)=(\xi,\eta+1)$, you see that for any $P\in E(\Bbb F_{16})$, you have $[4](P)=[-1](P)$. I did the computation by hand in about 10 minutes. From the fact that the curve has five points over the prime field, you use the Zeta function to see that it has $25$ points over $\Bbb F_{16}$, and that does it.

EDIT - addendum:
But the right way to do it is this: Since the curve has five points over $\Bbb F_2$, the characteristic polynomial for Frobenius $\mathbf f$ is $X^2+2X+2$, eigenvalues $-1\pm i$. Thus the characteristic polynomial of $\mathbf f^4$ has both eigenvalues equal to $-4$. That shows immediately that for $P\in E(\Bbb F_{16})$, $P=[-4](P)$.