$E$ - elliptic curve on field $F_{2^4}$ with equetion $y^2 + y = x^3$.
I need to show, that for any $P \in E$, $3P = 0$.
Here is list of 16 field elements:
$0,1,t,t+1,$
$t^2,t^2+1,t^2+t,t^2+t+1,$
$t^3,t^3+1,t^3+t,t^3+t+1,t^3+t^2,t^3+t^2+1,t^3+t^2+t,t^3+t^2+t+1$
Let's define multiplication in the field as $a*b=c,$ where c is a remainder of division of $a*b$ by $t^4+t+1$
Now, we can clculate all elements of the field in cube:
$(0)^3 = 0$
$(1)^3 = 1$
$(t)^3 = t^3$
$(t+1)^3 = t^3+t^2+t+1$
$(t^2)^3 = t^3+t^2$
$(t^2+1)^3 = t^3+t$
$(t^2+t)^3 = 1$
$(t^2+t+1)^3 = 1$
$(t^3)^3 = t^3+t^2$
$(t^3+1)^3 = t^3+t^2+t+1$
$(t^3+t)^3 = t^3+t^2+t+1$
$(t^3+t+1)^3 = t^3+t^2$
$(t^3+t^2)^3 = t^3$
$(t^3+t^2+1)^3 = t^3+t$
$(t^3+t^2+t)^3 = t^3$
$(t^3+t^2+t+1)^3 = t^3+t^2$
We also can look at the curve equation like this: $y^2+y=y*(y+1)=x^3$.Thus, we can take any element from field, add $1$, multiply this two polynomials and check wheter it equals to any cube element or not.
$y=0 \Rightarrow 0*1=0$
$y=1 \Rightarrow 1*0=0$
$y=t \Rightarrow t*(t+1)=t^2+t$
$y=t+1 \Rightarrow (t+1)*t=t^2+t$
$y=t^2 \Rightarrow t^2*(t^2+1)=t^2+t+1$
$y=t^2+1 \Rightarrow (t^2+1)*t^2=t^2+t+1$
$y=t^2+t \Rightarrow (t^2+t)*(t^2+t+1)=1$
$y=t^2+t+1 \Rightarrow (t^2+t+1)*(t^2+t)=1$
$y=t^3 \Rightarrow t^3*(t^3+1)=t^2$
$y=t^3+1 \Rightarrow (t^3+1)*t^3=t^2$
$y=t^3+t \Rightarrow (t^3+t)*(t^3+t+1)=t$
$y=t^3+t+1 \Rightarrow (t^3+t+1)*(t^3+t)=t$
$y=t^3+t^2 \Rightarrow (t^3+t^2)*(t^3+t^2+1)=t+1$
$y=t^3+t^2+1 \Rightarrow (t^3+t^2+1)*(t^3+t^2)=t+1$
$y=t^3+t^2+t \Rightarrow (t^3+t^2+t)*(t^3+t^2+t+1)=t^2+1$
$y=t^3+t^2+t+1 \Rightarrow (t^3+t^2+t+1)*(t^3+t^2+t)=t^2+1$
That's how I found points from curve:
$(0,0),$
$(0,1),$
$(1,t^2+t),$
$(1,t^2+t+1)$
$(t^2+t,t^2+t),$
$(t^2+t,t^2+t+1)$
$(t^2+t+1,t^2+t),$
$(t^2+t+1,t^2+t+1)$
Now, I'm tpying to calculate $3P$ using formulas:
$P(x_1,y_1) + Q(x_2,y_2) = R(x_3,y_3)$
if $(x_1 = x_2)$
$x_3 = \left(\frac{x_1^2+y_1}{x_1}\right)^2 + \left(\frac{x_1^2+y_1}{x_1}\right)$
$y_3=(\left(\frac{x_1^2+y_1}{x_1}\right)+1)*x_3 + x_1^2$
if $(x_1 \ne x_2)$
$x_3 = \left(\frac{y_1+y_2}{x_1+x_2}\right)^2 + \left(\frac{y_1+y_2}{x_1+x_2}\right)+x_1+x_2$
$y_3=(\left(\frac{y_1+y_2}{x_1+x_2}\right)+1)*x_3 + \left(\frac{y_1*x_2+y_2*x_1}{x_1+x_2}\right)$
I understand that $3P = P+P+P$. But when I put actual numbers in them, I have, for example for point $(0,0)$ $x_3=\left(\frac{0}{0}\right)^2$. What is that? 0?
The doubling formula on this curve is particularly simple. Indeed, if $(\xi,\eta)$ is a point on the curve, then $[2](\xi,\eta)=(\xi^4,\eta^4+1)$.
Here’s why: Take your point $(\xi,\eta)=P$ on the curve, so that $\xi^3=\eta^2+\eta$. The derivative there, as @Somos has pointed out, is $\xi^2$, so that the line through $P$ with this slope is $Y=\xi^2X+?$, where you adjust “?” for the line to pass through $P$. You get the equation of the line to be $Y=\xi^2X+\eta^2$. Call the line $\ell$.
Now, what is the third intersection of $\ell$ with our curve, beyond the double intersection at $P$? Substitute $Y=\xi^2X+\eta^2$ into $Y^2+Y=X^3$ to get a cubic in $X$ with a double root at $\xi$ and one other: $$ (\xi^2X+\eta^2)^2+\xi^2X+\eta^2+X^3=(X+\xi)^2(X+\xi^4)\,, $$ as you can check. Thus the other intersection of $\ell$ with our curve is $(\xi^4,\xi^6+\eta^2)=(\xi^4,\eta^4)=[-2](\xi,\eta)$. Now, in all cases, for a point $(a,b)$ on our curve, its negative is $(a,b+1)$: the other intersection of the curve with the vertical line through $(a,b)$.
Since $[-2](\xi,\eta)=(\xi^4,\eta^4)$, we have $[2](\xi,\eta)=(\xi^4,\eta^4+1)$.
Now, I say that for every point $(\xi,\eta)$ with coordinates in the field with four elements, we get $[3](\xi,\eta)=\Bbb O$, the neutral point at infinity, the one with projective coordinates $(0,1,0)$. That is, every one of the nine points with coordinates in $\Bbb F_4$ is a three-torsion point. Indeed, for $\xi,\eta\in\Bbb F_4$, $\xi^4=\xi$ and $\eta^4=\eta$. Thus for an $\Bbb F_4$-rational point $(\xi,\eta)$ on our curve, we have $[2](\xi,\eta)=(\xi,\eta+1)=[-1](\xi,\eta)$. Adding $(\xi,\eta)$ to both sides of this equation, we get $[3](\xi,\eta)=\Bbb O$.
It only remains to enumerate the nine points of the curve with $\Bbb F_4$-coordinates. Calling the elements of the field $0,1,\omega,\omega^2$, where $\omega^2+\omega+1=0$, we see that the only points on the curve are $$ \Bbb O,(0,0),(0,1)(1,\omega),(1,\omega^2),(\omega,\omega),(\omega,\omega^2),(\omega^2,\omega),(\omega^2,\omega^2)\,. $$ You can check that when you pass to the larger field $\Bbb F_{16}$, you don’t get any more points!