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I am given the elliptic curve with homogeneous equation $E: X^3+Y^3+Z^3=0$ over a field $K$ with a point $P_{\infty}=[1:-1:0]$. I am told that you can turn $E(K)$ into a group with identity element $P_{\infty}$ and I am asked to derive a formula for the addition of two points $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ and then I am asked to find the Weierstrass normal form of this curve.

I think that $P+Q = (-x_1-x_2-3\lambda^2b,-\lambda x_3-b)$ where $\lambda = \dfrac{y_2-y_1}{x_2-x_1}$ and $b=y_1-\lambda x_1$ and $x_3 = -x_1-x_2-3\lambda^2b$

(I basically did this using the usual addition in elliptic curves but I am not sure if this is correct.)

For the other part, I know that the Weierstrass Normal Form should be $y^2 = x^3+ax+b$ where $a,b \in \mathbb{R}$ but I don't know how to write this curve in such form.

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  • $\begingroup$ $F(X,Y,Z) = X^3+Y^3+Z^3$ if $F(a,b,c)=F(a+d,b+e,c+f)=0$ let $h(t)=F(a+dt,b+et,c+ft) = gt(t-1)(t-T)$ where $g = d^3+e^3+f^3$, $h(-1) = -2 g(T+1), T = \frac{-h(-1)}{2g}-1$ and the group law is $[a:b:c]+[a+d:b+e:c+f]+[a+dT:b+eT:c+fT] = [1:-1:0]$. To obtain the inverse let $[a:b:c]=[1:-1:0]$ $\endgroup$ – reuns Feb 17 at 22:01

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