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In this post (Substitutions that transform Fermat Equations to Elliptic Curves) it is proved that there exists a change of variables that trasform Fermat's curve $X^3+Y^3+Z^3=0$ into $y^2=x^3-432$, which is an elliptic curve in Weierstrass form. But this only work in characteristic $\neq 2,3$, that is becuase we have to divide by $6$ at some point.

My question is if we can do the same in the case of characteristic $2$ and $3$, because every ''standard'' change of variable I've tried needs multiplying or dividing by $2$ or $3$ at some point, which fails in our case.

Any hints or help will be appreciated.

Marcos
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    In characteristic $3$ it is $(X+Y+Z)^3$ whose vanishing set is $\cong\Bbb{P}^1$. In characteristic 2 it should be an elliptic curve, the Weirstrass form will be $y^2+y=f(x)$ or $y^2+xy=f(x)$ – reuns May 07 '22 at 20:46
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    @reuns thanks, the case of characteristic $3$ was easy... thanks. I know that the elliptic curves in characteristic $2$ is of one of these two ways, but I can't find the change that sends $X^3+Y^3+Z^3=0$ to this form. – Marcos May 07 '22 at 21:28

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