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How to rationally parametrize the cubic curve $x^3+y^3=1$? It's a slice of Fermat cubic, and the Fermat cubic does have rational parametrization, so I think it should also have rational parametrization.

I tried $u=x+y, v=x-y$ and got $u(u^2+3v^2)=4$,or $v=\pm\sqrt{\frac{4-u^3}{3u}}$, which is a bit closer to what I want, but still not good enough.

MaudPieTheRocktorate
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    If you could parametrize the curve rationally, Fermat's Last Theorem would be false. Only curves of genus $0$ admit a rational parametrization. –  Jun 14 '17 at 11:56

2 Answers2

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HINT.- You can do it however it is not "elementary".

The elliptic curve $E_1:X^3+Y^3=1$ can be rationally parametrized but with a trascendental parameter, a (corresponding) Weierstrass function $\wp (z)$. In order to do this, you can first put $V_1$ under its Weierstrass form which is $$E_2:y^2=x^3-432$$ which comes from the birational transformation $$(X,Y)\rightarrow \big(\frac{12}{X+Y},\frac{36(X-Y)}{X+Y}\big)$$ whose reciprocal is $$(x,y)\rightarrow\big(\frac{36+y}{6x},\frac{36-y}{6x}\big)$$ Now you have the classical formula linking $\wp(z)$ and its derivative $\wp'(z)$ $$\wp'^2=4\wp^3-g_2\wp-g_3$$ where $g_2$ and $g_3$ are the invariants of $\wp$ which must be calculated $$g_2=60\sum_{\omega\ne 0}\frac{1}{\omega^4}\text{ and}\space g_3=140\sum_{\omega\ne 0}\frac{1}{\omega^6}$$ where $\omega\in L$.

($L$ is the corresponding lattice for the doubly periodic function $\wp(z)$ which if I remember correctly is hexagonal).

This way we have the parametrization $$\begin{cases}x=\wp(t)\\y=\wp'(t)\end{cases}$$

Piquito
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  • I have a question? What even is $\wp$ function? Any link to info about it? – KKZiomek Feb 16 '19 at 03:16
  • It is known that if the complex function, $f$, is meromorphic and periodic, it is then necessarily simply periodic or doubly periodic (the elliptical functions are doubly periodic). Besides, it is shown that the order of all elliptical function is greater than or equal to 2. The Weierstrass function ℘(z) is a function that responds to the problem of constructing an L-elliptic function that takes exactly the minimum of 2 times every complex value, $u$, in any fundamental parallelogram $P$ for the lattice $L$. – Piquito Feb 17 '19 at 14:34
  • In Google you can find a lot of references about both Weierstrass function and L-elliptic function. – Piquito Feb 17 '19 at 14:37
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For $x\geq0$ and $y\geq0$ we have $x=\sqrt[3]{\sin^2\theta}$ and $y=\sqrt[3]{\cos^2\theta}$.

The similar for $x\leq0$ and $y\leq0$.

In the rest cases we can use $\sinh$ and $\cosh$.

Michael Rozenberg
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