# Sum of two cubes transformed to elliptic curve

Given $x^3+y^3=N$, we can perform some substitutions to obtain an elliptic curve $u^3-432N^2=v^2$, as given here, which are $x=\frac{36N+v}{6u}$, $y=\frac{36N-v}{6u}$. Here's the details: \begin{align*} \left(\frac{36N+v}{6u}\right)^3+\left(\frac{36N-v}{6u}\right)^3 &=N \\ u^3-432N^2 &=v^2\end{align*}

What's the inspiration for these substitutions? I see why linear substitutions won't work so perhaps I should have tried a simple rational substitution.

• I don't see how those substitutions give you the required form. I see nothing replacing the $x$ terms in any of the substitutions. – Ian Miller Jan 9 '16 at 6:57
• Yikes! I probably shouldn't have glazed over the last step. The other substitution mentioned in the webpage seems to work. I'll edit the question. – Kevin Jan 9 '16 at 7:18
• I corrected your typos. You can check it with this WA link. (The equation $u^3-432N^2=v^2$ is well-known.) – Tito Piezas III Jan 9 '16 at 8:27