I'm trying to understand the following famous arithmetical problem under a geometric point of view.
Find integers $a,b,c,d$ such that $a^3+b^3+c^3=d^3$.
Viete and Euler established algebraic identities that were used and developed by Ramanujan to obtain general solutions of this problem. Ramanujan introduced a real $x$ parameter and found a system of four quadratic equations related to $x$.
But, as I said, I'm looking for a more "graphic" solution. The classical surface parametrization of cubic surface doesn't seem to work here. Dividing the above equation by $d^3, \: d \neq 0$, we have (after substitution of variables) the cubic surface $(S)$: $x^3+y^3+z^3=1$ who has three real lines. One of which is $D:(x+y=0, z=1)$. In order to find a quadratic form defining the conic to be parametrized, I have to cut the surface $(S)$ by the real plane $P: \: \lambda(x+y)+z=1$ with line $D \in P$ and $\lambda \in \mathbb{R}$. But I'm lost in the calculus ! Please can someone explain how to determine the equation (in $x,y$) of the conic using this geometric method ?
I am not sure if your are still interested, and what are you really asking, but let me stress that, if you substitute $z=1-\lambda(x+y)$ into $x^3+y^3+z^3-1$, you get a polynomial whose factorization is
$$-(x + y)((\lambda^3-1)x^2 + (2\lambda^3 + 1)xy - 3\lambda^2x + (\lambda^3-1)y^2 - 3\lambda^2y + 3\lambda)$$
So may be the conic you are looking for is given by the equation $$(\lambda^3-1)(x^2 + y^2) + (2\lambda^3 + 1)xy - 3\lambda^2(x+y) + 3\lambda$$