It is easy to find binary quadratic form parameterizations $F(x,y)$ to,

$$a^3+b^3+c^3+d^3 = 0\tag{1}$$

(See the identity (5) described in this MSE post.) To solve,

$$x_1^3+x_2^3+x_3^3 = 1\tag{2}$$

in the integers, all one has to do is to check if one term of $(1)$ can be solved as a Pell-like equation $F_i(x,y) = 1$. For example, starting with the cubes of the taxicab number 1728 as $a,b,c,d = 1,-9,12,-10$, we get,

$$a,b = x^2-11xy+9y^2,\;-9x^2+11xy-y^2$$

$$c,d = 12x^2-20xy+10y^2,\;-10x^2+20xy-12y^2$$

We can then solve $a = x^2-11xy+9y^2 = \pm 1$ since it can be transformed to the Pell equation $p^2-85q^2 = \pm 1$, thus giving an infinite number of integer solutions to $(2)$.

Question: How easy is it to find a quadratic form parameterization to,

$$x_1^3+x_2^3+x_3^3 = Nx_4^3\tag{3}$$

for $N$ a non-cube integer? I'm sure one can see where I'm getting at. If one can solve,

$$x_4 = c_1x^2+c_2xy+c_3y^2 = \pm 1\tag{4}$$

as a Pell-like equation, then that would prove that,

$$x_1^3+x_2^3+x_3^3 = N\tag{5}$$

is solvable in the integers in an infinite number of ways. (So far, this has only been shown for $N = 1,2$). The closest I've found is a cubic identity for $N = 3$ in a 2010 paper by Choudhry,

$$\begin{aligned} x_1 &= 2x^3+3x^2y+3xy^2\\ x_2 &= 3x^2y+3xy^2+2y^3\\ x_3 &= -2x^3-3x^2y-3xy^2-2y^3\\ x_4 &= xy(x+y) \end{aligned}$$

which is a special case of eq.58 in the paper.

Anybody knows how to find a quadratic form parametrization to $(3)$? (If one can be found, hopefully $(4)$ can also be solved.)

  • $\begingroup$ Don't you need only $x_4$ quadratic and no restrictions on the other $x_i$? $\endgroup$ – joro Sep 20 '13 at 7:52

A parametrization with quadratic forms like you want defines a conic inside the projective cubic surface with equation (3). Such a conic is contained in a plane and the residual intersection of the plane with the surface is a line. Conversely any plane containing a line inside the cubic surface will have as residual intersection a conic that can be parametrized by quadratic forms. A smooth cubic surface such as (3) famously has 27 lines. If you want your quadratics to have rational coefficients, since you are interested in integral solutions,you need the line to be defined over the rationals. So you need to go through the 27 lines and see which are rational. The ones easy to eyeball ($x_i = \omega x_j, x_k = \alpha x_4, \omega^3=-1,\alpha^3 = N$ with $i,j,k$ a permutation of $1,2,3$. Did I get 27?) aren't defined over the rationals, so you are out of luck.


Perhaps if you start with my three-rational-cubes identity $$ ab^2 = \biggl(\frac{(a^2+3b^2)^3+(a^2-3b^2)(6ab)^2}{6a(a^2+3b^2)^2}\biggr)^{\!3} - \biggl(\frac{(a^2+3b^2)^2-(6ab)^2}{6a(a^2+3b^2)}\biggr)^{\!3} - \biggl(\frac{(a^2-3b^2)6ab^2}{(a^2+3b^2)^2}\biggr)^{\!3} $$ you might be able to find something?


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