Show: if $k = x^3 - y^3$, where $x, y \in \mathbb{Q}_{>0}$, then $k = u^3 + v^3$ with $u, v \in \mathbb{Q}_{>0}$

Question

Suppose we have a $k \in \mathbb{Q} \ $ ($k > 0$) such that $$k = x^3 - y^3, \textrm{ where } x, y \in \mathbb{Q} \ \textbf{ and } \ x > y > 0.$$

That is, some (rational) number $k$ is the difference of two cubes of positive rational numbers. I have to show that it is also $\textbf{the sum of two cubes of two positive rational numbers}$.

That is, there exist $u, v \in \mathbb{Q}$ where $u, v > 0$ such that $$k = u^3 + v^3.$$

I've only progressed as far as writing $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $u^3 + v^3 = (u + v)(u^2 - uv + v^2)$, but more advanced techniques are probably necessary. Can anyone help me?

Answer 1

Let us start by slighting translating your problem. The curve $C: x^3 + y^3 = k$ is a planar cubic curve (so it has genus 1, so it can be transformed into an elliptic curve, if you like this terminology more). Suppose it possesses a rational point $P = (x_0, y_0)$ in the fourth quadrant. You need to prove that the curve also admits a rational point in the first quadrant.

Now we see by some black magic that it is an elliptic curve, so we shall expect some group actions on it. We know that it passes through $-P = (y_0, x_0)$; and there is a point called $-2P$, which is acquired by taking the tangent line at $P$ and finding the intersection to the line.

My claim is, there exists a $l \in \mathbb Z_{>0}$ such that $2^lP$ is in the first quadrant. The equation of the tangent line at $(x_0, y_0)$ is

$$ y - y_0 = -\frac{x_0^2}{y_0^2}(x-x_0). $$

If the intersection is in the first quadrant, we are done. If not, we take $2P = (x_1, y_1)$ by finding the intersection of the tangent line and the curve, and then swap the $x$ and $y$ coordinate. Then we look at the tangent at $2P$.

Does this process halt eventually? Yes. Note that the slope of the tangent line is always less than $-1$. We know that the line between $(x_0, y_0)$ and $(y_0, x_0)$ has slope $-1$. If the slope is smaller, then the $y$-coordinate of the intersection is less than $x_0$, so we can promise that $x_1 < x_0$ after our iteration. It then suffices to show that this process does not converge to a fixed point in the fourth quadrant, which is true.

Remark: If you are interested in this elliptic curve, it can be transformed via a change of variable to the curve $Y^2 = X^3 - 432k^2$. For the curve $C$, when $k = 1$, there are nontrivial torsion points at $(1, 0)$ and $(0, 1)$; when $k = 2$, there is a nontrivial torsion point at $(1,1)$. Otherwise it is torsion free. (Reference: Dasgupta and Voight, Sylvester's Problem and Mock Heegner Point, Proc. Amer. Math. Soc.)

Answer 2

According to a Selmer's result, if the cubic $x^3+y^3=A$ has a non-torsion rational point then the set of rational points of the curve is dense in the curve. (Informally, this could be felt almost intuitively considering the method of chords and tangents that define the group law on the cubic: if the point is not of torsion, the procedure never gives a point previously calculated and you get infinitely many points all distinct). One can say that it happen with the cubic likely with the density of the set of rational points in the unit circle which is well known fact.

In our case $x^3-y^3=k$ and $u^3+v^3=k$ have both a rational point because $x^3-y^3=u^3+v^3=k$ for $(u,v)=(x,-y)$

The conclusion is that in all arc of the given cubics whose length is not null there are a lot of rational points.