Problem 1417

1. Define the curve and point:
Let $C$ be the curve defined by $x^3 - y^3 = n$, and let $P = (a, b)$ be a point on this curve, where $a$ and $b$ are rational numbers such that $a^3 + b^3 = n$.

2. **Find the tangent line at $P$:**
The tangent line $t$ to the curve $C$ at the point $P$ can be found using implicit differentiation:
$$x^3 - y^3 = n \implies 3x^2 \, dx - 3y^2 \, dy = 0 \implies \frac{dy}{dx} = \frac{x^2}{y^2}$$
Therefore, the equation of the tangent line at $(a, b)$ is:
$$y - b = \frac{a^2}{b^2}(x - a) \implies y = \frac{a^2}{b^2}x - \frac{a^3 - b^3}{b^2} = \frac{a^2}{b^2}x - \frac{n}{b^2}$$

3. Find the intersection of the tangent line with the curve:
Substitute $y = \frac{a^2}{b^2}x - \frac{n}{b^2}$ into the curve equation $x^3 - y^3 = n$:
$$x^3 - \left( \frac{a^2}{b^2}x - \frac{n}{b^2} \right)^3 = n$$
Simplify the equation:
$$x^3 - \left( \frac{a^6}{b^6}x^3 - \frac{3a^4n}{b^6}x^2 + \frac{3a^2n^2}{b^6}x - \frac{n^3}{b^6} \right) = n$$
$$x^3 - \frac{a^6}{b^6}x^3 + \frac{3a^4n}{b^6}x^2 - \frac{3a^2n^2}{b^6}x + \frac{n^3}{b^6} = n$$
$$\left( 1 - \frac{a^6}{b^6} \right)x^3 + \frac{3a^4n}{b^6}x^2 - \frac{3a^2n^2}{b^6}x + \frac{n^3}{b^6} - n = 0$$
Since $a^3 + b^3 = n$, we have $a^6 + b^6 = n^2$. Thus, the leading coefficient simplifies to:
$$1 - \frac{a^6}{b^6} = \frac{b^6 - a^6}{b^6} = \frac{n^2 - a^6}{b^6}$$
The independent term simplifies to:
$$\frac{n^3}{b^6} - n = \frac{n^3 - n b^6}{b^6} = \frac{n(n^2 - b^6)}{b^6}$$

4. Solve for the other intersection point:
Since $a$ is a double root, the other root can be found using Vieta's relations:
$$x = -\frac{a^3(2b^3 - a^3)}{a^2(a^3 + b^3)} = -\frac{a(2b^3 - a^3)}{a^3 + b^3}$$
Plugging this back into the tangent line equation, we get:
$$y = \frac{a^2}{b^2} \left( -\frac{a(2b^3 - a^3)}{a^3 + b^3} \right) - \frac{n}{b^2} = -\frac{b(2a^3 - b^3)}{a^3 + b^3}$$

5. Find the reflection of the intersection point:
The reflection of the point $\left( -\frac{a(2b^3 - a^3)}{a^3 + b^3}, -\frac{b(2a^3 - b^3)}{a^3 + b^3} \right)$ with respect to the line $x + y = 0$ is:
$$\left( \frac{b(2a^3 - b^3)}{a^3 + b^3}, \frac{a(2b^3 - a^3)}{a^3 + b^3} \right)$$

6. Conclusion:
The point $\left( \frac{b(2a^3 - b^3)}{a^3 + b^3}, \frac{a(2b^3 - a^3)}{a^3 + b^3} \right)$ is a rational point on the curve $x^3 - y^3 = n$ and both coordinates are non-negative rational numbers. Therefore, $n$ can be represented as the sum of the cubes of two non-negative rational numbers.

$\blacksquare$