Prove that $x^3 + y^3 = 9$ has infinitely many solutions $(x,y) \in \Bbb Q^2$.
My approach was the obvious one: writing $x$ and $y$ in terms of coprime integers and then turning the rational numbers problem into an integer numbers problem. But I couldn't advance much.
Also, in chat, a few days ago, some guys helped me using elliptic curves and geometric constructions but, as an undergrad olympiad problem, it was an overkill so I couldn't actually understand it.
Here is the 'tangent-chord' construction of infinitely many rational points from a single such point.
Suppose we have a point $p_0=(x_0,y_0)$ on the cubic curve $f(x,y)=x^3+y^3=9$. The tangent vector at this point is given by $$\left(\frac{\partial f}{\partial y},-\frac{\partial f}{\partial x}\right)_{p_0}=(3y_0^2,-3x_0^2)$$ and the tangent line at $p_0$ has the parametric form $$(x(t),y(t))=(x_0+3y_0^2 t,y_0 -3x_0^2 t).$$ The condition for this tangent to intersect the cubic is then $$f(x(t),y(t))=(x_0+3y_0^2 t)^3+(y_0-3x_0^2)^3=9.$$ If we use the fact that $f(x_0,y_0)=x_0^3+y_0^3=9$ and simplify the above accordingly, this becomes $243t^2(x_0 y_0-t(x_0^3-y_0^3))=0$. Two of the three solutions are $t=0$, corresponding to the line being tangent at $t=0$; the remaining case, corresponding to $t=\dfrac{x_0 y_0}{x_0^3-y_0^3}$, gives a new intersection with the cubic at the point $$(x(t),y(t)) = \left(x_0+\frac{3x_0 y_0^3}{x_0^3-y_0^3},y_0-\frac{3x_0^3 y_0}{x_0^3-y_0^3}\right)$$ By construction, this is a solution of $f(x,y)=9$. Moreover, if $(x_0,y_0)$ is a rational point then this new point on the curve is also rational, and this can be iterated to produce a family of such rational points. As an example, take $(x_0,y_0)=(1,2)$ and iterate the above tangent-chord map to obtain the infinite family of solutions \begin{align} (1,2) &\mapsto \left(-\frac{17}{7},\frac{20}{7}\right)\\ &\mapsto\left(\frac{188479}{90391},-\frac{36520}{90391}\right)\\ &\mapsto \left(\frac{1243617733990094836481}{609623835676137297449},\frac{487267171714352336560}{609623835676137297449}\right)\mapsto\cdots \end{align} In this way we generate a sequence of rational solutions to $x^3+y^3=9$ from one such solution.