Describe the rational points on $y^2 = 11 - 2x^3$ Describe the rational points on the title curve.
My attempt: Consider the line $L$ with slope $m$ that passes through the point $(1, 3)$. To obtain rational solutions, we need to have $m$ rational. Then $L$ will be of the form $y = m(x-1) + 3$. Now consider the intersection of $L$ with the title curve. 
Note that, although the resulting equation will be a cubic in $x$, we already know that $x=1$ is a root, so it may be somewhat easier to factor the equation and possibly derive further information. But this seems not to be getting us any further ?
 A: It seems to me that $(-\frac52,\frac{13}2)$ is a rational solution.
Additionally:
This is an elliptic curve, whose points (including a point “at infinity” in the $y$-direction) form a group, with the infinite point being the identity. The famous chord-and-tangent process allows addition on the curve. By knowing the point $(1,3)$, you can try doubling it: draw the tangent to the curve at this point, and by Bézout, there is one additional intersection with the tangent, beyond the double contact at $(1,3)$. The symmetric point to this third intersection will be twice the original point. Since the derivative is $-1$ there, the line $Y=-X+4$ must have a third intersection. Of course, it’d have been possible for the third intersection to be at $(1,3)$ again: this would happen if the point was an inflection point. It didn’t happen, though, and $(\frac52,-\frac{13}2)$ is the other intersection. So $(-\frac52,-\frac{13}2)=[2](1,3)$, the doubled point. To find more points, draw the line from $(1,3)$ to $(-\frac52,-\frac{13}2)$ and look for the third intersection.
An expert will tell you that the real question is what the structure of the group of all rational points on the curve may be. Although I haven’t checked that $(1,3)$ is not a torsion point, this does look likely, in which case the rank of the group will be $\ge1$. Good luck in trying to find points that are not in the group spanned by $(1,3)$.
