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Let $E$ be an elliptic curve defined over $\mathbb{Q}$. Is there an example of such an $E$ such that the only rational point in $E(\mathbb{Q})$ is the point at infinity?

In other words, consider the following Weierstrass equation: $$y^2 = x^3 + ax + b$$

Are there values of $a, b \in \mathbb{Q}$ (preferably in $\mathbb{Z}$) such that there is no $(x, y) \in \mathbb{Q}^2$ satisfying the equation?

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  • $\begingroup$ Keith Conrad's notes might be of interest to you (in particular, see pages 10-11): math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf $\endgroup$ – Alex Wertheim Jul 15 '16 at 8:25
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    $\begingroup$ @AlexWertheim: why not upgrade your comment to an answer that spells out what is on p.10? It seems to settle this question completely. $\endgroup$ – Nefertiti Jul 15 '16 at 8:36
  • $\begingroup$ The projective elliptic curve $X^3+Y^3=Z^3$ famously has only three rational points. I just wonder what happens, if we mod out that cyclic subgroup of order three by an isogeny. May be the resulting curve has only one? $\endgroup$ – Jyrki Lahtonen Jul 17 '16 at 19:04
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As requested in the comments, I will upgrade my comment to an answer. Keith Conrad's excellent article settles the question. On page $10$, it is noted that for the values $k = -5, -6, 6, 7, -24, 45$, the Mordell curve $y^{2} = x^{3}+k$ has exactly one rational point. Proof of this fact is not given, so the references may be helpful in this regard. He does prove in each case that there are no integral points, however.

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  • $\begingroup$ Is the only rational solution mentioned in each case ? I could not find any of them yet. $\endgroup$ – Peter Mar 8 '17 at 8:33
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    $\begingroup$ @Peter: the only rational point in each case is the point at infinity, meaning there are no pairs $(x, y) \in \mathbb{Q}^{2}$ which are solutions to the associated Mordell curve. $\endgroup$ – Alex Wertheim Mar 8 '17 at 8:48

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