I understand that when given a rational point on an elliptic curve, one can find more by method of secants and tangents. This also creates an abelian group of rational points on the curve.

When you generate such a group of points, will it contain all the rational points? I'd appreciate answers, other posts, or good reads on the question.

I have in mind specifically elliptic curves of the form: $$y^2=x^3-n^2x$$


No, in general one single point will not generate all the rational points. If $E$ is an elliptic curve defined over $\mathbb{Q}$, then the rational points form a finitely generated abelian group $E(\mathbb{Q})$. The classification of finitely generated abelian groups tells us that $$E(\mathbb{Q}) \cong E(\mathbb{Q})_\text{tors} \oplus \mathbb{Z}^{R_{E/\mathbb{Q}}},$$ i.e., $E(\mathbb{Q})$ is generated by a finite number of points, some of finite order (those in the torsion subgroup $E(\mathbb{Q})_\text{tors}$), and some of infinite order. In particular, if $R_{E/\mathbb{Q}}>1$, then a single point $P$ in $E(\mathbb{Q})$ will only generate part of the group, but not all. It also can happen that $R_{E/\mathbb{Q}}=0$, but the torsion subgroup is bicyclic, e.g., $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, and in this case although there are only finitely many rational points, you can't generate them all from one single rational point.

In the case you are interested in, for a curve $E_n:y^2=x^3-n^2x$ (known as the curves related to the congruent number problem) the group of rational points is of the form $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}^{R_n}$. So it is never generated by a single point. If you mean to ask whether $R_n$ is always $1$ when it is positive, then that is not true either. For instance, $R_5=1$, but $R_{34}=R_{41}=R_{65}=2$ (here $34,41,65$ are the only values of $n\leq 100$ such that $R_n\geq 2$).

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