I am trying to find the torsion subgroup $E(\mathbb{Q})$ of the elliptic curve $E: y^2=x^3+4x$ over $\mathbb{Q}$ which apparently is $\mathbb{Z}/4\mathbb{Z}$ according to exercise 4.9 of the book "Rational Points on Elliptic curves" by Silverman and Tate. But the only integer points of $E$ that I could find are $(0,0), (2,4)$ and $(2,-4)$. Apparently, $P=(2,4)$ is the generator of $E(\mathbb{Q})$ but the $x$-coordinate of $P+P$ is $x([2]P)=0$. So, how is $P$ a generator of the torsion subgroup of $E(\mathbb{Q})$ as mentioned here? If someone could point out what I am missing here and what is the fourth point of the torsion subgroup of $E(\mathbb{Q})$ I would really appreciate it.
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2The fourth point is the point at infinity! I.e. the identity. You have $[2]P = (0, 0)$, which is a 2-torsion point, but is not the identity. – Mathmo123 Sep 29 '22 at 14:36
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@Mathmo123 Then why is $P$ mentioned as a generator? – Anish Ray Sep 29 '22 at 14:39
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6Because it is a generator! $(0, 0) = [2]P$, $(2,-4) = [3]P$ and $\mathcal O = [4]P$. – Mathmo123 Sep 29 '22 at 14:41
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2@Mathmo123 could you convert that comments as an answer? – kelalaka Oct 03 '22 at 20:21