# Torsion group of $y^2 = x^3 \pm nx$

Let $$n\in \mathbb{Z}$$ be not divisible by $$15$$ and suppose $$n$$ and $$-n$$ are both not perfect squares. Prove that at least one of the elliptic curves $$Y^2 = X^3 + nX$$ and $$Y^2 = X^3 - nX$$ has torsion group given by $$\{{\bf o},(0,0)\}$$ (where $${\bf o}$$ is the point at infinity).

Here is my progress:

The discriminant is $$\Delta = \pm 4n^3$$ so we may use reduction with any odd prime $$p \nmid n$$. Since $$n$$ is not divisible by $$15$$, at least one of $$3\nmid n$$ and $$5\nmid n$$ holds.

• Suppose $$3\nmid n$$ - then we can use $$p=3$$. Since $$x^3 \equiv x \pmod 3$$, for the reduced curves we can equivalently consider $$Y^2 = (n+1)X$$ and $$Y^2 = (1-n)X$$. If $$n\equiv 1 \pmod 3$$, then the former has only $${\bf {\underline{o}}}, (0,0), (2,\pm 1)$$ as points, while the ones on the latter are $${\bf {\underline{o}}}, (0,0), (1,0), (2,0)$$. The points are exchanged when $$n\equiv 2 \pmod 3$$.

• Suppose $$5\nmid n$$ - then we can use $$p=5$$. For $$n\equiv 1,2,3,4 \pmod 5$$ a direct verification shows that $$Y^2 = X^3 + nX$$ has $$4,2,6,8$$ points respectively; hence $$Y^2 = X^3 - nX$$ has $$8,6,2,4$$ points, respectively.

Hence currently I have that at least one of the curves has $$1, 2$$ or $$4$$ torsion points. It is also easy to check that $$(0,0)$$ is always a point of order $$2$$. So how to rule out the case of $$4$$ torsion points?

Any help appreciated!

• Both curves have only a single rational point of order two. So you only need to rule out the case of a cyclic torsion group of order four. I might try and check whether it is possible that $(0,0)=P$ for a rational point $P$. In other words, can the tangent at a rational point pass through the origin? I don't know if this can be done though. Feb 1, 2020 at 14:44

Since $$n$$ and $$-n$$ are not perfect squares, the point $$(0, 0)$$ is the only point of order $$2$$ on both curves.
Now if $$P = (x, y)$$ is a rational point of order $$4$$ on the curve $$Y^2 = X^3 - nX$$, then by the doubling formula, we have $$X(2P) = \frac{x^4 + 2nx^2 + n^2}{4y^2}.$$ However, $$2P$$ must be the point $$(0, 0)$$, hence $$x^4 + 2nx^2 + n^2 = (x^2 + n)^2$$ must be zero, which means $$-n = x^2$$ is the square of a rational number, and therefore a perfect square. This contradicts the assumption.
Changing $$n$$ to $$-n$$ shows that the curve $$Y^2 = X^3 + nX$$ has no point of order $$4$$ either.