<s>[INST] 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.


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*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!
 [/INST] 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.
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