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Could you please tell me what is the points of order 3 (or 6) Hon the elliptic curve $y^2=x^3+sx^2-x$ where $$s = -\frac{1}{432}\frac{(81k^8-2592k^4-6912)}{k^6}$$ and $k$ is rational?

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4  
And the context is? – Anthony Quas Feb 26 '14 at 0:53

Two points of order $6$ are $(-4/k^2, \pm (k^4+16)/k^5)$, and two points of order $3$ are $(k^2/4, \pm (1+k^4/16)/k)$.

This can be verified via the following Magma code, which can be used on the free Magma online calculator:

K<k>:=FunctionField(Rationals());
s:=-1/432*(81*k^8-2592*k^4-6912)/k^6;
E:=EllipticCurve([0,s,0,-1,0]);
A,f:=TorsionSubgroup(E);
P:=f(A.1);
for n in [1..6] do n,n*P; end for;

which outputs

1 (-4/k^2 : (-k^4 - 16)/k^5 : 1)
2 (1/4*k^2 : (1/16*k^4 + 1)/k : 1)
3 (0 : 0 : 1)
4 (1/4*k^2 : (-1/16*k^4 - 1)/k : 1)
5 (-4/k^2 : (k^4 + 16)/k^5 : 1)
6 (0 : 1 : 0)
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