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I was trying to compute the torsion of $C$ : $y^2=x(x+1)(x-8)$ over $\mathbb{Q}$ by using the fact that the order of the torsion divides the order of $C$ reduced modulo any $p\nmid 2\Delta$.

For any prime $p$ I tried, I got $8\mid\#C(\mathbb{F}_p)$, but in fact the torsion turned out to be 4 (by Nagell-Lutz). However, I don't see how to prove that 8 divides $\#C(\mathbb{F}_p)$ for any prime $p$.

Any ideas would be appreciated.

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We already know that $C(\Bbb Q)$ has full $2$-torsion and no point of order $4$, so $Gal(C[4]/\Bbb Q)$ is a subgroup $G$ of $\{M \in Aut((\Bbb Z/4\Bbb Z)^2) \mid M = I_2 \pmod 2 \}$, with no other point fixed by $G$ than the elements of order $2$.

If $8$ divides $C(\Bbb F_p)$ for almost all $p$, then this is equivalent, by the Cebotarev density theorem, that the elements of $G$ all fix at least one element of order $4$.

After looking at the possibilities, one finds that $G$ has to be isomorphic to $(\Bbb Z/2\Bbb Z)^2$ and there is some basis where $G = \left\{I_2;\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} ; \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} ; \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} \right\}$

This tells us about a few arithmetic properties of $C$, in terms of divisors that are now rational (and that aren't in the general case) :

Let $(A,B)$ be a $\Bbb Z/4\Bbb Z$-basis of $C[4]$ in which $G$ as the form as described above.

$\{B; -B\}$ is stable by $G$, so $[B] + [-B] - 2[O]$ is a rational principal divisor. Given its form it must be the divisor of the function $(x-x_B)$ for some rational $x_B$ : $B$ and $-B$ must share a rational $x$ coordinate. The exact same thing happens with $2A+B$ and $2A-B$, and so they also must share a rational $x$ coordinate. (in general, those two pairs can be conjugates, so the Galois group would swap the two $x$-coordinates)

$\{A; A+2B\}$ is stable by $G$, so $[A] + [A+2B] - [O] - [2(A+B)]$ is also a rational principal divisor. This time it must be the divisor of a function of the form $y/(x-x_{2(A+B)}) - k$ for a rational number $k$, that is, $A$ and $A+2B$ are the other 2 intersections of the line through $2(A+B) \in C[2]$ whose slope is $k \in \Bbb Q$.
Applying the $(P \mapsto -P) (y \mapsto -y)$ automorphism, the line going $2(A+B)$ with slope $-k$ intersects the elliptic curve at $3A$ and $3A+2B$. (in general, those two pairs can be conjugates, and then the slope would be an irrational square root)

Finally, the exact same thing happens with $A+B$ and $A+3B$ (and the opposite pair), giving another $2$ lines through $2A$ with opposite rational slopes.


Summing up, it looks like the thing to do is to compute enough about the points of order $4$ of $C(\Bbb Q)$ to show that :

  • there is a point $P \in C[2]$ whose halves have rational $x$-coordinates

  • if $Q,R$ are the other two points of order $2$, the halves of $Q$ have a rational slope with $R$ and vice versa.

I haven't looked at all (I feel completely blind) at what happens modulo $p$, but if you look at the points in $\Bbb F_{p^2}$ you should be able to identify who $P$ is by looking at the $A^p-A$ (they are either $O$ or $P$). Then you should be able to identify all the rationals numbers involved, which speeds up the computations.

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Taking the low road for it's the only region where I can find my way.

Aided by Mathematica I got for the fourth division polynomial of your curve the following $$ \psi_4(x,y)=4y(x-2)(x+4)(x^2+8)(x^2-16x-8). $$ This shows a lot of potential for order four points modulo $p$. Disregarding the 2-torsion (zeros of $y$) the obvious zeros $x=2$ (resp. $x=-4$) lead to $\Bbb{F}_p$-rational points if the equations $y^2=-36$ (resp. $y^2=-144$) have solutions. Clearly this happens, iff $p\equiv1\pmod4$, because the 4-torsion points $(2,\pm6i)\in C$ (resp. $(-4,\pm12i)\in C$) then have reductions with coordinates in $\Bbb{F}_p$.

OTOH, if $p\equiv-1\pmod4$, then we know that exactly one of $\pm2$ is a quadratic residue modulo $p$. This is just as well, because the quadratic factors $x^2+8$ and $x^2-16x-8$ have respective zeros $x=\pm2i\sqrt2$ and $x=2(4\pm3\sqrt2)$. This produces $\Bbb{F}_p$ rational 4-torsion points as follows:

  • The 4-torsion point $(x,y)=(2i\sqrt2,8-2i\sqrt2)\in C$ reduces to a $\Bbb{F}_p$-rational 4-torsion point whenever $-2$ is a quadratic residue modulo $p$.
  • The 4-torsion point $(x,y)=(2(4+3\sqrt2),24+18\sqrt2)\in C$ reduces to a $\Bbb{F}_p$-rational 4-torsion point whenever $2$ is a quadratic residue modulo $p$.

From the above points and their conjugates it follows that $C(\Bbb{F}_p)$ has

  • the maximum of 12 points of order four, when $p\equiv1\pmod8$,
  • four 4-torsion points gotten by reducing $(2,\pm6i)$ and $(-4,\pm12i)$, when $p\equiv5\pmod8$,
  • four 4-torsion points with $x$-coordinate a zero of $x^2+8$, when $p\equiv3\pmod8$, and
  • four 4-torsion points with $x$-coordinate a zero of $x^2-16x-8$, when $p\equiv7\pmod8$.

Altogether these observations imply that the subgroup $C(\Bbb{F}_p)\cap C[4]$ has size $16,8,8,8$ all according to whether $p$ is congruent to $1,3,5,7\pmod8$. Therefore the size of the group $C(\Bbb{F}_p)$ is a multiple of $8$ in all the cases.

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