Finding $K$-rational points on $X_0(35)$ Let $K=\mathbb{Q}(\sqrt{-2})$. How can I compute the $K$-rational points on the modular curve $X_0(35)$?
Recall that $X_0(35)$ is a hyperelliptic curve of genus 3 and has the simplified affine model:
\begin{align*}
 y^2&=x^8-4x^7-6x^6-4x^5-9x^4+4x^3-6x^2+4x+1\\
    &=(x^2+x-1)(x^6-5x^5-9x^3-5x-1)
\end{align*}
My attempt at finding $K$-rational points on $X_0(35)$ is as follows: First I find a rational map $f$ from $X_0(35)$ to a quotient curve $E$ of $X_0(35)$ with $E$ an elliptic curve (which is induced by the involution $w_5$ by Kubert). Second, I determine the preimages of $E(K)$ under $f$. If $E$ is of rank 0, $E(K)$ is finite. However, in my case $E(K)$ is of rank 1.  As a result it is computationally infeasible to determine the preimages of the infinitely many points of $E(K)$. Is there a way to work-around this issue?
Any help in finding $K$-rational points on $X_0(35)$ would be appreciated.
 A: The group $J_0(35)(\mathbb Q)$ (where $J_0(35)$ is the Jacobian
of $X_0(35)$) has rank 0 (as shown for example by a 2-descent
computation in Magma); it is isomorphic to
${\mathbb Z}/24{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$,
with generators the difference of the two points at infinity
on $X_0(35)$ and the 2-torsion point corresponding to the
factorization of the polynomial on the right hand side of
the hyperelliptic equation. (One can check that the two points
generate a group of the given isomorphism type, and the
reduction of $J_0(35)$ mod 3 has a group of this type as its
group of ${\mathbb F}_3$-points.)
If $P$ is a point in $X_0(35)(K)$ with $x$-coordinate not
in $\mathbb Q$, then the sum of $P$ with
its Galois conjugate, minus the sum of the two points at
infinity, gives rise to a nonzero $\mathbb Q$-rational point on
$J_0(35)$. One can check that none of the points is of this
form.
This leaves the case when $x(P) \in \mathbb Q$. There are the
$\mathbb Q$-rational points (the two points at infinity
and the two points with $x(P) = 0$); for all other such points,
the Galois conjugate of $P$ must be the image of $P$ under
the hyperelliptic involution, so $y(P)$ must be $\sqrt{-2}$
times a rational number. Put differently, $P$ gives rise
to a $\mathbb Q$-rational point on the quadratic twist of
$X_0(35)$ by $-2$. But this twist does not even have
${\mathbb Q}_2$-rational points, let alone $\mathbb Q$-rational
ones.
The conclusion is that
$$ X_0(35)(K) = X_0(35)({\mathbb Q}) = \{\infty_+, \infty_-, (0,1), (0,-1)\} . $$
