Let $K=\mathbb{Q}(\sqrt{6})$. I am looking to determine all $K$-rational points on the curve
$$C: y^{2}=3x^6-24x^5+24x^4-54x^3+24x^2-24x+3.$$
More precisely, $C$ is a twist of the modular curve $X\_{0}(26)$. I know that Bruin and Najman (<https://arxiv.org/abs/1406.0655>) have determined all quadratic points on $X\_{0}(26)$ using the finiteness of the Jacobian of $X\_{0}(26)$ over $\mathbb{Q}$.


Let $J$ denote the Jacobian of $C$. Then $J$ has rank 1 over $\mathbb{Q}$. It's also interesting that $C(\mathbb{Q})=\emptyset$ (this can be seen using the TwoCoverDescent function on Magma).


I would really appreciate any pointers on how to proceed.



Finding the $K=\mathbb{Q}(\sqrt{6})$-rational points on the twist of $X_{0}(26)$
I used Magma to point search on $C/K$ up to a height of $1000$ and it appears that $C(K) = \emptyset$. If that's true, then one can probably use the Mordell-Weil sieve to prove it. Here's a bit more detail.


The curve $C$ has four automorphisms defined over $\mathbb{Q}$ and for one of these (the map $(x,y) \mapsto (1/x,y/x^{3})$), the quotient curve has genus $1$. In particular, there is a map from $\phi : C \to E$, where $E : y^{2} = x^{3} - 651x - 12742$. This curve $E$ has rank $1$ over $K$.


For a finite set $S$ of prime ideals of $\mathcal{O}\_{K}$ (all of which are primes of good reduction for $E/K$), one can write down the commutative diagram
$$
\require{AMScd}
\begin{CD}
C(K) @>>> E(K)\\
@VVV @VV{\beta}V\\
\prod\_{\mathfrak{p} \in S} C(\mathbb{F}\_{\mathfrak{p}}) @>\alpha>> \prod\_{\mathfrak{p} \in S} E(\mathbb{F}\_{\mathfrak{p}})\\
\end{CD}
$$
The horizontal maps in this diagram use the map from $C$ to $E$, while the vertical maps use reduction modulo the prime ideals in $S$.
If $P \in C(K)$ is a point, then the image of $\alpha$ and the image of $\beta$ (as subsets of $\prod\_{\mathfrak{p} \in S} E(\mathbb{F}\_{\mathfrak{p}})$) have a non-trivial intersection. So if one can find a set $S$ of primes for which the image of $\alpha$ and the image of $\beta$ are disjoint, this proves that $C(K)$ is empty.


For more about this, I recommend [The Mordell-Weil sieve: Proving non-existence of rational points on curves](https://arxiv.org/abs/0906.1934) by Nils Bruin and Michael Stoll.


