An elliptic curve $E$ over $\overline{\mathbb{Q}}$ is called a *$\mathbb{Q}$-curve* if it is isogenous (over $\overline{\mathbb{Q}}$) to all its Galois conjugates -- see [Are Q-curves now known to be modular?](https://mathoverflow.net/questions/3927/are-q-curves-now-known-to-be-modular) for example.


If I take a finite Galois extension $K / \mathbb{Q}$ and an elliptic curve $E / K$ whose base-extension to $\overline{\mathbb{Q}}$ is a $\mathbb{Q}$-curve, then all the Galois conjugates $E^{\sigma}$ are also defined over $K$, but the isogenies between them might not be. Supposing $E$ to be non-CM for simplicity, then what you get instead is a $K$-isogeny from each conjugate $E^{\sigma}$ to some possibly non-trivial quadratic twist of $E$. Let me say $E$ is a **strong $\mathbb{Q}$-curve** over $K$ if it's non-CM and it's actually $K$-isogenous to all its Galois conjugates. (Clearly any $\mathbb{Q}$-curve over $K$ becomes a strong $\mathbb{Q}$-curve over some finite extension $L / K$, but I want to keep $K$ fixed here.)


It's easy to produce examples of $\mathbb{Q}$-curves which aren't strong $\mathbb{Q}$-curves, by taking a strong $\mathbb{Q}$-curve and applying a quadratic twist by an element of $K^\times / K^{\times 2}$ that's not stable under $Gal(K / \mathbb{Q})$. However, I can't find any examples which aren't of this form.



> 
> Are there $\mathbb{Q}$-curves which are *not* twists of strong $\mathbb{Q}$-curves?
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(I'm chiefly interested in the case when $K$ is a real quadratic field here.)



Q-curves and twisting
I think there are examples of $\mathbb{Q}$-curves defined over a quadratic field $K$ which are not strong $\mathbb{Q}$-curves over $K$ in Jordi Quer's paper "$\mathbb{Q}$-Curves and Abelian Varieties of $\mathrm{GL}\_2$-Type". He uses the term $\mathbb{Q}$-curves completely defined over $K$ instead of strong $\mathbb{Q}$-curves (I will stick to Quer's terminology). The key result is Corollary 3.3, which then can be applied for instance to the family of $\mathbb{Q}$-curves with an isogeny of degree $3$ that he writes in Section 6:
$$
C^{(a)}\colon Y^2 = x^3 -3\sqrt{a}(4+5\sqrt{a})X+2\sqrt{a}(2+14\sqrt{a}+11a).
$$
Suppose that $C^{(a)}$ is not CM (the curve is CM only for 9 values of $a$). In the terminology of the article, the sets {a} and {3} are dual bases with respect to the degree map, and $K\_d = \mathbb{Q}(\sqrt{a})$. The curve $C^{(a)}$ is defined over $K\_d$ and by Corollary 3.3 if the quaternion algebra $(a,3)\_\mathbb{Q}$ is different in the Brauer group from $(-1,a)\_{\mathbb{Q}}^x$ for all $x\in\{0,1\}$ then there is no curve $\overline{\mathbb{Q}}$-isogenous to $C^{(a)}$ completely defined over $K\_d$ (when he writes isogenous in Corollary 3.3 he means over $\overline{\mathbb{Q}}$, not just over $K\_d$).


