Rational curves on the Fermat quartic surface Let $X$ be the Fermat quartic $x^4+y^4+z^4+w^4=0$ in $\mathbb P^3$. It is known that $X$ contains infinitely many $(-2)$-curves, that is, smooth rational curves. (One way to obtain in infinitely many is to use the various elliptic fibrations on $X$, and use translations in the fibers.) Note however that these curves are typically not defined over $\mathbb Q$, even though $X$ is. My question is the following: 
Is there a finite extension $K$ of $\mathbb Q$ such that all of the smooth rational curves are defined over $K$ and have a $K$-rational point?
I suspect that the answer is no, but I don't see how to prove it.
 A: I am posting my comments as an answer.  I am concerned that I misunderstand the OP, so let me state first the result.  There exists a finite field extension $K/\mathbb{Q}$ such that for every closed immersion $\mathbb{P}^1_\mathbb{C} \hookrightarrow X\otimes_{\mathbb{Q}}\mathbb{C}$, there exists a closed immersion $\mathbb{P}^1_K \hookrightarrow X \otimes_{\mathbb{Q}} K$ with the same image after base change to $X\otimes_{\mathbb{Q}}\mathbb{C}$.  
To prove this, first observe that the geometric Picard group $\text{Pic}(X\otimes_{\mathbb{Q}}\mathbb{C})$ is finitely generated.  Thus, there exists a finite field extension after which a finite generating set of the geometric Picard group is defined over that field.  
Second, the geometric automorphism group of the complex scheme, $\text{Aut}_{\mathbb{C}}(X\otimes_{\mathbb{Q}}\mathbb{C})$, is a discrete group that is finitely generated by Corollary 2.4 of Chapter 15, p. 315 of the following; result due to H. Sterk -- Finiteness results for algebraic K3 surfaces. Math. Z., 189(4):507–513, 1985.   
Daniel Huybrechts 
Lectures on K3 Surfaces 
Part of Cambridge Studies in Advanced Mathematics 
September 2016 
ISBN: 9781107153042 
http://www.math.uni-bonn.de/people/huybrech/K3Global.pdf
Thus, after a further finite field extension, there exists a finite subset of $\text{Aut}_K(X\otimes_{\mathbb{Q}} K)$ whose images generate the geometric automorphism group.  Thus, $\text{Aut}_K(X\otimes_{\mathbb{Q}} K)$ is already the full geometric automorphism group.
Third, by Corollary 4.6 of Chapter 8, p. 161 -- result again proved by Sterk in the article cited above -- the action of the geometric automorphism group on the set of smooth rational curves has only finitely many orbits.  For each of those finitely many orbits, after a degree $2$ field extension, every curve in that orbit acquires a rational point.  Thus, after a further finite field extension, every smooth, genus $0$ curve on $X$ acquires a rational point.
Please note, all of this is valid for every K3 surface over every characteristic $0$ field, not only for the Fermat quartic surface over $\mathbb{Q}$.  Using extensions of Sterk's results by Lieblich and Maulik, this should also be true in positive characteristic.
