There are quadratic solutions to $x^4+y^4 = z^4$ in $\mathbb{Q} (\sqrt{-7})$. But for equations such as $x^4+y^4 = nz^4$ where $n \in \mathbb{N}, \ n \neq 1$ do there still exist extension fields of $\mathbb{Q}$ that contain quadratic solutions?
I think what you are asking is, given $n\ne1$, must there exist an integer $d$ such that $x^4+y^4=nz^4$ has a solution in ${\bf Q}(\sqrt d)$, ignoring any solutions where all the variables take rational values. Is that the question? – Gerry Myerson Jul 29 '10 at 6:05
Maybe I'm confused. Take any integers $X$, $Y$ and $Z$, with $Z$ not square. Compute $N=(X^4+y^4)/Z^2$. Let $d$ be the denominator of $N$, so $n:=d^4 N$ is an integer. Set $(x,y,z)=(d^2 X, d^2 Y, d \sqrt{Z})$ is a solution to $(x^4+y^4)/z^4=n$ in $\mathbb{Q}(\sqrt{Z})$. Is this what you meant to ask? – David Speyer Jul 29 '10 at 11:20
@David, only Steven knows what Steven means, but I'm guessing Steven means to fix $n$ and then look for suitable quadratic extensions. – Gerry Myerson Jul 29 '10 at 13:02