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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?

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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
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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
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@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
    
@steven, i think this paper can help you:unomaha.edu/numtheory/UGresearch/manley.pdf –  Hashem sazegar Jul 29 '10 at 19:48
    
Thank you for the reference Hashem. –  Steven Jul 30 '10 at 3:47
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