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Timeline for Quadratic Solutions

Current License: CC BY-SA 2.5

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Jul 29, 2010 at 19:48 comment added Hashem sazegar @steven, i think this paper can help you:unomaha.edu/numtheory/UGresearch/manley.pdf
Jul 29, 2010 at 13:02 comment added Gerry Myerson @David, only Steven knows what Steven means, but I'm guessing Steven means to fix $n$ and then look for suitable quadratic extensions.
Jul 29, 2010 at 11:20 comment added David E Speyer 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?
Jul 29, 2010 at 6:05 comment added Gerry Myerson 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?
Jul 29, 2010 at 5:14 history asked Steven CC BY-SA 2.5