Timeline for Solutions of the equation $X^4-DY^4=z^4$
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2014 at 12:15 | comment | added | Joe Silverman | @NoamD.Elkies Good point, the 3-fold is rational. So certainly the solutions in $\mathbb{Q}^4$ are dense. And by modifying your example to, say, $Z=X+cY^n$ with $c\in\mathbb{Z}$, you'll get a Zariski dense set of solutions $(W,[X,Y,Z])$ that are Zariski dense in $\mathbb{Z}\times\mathbb{P}^2(\mathbb{Q})$. | |
| Jan 26, 2014 at 3:52 | comment | added | Vassilis Parassidis | If we set $k=a/2$ where $a$ an integer, then in the equation $4k^3+k=±m$ $m$ is always an integer and so the equation $X^4 –DY^4=Z^4$ is solvable. | |
| Jan 26, 2014 at 3:18 | comment | added | Noam D. Elkies | But the variety $X^n - W Y^n = Z^n$ is clearly rational because the equation can be solved for $W$. We can even make $\gcd(X,Y,Z) = 1$ by fixing $Y \neq 0$ and (for example) taking $X$ coprime to $Y$ and $Z = X + Y^n$. | |
| Jan 26, 2014 at 1:23 | comment | added | Vassilis Parassidis | The filled circle does mean multiplication. Thanks for your suggestions. I was hoping for a technique that yields infinite solutions, not just a lot. | |
| Jan 26, 2014 at 1:13 | history | answered | Joe Silverman | CC BY-SA 3.0 |