Timeline for Rational points on surfaces
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2021 at 20:15 | comment | added | Jérémy Blanc | Your surface is a conic bundle over $\mathbb{A}^1$ with seven singular fibres. You may see it as an open surface of a projective surface with a $\mathbb{P}^1$ with a conic bundle having $7$ singular fibres (or $8$ if the point at infinity is singular). I would guess that if the $a_i$ are general, then the surface is a del Pezzo surface of degree $1$ and in this case is always has a k-rational point. But maybe the point is outside of your open subset? Or maybe the degree of the $a_i$ impose bad conditions so that the surface is not del Pezzo? | |
| Nov 22, 2021 at 5:40 | comment | added | individ | It is possible to reduce the search for solutions to the Pell equation. But for this you will have to use formulas that are quite long. The representation of coefficients in some form further complicates the calculations. math.stackexchange.com/questions/2773097/… When looking at the formula, usually all the desire to solve the equation in this form disappears. And the question constantly arises how to solve this equation simply. But as you have already been told, there are no simple solutions. | |
| Nov 22, 2021 at 0:25 | comment | added | Kevin Buzzard | Now my answer is "almost certainly not, because unfortunately number theory isn't that easy" :-( | |
| Nov 21, 2021 at 22:26 | comment | added | user168611 | I made my question more precise. | |
| Nov 21, 2021 at 22:26 | history | edited | user168611 | CC BY-SA 4.0 |
added 114 characters in body
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| Nov 21, 2021 at 16:10 | comment | added | Kevin Buzzard | any field of char 0? x^2+y^2+t^2+1=0 has no rational (i.e. \Q) points and neither do a gazilion variants. | |
| Nov 21, 2021 at 15:47 | history | asked | user168611 | CC BY-SA 4.0 |