Timeline for What is the smallest unsolved Diophantine equation?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2020 at 19:14 | comment | added | JoshuaZ | @HarryAltman Oh, hmm, Not sure what I meant then. I'll have to think about this again. | |
| Nov 1, 2020 at 18:15 | comment | added | Harry Altman | That results in $x^2 + 1=0$, though, i.e., just again covers the $p\equiv1 \pmod{4}$ case. | |
| Nov 1, 2020 at 12:36 | comment | added | JoshuaZ | @HarryAltman, I think that should be $x \equiv -y$ (mod $p$). | |
| Oct 31, 2020 at 23:37 | comment | added | Harry Altman | I don't follow the part about $x=yy$; did you mean something else? | |
| Sep 4, 2020 at 11:44 | comment | added | JoshuaZ | This equation is solvable mod p for any prime p. $z \equiv 1$ (mod p), $x \equiv y $ (mod p) has a solution when $p \equiv 1$ (mod 4). $z \equiv 0$ (mod p), and $x \equiv y y$ has a solution when -2 is a QR mod $p$. Similarly, $z \equiv -2$ (mod p) and $x \equiv y$ has a solution when $2$ is a QNR mod $p$. So that covers all of them. I haven't checked lifting for powers of $p$ but it should go through also. So if this one has no solutions it isn't just due to a simple modulus argument. | |
| Aug 31, 2020 at 22:04 | comment | added | Will Sawin | Let me comment more generally that the heuristic of Heath-Brown on the sum of three cubes problem suggests, I believe, that broadly similar equations (i.e. cubics in three variables) that are sufficiently generic (don't have quadratic-recriprocity-based obstructions, Vieta jumping structure, or maybe a few other similar things), should have a constant times $\log n$ solutions of height up to $n$, so that if this constant is small, might have no solutions found by an easy computer search. So I think there is an unsolved problem somewhere among the three-variable cubics. | |
| Aug 31, 2020 at 21:54 | comment | added | Will Sawin | Your equation looks a bit like the Markoff surface $x^2+y^2+z^2=3xyz$. One can consider the phenomenon (Vieta jumping) If $(x,y,z)$ is a solution then $(yz-x, y,z), (x,xz-y,z),$ and $(x,y,-xy-z)$ are all solutions. Maybe one can show that, for all solutions, at least one of these must have smaller height? | |
| Aug 31, 2020 at 21:21 | history | answered | Bogdan | CC BY-SA 4.0 |