Timeline for What is the smallest unsolved Diophantine equation?
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
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| Sep 10 at 3:26 | comment | added | Lucenaposition | Maybe make an equation that has no solutions iff ZFC is consistent. Then if ZFC is consistent, you can't prove it has a solution (because it has none) but you can't prove it has no solutions in ZFC (by incompleteness). | |
| Sep 3 at 13:40 | answer | added | Bogdan Grechuk | timeline score: 9 | |
| S Feb 5, 2022 at 8:46 | history | suggested | The Amplitwist | CC BY-SA 4.0 |
used Markdown formatting for bullet points instead of the MathJax workaround
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| Feb 4, 2022 at 18:58 | review | Suggested edits | |||
| S Feb 5, 2022 at 8:46 | |||||
| Jan 3, 2022 at 10:42 | answer | added | Bogdan Grechuk | timeline score: 1 | |
| Nov 5, 2021 at 13:09 | answer | added | Bogdan Grechuk | timeline score: 8 | |
| Jul 26, 2021 at 19:56 | comment | added | Bjorn Poonen | In contrast to what is claimed above, deciding whether a two-variable polynomial equation has an integer solution is still an open problem. Runge's method (used in the Hilliker-Straus paper mentioned above) and Baker's method are effective only for special equations. | |
| Jul 7, 2021 at 17:30 | answer | added | Bogdan | timeline score: 6 | |
| May 18, 2021 at 2:53 | comment | added | LSpice | Re the opening bid, $x^3 + y^3 + z^3 = 33$ is now (well, for some time, but this question just recently came back to the front page) known to be soluble. | |
| May 17, 2021 at 19:10 | answer | added | Will Sawin | timeline score: 21 | |
| Aug 31, 2020 at 21:21 | answer | added | Bogdan | timeline score: 14 | |
| Aug 31, 2020 at 13:46 | answer | added | Bogdan | timeline score: 28 | |
| Dec 4, 2018 at 19:29 | comment | added | user44143 | Given the logic tag, I’d like to advertise $a^5+b^5=c^5+d^5$, a simple equation that would have lower height than the $33$ equation if we used $3/2$ instead of $2$. If writing a number as a sum of fifth powers is unique, then we can use $x^5+y^5$ for a pairing function on the rationals, which would be an advance on Hilbert’s 10th problem over $\mathbf{Q}$. | |
| Dec 3, 2018 at 15:24 | history | edited | Zidane |
edited tags
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| Dec 3, 2018 at 14:46 | comment | added | Wojowu | I agree such theories are unnatural (and, arguably, dumb). I just meant to imply there is clarification needed for this to become a question which can be answered. | |
| Dec 3, 2018 at 14:31 | comment | added | Zidane | @Wojowu I agree that, once you prove that deciding the existence of an integral solution is undecidable in a system, you can add an axiom saying there is no solution and hence get a new system in which the problem is not undecidable. I was more interested in a natural system (ZFC say) and have an idea of how far one has to go to run into real problems: I think that there are explicit examples of undecidable diophantine equations, but they are probably quite big. | |
| Dec 3, 2018 at 14:29 | comment | added | Zidane | @TerryTao Baker's bounds make the determination of integral solutions in $2$ variables deterministic, but they are usually much too big to allow you to actually certify that you have found all solutions. It's like for the ternary Glodbachh conjecture: there was a time gap between Vinogradov's result and the complete solution. | |
| Dec 3, 2018 at 14:25 | comment | added | Zidane | @Chris This was the simplest formule I could think of that makes sure that the number of $P$ with $h(P)\leq N$ is finite (not fixing degrees or number of variables (up to obvious isomorphisms)). If you want to put more emphasis on degree $d$ and number of variables $n$ you can replace $2$ by $(d+C)^n$ or anything that suits your purposes. | |
| Dec 2, 2018 at 20:57 | comment | added | François Brunault | @Chris I guess this definition is used to ensure the degree of the polynomial cannot be arbitrarily large. It is a bit unusual since finiteness properties of height functions are usually stated for bounded degree. | |
| Dec 2, 2018 at 20:30 | comment | added | Chris Wuthrich | $x^3-1-z^2(y^3-1)=0$ mentioned here: mathoverflow.net/questions/54857/… has $h(P) = 45$. I don't know if it has been solved. Not sure why $x$ are taken to be $2$ in the definition of $h$, though. | |
| Dec 2, 2018 at 18:44 | comment | added | Terry Tao | I believe this paper of Hilliker and Strauss jstor.org/stable/1999638?seq=1#metadata_info_tab_contents implies that the first question is decidable in two variables, so one needs at least three variables. I believe all quadratic Diophantine equations are known to be decidable also, so this makes Poonen's example likely to be almost the best possible. | |
| Dec 2, 2018 at 17:51 | comment | added | Terry Tao | To start the bidding: according to Poonen at www-math.mit.edu/~poonen/papers/h10_notices.pdf , it is not known whether there are integer solutions to $x^3+y^3+z^3-33 = 0$ (so $h(P) = 57$ in this case). | |
| Dec 2, 2018 at 17:04 | comment | added | Wojowu | Regarding fourth and fifth point: if you mean algorithmically undecidable, there is none - the solution set is trivially decidable for any integer polynomial. If you mean undecidable in the sense of it being unprovable what the solutions are, the answer depends on background theory (Peano arithmetic, ZFC, etc.) There isn't a single polynomial which could work for all theories. | |
| Dec 2, 2018 at 17:00 | comment | added | Wojowu | Not a duplicate, but closely related: mathoverflow.net/questions/54857/… | |
| Dec 2, 2018 at 11:25 | review | First posts | |||
| Dec 2, 2018 at 11:39 | |||||
| Dec 2, 2018 at 11:22 | history | asked | Zidane | CC BY-SA 4.0 |