Timeline for a family of Pellian equations
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 2, 2022 at 21:39 | comment | added | duje | The mentioned related conjecture that there does not exist a set of four positive integers with the property that the product of any two of them is 1 greater than a square was proved recently in the paper N. C. Bonciocat, M. Cipu, M. Mignotte, There is no Diophantine D(-1)-quadruple, J. London Math. Soc. 105 (2022), 63-99. | |
| Oct 2, 2022 at 21:23 | history | edited | duje | CC BY-SA 4.0 |
added information on a recent paper by Le and Srinivasan concerning a special case of the conjecture
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| Mar 12, 2016 at 19:26 | answer | added | Will Jagy | timeline score: 2 | |
| Mar 12, 2016 at 1:28 | comment | added | Will Jagy | found slides from a nice talk by Keith Matthews numbertheory.org/pdfs/dujella_slides.pdf | |
| Feb 13, 2013 at 6:57 | vote | accept | duje | ||
| Jan 16, 2013 at 6:57 | answer | added | Jim White | timeline score: 1 | |
| Jan 12, 2013 at 9:20 | answer | added | Jim White | timeline score: 3 | |
| Nov 13, 2012 at 23:41 | comment | added | Max Alekseyev | Just a simple observation: this equation is equivalent to $x^2+1=(y^2+1)(k^2+1)$, i.e., when the product of two numbers of the form $m^2+1$ is again a number of this form. | |
| Feb 24, 2012 at 18:15 | comment | added | duje | Let me mention that I have checked "the conjecture" for $k\leq 1000000$. In that range, there are 1045 $k$'s for which this equation has 5 classes of solutions, while for all other $k$'s there are 3 classes. It seems that the number of $k$'s such that $k \leq N$ and the equation has 5 classes of solutions is $O(\sqrt{N})$ (it seems that, asymptotically, the most on such cases comes from $k=2t^2$). | |
| Feb 24, 2012 at 17:54 | answer | added | Franz Lemmermeyer | timeline score: 8 | |
| Feb 19, 2012 at 10:51 | history | asked | duje | CC BY-SA 3.0 |