Timeline for Are the nonnegative rationals diophantine with only two quantifiers?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 18, 2021 at 16:43 | comment | added | user44143 | Your formula with three squares is creative, and I see that no simple variant will work with two squares or two triangular numbers: for any finite list of functions $f_i$, the set $\{x: \exists y\,\exists z\, (f_1(x)=y^2+z^2) \vee \cdots \vee (f_n(x)=y^2+z^2)\}$ has density 0 and must miss many positive rationals. | |
| Jun 18, 2021 at 16:05 | comment | added | Sergey Kiselev | Not sure if it can help, but it is known that the subset $\mathbb N$ of $\mathbb Z$ can be defined with two variables and can't be defined with one: mathoverflow.net/a/249525 | |
| May 4, 2021 at 16:52 | comment | added | Apjoo | sorry I got it now | |
| May 4, 2021 at 16:52 | comment | added | Arno Fehm | @Apjoo: Euler's four squares theorem for the integers implies Lagrange's four squares theorem for the rationals: Write $ab$ as a sum of four integer squares, then divide by $b^2$. | |
| May 4, 2021 at 16:47 | history | edited | Arno Fehm | CC BY-SA 4.0 |
added 27 characters in body
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| May 4, 2021 at 16:34 | history | asked | Arno Fehm | CC BY-SA 4.0 |