Timeline for A question on maps from $\mathbb{Z}/p\mathbb{Z}$ to itself
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 27, 2013 at 15:26 | history | edited | user9072 |
edited tags
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| Jul 9, 2013 at 8:52 | answer | added | Peter Mueller | timeline score: 24 | |
| Jul 8, 2013 at 11:17 | comment | added | Benjamin Dickman | The same language in your link is used at MSE, as well: math.stackexchange.com/questions/376931/… | |
| Jul 8, 2013 at 10:50 | comment | added | user36734 | This problem was posted on MathLinks, so presumably the OP knows an elementary proof. | |
| Jul 8, 2013 at 8:57 | vote | accept | Jean-Marc Schlenker | ||
| Jul 8, 2013 at 5:02 | answer | added | Noam D. Elkies | timeline score: 41 | |
| Jul 7, 2013 at 18:19 | comment | added | Jean-Marc Schlenker | Thanks Ian. I had actually tried (very superficially) to look in this direction, also based on wikipedia. But it's hard to get the permutation condition to work for all difference polynomials (except of course for a polynomial of degree 2. But I'll try more of that. Or maybe try blindly a large set of polynomials, for a given $p$, to see if I can find one which works. | |
| Jul 7, 2013 at 18:17 | comment | added | Greg Martin | I record the following equivalent statement. Suppose that $g\colon \mathbb Z/p\mathbb Z \to \mathbb Z/p\mathbb Z$ is a permutation such that each of the following functions are also permutations: $g(x)+g(x+1)$ and $g(x)+g(x+1)+g(x+2)$ and so on. Must $g$ be a linear polynomial? (The connection is that $g(x) = u(x+1)-u(x)$.) | |
| Jul 7, 2013 at 16:11 | comment | added | Ian Agol | If $u$ is a polynomial, then $u(k+l)-u(k)$ will be a polynomial of one lower degree. So one could try to analyze which polynomials give permutations, and see if one of these can arise as such a difference. en.wikipedia.org/wiki/Permutation_polynomial | |
| Jul 7, 2013 at 12:48 | comment | added | Jean-Marc Schlenker | Igor's comment on the $p$ queens problem made me think that this question has a simple geometric statement. Consider $p$ prime and a $p\times p$ toroidal blackboard. Is it possible to put one queen (say) on each line, so that no four of them are at the vertices of a parallelogram, other than by placing them according to a polynomial of degree 2? | |
| Jul 7, 2013 at 6:04 | comment | added | Yemon Choi | @IgorRivin Touché - I meant, and should have said, "the corner of the mathematical community that I encounter online". | |
| Jul 7, 2013 at 5:43 | comment | added | Igor Rivin | @YemonChoi: your concept of "world" is an interesting one, since most of the denizens of the planet would not know \mathbb if it came up and bit them on the tender parts. | |
| Jul 7, 2013 at 5:12 | comment | added | Yemon Choi | Off-topic, but sometimes I think the world's fetish for blackboard bold over actual bold has become the catechism of a cult... (This comment directed at the recent edit, not at the original post.) | |
| Jul 7, 2013 at 5:09 | comment | added | Igor Rivin | Requiring this for two $\ell$ (I prefer $\pm 1$) I think makes it equivalent to toroidal $p$ queens problem (which requires you to find a permutation $\sigma$ of $Z/p Z$ such that $\sigma \pm Id$ are both permutations also. There are lots of solutions, but not clear exactly how many (see old paper of Rivin/Vardi/Zimmerman in the Monthly). | |
| S Jul 7, 2013 at 4:56 | history | suggested | José Hdz. Stgo. | CC BY-SA 3.0 |
improved formatting
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| Jul 7, 2013 at 4:50 | review | Suggested edits | |||
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| Jul 7, 2013 at 3:15 | comment | added | Noam D. Elkies | There must be lots of examples that work for just $l=1,2$: there are $p^p$ functions from ${\bf Z}/p{\bf Z}$ to itself, about $e^{-p}$ of which are permutations; so we expect (very roughly) $e^{-2p} p^p$ random examples, which for large $p$ is way more than the number of quadratic functions. A quick numerical search finds that this first occurs for $p=7$, with $336$ functions of which only $42$ are quadratic. One that isn't has $f(n)=0,0,1,3,0,3,2$ for $n=0,1,2,3,4,5,6$. (But none of them extends to a counterexample for $l=3$, which would refute the conjecture at hand.) | |
| Jul 6, 2013 at 21:43 | comment | added | Steven Landsburg | As far as you're aware, is it possible to weaken the hypothesis by requiring your map to be a permutation only for $\ell=1,2$? | |
| Jul 6, 2013 at 19:08 | history | asked | Jean-Marc Schlenker | CC BY-SA 3.0 |