Timeline for Smallest solution to $x^2 \equiv x\pmod{n}$

14 events

when toggle format	what		by	license	comment
Jun 21, 2016 at 16:18	vote	accept	Xiaosheng Mu
Jun 21, 2016 at 16:18	vote	accept	Xiaosheng Mu
Jun 21, 2016 at 16:18
Jun 21, 2016 at 15:14	vote	accept	Xiaosheng Mu
Jun 21, 2016 at 16:18
Jun 20, 2016 at 19:46	answer	added	David E Speyer		timeline score: 6
Jun 20, 2016 at 18:47	answer	added	Fedor Petrov		timeline score: 6
Jun 20, 2016 at 18:11	history	edited	Mikhail Katz	CC BY-SA 3.0	added 2 characters in body
Jun 20, 2016 at 17:19	history	edited	Arturo Magidin	CC BY-SA 3.0	fix mod latex
Jun 20, 2016 at 17:03	comment	added	Emil Jeřábek		There is a nontrivial solution by the Chinese Remainder Theorem. Moreover, each solution $x$ is paired with another solution $1-x$, that is, $n+1-x$, so one of those satisfies $1<x<n+1-x$, i.e., $x\le n/2$.
Jun 20, 2016 at 16:55	comment	added	Kimball		Based on a small amount of numerical calculations, this may indeed be possible. What is your argument to get $x \le \frac n2$ when $n$ has at least 2 prime factors?
Jun 20, 2016 at 15:59	answer	added	David E Speyer		timeline score: 6
Jun 20, 2016 at 14:22	comment	added	Emil Jeřábek		On the other hand, for arbitrary $n$ we have the lower bound $x\ge1/2+\sqrt{n+1/4}$.
Jun 20, 2016 at 14:06	comment	added	Karl Fabian		For $n=k(k+1)$ one evidently has $x\leq 1/2+\sqrt{n+1/4}$ for $x=k+1$.
Jun 20, 2016 at 12:31	answer	added	Mikhail Katz		timeline score: 3
Jun 20, 2016 at 2:24	history	asked	Xiaosheng Mu	CC BY-SA 3.0