Timeline for A question on fixed point theory

7 events

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Mar 11, 2015 at 23:04	comment	added	Eric Wofsey		In case anyone cares, I believe I've worked out that the necessary and sufficient conditions for this argument to work when $k$ is odd are: (1) $n$ is odd, (2) $n>k+2^d$ if $2^d$ is the least power of $2$ dividing $k+1$, and (3) $n$ is not the least integer greater than $k$ which is $1$ less than a multiple of $2^d$ for any $d$. The proof is a messy but straightforward induction on $n$ with $k$ fixed.
Dec 16, 2014 at 15:01	comment	added	Ali Taghavi		@WłodzimierzHolsztyński thank you very much for your communication in my question.
Dec 16, 2014 at 15:00	comment	added	Ali Taghavi		@EricWofsey thank you so much for your very interesting question.
Dec 16, 2014 at 14:06	vote	accept	Ali Taghavi
Dec 16, 2014 at 7:25	comment	added	Włodzimierz Holsztyński		Eric, you're much too kind to me. My answer (which I have just removed) was not slightly but grossly incorrect. I am very glad for your meaningful post.
Dec 16, 2014 at 4:02	comment	added	Eric Wofsey		Actually, the hypothesis that $k+1$ and $n+1$ are divisible by $2$ the same number of times is far stronger than what is needed for this argument to work, though the actual necessary and sufficient condition seems a lot more complicated to state. For instance, if I'm not mistaken, if $k+1=2^d$ for some $d>0$ then $n+1$ can be any even number greater than $2^{d+1}$ that is not a power of $2$.
Dec 16, 2014 at 2:21	history	answered	Eric Wofsey	CC BY-SA 3.0