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I was trying to generalize a solution to a problem (if $P\in \mathbb Z[x], n\in\mathbb Z,$ then $P(P(P(n)))\neq n$ unless $P(n)=n$), and I found a sufficient condition to replace $\mathbb Z$ with a domain $R$: $u$ and $1-u$ are never simultaneously units. Equivalently, $u(1-u)$ is never a unit in $R$ for $u\in R$. Unfortunately, the only examples I can find of rings with this property are the integers and the Gaussian integers.

Are there other rings with this property, or perhaps a more general condition that implies this property? What if we restrict ourselves to rings of integers of number fields?

Aaron
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  • Could you explain (or give a reference) for "if $P\in \mathbb Z[x], n\in\mathbb Z,$ then $P(P(P(n)))\neq n$"? ;-) – Alphonse Jan 21 '17 at 11:26
  • I suppose that at least $P=x$ (or any polynomial with zero constant term in case $n=0$) s to be excluded for your original problem ... – Hagen von Eitzen Jan 21 '17 at 11:27
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    @Alphonse Assume for contradiction that $P(a)=b, P(b)=(c), P(c)=a$, for $a,b, c$ distinct integers (the actual setup of the problem, what I wrote above allows fixed points). Then $(a-b)|P(a)-P(b)=b-c$, and similarly for other pairs. This yields a contradiction if $a-b$ is nonzero. – Aaron Jan 21 '17 at 11:32
  • Ah yes, thank you. It was a question here, I think. – Alphonse Jan 21 '17 at 11:34
  • @Alphonse I didn't realize it had been posted here, I found it elsewhere, but regardless, I was only mentioning it for context. – Aaron Jan 21 '17 at 11:37
  • @Barto Interesting. I did use the assumption that my ring was a domain to reach the condition ($u(1-u)$ never a unit), but it's nice to see that things fail so easily and spectacularly without this assumption. – Aaron Jan 21 '17 at 11:54
  • @Aaron I deleted my comment because I realized you're looking for domains only. Anyway: here it is again: $\mathbb Z/2n\mathbb Z$, and if $R$ is such a ring, then so is $R\times S$ for any $S$. – Bart Michels Jan 21 '17 at 11:59
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    Note that $\forall u\in R:u^2-u\notin R^\times$ is equivalent to $\forall u,v\in R^\times: u+v\notin R^\times$; perhaps it's easier to think of it that way – Bart Michels Jan 21 '17 at 12:13
  • OK, very many rings do have two units that sum up to a unit, e.g., all fields except $\Bbb F_2$. On the other hand, if $R$ is the ring of integers of a quadratic number field $\Bbb Q(\sqrt d)$ with $d<0$, then there are only few candidates for units (3rd, 4th, 6th roots of unity) and indeed we have $\frac{-1+i\sqrt 3}2+1=\frac{1+i\sqrt 3}2$, if applicable, but essentially nothing else. Even if we drop the condition $d<0$, and assume $a+b\sqrt d$ and $(a+1)+b\sqrt d$ are units, then $a^2-db^2$ and $a^2+2a+1-db^2$ must both be $\pm1$, so $a\in{-\frac32,-\frac12,\frac12}$ – Hagen von Eitzen Jan 21 '17 at 12:54

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