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Let $A$ be a finite subset of $\mathbb{C}$ with at least two elements such that $f:z\to z^2$ induces a bijection from $A$ to $A$.

If $1\notin A$, how can I show that $\displaystyle\prod_{a\in A}(1+a)=1$ ?

What are all the subsets $A$ such that $\displaystyle\prod_{a\in A}(1+a)=1$ ?


What I have tried so far :

I have tried to see what happened in simple case; I considered the simplest case with only two elements, like $A=\{e^{2i\pi/3},e^{4i\pi/3}\}$. We do get the right result, but it doesn't give much of a hint to solve the general problem...

For the second part I have only found that all of $A$'s elements must be unit roots.

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  • $\begingroup$ In the second part, do these A still have to have the property that $f$ is a bijection on them? $\endgroup$ – Tim B. May 12 '15 at 20:15
  • $\begingroup$ @LeBtz That's correct. (otherwise there would be too many solutions) $\endgroup$ – Hippalectryon May 12 '15 at 20:17
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We have $\displaystyle\prod_{a\in A}(1-a) \prod_{a\in A}(1+a) = \prod_{a\in A}(1-a^2) = \prod_{a\in A}(1-a)$, hence $\displaystyle\prod_{a\in A}(1+a)$ must be $1$, since $\displaystyle\prod_{a\in A}(1-a) \neq 0$.

For the second question:

Take any primitive root of unity $x$ of a degree $n$ for some $n>1$ which has the property that the sequence $x,x^2,x^4,...$ reaches $x$ again. This is equivalent to $2^j \equiv 1 \mod n$ for some $j>0$ (which is the the case iff $n | 2^j-1$ for some $j$. This however is exactly the case if $n$ is odd as i will show at the end). Then $A = \{x^k ~\vert~ k\in\mathbb N^\ast\}$ has the above property since $f$ is bijection of $A$ by construction. Take now arbitrary finite unions of sets as constructed above and then take all these sets and add $0$ to it or don't do it. What you are then left with are all possible such sets. Taking unions and adding $0$ obviously doesn't disturb the wanted property.

There can't be any other such sets because for each $x\in A$, $x^{2^n}$ has to be $x$ for some $n\in \mathbb N$, hence $x^{2^n}-x = 0$, which leads to $x=0$ or $x$ is a root of unity. Then $x$ is a primitve root of unity for some degree $n$. But then $2^j\equiv 1 \mod n$ must hold for some $j>0$, because otherwise the sequence $x,x^2,x^4,...$ would only contain $x$ once. But then by injectivity of $f$, it only contains $x^2$ once aswell. Inductively we obtain, that this sequence indeed has no repitition which is a contradiction to $A$ being finite. But if $x$ is an element of $A$, then of course $x,x^2,x^4,...$ have to be elements of $A$ aswell, hence it is one of the sets construced above.

It remains to prove that $2^j \equiv 1 \mod n$ for some $j>1$ iff $n$ is odd. If $n$ is even then $2^j \equiv 1 \mod n$ is impossible of course. If $n$ is odd than $gcd(2,n) = 1$, hence $2$ is a unit in $\mathbb Z/n\mathbb Z$. Since the units are a subgroup, applying Lagrange theorem, $2^j = 1$ in $\mathbb Z/n\mathbb Z$ for some $j>1$.

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  • $\begingroup$ Very smooth indeed! Endorsed!!! $\endgroup$ – Robert Lewis May 12 '15 at 17:07
  • $\begingroup$ This is very nice ! Do you have an idea on the second part ? $\endgroup$ – Hippalectryon May 12 '15 at 17:17
  • $\begingroup$ For $n=1$, the 1st root of unity would be $1$, which does disturb the property. Am I wrong ? $\endgroup$ – Hippalectryon May 12 '15 at 20:38
  • $\begingroup$ Yes, that's right. 1 has to be excluded. $\endgroup$ – Tim B. May 12 '15 at 20:41
  • $\begingroup$ And I suppose that the contradiction if $1\in A$ comes from the fact that $\displaystyle\prod_{a\in A,a\neq1}(1+a)=1$ hence $\displaystyle\prod_{a\in A}(1+a)=2$ ? $\endgroup$ – Hippalectryon May 12 '15 at 20:44

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