Silk Road Mathematics Competition

For real numbers $r$ and $s$ we call them equivalent (and write $r\equiv s$), if $r-s\in \mathbb{Z}$. Then, $$g(x)\equiv \{\begin{array}{ll}x-a,& \text{if }x\not\equiv a,b,\\ b-a,& \text{if }x\equiv a,\\ -a,& \text{if }x\equiv b.\end{array}$$ Note that $g$ is a bijection on $(0,1)$. Let $$S=\{g^{-i}(a):0\le i\le n-1\}\cup \{g^{-i}(b):0\le i\le n-1\}.$$ Then $g^n(x)=x-na$ for $x\notin S$. Case 1. $g^n$ has a fixed point $x^{\prime }$ outside $S$. Then $x^{\prime }\equiv g^n(x^{\prime })\equiv x^{\prime }-na$, and therefore, $na\equiv 0$ and $g^n(x)=x$ for all $x\notin S$. On the other hand, $g^n$ is a permutation of $S$, and for some positive integer $m$, $(g^n{)}^m$ be the identity in $S$. So we can take $N=mn$ in this case. Case 2. All fixed points of $g^n$ are in $S$. If $g(a)\notin S$, then none of $g^{-i}(a)$ for $0\le i\le n-1$ can be a fixed point of $g^n$, and $g^n$ can not have more than $n$ fixed points. Similarly, $g(b)\notin S$ is impossible. Therefore, $g(S)=S$. In particular, as in Case 1, there is a positive integer $m$ such that $g^{nm}(x)=x$ for all $x\in S$. Since $g(S)=S$, $g^n$ is a bijection on $(0,1)\setminus S$. As $g^n$ is defined by $x\to x-na$ on $(0,1)\setminus S$, it means that $x\to x-na$ is also bijection on $S$. It follows that $lna\equiv 0$ for some positive integer $l$, and therefore $g^{nl}(x)=x$ for all $x\notin S$. Hence we can choose $N=nml$ in this case.