Saudi Arabia Mathematical Competitions 2012

Since $n$ is odd, $$|S|=\varphi (n)=n\prod _{p|n}(1-1/p)$$ is even. Given an element $k$ of $S$, write $k^{|S|}\equiv 1+n{r}_k(\mathrm{mod}{n}^2)$ and $(n-k{)}^{|S|}\equiv 1+n{r}_{n-k}(\mathrm{mod}{n}^2)$, and notice that $$(n-k{)}^{|S|}\equiv k^{|S|}\cdot n\cdot k^{|S|-1}(\mathrm{mod}{n}^2),$$ to get $$r_k-r_{n-k}\equiv |S|\cdot k^{|S|-1}(\mathrm{mod}n).$$ Hence $$\prod _{k\in T}(r_k-r_{n-k})\equiv |S{|}^{|T|}{\left(\prod _{k\in T}k\right)}^{|S|-1}(\mathrm{mod}n).$$ Finally, notice that the product in the right-hand member is congruent to 1 modulo $n$. To see this, let $k^{\prime }$ denote the modulo $n$ multiplicative inverse of an element $k$ of $S$, and notice that $$k^{\prime }+1\equiv k^{\prime }(k+1)(\mathrm{mod}n)\text{and}(k-1)(k^{\prime }+1)\equiv k-k^{\prime }(\mathrm{mod}n).$$ The first congruence shows that if $k$ belongs to $T$, then so does $k^{\prime }$, and the second shows that $k=k^{\prime }$ if and only if $k=1$. Consequently, if $k\ne 1$, the factors $k$ and $k^{\prime }$ in the product can be paired off and the conclusion follows.