37th Iranian Mathematical Olympiad

Take a sufficiently large $m$ ($m>2^{3000}$) and put $x=m+\frac{1}{k-1}$. Note that $$\left\{{\left(m+\frac{1}{k-1}\right)}^n\right\}=\sum _{k_i>n}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\frac{m^{n-i}}{m^{(k-1)i}}$$ Because $$\sum _{k_i>n}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\frac{m^{n-i}}{m^{(k-1)i}}<\frac{1}{m}\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)<\frac{2^{n+1}}{m}<1,$$ and the remaining terms of ${\left(m+\frac{1}{k-1}\right)}^n$ are all integers. Now if $n=kt+r$ such that $k-1\ge r\ge 0$, we have two cases:

Case 1 $r\ne 0$. $$\begin{gathered} \left\{{\left(m+\frac{1}{k-1}\right)}^n\right\}=\sum _{k_i>n}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\frac{1}{m^{k_i-n}}\stackrel{i=t+1}{>}\frac{1}{m^{k-r}}, \\ \left\{{\left(m+\frac{1}{k-1}\right)}^{n-1}\right\}=\sum _{\begin{array}{c}{k}_i>n-1\\ i\ge t+1\end{array}}\left(\genfrac{}{}{0ex}{}{n-1}{i}\right)\frac{1}{m^{k_i-n+1}}<2^n\times \frac{n}{m^{k-r+1}}, \\ \frac{1}{m^{k-r}}>2^n\times \frac{n}{m^{k-r+1}}⟹\left\{{\left(m+\frac{1}{k-1}\right)}^n\right\}>\left\{{\left(m+\frac{1}{k-1}\right)}^{n-1}\right\}. \end{gathered}$$

Case 2 $r=0$. $$\begin{gathered} \left\{{\left(m+\frac{1}{k-1}\right)}^n\right\}=\sum _{\begin{array}{c}{k}_i>n\\ i\ge t+1\end{array}}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\frac{1}{m^{k_i-n}}<2^n\times \frac{n}{m^k}, \\ \left\{{\left(m+\frac{1}{k-1}\right)}^{n-1}\right\}=\sum _{\begin{array}{c}{k}_i>n-1\\ i\ge t\end{array}}\left(\genfrac{}{}{0ex}{}{n-1}{i}\right)\frac{1}{m^{k_i-n+1}}\ge \frac{1}{m}, \\ \frac{1}{m}\stackrel{k\ge 2}{>}{2}^n\times \frac{n}{m^k}⟹\left\{{\left(m+\frac{1}{k-1}\right)}^{n-1}\right\}>\left\{{\left(m+\frac{1}{k-1}\right)}^n\right\}. \end{gathered}$$ Therefore, $$\begin{gathered} \left\{{\left(m+\frac{1}{k-1}\right)}^{n-1}\right\}>\left\{{\left(m+\frac{1}{k-1}\right)}^n\right\}⟺k\mid n, \\ \{x^{n-1}\}>\{x^n\}⟺k\mid n. \end{gathered}$$