45th Mongolian Mathematical Olympiad

For every $m\in \mathbb{N}$ we can choose that way: $$m=p((p-1)m+3)=(p-1)(pm+3)+3$$ Therefore $$2S_n=\sum _{k=1}^{p-1}(k^n+(p-k{)}^n)=\sum _{k=1}^{p-1}\sum _{m=0}^{n-1}{C}_n^m(p-k{)}^m{p}^{n-m}\equiv$$ $$\begin{array}{rl}& \equiv \sum _{k=1}^{p-1}(-1{)}^{n-2}\cdot C_n^{n-2}\cdot (p-k{)}^{n-2}{p}^2+(-1{)}^{n-1}{C}_n^{n-1}\cdot (p-k{)}^{n-1}\cdot p\equiv \\ & \equiv (-1{)}^{n-2}\cdot C_n^{n-2}{S}_{n-2}{p}^2+(-1{)}^{n-1}{C}_n^{n-1}{S}_{n-1}p(\mathrm{mod}{p}^3)\end{array}$$ Moreover, $$S_{n-2}=1^{n-2}+2^{n-2}+\cdots +(p-1{)}^{n-2}\equiv S_1=\frac{p(p-1)}{2}\equiv 0(\mathrm{mod}p)(\ast )$$ $$S_{n-1}=1^{n-1}+2^{n-1}+\cdots +(p-1{)}^{n-1}\equiv S_2=\frac{p(p-1)(2p+1)}{6}\equiv 0(\mathrm{mod}p)(\ast \ast )$$ $p$ is odd prime number hence $(p,6)=1$. Thus $2S_n\equiv 0(\mathrm{mod}{p}^3)$. Because, we chose above $p\mid n$ and $n=(p-1)(pm+3)+3$. By Fermat's theorem, we get $k^{n-1}\equiv k^2(\mathrm{mod}p)$, $k^{n-2}\equiv k(\mathrm{mod}p)$ then $S_{n-2}\equiv S_1(\mathrm{mod}p)$, $S_{n-1}\equiv S_2(\mathrm{mod}p)$. Finally, by () and (*) proof is completed.