Mongolian Mathematical Olympiad

If rational numbers $\frac{p_1}{q_1},\frac{p_2}{q_2}$ satisfy the conditions $q_1\cdot q_2\ne (\text{mod }p)$ and $p_1{q}_2-p_2{q}_1\equiv 0(\mathrm{mod}p)$, then we write $\frac{p_1}{q_1}\equiv \frac{p_2}{q_2}(\mathrm{mod}p)$. Then our task is to show $S_{2n}+S_{2n-1}-2S_n-S_2\equiv 0(\mathrm{mod}p)$.

$$\begin{array}{rl}\frac{1}{p}{C}_p^k& \equiv \frac{(p-1)\cdot (p-2)\cdots (p-k+1)}{k!}\\ & \equiv \frac{(-1)\cdot (-2)\cdots (-k+1)}{k!}\equiv \frac{(-1{)}^{k-1}}{k}(\mathrm{mod}p).\end{array}$$

Proposition 2: $$S_a\equiv \frac{(a-1{)}^p-a^p+1}{p}(\mathrm{mod}p)$$ $$\text{** *Proof:* **}▲S_a=-\sum _{k=1}^{p-1}\frac{(-a{)}^k(-1{)}^{k-1}}{k}\equiv \sum _{k=1}^{p-1}(-a{)}^k\frac{1}{p}{C}_p^k\equiv -\frac{1}{p}\left(-1-(-a{)}^p+\sum _{k=0}^{p-1}(-a{)}^k{C}_p^k\right)\equiv \frac{(a-1{)}^p-a^p+1}{p}(\mathrm{mod}p)▲$$

By proposition 2, it follows $S_{2n}+S_{2n-1}-2S_n-S_2\equiv \frac{1}{p}((2n-1{)}^p-(2n{)}^p+1+(2n-2{)}^p-(2n-1{)}^p+1-2(n-1{)}^p+2n^p-2-1+2^p-1)\equiv -\frac{1}{p}(2^p-2)\cdot ((n^p-n)-((n-1{)}^p-(n-1)))(\mathrm{mod}p)$.

By Fermat's theorem we can write $p\mid 2^p-2$, $p\mid n^p-n$, $p\mid (n-1{)}^p-(n-1)$. Thus we have $S_{2n}+S_{2n-1}-2S_n-S_2\equiv 0(\mathrm{mod}p)$ and the problem is solved.