Mongolian Mathematical Olympiad

If rational numbers $\frac{p_1}{q_1}, \frac{p_2}{q_2}$ satisfy the conditions $q_1 \cdot q_2 \neq (\text{mod } p)$ and $p_1q_2 - p_2q_1 \equiv 0 \pmod{p}$, then we write $\frac{p_1}{q_1} \equiv \frac{p_2}{q_2} \pmod{p}$. Then our task is to show $S_{2n} + S_{2n-1} - 2S_n - S_2 \equiv 0 \pmod{p}$.

$$\begin{align*} \frac{1}{p} C_p^k &\equiv \frac{(p-1) \cdot (p-2) \cdots (p-k+1)}{k!} \\ &\equiv \frac{(-1) \cdot (-2) \cdots (-k+1)}{k!} \equiv \frac{(-1)^{k-1}}{k} \pmod{p}. \end{align*}$$

Proposition 2:
$$S_a \equiv \frac{(a-1)^p - a^p + 1}{p} \pmod{p}$$
$$\text{** *Proof:* **} \blacktriangle S_a = - \sum_{k=1}^{p-1} \frac{(-a)^k (-1)^{k-1}}{k} \equiv \sum_{k=1}^{p-1} (-a)^k \frac{1}{p} C_p^k \equiv - \frac{1}{p} \left( -1 - (-a)^p + \sum_{k=0}^{p-1} (-a)^k C_p^k \right) \equiv \frac{(a-1)^p - a^p + 1}{p} \pmod{p} \blacktriangle$$

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

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 $S_{2n} + S_{2n-1} - 2S_n - S_2 \equiv 0 \pmod{p}$ and the problem is solved.