Taiwan IMO Selection Camp

For rational numbers $p_1/q_1$ and $p_2/q_2$ where $p \nmid q_1, p \nmid q_2$. If $p|(p_1q_2 - p_2q_1)$, we write $p_1/q_1 \equiv p_2/q_2 \pmod p$.
We begin with $S_a$ modulo $p$. Observe: $p|(\binom{p}{k})$, $k = 1, \cdots, p-1$ and
$$\begin{aligned} \frac{1}{p}\binom{p}{k} &= \frac{(p-1)(p-2)\cdots(p-k+1)}{k!} \\ &\equiv \frac{(-1)\cdot(-2)\cdots(-k+1)}{k!} = (-1)^{k-1} \pmod{p}. \end{aligned}$$
Then, we have
$$S_a = - \sum_{k=1}^{p-1} \frac{(-a)^k (-1)^{k-1}}{k} \equiv - \sum_{k=1}^{p-1} (-a)^k \cdot \frac{1}{p} \binom{p}{k} \pmod{p}.$$
The right-hand side of the above expression is an integer. By the binomial theorem, we obtain
$$- \sum_{k=1}^{p-1} (-a)^k \cdot \frac{1}{p} \binom{p}{k} = - \frac{1}{p} \left( -1 - (-a)^p + \sum_{k=1}^{p} (-a)^k \binom{p}{k} \right) = \frac{(a-1)^p - a^p + 1}{p}$$
since $p$ is odd. Hence
$$S_a \equiv \frac{(a-1)^p - a^p + 1}{p} \pmod{p}.$$
Finally, we obtain
$$\begin{aligned} S_3 + S_4 - 3S_2 &\equiv \frac{(2^p - 3^p + 1) + (3^p - 4^p + 1) - 3(1^p - 2^p + 1)}{p} \\ &= \frac{4 \cdot 2^p - 4^p - 4}{p} = -\frac{(2^p - 2)^2}{p} \pmod{p}. \end{aligned}$$
By Fermat's theorem, $p|(2^p - 2)$, so $p^2|(2^p - 2)^2$. Therefore
$$S_3 + S_4 - 3S_2 \equiv 0 \pmod{p}.$$