Taiwan IMO Selection Camp

We need the partial fraction decomposition
$$\frac{2}{(k+1)(k+2)(k+3)} = \frac{1}{k+1} - \frac{2}{k+2} + \frac{1}{k+3}$$
Sum over $q = 1,4,7, \dots$, we get
$$\begin{aligned} 2S_q &= \left(\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} - \frac{2}{6} + \frac{1}{7}\right) + \cdots + \left(\frac{1}{q+1} - \frac{2}{q+2} + \frac{1}{q+3}\right) \\ &= \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \cdots + \frac{1}{q+1} + \frac{1}{q+2} + \frac{1}{q+3} \\ &\quad - \left(\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{q+2}\right) \end{aligned}$$
Also we have
$$\frac{1}{p+1} + \frac{1}{p+2} + \cdots + \frac{1}{q+3} \equiv \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{(q-p)+3} \pmod{p}$$
Note that
$$q - p + 3 = \frac{q + 2}{3} \iff q = \frac{3p - 7}{2}$$
In the final, we have
$$\begin{aligned} 1 + 2S_q - \frac{1}{p} &\equiv 1 + \frac{1}{2} + \cdots + \frac{1}{p-1} \pmod{p} \\ &\equiv 0 \pmod{p} \end{aligned}$$