For an odd prime number p, let S denote the following sum taken modulo p:
S \equiv \frac{1}{1 \cdot 2}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{(p-2) \cdot(p-1)} \equiv \sum_{i=1}^{\frac{p-1}{2}} \frac{1}{(2 i-1) \cdot 2 i} \quad(\bmod p)
Prove that p^{2} \mid 2^{p}-2 if and only if S \equiv 0(\bmod p).