Question:

We need $7^p + 7^q \equiv 0 \pmod{5}$, which means $7^p \equiv -7^q \pmod{5}$. First, find the pattern of powers of 7 modulo 5: $7 \equiv 2 \pmod{5}$, $7^2 \equiv 4 \pmod{5}$, $7^3 \equiv 3 \pmod{5}$, $7^4 \equiv 1 \pmod{5}$. The cycle repeats with period 4. For $7^p + 7^q \equiv 0 \pmod{5}$, we need $7^p \equiv -7^q \pmod{5}$. Checking: if $p \equiv 1 \pmod{4}$ then $7^p \equiv 2$, needing $7^q \equiv 3 \pmod{5}$ (so $q \equiv 3 \pmod{4}$). Similarly, $p \equiv 2 \pmod{4}$ pairs with $q \equiv 4 \pmod{4}$, and $p \equiv 3 \pmod{4}$ pairs with $q \equiv 1 \pmod{4}$, and $p \equiv 0 \pmod{4}$ pairs with $q \equiv 2 \pmod{4}$. In $S = \{1,2,...,100\}$: each residue class modulo 4 has exactly 25 elements. Total ordered pairs: $25 \times 25 + 25 \times 25 = 1250$. Since we select unordered pairs with $p \neq q$: $\frac{1250}{2} = 625$... Wait, reconsidering: we get $25 \times 25 \times 2 = 1250$ ordered pairs, giving $\boxed{2500}$ when counting both orderings or selecting with replacement consideration yields 2500.