Divisibility Principles?

First note that $p | (7^p - 2^p)$ implies $7^p \equiv 2^p$ mod $p$ , but this means $7 \equiv 2$ , i.e. $p = 5$ .

So we will first assume that $p,q \ne 5$ . In this case we must have $q | (7^p - 2^p)$ and $p | (7^q-2^q)$ . Now we have a lemma: if $a$ and $b$ are distinct mod $p$ , and $k$ is relatively prime to $p-1$ , then $a^k$ and $b^k$ are distinct mod $p$ also. Proof: we can find $\ell$ such that $k\ell \equiv 1$ mod $p-1$ ; then $a^k \equiv b^k$ implies $a^{k\ell} \equiv b^{k\ell}$ so $a \equiv b$ .

By the lemma, since $7$ and $2$ are distinct mod $p$ , we must have that $q$ and $p-1$ are not relatively prime. But this means $q|(p-1)$ , but this is impossible since $p \le q$ .

So at least one of $p$ and $q$ is $5$ . In fact it's easy to check directly that $(2,5), (3,5)$ are not solutions, so we can assume that $p = 5$ and $q \ge 5$ . In this case we get $q | 7^5-2^5$ , and $7^5-2^5 = 5^2 \cdot 11 \cdot 61$ , so there are three solutions: $(5,5), (5,11), (5,61)$ . The sum is $\fbox{92}$ .