Prime Pairs

Rearranging, we have:

$p^q - p = q^p + q$

$p(p^{q-1} - 1) = q(q^{p-1} + 1)$

This means that either $p|q$ or $p|q^{p-1} + 1$

However, it is easily seen that if $p|q$ , then $p = q$ , which would yield

$0 = 2p$ or $p = q = 0$

then, we have $p|q^{p-1} + 1$ . Since $gcd(p,q) = 1$ , then by using Fermat's Little Theorem, we have:

$q^{p-1} \equiv 1 \mod p$

$q^{p-1} + 1 \equiv 2 \mod p$ , hence

$2 \equiv 0 \mod p$ , therefore $p = 2$

however, 2 is not an odd prime, so we're left with the conclusion that there are no odd primes that satisfy the given condition.

Therefore $a = 0$ , giving us the value of $\fbox{27}$ .