First, $p>q$. Note that $p$ or $q$ have to be less than $5$, as otherwise either $6 \mid p-q$, $6 \nmid p+q$ or $6 \mid p+q$, $6\nmid p-q$, which both give a contradiction. Thus, we casebash with $p=2$ (reject since $p>q$), $p=3$ (so $p=2$ which fails), $q=2$, $q=3$. $q=2\implies p^3-6p^2+11p-10=0$ which has no rational roots by RRT. $q=3 \implies p^3-9p^2+26p-30=0 \implies (p-5)(p^2-4p+6)=0 \implies p=5$ is the only solution. Thus, only $(5,3)$ works.

Transforming the given equation, we have \[ 2q = (p-q)\left((p-q)^2 - 1\right). \]Since \( p \ne q \), \( p-q \) and \( q \) are coprime, so \( p-q \) must be a divisor of \( 2 \). Here, we consider the possible cases: - When \( p-q = 1 \), it follows that \( 2q = 0 \), which is invalid. - When \( p-q = 2 \), substituting gives \( 2q = 6 \), which leads to \( (p, q) = (5, 3) \). Thus, we conclude that \( (p, q) = (5, 3) \).

See Modern Olympiad Number Theory, page 49.