, 2012

The only such pair is $(p, q) = (37, 11)$.
We have $p(p-1) = (q+1)(q^2-q+1)$. Since $p$ is prime and $p > p-1$ and $q^2-q+1 > q+1$, there exists an integer $m > 0$ such that
$$q^2 - q + 1 = mp, \quad (1)$$
$$p-1 = m(q+1). \quad (2)$$
Since $q^3 = p^2 - p + 1 > (p-1)^2 = m^2(q+1)^2 > m^2q^2$ we deduce that $q > m^2$. We have from (1) and (2) that
$$mp \equiv 1 \pmod{q} \qquad p \equiv m+1 \pmod{q}$$
and so we obtain that $m^2 + m - 1$ is divisible by $q$.
Since $m^2 + m - 1 < 2m^2 < 2q$ this means that
$$q = m^2 + m - 1. \quad (3)$$
Applying the substitution from (3) to (1) and (2), we obtain
$$q^2 - q + 1 = mp = m^2(q+1) + m = (q - m + 1)(q + 1) + m$$
which after cancellations gives $(m-3)q = 0$ and therefore $m = 3$. Replacing in (3) and then in (2) we obtain $(p, q) = (37, 11)$, the only solution.