8.3. Find the prime solutions of the equation $\left[\frac{p}{2}\right]+\left[\frac{p}{3}\right]+\left[\frac{p}{6}\right]=q$ (where $[x]$ denotes the greatest integer not exceeding $x$).
8.3. If \( p > \)
3, then \( p = 6k + 1 \) or \( p = 6k - 1 \). Substituting into the equation, we get \( q = 6k \) or \( q = 6k - 3 \). Since \( q \) is a prime number, \( q \neq 6k \); and the latter case is only possible when \( k = 1 \), hence \( p = 2, 3, 5 \). Substituting them into the equation, we get \( q = 1 \) (discard), \( q = 2, q = 3 \). Therefore, the original equation has two solutions: \( p = 3, q = 2 \); \( p = 5, q = 3 \).