1. The problem is to understand why $q^2$ can divide the expression $3p(p-1)$. 2. To say that $q^2$ divides $3p(p-1)$ means that when you divide $3p(p-1)$ by $q^2$, the result is an integer with no remainder. 3. This implies that $3p(p-1)$ contains at least two factors of $q$ in its prime factorization. 4. Since $3p(p-1)$ is the product of three terms: 3, $p$, and $(p-1)$, for $q^2$ to divide it, $q$ must appear at least twice among these factors. 5. Note that $p$ and $p-1$ are consecutive integers, so they share no common factors other than 1. 6. Therefore, the two factors of $q$ must come from either $p$ or $p-1$ alone, or one factor from each if $q$ is 1 (which is trivial). 7. In summary, $q^2$ divides $3p(p-1)$ if and only if $q^2$ divides either $p$ or $p-1$ (since 3 is constant and usually does not contribute factors of $q$ unless $q=3$). Final answer: $q^2$ divides $3p(p-1)$ if $q^2$ divides either $p$ or $p-1$.