4. For a positive integer $n$, we denote by $P(n)$ the product of the positive divisors of $n$. For example, $P(20)=8000$. The positive divisors of 20 are 1, 2, 4, 5, 10, and 20, with the product $1 \cdot 2 \cdot 4 \cdot 5 \cdot 10 \cdot 20=8000$.

(a) Find all positive integers $n$ for which $P(n)=15 n$.

(b) Show that there are no positive integers $n$ for which $P(n)=15 n^{2}$.
4. a) Since $P(n)=15 n$ is the product of the positive divisors of $n$, the prime divisors 3 and 5 of $P(n)$ must also be divisors of $n$. It follows that $n$ is a multiple of 15. If $n>15$, then $3,5,15$ and $n$ are different divisors of $n$, so $P(n) \geqslant 3 \cdot 5 \cdot 15 \cdot n=225 n$. This contradicts the given that $P(n)=15 n$. The only remaining possibility is thus $n=15$. This is indeed a solution, as $P(15)=1 \cdot 3 \cdot 5 \cdot 15=15 \cdot 15$.

b) Suppose $P(n)=15 n^{2}$. Then $n$ must again be a multiple of 15. Clearly, $n=15$ does not satisfy this, so $n \geqslant 30$. We see that $\frac{n}{5}>5$, so $115$ holds $P(n) \geqslant n \cdot n^{2}>15 n^{2}$, a contradiction. Therefore, there does not exist any $n$ such that $P(n)=15 n^{2}$.