
4. For a positive integer $n$ the number $P(n)$ is the product of the positive divisors of $n$. For example, $P(20)=8000$, as the positive divisors of 20 are $1,2,4,5,10$ and 20 , whose product is $1 \cdot 2 \cdot 4 \cdot 5 \cdot 10 \cdot 20=8000$.

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

(b) Show that there exists no positive integer $n$ such that $P(n)=15 n^{2}$.


4. a) Because $P(n)=15 n$ 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 distinct divisors of $n$, yielding $P(n) \geqslant 3 \cdot 5 \cdot 15 \cdot n=225 n$. This contradicts the fact that $P(n)=15 n$. The only remaining possibility is $n=15$. This is indeed a solution, because $P(15)=1 \cdot 3 \cdot 5 \cdot 15=15 \cdot 15$.

b) Suppose that $P(n)=15 n^{2}$ holds. Again, we find that $n$ is a multiple of 15 . It is clear that $n=15$ is not a sulution, hence $n \geqslant 30$. We observe that $\frac{n}{5}>5$. It follows that $115$ holds, we have $P(n) \geqslant n \cdot n^{2}>15 n^{2}$, which contradicts the assumption of this problem. We conclude that no $n$ exists for which $P(n)=15 n^{2}$.
