China Southeastern Mathematical Olympiad

If $n$ is odd, then all factors of $n$ are odd. So $f(n)$ is odd and $g(n)$ is even. $g(n)$ cannot be $f(n)$ to some power of positive integer. Therefore $n$ is even. The smallest two divisors of $n$ are $1$ and $2$, and the largest two divisors of $n$ are $n$ and $n/2$.

Let $d$ be the third smallest divisor of $n$. If there exists $k\in {\mathbb{N}}^{\ast }$ such that $g(n)=f^k(n)$, then

$$\frac{3n}{2}=(1+2+d{)}^k=(3+d{)}^k\equiv d^k(\mathrm{mod}3).$$

Since $3\nmid \frac{3n}{2}$, we see that $3\mid d^k$. So $3\mid d$, and since $d$ is the third smallest, we see that $d=3$.

Thus, $\frac{3}{2}n=6^k$, we get $n=4\times 6^{k-1}$. Since $3\nmid n$, we see that $k\ge 2$.

Summing up, $n=4\times 6^l(l\in {\mathbb{N}}^{\ast })$.