Let $n=p \cdot q .......... $ where , $p<q$ , $p$(prime) is least divisor after $1$ and $q$ is least divisor after $p$ . Then , $f(n) = 1+p+q$ and , $g(n)= n+\frac{n}{p}$ . Now , you get , $n+\frac{n}{p} = (1+p+q)^m \implies n(p+1)=p(1+p+q)^n \implies p+1 | 1+p+q \implies p+1|q$ Now there are two cases , if $ q=p^2 \implies p+1|p $ which is impossible .If $q$ is prime then , $p+1|q \implies p+1=q$ which means $p=2$ and $q=3$ . Which gives $n= 2^{m+1}3^{m-1} \Box $