Fall Mathematical Competition

If $m>2$ then the positive integers which are less than $m$ and co-prime to $m$ can be partitioned into pairs $(k,m-k)$, $m-k\ne k$. The number of these pairs is $\frac{\phi (m)}{2}$ and the sum of the numbers in every pair is $m$. Therefore $f(m)=\frac{m\phi (m)}{2}$, which is true for $m=2$ as well. Since $\phi (n^k)=n^{k-1}\phi (n)$, we obtain $$f(n^k)=\frac{n^k\phi (n^k)}{2}=\frac{n^{2k-1}\phi (n)}{2}.$$ Hence the given condition becomes $n^{2k-1}\phi (n)=2n^l$. If $l>2k-1$, then we obtain $\phi (n)=2n^{l-2k+1}\ge 2n$, which is impossible. If $l=2k-1$ we get $\phi (n)=2$, which gives $n=3,4$ or $6$. If $l<2k-1$, then $n^{2k-1-l}\phi (n)=2$, which is satisfied only by $n=2$ and $2k-1-l=1$, i.e. $l=2k-2$. Finally, the solutions are $n=2,3,4$ and $6$.