Let $\phi$ be Euler's totient function, find all $n$ such that $\phi(n) = \frac{1}{3} n$. Let $\phi$ be Euler's function, find all $n$ such that $\phi(n) = \frac{1}{3} n$.
Now while yes I've seen this proof asked before, I'm looking for a specific approach using this theorem.
That is if $n = \prod p^c$, then $\phi(n) = n\prod_{p \mid n} (1 - \frac{1}{p})$.
Not really sure how to continue here. Consider the largest prime divisor of $n$?

We need $n\prod_{p|n}(1 - \frac 1p) = \frac 13 n$ so
$\prod (1-\frac 1p) = \frac 13$
Now to get $3$ in the denominator we must have $3|n$.
So $(1 - \frac 13)\prod_{p|n; p\ne3}(1-\frac 1p)= \frac 13$ so
$\prod_{p|n;p\ne =3} (1 - \frac 1p) = \frac 12$.
To have $2$ in the denominator we must have $2|n$ so 
$(1 - \frac 13)(1-\frac 12)\prod_{p|n;n\ne 2; n\ne 3} (1- \frac 1p) = \frac 13$ so  $\prod_{p|n;n\ne 2; n\ne 3} (1- \frac 1p) =1$ and the only prime factors of $n$ are $2$ and $3$.
So $n$ may be any number of the form $2^a3^b; a\ge 1; b\ge 1$.
