Show that there does not exist an integer $n\in\mathbb{N}$ s.t $\phi(n)=\frac{n}{6}$ Show that there does not exist an integer $n\in\mathbb{N}$ s.t $$\phi(n)=\frac{n}{6}$$.
My solution:
Using the Euler's product formula:
$$\phi(n)=n\prod_{p|n}\Bigl(\frac{p-1}{p}\Bigr)$$
We have:
$$\frac{\phi(n)}{n}=\prod_{p|n}\Bigl(\frac{p-1}{p}\Bigr)=\frac{1}{6}$$
But  $6=3\cdot 2$, hence 
if $p=2,\;\;\;\;\; \Bigl(\frac{p-1}{p}\Bigr)=\Bigl(\frac{2-1}{2}\Bigr)=\frac{1}{2}$
if $p=3,\;\;\;\;\; \Bigl(\frac{p-1}{p}\Bigr)=\Bigl(\frac{3-1}{3}\Bigr)=\frac{2}{3}$
But $\Bigl(\frac{2-1}{2}\Bigr)\Bigl(\frac{3-1}{3}\Bigr)\neq \frac{1}{6}$
Is this correct?
Thanks

If there is such $n$ , then as $6|n$ , so let $n=6k$ for some positive integer $k$ and let $k=2^m k'$ , where $k'$ is odd and $m \ge 0$ .
Then $\phi(n)=\phi(6k)=\phi(2^{m+1}.3.k')$ as $3k'$ is odd , so 
$\phi(2^{m+1}.3.k')=\phi(2^{m+1})\phi(3k')=2^m\phi(3k')$ hence $2^m\phi(3k')=\phi(n)=\dfrac n 6=k=2^mk'$ , so
$\phi(3k')=k'$  , now as $k' \ge 1$ , so $3k' \ge 3$ , so
 $2|\phi(3k')$ , hence we should have $2|k'$ , 
contradicting $k'$ is odd .
