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

Let $n=p_{1}^{a_1}\cdots p_{r}^{a_r}$, so $\;\;\phi(n)=p_{1}^{a_{1}-1}(p_{1}-1)\cdots p_{r}^{a_{r}-1}(p_{r}-1)$.
If $n=6\phi(n)$, then $\;\;p_{1}^{a_1}\cdots p_{r}^{a_r}=6\big[p_{1}^{a_{1}-1}(p_{1}-1)\cdots p_{r}^{a_{r}-1}(p_{r}-1)\big]$, so
$\;\;p_{1}\cdots p_{r}=6(p_1-1)\cdots(p_{r}-1)=2\cdot3(p_1-1)\cdots(p_{r}-1)$.
$\;\;$We can take $p_1=2$ and $p_2=3$, so 
$\;\;p_{3}\cdots p_{r}=1\cdot2 (p_3-1)\cdots(p_{r}-1)$; and this gives a contradiction since $p_i\ne2$ for $i=3,\cdots,r$.
