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
 A: What if other primes divide $n$?  Consider that $\phi(12)/12 = 1/3$, but 12 is even.  So, $n$ can be divisible by more primes than appear in the denominator of $\phi(n)/n$.
Here are some hints.
Argue that $n$ must be of the form $2^k 3^m z$, $k>0$, $m>0$, where $z$ is not divisible by 2 or 3.
Then argue that $\displaystyle \frac{\phi(z)}{z}=\frac{1}{2}$, and show this is a contradiction.
A: $$\phi(n)=\frac{n}{6}$$
$$\iff \prod_{p\mid n}\frac{p}{p-1}=6 $$
$$\text{ Taking the 2-adic order of both sides}$$
$$v_2(\prod_{p\mid n}\frac{p}{p-1})=v_2(6)$$
$$\sum_{p\mid n}v_2(\frac{p}{p-1})=1$$
$$1=\sum_{p\mid n} v_2(p)-v_2(p-1)$$
$$1\leq \sum_{p\mid n} v_2(p)$$
$$\text{ But }v_2(p)=0, \text{ For all primes } p  \text{ unless } p=2$$
$$\text{Thus we must have that one of the primes is } 2$$
$$\text{ Now going back we have that}$$
$$1=\sum_{p\mid n} v_2(p)-v_2(p-1)$$
$$1=v_2(2)-v_2(1)+\sum_{p\mid n}_{p\ne 2} v_2(p)-v_2(p-1)$$
$$0=\sum_{p\mid n}_{p\ne 2} v_2(p)-v_2(p-1)$$
$$\text{ But since } p\ne 2 \text{ we know that } v_2(p)=0 \text{ for all other primes } p$$
$$\text{ So we get that}$$
$$0=\sum_{p\mid n}_{p\ne 2} v_2(p-1)$$
$$\text{But for every odd prime } p \text{ we know that } p-1 \text{ has at least one factor of } 2$$
$$\text{ So the only way the right hand side can be zero is if no other primes divide } n \text{ other then } 2$$
$$\text{ Which leaves us with}$$
$$\prod_{p\mid n}\frac{p}{p-1}=\frac{2}{2-1}=2=6$$
$$\text{ A contradiction and thus there can be no solutions to } \phi(n)=\frac{n}{6}$$
A: 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 $i=3,\cdots,r$.
A: 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 .
