Show that there does not exist an integer $n\in\mathbb{N}$ s.t $\phi(n)=\frac{n}{6}$

Question

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

Answer 1

$$\phi(n)=\frac{n}{6}$$ $$\iff \prod_{p\mid n}\frac{p}{p-1}=6 $$ $$\text{ Taking the 2-adic order of both sides}$$ $$v_2(\prod_{p\mid n}\frac{p}{p-1})=v_2(6)$$ $$\sum_{p\mid n}v_2(\frac{p}{p-1})=1$$ $$1=\sum_{p\mid n} v_2(p)-v_2(p-1)$$ $$1\leq \sum_{p\mid n} v_2(p)$$ $$\text{ But }v_2(p)=0, \text{ For all primes } p \text{ unless } p=2$$ $$\text{Thus we must have that one of the primes is } 2$$ $$\text{ Now going back we have that}$$ $$1=\sum_{p\mid n} v_2(p)-v_2(p-1)$$ $$1=v_2(2)-v_2(1)+\sum_{p\mid n}_{p\ne 2} v_2(p)-v_2(p-1)$$ $$0=\sum_{p\mid n}_{p\ne 2} v_2(p)-v_2(p-1)$$ $$\text{ But since } p\ne 2 \text{ we know that } v_2(p)=0 \text{ for all other primes } p$$ $$\text{ So we get that}$$ $$0=\sum_{p\mid n}_{p\ne 2} v_2(p-1)$$ $$\text{But for every odd prime } p \text{ we know that } p-1 \text{ has at least one factor of } 2$$ $$\text{ So the only way the right hand side can be zero is if no other primes divide } n \text{ other then } 2$$ $$\text{ Which leaves us with}$$ $$\prod_{p\mid n}\frac{p}{p-1}=\frac{2}{2-1}=2=6$$ $$\text{ A contradiction and thus there can be no solutions to } \phi(n)=\frac{n}{6}$$

Answer 2

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.

Answer 3

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 $i=3,\cdots,r$.