Is there $\phi(n)=n/6$ I know how to find for which $n$ $\phi(n)=n/2$ or $\phi(n)=n/3$, my method for finding those was simply to find primes $p$ that satisfy $\Pi_p$$_|$$_n$$1-1/p$ $  = 1/2$ or $1/3$.
However, I don't know how to find $\Pi_p$$_|$$_n$$1-1/p = n/6$. Intuitively it seems that if I combine results for both $\phi(n) = n/2$ and $\phi(n) = n/3$ I'll get $\phi(n) = n/6$ but it does not work, cause I get number of the form $2^a3^b$ which gives $\phi(n) = n/3$ again.
Is there a way to find $n$ for which $\phi(n)=n/6$? Or do such numbers exist at all?
 A: We have:
$$\nu_2\left(\frac{\varphi(n)}{n}\right)=\sum_{p\mid n}\nu_2\left(\frac{p-1}{p}\right)\geq \omega(n)-1$$
hence it is not possible that $\frac{\varphi(n)}{n}$ equals a rational number $\frac{p}{q}$  with an even $q>2$.
Here, $\nu_2$ is the $2$-adic height: $$\gcd(p,q)=1,\quad \nu_2\left(\frac{p}{q}\right)=\max\{m\in\mathbb{N}:2^m\mid p\}-\max\{m\in\mathbb{N}:2^m\mid q\}$$ and $\omega(n)$ is the number of prime divisors of $n$.
A: I think this is valid:
Suppose $ϕ(n)=\frac{n}{6}$ then since the Euler's totient function is by definition the size of a set, it must be an integer hence $\frac{n}{6}$ is an integer hence $6$ must be a factor of $n$. Rewrite $n$ to $n=6k$ and we get:
$\rightarrow ϕ(6k)=k$
Now we rewrite k so that it's free from all it's factors of $2$ and $3$ , i.e. ,
$k=2^p*3^q*m$ where $m$ is not divisible by $2$ or $3$. So we now get :-
$$ϕ(6*2^p*3^q*m)=ϕ(2^{(p+1)}*3^{(q+1)}*m) = 2^p*3^q*m$$
So now using multiplicativity:
$$ϕ(2^{(p+1)}*3^{(q+1)}*m)$$
$$\rightarrow ϕ(2^{(p+1)})ϕ(3^{(q+1)})*ϕ(m)$$
$$\rightarrow 2^p(2-1)*3^q(3-1)ϕ(m)$$
$$\rightarrow 2^p*3^q*m$$ (last equality coming from the "equalling $k$" part)
On cancelling/comparing the last two equalities we get:
$$2*ϕ(m)= m$$
Hence $m$ has a factor of $2$ which is a contradiction.
