Are there any positive integers n for which $\phi(n) = \frac{n}{4}$?

Question

For starters, I established that n cannot be prime, as $\phi(n) = n - 1$ for prime numbers. So, if $n$ is composite, I wrote $n$ as its prime factorization: $n = p_1^{r_1}...p_k^{r_k}$ and thus the phi-totient of n is $\phi(n) = n (\frac{p_1 - 1}{p_1})...(\frac{p_k - 1}{p_k})$. However, now I'm stuck. I know since $\phi(n) = \frac{n}{4}$ then $(\frac{p_1 - 1}{p_1})...(\frac{p_k - 1}{p_k}) = \frac{1}{4}$ but I'm not really sure how to prove this, or if I can. Where do I go from here?

Answer 1

We have $4(p_1 - 1)...(p_k - 1) = p_1...p_k$.
Hence one of the $p_i$, say $p_1 = 2$.
But then dividing by $p_1$ you get $2(p_1 - 1)...(p_k - 1) = p_2...p_k$.
The LHS is even, the RHS is odd, so this is impossible (if $k = 1$ then the RHS is the empty product which equals $1$, and this is also odd).

Answer 2

$$\frac{\varphi(n)}{n}=\prod_{p\mid n}\left(1-\frac{1}{p}\right) $$ but there is a unique even prime, so $\nu_2\left(\frac{\varphi(n)}{n}\right)\geq -1$.