Find all solutions of $\phi(n) = 4$ and prove there are no more. I have found that when n = 5, 8, 10, & 12 are the only solutions where $\phi(n) = 4$. I also know that I can spilt up the solutions to say that $\phi(n) * \phi(m) = 2 *2$. I am just not sure where to start with the proof.
Help would be appreciated.
Since $\phi(n)$ is a product of $p^e(p-1)$, the only prime factors of $n$ are such that $p-1$ divides $4$, that is, $p-1=1$ or $p-1=2$ or $p-1=4$.
If $n$ is a prime power, $n=p^e$, then $p^{e-1}(p-1)=4$, which implies $p=2$ or $p=5$.
Otherwise, $n=ab$ with $\gcd(a,b)=1$ and so $4=\phi(n)=\phi(a)\phi(b)$, which implies $\phi(a)=1$ and $\phi(b)=4$, or $\phi(a)=\phi(b)=2$.
