Find all positive integers $n$ such that $\varphi(\varphi(n)) = 2.$
Let $\varphi(n)=m$ so I need the integers $m$ such that $\varphi(m)=2$. I know that $3,4,6$ work but I need to see if there are no others.
If $m$ is prime then $m=3$
Suppose that $m$ is even then $m=2^a(x)$ such that $x$ is odd.
$\varphi(2^a(x))$ since the $gcd(2^a,x)=1$ you have that $\varphi(2^a(x))=2^{a-1}\varphi(x)$. So if $a$ is $1$ then you have that $\varphi(m)=\varphi(x)$ so $x$ is composite let $x=uv$ where $u$ and $v$ are relatively prime and greater then one. So I get that $\varphi(x)=\varphi(u)\varphi(v)$ which $\varphi(u)\geq 2$ and $\varphi(v)\geq(2)$ then $\varphi(x)\geq4$. Also if $m$ is odd you can use the same argument to show that $\varphi(m) \geq 4$
I'm having trouble tying it together to show that you will only get $4$ and $6$ as the only solutions in the case where $m$ is not prime.