Find all positive integers $n$ such that $\phi(\phi(n))=2$ 
Find all positive integers $n$ such that $\phi(\phi(n))=2$.

First I want to find all positive integers $n$ such that $\phi(n)=2$ then solve for all $m$ such that $\phi(m)=n$.
Knowing that none of my $n$ can be odd should limit my choices to pick but I'm not sure how to go about finding them. 
 A: This is by all means not the most economical way of checking the cases.
We recall that $\phi(p_1^{a_1}p_2^{a_2}...p_k^{a_k})=(p_1-1)(p_2-1)...(p_k-1)p_1^{a_1-1}p_2^{a_2-1}...p_k^{a_k-1}$.
So $m$ is a product of at most two prime factors since if there are three prime factors in $m$ (not necessarily distinct), then we are guaranteed to get two or more prime factors(counting multiplicities).
If $m=p$, then $\phi(m)=\phi(p)=p-1=2$ implies, $p=3$.
If $m=p^2$, then $\phi(m)=p(p-1)=2$ implies $p^2-p-2=0$ so $p=2$(As $p=-1$ is not a valid solution.)
If $m=pq$ where $p>q$, then, $(p-1)(q-1)=2$. Then, $p-1=2, q-1=1$ is the only valid factorisation and incidentally, this is valid as $p=3, q=2$ are both primes.
Therefore, $\phi(m)=2$ implies $m=3,4,6$. 
So $\phi(n)=3,4,6$
With $\phi(n)=3$ you get nothing as $\phi(w)$ is even except for when $w=2$. (Looking at the formula, $\phi(p_1^{a_1}p_2^{a_2}...p_k^{a_k})=(p_1-1)(p_2-1)...(p_k-1)p_1^{a_1-1}p_2^{a_2-1}...p_k^{a_k-1}$ and noting that $p_i-1$ is even except when $p_i=2$ immediately tells us why.)
For the case $\phi(n)=6$, we first note that $n=7$ is a solution. With the above remark, we note that if $\phi(n)=6$, then there does not exist a $r|n$ such that $\phi(r)=3$. So as a result, we know that if $n$ is not prime and $n=st$ for some positive coprime integers $s,t\neq 1$ then, WLOG, $\phi(s)=1$ and $\phi(t)=6$. So $s=2$. $t=7$ is a valid solution. If $t$ is not prime, then we can repeat the argument and say $t=ab$ for some coprime $a,b$ with $\phi(a)=1$, $\phi(b)=6$. But then, that tells us again that $a=2$ and then that contradicts the fact that $s$ and $t=ab$ are coprime. So we have exhausted all the possible cases for when $\phi(n)=6$. In this case, $\underline{n=7,14}$
We consider what happens, when $n$ can be written as a product of two non-coprime factors and $\phi(n)=6$. Then, WLOG, $n=p^2 c$. So in particular, $\phi(p^2)=p(p-1)$ divides $6$. $p$ divides, $p(p-1)$ which divides $6$. So $p=2,3$. If $n=3^2 c$ where $c>1$, then $c$ has to be coprime to 3 as otherwise, $\phi(3^3)=12$ divides $\phi(n)=6$. Thus, since $c$ is coprime to $3$,  $\phi(c)=1$. This tells us $c=2$. So we have two further solutions, $\underline{n=9,18}$.
With the case when $\phi(n)=4$, we note that if $n$ is a product of 4 or more prime factors(not necessarily distinct) then, $\phi(n)$ has 3 or more prime factors(not necessarily distinct). 
If it is a product of one prime factor, (i.e. $n=p$) then $\underline{n=5}$.
If it is a product of two prime factors, (i.e $n=pq$ or $n=p^2$) then we know that $n=10$ as the latter case when $n=p^2$ gives us $p^2-p-4=0$ which has non-integral solutions.
If it is a product of three prime factors, (i.e. $n=p^3$, $n=p^2q$ or $n=pqr$)  we note that the first case gives us $p^3-p^2-4=0$. This factorizes as $(p-2)(p^2+2)=0$ and the quadratic term has no integral factors. So $p=2$ and $\underline{n=8}$.
The second case gives us $p(p-1)(q-1)=4$. If $p,p-1,q-1$ are distinct integers, then, we get a contradiction as $4$ cannot be written as a product of three distinct positive integers. $p$ and $q-1$ are the only pair that might not be distinct. This tells us that $p=2$ and $q=3$ as $p,q$ are both primes. 
If $n=2^2c$ similarly, we can say that $c$ has to be coprime with $2$ as otherwise $8$ divides $n$ and so $\phi(8)=4$ divides $\phi(n)$. Thus, $6=\phi(2^2)\phi(c)$ gives us $\phi(c)=3$ which is a  contradiction. 
This gives us the case $\underline{n=2^2\cdot 3=12}$ and $\phi(12)=4$ as required.
The last case when $n=pqr$ can be ruled out as $\phi(n)$ is a product of three distinct positive integers.
That was a lot of checking rather than actual maths. 
