Computing $n$ such that $\phi(n) = m$ Is there a general procedure for computing an inverse of the euler totient function? I did find an old SE post that seemed to have some pointers -How to solve the equation $\phi(n) = k$? 
However, I am interested in the case $\phi(n) = 72$. The procedure outline in the link above leads me to write out $n = 2^\alpha3^\beta\prod_{p_i|n}p_i$. However, since 72 + 1 = 73 is a prime, this clearly fails to find all n. 
Would someone be willing to point in the right direction for this case?
 A: If $\phi(n)=72$, and $p$ is a prime dividing $n$, then $p-1$ divides $72$ (why?), so the first thing to do is write down all the divisors of $72$ and see which ones are $1$ less than a prime. 
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. 
2, 3, 4, 5, 7, 9, 10, 13, 19, 25, 37, 73. 
The primes in this second list are 2, 3, 5, 7, 13, 19, 37, and 73, so these are the only possible prime divisors of $n$. 
If $p^r$ divides $n$ for some $r\ge2$, then $p^{r-1}$ divides $72$ (why?), so $p^r$ is one of the numbers 4, 8, 16, 9, 27 (why?). 
So now you have the complete list of possible prime and prime-power divisors of $n$. You test them systematically to see which combinations work. 
For example, if $73$ is a divisor of $n$, that doesn't leave much room for anything else --- $n$ must be $73$ or $2\times73$. 
If $37$ is a divisor of $n$, then you're just looking for a factor of $2$, so you can use $3\times37$, or $4\times37$, or $6\times37$. 
And so on. 
A: There are a total of 17 such $n$ and they are: 
73, 91, 95, 111, 117, 135, 146, 148, 152, 182, 190, 216, 222, 228, 234, 252,  270.
Regards
