Solving equation involving Euler's totient function

Question

Given $m\in\mathbb{N}$, is there a general way to solve the equation \begin{equation} \varphi(n)=m \end{equation} where $\varphi$ is the Euler's totient function? For example, I want to solve for all the $n$'s such that $\varphi(n)=4$. Is there any way to solve it that are also applicable to many (if not general) cases?

Answer 1

Let $n=p_1^{k_1}\cdot p_2^{k_2}\cdots p_r^{k_r}$, where $p_i$ are disntinct prime numbers.

Then, $\varphi(n)=p_1^{k_1-1}(p_1-1)p_2^{k_2-1}(p_2-1)\cdots p_r^{k_r-1}(p_r-1)=2^2=1\cdot 4=2\cdot 2$.

Now it is easy to conclude $r=1$, $p_1=2$, $k_1=3$ or $p_1-1=4$, $p_1=5$, $k_1=1$ (remain case $p_1-1=2=p_2-1$, isn't possible since $p_1\neq p_2$).

Solutions are $n=8$ and $n=5$.

Answer 2

Consider the wiki

https://en.wikipedia.org/wiki/Euler%27s_totient_function#Unsolved_problems

and look up "totient numbers". There are many nontotient numbers. The page as a whole is very informative.