Finding $n$ such that $\phi(n)=34$ (where $\phi$ is Euler's totient)

Question

How can I find $n$ such that $\phi(n)=34$ (where $\phi$ is Euler's totient) or prove that it does not exist?

And how can I find $c$ for which $\phi(n)=c$ if $n$ does exists for $c$?

Answer 1

If $p\mid n$ then $p-1\mid\phi(n)$, i.e. $p-1\in\{1,2,17,34\}$. This implies $p=2$ or $p=3$ as $18$ and $35$ are not prime. But $\phi(2^a3^b)$ can never be a multiple of $17$.

Answer 2

For a (partial) answer to the second question, if $n = p_1^{a_1} \cdots p_r^{a_r}$ for $p_i$ different primes then $$ \phi(n) = n \left( 1 - \frac{1}{p_1} \right) \cdots \left( 1 - \frac{1}{p_r} \right) = p_1^{a_1 - 1} \cdots p_r^{a_r - 1} (p_1 - 1) \cdots (p_r - 1) $$