Does there exist a number for which euler phi of $9n$ equals $91$?

Question

Does there exist such $n$ that euler phi for $9n$ is equal $91$?

Answer 1

No there doesn't because as Joffan rightly comments, $\phi(n)$ is even for $n > 2$.

Trivially, $\phi(2)=1$. For any $n > 2$, all values taken on by $\phi$ will be even.
Consider any number which has an odd prime $p$ as a factor. Suppose $p^k$ is the highest power of $p$ which divides $n$.
Then $\phi(n) = \phi(p^k)\phi(n/p^k) = (p^k - p^{k-1})\phi(n/p^k).$
Since $p$ is odd, so are both $p^k$ and $p^{k-1}$, thus their difference is even. Thus there is an even factor, so the value taken by the totient function is even. The only remaining case is $n=2^k$. But then we have $\phi(2^k) = 2^k - 2^{k-1},$ and $2$ clearly divides this difference if $k-1 \ge 1\Rightarrow k \ge 2$.