Does there exist such $n$ that euler phi for $9n$ is equal $91$?
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1Euler's totient $\phi(n)$ is even for $n>2$, so - no. – Joffan Mar 18 '17 at 14:19
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Did you look at the prime factorization of $91$ and the definition of the $\varphi$-fubction? – martin.koeberl Mar 18 '17 at 14:20
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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 $k-1 \ge 1\Rightarrow k \ge 2$.
Shraddheya Shendre
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