# Euler Phi Function Equality

I am trying to prove that $\phi(1)=1$ is the only case where $\phi (n)=n$

I can obviously discount all prime numbers due to the fact that $\phi(p) = (p-1)$

Whenever I try and prove the rest of the cases (for example, using $\phi(mn)=\phi(m)\phi(n)$), I keep falling into an endless loop of needing to prove that the only case where $\phi(n)=n$ only holds for 1.

• For $n>1$, $\varphi(n)$ counts those elements between $1$ and $n-1$ which are relatively prime to $n$. Hence $\varphi(n)≤n-1$ (note that $n>1$ implies $\gcd(n,n)=n>1$). – lulu Apr 2 '17 at 0:05
• @lulu $\varphi(n) = \# \{k \in \{1 \ldots n \}, gcd(k,n) = 1\}$. As you said $gcd(n,n) = n$ so $\varphi(n) \le n-1$ for $n > 1$ – reuns Apr 2 '17 at 0:49

$1$ is the only number relatively prime to itself. Any other $n>1$ will have a maximum $n-1$ numbers from $1$ to $n$ itself that can be relatively prime to it, having to rule out at least $n$