How to prove that $ φ(n^2) = n * φ(n) $? I Want to prove that:  
$$ φ(n^2) = n * φ(n) $$
the direction is :  
$$ φ(nm) = φ(n)φ(m) $$
or I'm not even close?
 A: HINT: Suppose that $m_1,\dots,m_{\varphi(n)}$ are the positive integers in $\{1,\dots,n\}$ that are relatively prime to $n$. Which integers in $\{n+1,n+2,\dots,2n\}$ are relatively prime to $n$? How about in $\{2n+1,2n+2,\dots,3n\}$? In $\{kn+1,kn+2,\dots,(k+1)n\}$?
It doesn’t hurt to start by looking at numerical examples. For instance, with $n=6$:
$$\begin{array}{c}
\color{red}1&2&3&4&\color{red}{5}&6\\
\color{red}{7}&8&9&10&\color{red}{11}&12\\
\color{red}{13}&14&15&16&\color{red}{17}&18\\
\color{red}{19}&20&21&22&\color{red}{23}&24\\
\color{red}{25}&26&27&28&\color{red}{29}&30\\
\color{red}{31}&32&33&34&\color{red}{35}&36
\end{array}$$
A: Well this formula works for $\gcd(n,m)=1$ if your function is the Euler totient function.
But note that $\phi(p^k)=p^k(1-\frac 1p)$ so that $\phi(p^{2k})=p^{2k}(1-\frac 1p)=p^k\phi(p^k)$ (a proof is here).
From this you should deduce that indeed $\phi(n^2)=n\phi(n)$.
A: Here is a proof based on Euler's product formula:
$$
\frac{\varphi(n^2)}{n^2} = \prod_{p\mid n^2} 1-\frac1p = \prod_{p \mid n} 1-\frac1p = \frac{\varphi(n)}{n}
$$
It does not really matter what is in the product, just that the set of primes dividing $n^2$ is the same as the set of primes dividing $n$.
