Showing $\phi(n^2) = n\, \phi(n)$

Question

How can I show this for Euler's $\phi$ function? $$\phi(n^2) = n\,\phi(n)$$

Ive been struggling a lot with this problem, I would appreciate some help.

Answer 1

$\begin{align}\frac{\varphi(n^2)}{n^2}&=\Pi_{p|n^2} (1-\frac{1}{p})\\&=\Pi_{p|n} (1-\frac{1}{p}) \\&=\frac{\varphi(n) }{n}\end{align} $

Note : $p|n^2=n\cdot n$ implies $p|n$ . Hence $p|n^2$ iff $p|n$

Answer 2

Remember, $\phi(n)$ counts the integers $1\le k\le n$ coprime to $n$.

Consider $\phi(n^2)$. For any $1\le k\le n^2$, we can divide $k$ by $n$ to write $k=nq+r$ (with $q$ the quotient $0\le q<n$ and $r$ the remainder $0\le r<n$). Note $k$ coprime to $n^2$ iff it's coprime to $n$, iff $k-qn=r$ is coprime to $n$. Thus, there are $n\phi(n)$ different $k$s we're counting, where $n$ is the number of options for $q$ and $\phi(n)$ is the number of options for $r$. Therefore, $\phi(n^2)=n\phi(n)$.