I'm currently working in the following Euler's theorem exercise:
Let $n$ be a positive integer. Show that $n\phi(n)=\phi(n^2).$
Here are my thoughts:
If $n$ is prime, then
$$\phi (n^2) = n^2-n = n(n-1) = n \phi (n).$$
If $n$ is not prime and $p$ and $q$ are prime factors of $n,$ then
$$\phi (n^2) = (p^2-p)*(q^2-q)$$ $$\phi (n^2) =p(p-1)*q(q-1)$$ $$\phi (n^2) =p\phi(p)*q\phi(q)$$ $$\phi (n^2) =pq \phi(pq)$$ $$\phi (n^2) = n \phi(n)$$
Is that enough proof? Is there a more general one? Am I missing something? Any hint will be really appreciated.