How to prove $ \phi(n) = n/2$ iff $n = 2^k$? How can I prove this statement ? $ \phi(n) = n/2$ iff $n = 2^k $
I'm thinking n can be decomposed into its prime factors, then I can use multiplicative property of the euler phi function to get the $\phi(n) = \phi(p_1)\cdots\phi(p_n) $. Then use the property $ \phi(p) = p - 1$. But I'm not sure if that's the proper approach for this question.

Edit: removed my full answer to be more pedagogical.
You know that $\varphi(p) = p-1$, but you need to remember that $\varphi(p^k) = p^{k-1}(p-1).$ Can you take it from here?
