if $p \equiv 3 \pmod 4$, does that automatically imply that 4 does not divide $\phi(p^k)$?

Question

Problem:

Describe all $m$ such that $\phi(m)$ is not divisible by $4$.

The solution for this is this set:

Let $p$ be a prime number, then $$T = \{p^k\text{ where }p \equiv 3\pmod 4\} \cup\{1, 2, 3\} \cup \{2m\mid m\in T\}$$

My main question is this:

We know that for $p \equiv 3 \pmod 4 $, $\phi(p)$ does not divide 4, because $\phi(p)$ is $p-1$. But why does 4 also not divide $\phi(p^k)$, when p is congruent to 3 mod 4?

Answer 1

If $p$ is prime, then $$ \varphi\left(p^k\right)=(p-1)\,p^{k-1} $$ so $4\mid\varphi\left(p^k\right)\iff p=2\text{ and }k\ge3\text{ or }p\equiv1\pmod{4}\text{ and }k\ge1$