If $n$ has $r$ odd prime divisors then $2^r | \phi(n)$

Question

Let $\phi (n)$ be the Euler totient function $n\in \Bbb N$. If $n$ has $r$ odd prime divisors then $2^r | \phi(n)$

If $n$ is prime then $\phi(n)= n-1$ except $2$ all primes are odd therefore $n-1$ would be even and $2^r$ a divisor.

What should I do if $n$ is not prime?

Answer 1

If $n$ is not prime, then $n=p_1^{\alpha_1}\cdots p_r^{\alpha_r}$, and $\phi(n)=n\prod_{i=1}^r\left(1-\frac{1}{p_i}\right)$. Rewriting the parenthetical as $\frac{p_i-1}{p_i}$,we see that $p_i$ will divide $n$ (since each $\alpha_i\geq 1)$ and there are $r$ terms of $p_i-1$ each of which is even (since $p_i$ are odd). Hence $2^r$ will divide the numerator.

Answer 2

$\phi()=(p_1^{k_1}\cdots p_r^{k_r})[(p_1-1)\cdots(p_r-1)]$.
For every $i$ such that $1\leq i\leq r$, $p_i-1$ is even. Therefore you can re-write the equation replacing the $(p_i-1)$ portions with $2$ times an integer:
$\phi()=(p_1^{k_1}\cdots p_r^{k_r})[(2)(m_1)\cdots (2)(m_r)]=(p_1^{k_1}\cdots p_r^{k_r})[(2^r)(m_1\cdots m_r)]$, where $m_1\dots m_r$ are integers.
Thus $2^r|\phi()$.