[Math] Show that if $n$ is a positive integer with $r$ distinct odd prime factors, then $2^r \mid \varphi(n)$

Question

Show that if $n$ is a positive integer with $r$ distinct odd prime factors, then $2^r \mid \varphi(n)$.

So I know that
$$\varphi(n) = \varphi(p_1^{a_1}\cdot p_2^{a_2} \cdots p_r^{a_r}) = (p_1^{a_1}-p_1^{a_1-1})(p_2^{a_2}-p_2^{a_2-1})\cdots(p_r^{a_r}-p_r^{a_r-1})=p_1^{a_1}(1-\frac{1}{p_1})p_2^{a_2}(1-\frac{1}{p_2}) \cdots p_r^{a_r}(1-\frac{1}{p_r}) = n\cdot (1-\frac{1}{p_1})(1-\frac{1}{p_2})(1-\frac{1}{p_r}).$$

and then I'm stuck. Any help would be appreciated!

Answer 1

Let $n=p_1\cdots p_r$ then $\varphi(p_i) = p_i-1 = 2k_i$ thus $\varphi(n)=2k_1\cdots 2k_r = 2^r\cdot(k_1\cdots k_r)$

So $\frac{2^r\cdot(k_1\cdots k_r)}{2^r} = (k_1\cdots k_r) \Rightarrow 2^r \mid \varphi(n)$