Let $F(n) = \sum_{d^2|n} \phi(d)$. We must show that if $F(1) = 1$, and if $n>1$ factors as $n=p^{a_1}_1p^{a_2}_2...p^{a_m}_m$, then $$ F(n)=\prod_{i=1}^{m} p^{[a_i/2]}_i. $$ If I understood correctly, we must first use Mobius inversion to get $ \sum_{d^2|n} \phi(d) = \sum_{\sigma |n} g(\sigma) $ , where
$ g(\delta) = f(d) = f(\sqrt{\delta}) $ . But I'm confused about what follows after that.