Let $R(n)$ denote the number of ways of writing $n$ as a sum of a prime and a square-free number. Prove that for every $A >0 $ we have $$ R(n) = \operatorname{Li}(n) \prod_{ p \nmid n} \left( 1 - \frac{1}{p(p-1)} \right) + O\left( \frac{n}{(\log n)^A} \right)$$
Hint: Prove first that $$ R(n) = \sum_{d \leq \sqrt{n} } \mu(d) \pi(n-1; d^2,n) $$
Trying to follow what was done in "On the representations of a number as the sum of a prime and a quadratfrei number" by Estermann we actually have that $$ R(n) = \sum_{d \leq \sqrt{n} } \mu(d) \pi(n; d^2,n) $$ anyway now what I want to apply is the Siegel-Walfisz Theorem, soo defining $$ R_1(n) = \sum_{ d, (d,n)=1} \mu(d) \pi(n; d^2,n) $$ and noticing that $ \pi(n; d^2,n)= 0 $ if $ d > \sqrt{n} $ (clearly), and noticing that $\pi(n; d^2,n) \leq 1$ if $(d,n)>1$ (why?) then $$ \left| R(n) - R_1(n) \right| \leq \sqrt{n} $$ hence using Siegel-Walfisz Theorem, we deduce that $$ R(n) = R_1(n) + O(\sqrt{n}) = \operatorname{Li}(n) \sum_{d, (n,d)=1} \frac{\mu(d)}{\varphi(d^2)} + O \left( \sqrt{n}+ \sum_{d, (n,d)=1} \mu(d) n \exp( -C_A \sqrt{\log n}) \right) $$ Now I read it but I don't understand why is true that $$\sum_{d, (n,d)=1} \frac{\mu(d)}{\varphi(d^2)} = \prod_{ p \nmid n} \left(1 - \frac{1}{\varphi(p^2)} \right) = \prod_{ p \nmid n} \left( 1 - \frac{1}{p(p-1)} \right)$$ Hence I have to prove that the error term $$ O \left( \sqrt{n}+ \sum_{d, (n,d)=1} \mu(d) n \exp( -C_A \sqrt{\log n}) \right) = O\left( \frac{n}{( \log n )^A} \right) $$ but I don't see how Maybe I have to do some more work on the summation before to be able to apply Siegel-Walfisz, some help?