how to find an upper bound for this series

Question

I don't understand why $$\left|\sum_{b<y}\frac{\mu(b)\rho(bc)}{b}\right|\ll\tau(c)log(y)^{-A}$$

with $A>1$, where $\mu(b)$ is the mobius function and $\rho(n)$ is the number of solution of $x^2+1 \text{ mod }n$. Thank you in advance

Answer 1

Bear,

Here I think is one way to show this.

From Appendix A.5 of Friedlander's and Iwaniec's Opera de Cribro, I think we have the following proposition (implicitly not explicitly stated):

Proposition: Let $g(d)$ be a multiplicative function such that for $0 \leq g(p) < 1$ for all primes $p$ and satisfies $$\sum_{p < x} g(p) \log p = \log x + C + O\left((\log x)^{-A}\right),$$ for some constant $C$ and for any $A > 0$. Then $$\sum_{n < x} \mu(n)g(n) \ll (\log x)^{-A}, \quad \text{for any } A > 0.$$

We wish to show that $$\sum_{b < y} \frac{\mu(b)\rho(cb)}{b} \ll \tau(c)(\log y)^{-A}$$ for any arbitrary $A > 0$, where $\tau(c) = \sum_{d|c} 1$, and $\rho(d)$ is the number of solutions to $x^2 + 1 \equiv 0 (\bmod d)$.

Observe that $\rho(d)$ is multiplicative, and on the primes is equal to $\rho(2) = 1$, $\rho(p) = 2$ for $p \equiv 1 (\bmod 4)$ and $\rho(p) = 0$ for $p \equiv 3 (\bmod 4)$.

Let $g(d) = \rho(d)/d$. I think that, by the prime number theorem for arithmetic progressions, we have that $g(p)$ satisfies the conditions of the above proposition, proving our statement for $c=1$.

More generally, for $c$ a fixed positive integer, $g(d) = \rho(cd)/(\rho(c)d)$ also satisifies the conditions of the proposition. Since $\rho(c) \ll \tau(c)$, we have our desired result.

Please let me know if you have any questions or would like further details.