Let $\mu(\cdot)$ be the Mobius function, defined on the natural numbers by

$$\displaystyle \mu(n) = \begin{cases} (-1)^{\omega(n)} & \text{if } n \text{ is square-free} \\ 0 & \text{otherwise}.\end{cases}$$

Here $\omega(n)$ is the number of distinct prime divisors of $n$. It is well-known that the Mertens function $M(x)$ defined by

$$\displaystyle M(x) = \sum_{n \leq x} \mu(n)$$

satisfies the asymptotic

$$\displaystyle M(x) = O\left(x \exp(-c \sqrt{\log x}) \right)$$

for some positive number $c$; this is a consequence of the prime number theorem. The Riemann hypothesis is equivalent to the assertion that for any $\varepsilon > 0$ the Mertens function satisfies $M(x) = O_\varepsilon \left(x^{1/2 + \varepsilon}\right)$.

Let $a, q$ be co-prime positive integers, and let $P_{a,q}$ denote the set of primes $p \equiv a \pmod{q}$. Let $N_{a,q}$ be the set of positive integers such that $p | n \Rightarrow p \in P_{a,q}$. Let

$$M_{a,q}(x) = \sum_{\substack{n \leq x \\ n \in N_{a,b}}} \mu(n).$$

Does $M_{a,q}(x)$ satisfy a similar asymptotic as $M(x)$?

Next, let $f \in \mathbb{Z}[x]$. Put $\rho_f(n) = \# \{m \in \mathbb{Z}/n \mathbb{Z} : f(m) \equiv 0 \pmod{n}\}$. Define

$$\displaystyle M_f(x) = \sum_{n \leq x} \mu(n) \rho_f(n).$$

Does $M_f(x)$ satisfy a similar asymptotic upper bound as $M(x)$?