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Let $\mathcal{F}_{Q;r,q}=\{\gamma=\frac{m}{n} | 0\leq m \leq n \leq Q, \gcd(m,n)=1, n \equiv r \mod q, \gcd(r,q)=1\}$. Usually, with no condition on arithmetic progression, then $\# \mathcal{F}_{Q}$ (cardinality) $=\varphi(1)+\varphi(2)+\varphi(3)+\cdots+\varphi(Q)$ $=\frac{3Q^2}{\pi^2}+O(Q\log Q)$ as $Q\rightarrow \infty$, but now how to get the cardinality for $\mathcal{F}_{Q;r,q}$ as $Q \rightarrow \infty$?

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I suppose $\phi(\cdot)$ refers to the Euler phi (totient) function, sometimes denoted $\varphi(\cdot)$. –  Srivatsan Oct 30 '11 at 3:18
    
Have you tried looking at a proof of the estimate for the sum of the phi-values and adapting it to the sum over an arithmetic progression? –  Gerry Myerson Oct 30 '11 at 5:44
    
So I have tried $$\#\mathcal{F}_{Q;r,q}=\frac{1}{\varphi(q)}\sum_{n \leq Q} \varphi(Q) \sum_{\chi \mod q}\chi(n)\bar{\chi(r)}$$, which use $\varphi(n)=\sum_{d|n} \frac{\mu(d)}{d}$ becomes $$\frac{1}{\varphi(q)}\sum_{\chi \mod q} \bar{\chi(r)}\sum_{d \leq Q}\mu(d)\chi(d)\sum_{m \leq \frac{x}{d}}m \chi(m)$$, but then I don't know how to continue. –  Rob Oct 31 '11 at 21:44
    
Are we given that $r\in(\frac{Z}{qZ})^{\times}$? If so we could use symmetry arguments. –  pre-kidney Jan 23 '13 at 19:46

2 Answers 2

For all $b$ coprime to $a$, with $1 \leq b \leq a$ $$\sum_{n\leq x}_{n\equiv b \text{ mod a}}\phi(n)=(\frac{3}{\pi^2}\prod_{p\mid a}\frac{p^2}{p^2-1})x^2+O(x\ln(x)).$$ Can be shown pretty easily with out characters, though with some additional tools I think the O term could be improved.

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You can find a proof in the book Postnikov, A. G. Introduction to analytic number theory American Mathematical Society, 1988, (see section 4.2).

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