$\displaystyle \sum_{n\:\equiv\: 0 \mod k} \chi (n) = \sum_{l|k} \mu (l) \sum_{(n,l)=1} \chi (n)$
where $\mu$ is the Mobius function. It's clearly some application of Mobius inversion. Can anyone show this rigorously?
2. Hint: $\displaystyle \sum_{d\mid n} \mu(d) = \begin{cases} 1 \text{, if } n = 1 \\ 0 \text{, otherwise} \end{cases}$