Let $\chi$ be a primitive quadratic Dirichlet character of d modulus $m$, and consider the product
$$\prod\_{\substack{p \text{ prime} \\ \chi(p) = 1}} (1-p^{-2})^{-1}.$$


What can we say about the value of this product? Do we have good upper or lower bounds?


**Some observations, ideas, and auxiliary questions**


* When $\chi$ is trivial, it has value $\zeta(2)$.
* In general, since $\chi(p)$ is equidistributed in the limit, I would "want" the value to be something like


$$\Big(\zeta(2)\prod\_{p | m} (1-p^{-2})\Big)^{\frac{1}{2}}.$$


However, if I'm not mistaken, it seems that the error terms in effective forms of this result may cause this to be very far from the truth. We can't ignore what happens before we are close to equidistribution as the tail and the head are both $O(1)$. We can't even control the error term well (without GRH) because of Siegel zeroes.


* I don't think we can appeal to Dirichlet density versions of CDT since those only tell us things in the limit as $s$ goes to $1$ and here $s = 2$.
* Is there a way to "Dirichlet character"-ify a proof of $\zeta(2) = \pi^2/6$ to get a formula for this more general case? At least with Euler's proof via Weierstrass factorization, it seems that we would need some holomorphic function which has zeroes whenever $\chi(n) = 1$.


I had a few other ideas but they all seem to run into the same basic problem of "can't ignore the stuff before the limit"... am I missing something?



Sum of inverse squares of numbers divisible only by primes in the kernel of a quadratic character
No need to use Chebotarev or Dirichlet: set $Z(k)=\prod\_{p\nmid m}(1-p^{-k})^{-1}$ and $L(k)=\prod\_p(1-\chi(p)p^{-k})^{-1}$. Then if $P(k)=\prod\_{\chi(p)=1}(1-p^{-k})^{-1}$ we have the recursion
$$P(k)=(Z(k)L(k)/Z(2k))^{1/2}P(2k)^{1/2}$$
from which you can deduce bounds that you want.


