Krdytkyu

Let $$A_b= leftlbrace{p in mathbb{P}| left(frac{b}{p}right)=1 } rightrbrace $$ and $$ nu(A_b)=limlimits_{x to infty} frac{#lbrace {p in A_b|ple x}rbrace}{pi(x)}$$ the natural density.
For $b in mathbb{Z}$ square-free, prove that $$nu(A_b)=frac{1}{2}. $$
I do not have really an idea, but I know that $$ left(frac{b}{p}right)$$ is a Dirichlet-character modulo $4|b|$.
Can someone give me a hint ? Thanks for helping .

Write $b =p_{n_1}dots p_{n_k}$ a product of distinct primes. First assume each $p_{n_i}$ is odd. Then for $p not = p_{n_1},dots,p_{n_k}$, by quadratic reciprocity, $$(frac{b}{p}) = (frac{p_{n_1}}{p})dots(frac{p_{n_k}}{p}) = (frac{p}{p_{n_1}})dots (frac{p}{p_{n_k}})(-1)^{epsilon_p}$$ where $epsilon_p$ depends (at most) on the residue of $p$ mod $4$. By Dirichlet's theorem on primes in arithmetic progression, $p$ is equidistributed mod $4p_{n_1}dots p_{n_k}$ and thus, since $(frac{p}{p_{n_j}})$ is $+1$ for $frac{1}{2}$ of the nonzero residues mod $p_j$ and $-1$ for the others, we conclude that $(frac{b}{p})$ is $+1$ half of the time and $-1$ half of the time.

If some $p_{n_j} = 2$... (try to finish this case).