Show that the natural density is $1/2$.

Question

Let $$A_b= \left\lbrace{p \in \mathbb{P}| \left(\frac{b}{p}\right)=1 } \right\rbrace $$ and $$ \nu(A_b)=\lim\limits_{x \to \infty} \frac{\#\lbrace {p \in A_b|p\le x}\rbrace}{\pi(x)}$$ 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 .

Answer 1

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).