Proportion of primes that satisfies two legendre symbols? I have a problem in regards to a paper that I am reading and I hope someone can guide me a little bit. Assume that $3s \in \mathbb{Z}$ is not a square, the paper states that the proportion of primes such that the Legendre symbols $\left(\frac{-1}{p}\right) = \left(\frac{-3s}{p}\right) = 1$ is 1/4.
I'm not sure how he obtained this proportion, i understand for example:

*

*That the proportion of primes such that $\left(\frac{-1}{p}\right) = \left(\frac{3}{p}\right) = 1$, is in fact 1/4.

*The proportion of primes such that $\left(\frac{-1}{p}\right) = -\left(\frac{3}{p}\right) =1$ is also $ 1/4$.

Both of these propositions due to Dirichlet theorem.
Clearly, if $s$ is a square in $\mathbb{Z}$ we have what is claimed. Otherwise, i know that by Cheboratev's density theorem the amount of primes such that $s$ is a square mod $p$ is 1/2. But here I am stucked, because I'm missing a way to connect this two ideas, i can naively think that then a half of the primes with the first condition plus a half of the primes with the second condition will lead to the proportion that I'm searching for (which adds to 1/4) but I am not sure if this reasoning is correct because some proportions of these primes might be "overlapping" or this cannot be the case?.
I hope someone can give me some ideas.
 A: Let $p$ be a prime number, and let $a$ and $b$ be integers such that $a$ is not divisible by $p$. The Legendre symbol $(\frac{a}{p})$ is defined as $1$ if $a$ is a quadratic residue modulo $p$, $-1$ if $a$ is a quadratic non-residue modulo $p$, and $0$ if $a$ is divisible by $p$.
The proportion of primes that satisfy $(\frac{a}{p}) = 1$ and $(\frac{b}{p}) = 1$ is equal to the product of the proportions of primes that satisfy each condition individually, since the conditions are independent. Therefore, we have:
$$\frac{\text{number of primes }p\text{ such that }(\frac{a}{p}) = 1 \text{ and } (\frac{b}{p}) = 1}{\text{total number of primes}} = \frac{\text{number of primes }p\text{ such that }(\frac{a}{p}) = 1}{\text{total number of primes}} \cdot \frac{\text{number of primes }p\text{ such that }(\frac{b}{p}) = 1}{\text{total number of primes}}$$
By the quadratic reciprocity law, the proportion of primes that satisfy
$(\frac{a}{p}) = 1$ is equal to the proportion of primes that satisfy
$(\frac{p}{a}) = 1$. Therefore, the proportion of primes that satisfy $(\frac{a}{p}) = 1$ and $(\frac{b}{p}) = 1$ is equal to:
$$\frac{\text{number of primes }p\text{ such that }(\frac{a}{p}) = 1 \text{ and } (\frac{b}{p}) = 1}{\text{total number of primes}} = \frac{\text{number of primes }p\text{ such that }(\frac{a}{p}) = 1}{\text{total number of primes}} \cdot \frac{\text{number of primes }p\text{ such that }(\frac{p}{b}) = 1}{\text{total number of primes}}$$
Since $(\frac{a}{p})$ and $(\frac{p}{b})$ are either $1$ or $-1$, the proportion of primes that satisfy $(\frac{a}{p}) = 1$ and $(\frac{p}{b}) = 1$ is equal to $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$. Therefore, the proportion of primes that satisfy $(\frac{a}{p}) = 1$ and $(\frac{b}{p}) = 1$ is $\frac{1}{4}$.
For example, let $a = 2$ and $b = 3$. The proportion of primes that satisfy $(\frac{2}{p}) = 1$ and $(\frac{3}{p}) = 1$ is $\frac{1}{4}$, since there are 8 total primes (2, 3, 5, 7, 11, 13, 17, 19) and 2 of them (3 and 17) satisfy both conditions.
