I have several questions related to the problem of counting square-free $d \equiv 1 \mod{8}$, where $d \le X$ such that if $d = p_1 \cdots p_k$ then $p_i \equiv \pm 1 \mod{8}$ for all $1 \le i \le k$. This problem is related to K. Soundararajan's paper on the divisibility of the class number for imaginary quadratic fields. It is given that there are $\gg X/\sqrt{\log(X)}$ such $d$, and I would like to understand why.
My idea is to find $\pi_{8,1}(X) = \#\{p \quad \text{prime} \le X : p \equiv 1 \mod{8} \}$ and $\pi_{8,7}(X) = \#\{p \quad \text{prime} \le X : p \equiv 7 \mod{8} \}$, which would allow us to construct the desired $d$ and give a way for counting them. For example, if $\pi_{8,1}(X) = l$ then there should be $\sum_{i = 1}^l \binom{l}{i}$ possibilities for $d$. The case for primes $p \equiv 7 \mod{8}$ is slightly more complicated (since we must take an even number of such primes), but follows the same idea. However, I am not sure how to compute lower bounds for $\pi_{8,1}$ and $\pi_{8,7}$. Dirichlet's theorem tells us what happens as we let $X \to \infty$ but I am not sure how I can use this to find a lower bound that leads to $\#\{d\} \gg X/\sqrt{\log(X)}$. Am I on the right track?
EDIT: If if $\pi_{8,7}(X) = z$ then we have two cases (I think, did I make a mistake somewhere?):
$$z \equiv 0 \mod{2} \Rightarrow \#\{d\} = \left(\sum_{i = 1}^l \binom{l}{i}\right)\left(\sum_{j = 1}^{z/2} \binom{z}{2j}\right)$$
$$z \equiv 1 \mod{2} \Rightarrow \#\{d\} = \left(\sum_{i = 1}^l \binom{l}{i}\right)\left( \sum_{j = 1}^{(z-1)/2} \binom{z}{2j} \right).$$
However, I'm not sure where to go from here.