Consequence of Brun's Sieve

Question

I'm following a text that uses Brun's sieve (as described in https://pages.cs.wisc.edu/~cdx/Sieve.pdf in Theorem 2.2.2) to prove the following theorem: For any $\alpha\in\mathbb{Z},\alpha\neq0$, we have $$\#\{p\le x: |p+\alpha|\text{ is prime}\}<\frac{cx}{(\log x)^2}\prod_{p|\alpha}\left(1-\frac{1}{p}\right)^{-1}$$ for some absolute positive constant $c$.

They then use the set $A:=\{n+\alpha: n\le x\}$, $P$ is the set of all primes and $\omega(2)=1$ and $\omega(p)=2$, $p>2$. For this, I have 2 questions:

1. They don't show that the hypotheses of Brun's Sieve are satisfied. I've figured out the first 2 hypotheses, but the last one is giving me trouble. How do we know that $|R_d|\le \omega(d)$ for any square free $d$? (where $R_d = |A_d| - \frac{\omega(d)}{d}|A|$)

2. Further in the proof, we have the following: $$\prod_{p<z}\left(1-\frac{\omega(p)}{p}\right) = \prod_{\substack{p<z\\p|\alpha}}\left(1-\frac{1}{p}\right)\prod_{\substack{p<z\\p\not|\alpha}}\left(1-\frac{1}{p}\right)=\prod_{\substack{p<z\\p|\alpha}}\left(1-\frac{1}{p}\right)\left(1-\frac{2}{p}\right)^{-1}\prod_{\substack{p<z}}\left(1-\frac{2}{p}\right)\le \frac{c_1}{(\log z)^2}\prod_{p|\alpha}\left(1-\frac{1}{p}\right)^{-1}$$ The only thing that I'm stuck on here is the inequality at the end. I know I should be using Merten's Theorem here, but I'm not sure how to get the value that's shown.

Any help would be greatly appreciated.

EDIT

I've figured out my second problem, thanks to Theorem precedding Schnirelmann's Theorem, however, I'm still not sure about my first problem. Any help would still be appreciated

Answer 1

For practical purposes, one only needs to care about the case where $\alpha$ is even, which indicates that

$$ n+\alpha\equiv0\pmod p\Rightarrow \begin{cases} \text{1 solution} & p=2 \\ \text{2 solutions} & p>2 \end{cases}\Rightarrow\omega(p)\text{ solutions}. $$

By the Chinese remainder theorem, we know for sure that $n+\alpha\equiv0\pmod d$ has $\omega(d)$ solutions whenever $d$ is squarefree. If we were to divide the range $[1,x]$ into intervals of size $d$. Then

$$ \left||A_d|-\omega(d)\left\lfloor\frac xd\right\rfloor\right|\le\omega(d). $$

Because $|y-\lfloor y\rfloor|\le1$, we see immediately that

$$ |R_d|=\left||A_d|-{\omega(d)\over d}x\right|\le2\omega(d), $$

which still satisfies the conditions for Brun's sieve (In fact, $|R_d|\le L\omega(d)$ for any fixed $L>0$ will work).