A Selberg's sieve application I have a problem concerning Selberg's sieve: I have to prove that, given $k>1$ a fixed even integer, the number of primes $p\leq x$ such that $1+pk$ is still a prime is $O(\frac{x}{\log^2(x)})$. 
Basically, I'd like to use Selberg's method as suggested in  Narkiewicz's Number Theory (pg. 165 ex 2): There, I can see a reference to an article by Erdos.
Anyway, I can't find out the number $b(p_j)$ of classes mod. $p_j$ to sift for every $p_j\leq x$ and, then, proceed with the usual argument by Selberg (here I'm using Montgomery-Vaughan's notation). 
The question is, what is $b(p_j)$?
It is obvious that if $p\leq x$ is a prime such that $q=1+kp$ is a prime as well, then $q\equiv1$ mod. $p$, so I must not sift the class $1$ mod. $p$. So I guess the only admissible class mod. $p$ is $1$.
But, how do I know that a prime $q\equiv1$ mod. $p$ is not $\equiv h\neq1$ mod. $p_2$ for another prime $\leq x$?
 A: Let $z=\frac{\sqrt{x}}{\ln{x}}$ be the sifting parameter, and let $S(z,x)$ be the set of $y<x$ such that for every prime $q<z$, we have $q\nmid y$ and $q\nmid 1+yk$. Note that if $p$ is a prime of the form you're interested in, then either $p<z$ or $p\in S(z,x)$. Since $z=O(x/\ln^2x)$, it suffices to show that $|S(z,x)| = O(x/\ln^2x)$.
Now we define for each prime $q<z$ the set $b(q)$ consisting of those $y<x$ such that $q\mid y$ or $q\mid 1+yk$. We're interested in the set of numbers that are not in any of these $b(q)$.
Now suppose $q\neq 2$. Then clearly $q\mid y$ iff $y\equiv 0\mod q$ and $q\mid 1+yk$ iff $y\equiv k^{-1}\mod q$. Since these conditions are obviously disjoint, and each occurs for $\frac{x}{q}+O(1)$ values of $y<x$, we have $|b(q)| = \frac{2x}{q}+O(1)$.
For $q = 2$, we have that $q\nmid 1+ky$ for every value of $y$, and $q\mid y$ for $\frac{x}{2}+O(1)$ values of $y<x$. Thus $|b(2)| = \frac{x}{q}+O(1)$.
Using the Chinese remainder theorem you can show that up to small constant factors we also have that $|b(p_1\cdot...\cdot p_n)|=|b(p_1)|\cdot...\cdot|b(p_n)|$, so you can try to bound $V(z)$ and apply the Selberg sieve. I haven't run the explicit calculation, but I think you can bound $V(z)\geqslant \sum_{d<z} \frac{log_2(d)}{d}\geqslant \Omega(\log^2(z)) = \Omega(\log^2(x))$, which is exactly what you want.
