Show that number of primes $p\leq x$ so that $2p+1$ is also a prime... This question is from assignment 4 of the following sieve theory course: http://www.math.tau.ac.il/~rudnick/courses/sieves2015.html
Question : Show that the number of primes $p\leq x$ so that $2p+1$ is also a prime is atmost $O(x/(\log x)^2)$.
If I use the Brun Titchmarsh Inequality given in Lecture 13 with $q=2$ and $a=1$, I get that number of  such primes in $O( x/\log x)$.
I am not getting any ideas on what result I should use to improve the bound despite reading the notes again. Can you please give some hints?
 A: I think the question is asking you to construct a sieve for this problem.
In essence, let $\mathcal A=\{2p+1:p\le x\}$. Then the task will be to estimate the number of primes in $\mathcal A$, which is bounded by
$$
S(\mathcal A,\mathcal P,z)+z
$$
Now, our task is transformed into estimating the upper bound for $S(\mathcal A,\mathcal P,z)$.
To continue, we need to estimate the size of $\mathcal A_d$, the multiples of $d$ in $\mathcal A$. For $p\le x$, we know that the following congruence
$$
2p+1\equiv0\pmod d
$$
has unique solution in $\mathbb Z_d$ if and only if $d$ is not an even number. This suggests that if
$$
\omega(d)={d\over\varphi(d)}
$$
for odd $d$ and zero for even $d$, then prime number theorem in arithmetic progression suggests that
$$
|\mathcal A_d|={\omega(d)\over d}\operatorname{li}x+\mathcal O\{E(x,d)\},
$$
where
$$
E(x,d)=\max_{(a,d)=1}\left|\pi(x;d,a)-{\operatorname{li}x\over\varphi(d)}\right|.
$$
Consequently, applying the fundamental lemma, we see that when
$$
W(z)=\prod_{p<z}\left(1-{\omega(p)\over p}\right),
$$
there is
$$
S(\mathcal A,\mathcal P,z)\ll W(z){x\over\log x}+\sum_{\substack{d|P(z)\\d\le z^2}}3^{\nu(d)}E(x,d),
$$
where $P(z)$ denote the product of primes $<z$ and $\nu(d)$ denotes the number of distinct prime divisors of $d$. Now, set $z=x^{\frac14-\delta}$, so that Bombieri-Vinogradov theorem can help us cope with the error term:
$$
\sum_{\substack{d|P(z)\\d\le z^2}}3^{\nu(d)}E(x,d)\ll_A{x\over\log^Ax}
$$
For the main term, we plug in the definition and apply Mertens' second theorem so that
$$
\log W(z)\le-\sum_{p<z}{\omega(p)\over p}=-\sum_{2<p<z}{1\over p-1}=-\log\log z+O(1)
$$
Finally, putting everything together allows us to conclude
$$
\#\{p\le x:2p+1\text{ prime}\}\le
S(\mathcal A,\mathcal P,x^{\frac14-\delta})+x^{\frac14}\ll{x\over\log^2x}
$$
