Some Sieves problems

Question

I'm currently reading some lectures notes on Sieves Problems and I stumbled across a couple of problems I'm not able to solve. Honestly, since this is my first time I'm reading on analytic number theory, I'm not even sure how to attack them.

1. Given an odd positive integer $n$ we have: $$\#\{(p_1,p_2,p_3): \ p_i\text{'s are prime and } \ p_1+p_2+p_3=n \} \ll \frac{n^2}{\log(n)^3}$$ Now the bound intuitively makes sense, since we can choose the first two primes in about $\frac{n^2}{\log(n)^2}$ ways, but since it's clear that $n-p_1-p_2$ will not always be a prime, we need to multiply by $\frac{1}{\log(n)}$ factor, but I'm not sure how to prove this.

2. A positive proportion of integers can be written as the sum of two prime numbers. I can prove this without using Sieves, but I'd like to see a proof employing them. I guess the proof would go something like that: Given $n$, the number of ways in which $n$ can be written as the sum is bounded, and then we can use this bound and the fact that $\pi(n) \sim \frac{n}{\log(n)}$ to prove the result.

Answer 1

This answer is dedicated to OP's first question as the second one is basically hinted in Greg Martin's comment.

Let $r_k(n)$ be the number of ways to write $n$ as a sum of exactly $k$ primes. Then we have $$ r_k(n)=\sum_{p<n}r_{k-1}(n-p). $$ Thus, to estimate $r_3(n)$ it suffices to estimate $r_2(n)$, the number of ways to write $n$ as a sum of 2 primes. As this result is available in basically every sieve theory textbook, I am quoting it without a proof:

Theorem 1 (Schnirelman): For large $n$, we have $$ r_2(n)\ll{n\over\log^2n}\prod_{p|n}\left(1+\frac1p\right). $$

This indicates that $$ r_3(n) \ll{n\over\log^2n}\underbrace{\sum_{p_0<n}\prod_{p|n-p_0}\left(1+\frac1p\right)}_S $$ Therefore, it suffices to show that $S\ll n/\log n$. Using the properties of Dirichlet convolution, we have $$ S\le\sum_{p_0<n}\sum_{d|(n-p_0)}\frac1d=\sum_{d<n}\frac1d\color{red}{\sum_{\substack{p_0<n\\p_0\equiv n(d)}}1} $$

Trivially the red sum is $\ll n/d$, so we can pick $D<n$ and obtain $$ S\ll\underbrace{\sum_{d<D}\frac1d\color{red}{\sum_{\substack{p_0<n\\p_0\equiv n(d)}}1}}_{S_1}+\underbrace{\sum_{d\ge D}{n\over d^2}}_{\ll n/D} $$

To estimate the remaining red sum, we quote Brun-Titchmarsh inequality:

Theorem 2 (Brun-Titchmarsh): For all $x\ge q\ge1$ and $(a,q)=1$ we have $$ \#\{p\le x:p\equiv a\pmod q\}\ll{x\over\varphi(q)\log x/q}. $$

Therefore, we see that \begin{aligned} S_1 &\ll\sum_{\substack{d<D\\(d,n)=1}}{1\over d\varphi(d)}{n\over\log n/d}+\sum_{\substack{d<D\\(d,n)>1}}\frac1d \\ &\ll{n\over\log n/D}\sum_{d\ge1}{1\over d\varphi(d)}+\log D \end{aligned}

Finally, setting $D=\sqrt n$ gives $S\ll n/\log n$, completing the proof of $$ r_3(n)\ll{n^2\over\log^3n}. $$