$\mathbb N-\{1\}\subseteq\mid S+S\mid$ such that $d\left(S\right)=0$

Question

I am dealing with the set of integers $A=\{x:x\neq i+j+2ij, x\in\mathbb N, i\in\mathbb N, j\in\mathbb N\}$. I am trying to show that $\mathbb N-\{1\}\subseteq\mid A+A\mid$, $\mid A+A\mid=\{a_i+a_j: a_i\in A, a_j\in A\}$ as there is heavy numerical evidence that it is true; but it is really complicated, because the natural and Schnirelmann's density of $A$ equals zero (approximately, for some $a_j\in A$, $j\sim \frac{2a_j}{\ln\left(a_j\right)}$).

Do you know of some set of positive integers $B$ with natural and/or Schnirelmann's density equal to zero, such that has been proved that $\mathbb N-\{1\}\subseteq\mid B+B\mid$? Are there known conditions that some set of integers $S$ has to comply with in order to have that $\mathbb N-\{1\}\subseteq\mid S+S\mid$?

Thanks!

Answer 1

Note that $x=i+j+2ij$ iff $(2x+1)=(2i+1)(2j+1)$.

It follows that $x\in A$ iff $2x+1$ is prime. Thus $A=\lbrace \frac{p-1}{2} \ | \ p \ \textrm{is an odd prime}\rbrace$.

Your claim then amounts to saying that any $n\geq 2$ can be written $n=\frac{p_1-1}{2}+\frac{p_2-1}{2}$ for odd primes $p_1,p_2$, or equivalently $2n+2=p_1+p_2$.

Thus your claim is equivalent to the well-known Goldbach conjecture.

You are quite correct that there is heavy evidence supporting it : according to Wikipedia, it has been checked up that every even $m \leq 4 \times 10^{18}$ is a sum of two primes.

In answer to your second question, a well-known example of $B$ that works is the set of sums of two perfect squares. Lagrange's four-square theorem ensures that $N\setminus \lbrace 1 \rbrace \subseteq B+B$, and it is shown here that $B$ has density $0$.