The field is also known as additive number theory. I am interested in sums $z=x + y$ where $x \in S, y\in T$, and both $S, T$ are infinite sets of positive integers. For instance:

  • $S = T$ is the set of primes (leading to Goldbach's conjecture)
  • $S$ is the set of squares and $T$ is the set of primes, leading to the deeper Hardy and Littlewood's Conjecture $H$, see my previous question here

A possible approach to check if $S+T = \{x+y, x\in S, y \in T\}$ covers all sufficiently large integers is as follows.

Define $N_S(x)$ as the number of elements in $S$ that are smaller or equal to $x$, and $N_T(y)$ as the number of elements in $T$ that are smaller or equal to $y$. The $n$-th element of $S$ is $N_S^{-1}(n)$, and $n$-th element of $T$ is $N_T^{-1}(n)$. The number $r(z)$ of solutions to $$N_S^{-1}(x) + N_T^{-1}(y) \leq z$$ is asymptotically $$r(z) \sim \int_0^{N_S(z)} N_T(z-N_S^{-1}(x)) dx.$$

The number $t(z)$ of ways that an integer $z$ can be written as $x+y$ with $x\in S, y\in T$ is thus $$t(z) = r(z) - r(z-1) \sim \frac{dr(z)}{dz}$$ as $z$ becomes larger and larger. So in order to prove that for $z$ large enough, $z$ is the sum of an element of $S$ and an element of $T$, one "only" has to prove that $t(z) > 0$ for $z$ large enough.


Is it possible to solve this problem using extremely precise approximations in all the asymptotic derivations discussed here? For instance, if $S$ is the set of prime numbers, then $N_S(z) \sim z/\log z$ and $N_S^{-1}(z)=z\log z$, but this is not precise enough to prove that every large enough even integer is the sum of two primes. You need far better approximations. Likewise, if $S$ is the set of squares, then $N_S(z) \sim \sqrt{z}$ and $N_S^{-1}(z)=z^2$, but this is not enough to prove that every large enough non-square integer is the sum of a square and a prime.

One issue is with the integral, which is only the first term in an Euler - Maclaurin series expansion to approximate $r(z)$. You need to use more than just the first term. If $S=T$ are the sets of squares, rather precise formulas are available for $r(z)$: see the Gauss-circle problem, here (Wikipedia) and here (MSE).

Another question is whether my method is equivalent to the circle method.


Besides $N_S(x), N_S^{-1}(x), N_T(y), N_T^{-1}(y), r(z), dr(z)/dz$, another quantity of interest is the probability for an integer $z$ to belong to $S$: it is defined as $dN_S(z)/dz$, for instance, equal to $1/\log z$ if $S$ is the set of primes.


When $S$ is the set of squares and $T$ the set of primes, I made all the computations in my previous question: see here. I also added a lot of new material recently, for instance: among the first 750,000 integers, $z=78754$ is the last one to admit only one ($r(z) = 1$) decomposition as $z=x^2+y$ with $x$ integer and $y$ prime. That is, if $z>78754$ then $r(z) > 1$. Likewise:

  • $z=101794$ is the last one with $r(z) =2$
  • $z=339634$ is the last one with $r(z) =3$
  • $z=438166$ is the last one with $r(z) =4$
  • $z=383839$ is the last one with $r(z) =5$

The sequence of $z$'s with $r(z)=1$ is listed at the bottom of my previous question, see here. I searched for this sequence to see if it had been discovered, but could not find any reference.


If my approach (assuming it is new!) ever leads to a proof of some famous conjectures, the proof will be very technical, difficult and long. It is beyond my reach but some mathematicians with experience dealing with extremely precise (second- or third-order approximations) to the asymptotics involved might have an answer about the feasibility of my approach. Just to give an idea of the many problems, it may require excellent asymptotics about a function more complex than the Lambert function (again, this briefly outlined in my previous question).

Maybe the following is true for sums of two primes and sums of a prime and and a square: there only finitely many $z$'s that can be expressed as $z=x+y$ in less than $k$ different ways, with $x\in S, y \in T$, regardless of $k$. This would imply that all but a finite number of $z$'s can be expressed as the sum in question.



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