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I came across the following problem after a colleague and I were discussing nonregular languages whose concatenation is regular.

If $A, B \in P(\mathbb{N})$, we can define their Minkowski sum as

$$A + B = \{ m + n \ | \ m \in A \land n \in B \}\text.$$

For the purposes of this problem, assume $0 \in \mathbb{N}$.

Some sets $S \subseteq \mathbb{N}$ have the property that $S + S = \mathbb{N}$. For example:

  • The set $\mathbb{N}$ itself.
  • The set of all natural numbers except $137$.
  • The set of all natural numbers not congruent to $2$ modulo $3$.
  • The odd natural numbers, along with 0.
  • The set of all natural numbers that are the sums of two squares (a consequence of Lagrange’s Four Square Theorem).

I haven’t found a principled way to discover sets like this and have been approaching finding them largely as an exercise in trial and error.

Is there an alternative exact characterization of these sets that makes it quick and easy to discover them?

Caleb Stanford
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templatetypedef
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1 Answers1

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This is likely an unsolved problem; I will provide some keywords and related references. Such a set is sometimes known as an additive basis of order 2. Related questions on mathSE are 1, 2, 3 (unanswered), and 4 (unanswered).

There are some elementary properties you can give about additive bases; for example, for such a set $S$, it must contain at least $n$ of the first $n + \binom{n}{2} = \binom{n+1}{2}$ nonnegative integers, or else there wouldn't be enough pairs to cover all $0, 1, \ldots, \binom{n+1}{2} - 1$. But an exact characterization appears to be much harder -- there are many open conjectures.

For starters, finding an exact characterization would have implications for the strong Goldbach conjecture. To see this, consider the set $S$ of numbers of the form $p - 3$ or $p - 2$ where $p$ is an odd prime. If the strong Goldbach conjecture is true, then for any $n$ either $n + 5$ or $n + 6$ is an even number greater than or equal to $6$, and thus equal to the sum of two odd primes, so $S + S = \mathbb{N}$. On the other hand, if the strong Goldbach conjecture is false, then there is some even number $n > 6$ that can't be written as the sum of two primes, and it follows that the odd number $n + 5$ can't be written as a sum of two elements of $S$. In particular, since $n + 5$ is odd, it would have to be written as an odd number plus an even number, so it would equal $(p_1 - 2) + (p_2 - 3)$ for some odd primes $p_1, p_2$.

This means that a sufficiently "simple" exact characterization would lead to a simple solution to the strong Goldbach conjecture, and such a simple characterization is thought to be very difficult to obtain.

There are other relevant open problems. For example, the Erdős–Turán conjecture concerns the number of ways $f(n)$ to write $n$ as the sum of two elements of $S$, for sufficiently large $n$. It conjectures that for any set $S$ where $S + S = \mathbb{N}$, there is no upper bound on $f(n)$ (so $f(n)$ has arbitrarily large values).

I initially found these references by writing down the number of possibilities in such a set $S$ for the elements $0, 1, 2, \ldots n$. That is, defining the notation $[n] := \{0, 1, \ldots, n\}$, we can ask, what is the number of subsets $S \subseteq [n]$ such that $S + S$ covers $[n]$? We get the sequence $1, 1, 2, 3, 6, 10, 20, 37, 73, 139, 275, \ldots$, which turns up the following result in OEIS:

  • A066062: Number of distinct subsets S of T={0,1,2,...,n} such that each element of T is the sum of two elements of S.

There appear to be two other sequences in OEIS that are also equivalent (though the equivalence is not immediately obvious):

  • A008929: Number of increasing sequences of Goldbach type with maximal element n.

  • A164047: The number of symmetric numerical sets with odd Frobenius number and no small atoms.

Each of these contains some additional references on this problem and related problems.

Caleb Stanford
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  • Fantastic, thanks! For your first point about needing $n$ of the first $\binom{n+1}{2}$ natural numbers: should that be $n$ of the first $n^2$, since the number of pairs of $n$ naturals is $n^2$ because repetition is allowed? – templatetypedef May 15 '23 at 05:38
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    @templatetypedef I originally had $n^2$ also, but $\binom{n+1}{2}$ is a more precise bound. While it's true that repetition is allowed, order doesn't matter -- so it's $n + \binom{n}{2} = \binom{n+1}{2}$ possible pairs that can be chosen. – Caleb Stanford May 15 '23 at 16:40