Ostmann’s inverse Goldbach conjecture

The explanation

Take two small sets of whole numbers, A = {0, 2} and B = {3, 5, 11, 17, 29, 41}. Add every number in A to every number in B. You get 3, 5, 7, 11, 13, 17, 19, 29, 31, 41 and 43. Every one of those sums is prime. Yet the list skips three primes below 43: 2, 23 and 37. Could some cleverer pair of sets produce the primes, and nothing else?

The question is the inverse of Goldbach's. Goldbach asked which numbers are sums of primes. Here we ask whether the primes themselves are a set of sums. Some care is needed to make it fair. With A = {0} and B equal to all the primes, the answer is trivially yes. So both sets must hold at least two numbers. The question also forgives finitely many mistakes. A few missing primes, or a few stray non-primes, are allowed. Only the long run counts.

The missing prime 23 shows why this is hard to arrange. To make 23 with A = {0, 2}, B would need 21 or 23. But 21 + 0 = 21 is not prime, and 23 + 2 = 25 is not prime. Adding either one breaks the rule.

Hans-Heinrich Ostmann conjectured in 1956 that no such pair of sets exists. This is also called the inverse Goldbach problem. In 1964, Laffer and Mann showed that a pair would need both sets to be infinite. Elsholtz later ruled out splitting the primes into three summed sets. Elsholtz and Harper showed that each set would have roughly √x members up to x, apart from small factors.

Counting alone cannot settle it. Below one million there are 78,498 primes. Two sets with 1,000 members each produce up to a million sums. That is plenty of room. So the obstacle has to come from somewhere other than size.

The real pressure comes from remainders. Divide by 3, for example. Suppose a large number in A leaves remainder 1 and a large number in B leaves remainder 2. Then their sum is a multiple of 3, so it is not prime. The same holds for every prime divisor at once. Green and Harper showed in 2014 that an unproved conjecture about such sets would imply Ostmann's. That conjecture says these sets must look like the values of a quadratic formula. Later work found large overlaps with such formulas, but not enough to finish.

This paper takes a direct route. It uses the remainder restrictions at all primes together with the demand that every large prime appearSource 1. It does not need the unproved conjecture, and it does not prove it eitherSource 2. The paper spells out one consequence. Take two sets that each hold at least two numbers. If their sums include every large enough prime, they must also include infinitely many non-primesSource 3.

So no pair like our example can ever work, however clever. Forgiving finitely many mistakes does not help. Any two sets, each with at least two numbers, miss infinitely many primes or hit infinitely many non-primesSource 3.

How sure is this? A computer proof checker called Lean has a formal version of the full statement listed. It covers the main conjecture and the case where both sets are infinite. No specialist review of the paper has been published.

What it doesn't show. This does not prove the inverse-sieve conjecture of Green and Harper. It shows only that the primes are not a sum of two such sets.