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Source. Christian Elsholtz, The inverse Goldbach problem, Mathematika 48 (2001), 151-158, read in the author's version identified on the source card; labels and pages are that version's (pp. 1-8). The Theorem is unnumbered; it begins on p. 1 and its last sentence is on p. 2.

Statement

Setting (p. 1). $\mathcal{A}$ and $\mathcal{B}$ are sets of positive integers, $\mathcal{A}+\mathcal{B}=\{a+b:a\in\mathcal{A},b\in\mathcal{B}\}$, $\mathcal{P}$ is the set of primes, $\mathcal{P}'$ is a set of positive integers differing from $\mathcal{P}$ in only finitely many elements, and $A(x)=\sum_{a\in\mathcal{A},a\le x}1$, with $B(x)$ defined in the same way.

The paper states (pp. 1-2): "Suppose that there exist sets $\mathcal{P}',\mathcal{A},\mathcal{B}$ with $\mathcal{P}'=\mathcal{A}+\mathcal{B}$, where $|\mathcal{A}|,|\mathcal{B}|\geq 2$ and $\mathcal{P}'$ coincides with the set of primes for elements $p>x_0$. For sufficiently large $x\geq x_1$ the following bounds hold:

$$\frac{x^{1/2}}{(\log x)^5}\ll A(x)\ll x^{1/2}(\log x)^4.$$

The same bounds hold for $B(x)$."

In words: any such decomposition has both counting functions of size $x^{1/2}$ up to a bounded power of $\log x$. The theorem is conditional on the decomposition existing; it does not say whether one exists, and the paper calls Ostmann's question whether one exists still open (p. 1). The paper compares the bounds with earlier ones of Hornfeck, of Hofmann and Wolke, and its author's own, and with Wirsing's $A(x)B(x)=O(x)$ (p. 2).

Read depth. Claims checked: the setting and the statement were read clause by clause on the page images of the author's version. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Section 2 (pp. 2-7), proof of the Theorem pp. 3-7. The proof combines Montgomery's sieve (Lemma 2, p. 3, with Vaughan's lower bound for its denominator, Lemma 3, p. 3) applied to $\mathcal{A}$ with Gallagher's larger sieve (Lemma 4, p. 3) applied to $\mathcal{B}$: a residue class modulo $p$ met by $\mathcal{B}$ is forbidden for $\mathcal{A}$, and a class missed by $\mathcal{B}$ can be sifted from $\mathcal{B}$, so every class is used by one of the two sieves. A first round (Iteration A, pp. 4-6, with Lemmas 5-8, p. 5) proves the weaker Proposition (p. 4), $x^{1/2-\varepsilon}\ll A(x)\ll x^{1/2+\varepsilon}$ and the same for $B(x)$, for every $\varepsilon>0$ and $x\ge x_2$; a second round (Iteration B, pp. 6-7, with $m=2$, $y=x^{1/4}$, $c=20$) gives the upper bound $x^{1/2}(\log x)^4$, and the lower bound follows from $A(x)B(x)\gg\pi(x)$ and the symmetry between $\mathcal{A}$ and $\mathcal{B}$ (p. 7).

Dependencies

Montgomery's sieve and Vaughan's evaluation of it, Gallagher's larger sieve, the prime number theorem, and Hornfeck's earlier lower bound for $B(x)$ (p. 5), all of which the paper cites.

Bears on

• Problem 431: two infinite sets have at least two elements each, so the theorem applies to any pair of sets of positive integers the problem asks for, and shows that for such a pair $A(x)$ and $B(x)$ would both lie between $x^{1/2}(\log x)^{-5}$ and $x^{1/2}(\log x)^4$ up to constants for large $x$. It proves neither that such a pair exists nor that none does.