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Statement

Notation (pp. 410-411). $S=\{1^2,2^2,\ldots\}$, so the square $0$ is not in $S$, and $B$ is an additive complement of $S$ if every sufficiently large integer is $a+b$ with $a\in S$ and $b\in B$.

Theorem 1.2 (p. 413). Let $\alpha$ and $\beta$ be any positive constants with

$$0<\alpha<\sqrt{\frac{2}{\pi}}\,\frac{1}{\log4}=0.5755\cdots.$$

If $B=\{b_n\}_{n=1}^\infty$ satisfies

$$b_n\ \ge\ \frac{\pi^2}{16}n^2-\alpha n^{1/2}\log n-\beta n^{1/2}, \qquad n=1,2,\ldots,$$

then $B$ is not an additive complement of $S$.

The inequality is required for every $n\ge1$; the constant $\beta$ absorbs any finite initial segment, which is how Corollary 1.1 is deduced. The context is a question Ben Green put to the second author (p. 412): whether some additive complement $B$ of $S$ has $b_n=\frac{\pi^2}{16}n^2+o(n^2)$, displayed as (1.1). Green observes there that (1.1) gives $B(N)=\frac4\pi\sqrt N+o(\sqrt N)$ and $\lim_{N\to\infty}\frac1N\sum_{n=1}^{N}R_{S,B}(n)=1$. Theorem 1.2 excludes only lower-order deviations of size $\alpha n^{1/2}\log n+\beta n^{1/2}$ and does not answer Green's question.

Source. Yong-Gao Chen and Jin-Hui Fang, Additive complements of the squares, J. Number Theory 180 (2017), 410-422, doi:10.1016/j.jnt.2017.04.016: Green's question on p. 412, Theorem 1.2 on p. 413, its proof on pp. 417-421. The edition read is identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the printed page. The proof (pp. 417-421) was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pages 417-421. Suppose $B$ is a complement. Choose $\alpha_1$ with $\pi/4<\sqrt{\alpha_1}<\frac1\alpha\sqrt{\pi/8}\,\frac1{\log4}$ (3.1), possible by the bound on $\alpha$, and assume $b_n$ nondecreasing from some point on. Case 1, $b_n<\alpha_1n^2$ infinitely often: then $B(2\sqrt N)$ is at least of order $N^{1/4}$ along infinitely many $N$, and bounding $\sum_{n\le N}R_{S,B}(n)$ by $\sum_{b<N}\sqrt{N-b}$, comparing with an integral and using the hypothesis, the surplus is at most $\frac{\delta}{\log4}B(2\sqrt N)\log B(2\sqrt N)$ for some $\delta<1$, against Theorem 2.1. Case 2, $b_n\ge\alpha_1n^2$ for all large $n$: the same integral comparison gives $\sum_{n\le N}R_{S,B}(n)\le\frac{\pi}{4\sqrt{\alpha_1}}N+n_1\sqrt N$, whose leading coefficient is below $1$ by (3.1), so the complement property fails.

Dependencies

Theorem 2.1 of the same paper.

Bears on

• Problem 33: the problem asks for the smallest limsup, and whether the liminf exceeds $1$, of $\lvert A\cap\{1,\ldots,N\}\rvert/N^{1/2}$ over sets $A$ with every large integer $n^2+a$, $n\ge0$; every additive complement of $S$ is such a set, and the converse need not hold. By Green's observation the profile $b_n\approx\frac{\pi^2}{16}n^2$ is that of a set with counting function $\frac4\pi\sqrt N+o(\sqrt N)$, the known lower bound for both quantities. Theorem 1.2 says a complement of $S$ cannot lie above that profile up to the stated error; it does not raise the lower bound $4/\pi$ for either quantity and does not determine the smallest limsup.