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Statement

Notation (pp. 410-411). Two infinite sequences $A$ and $B$ of nonnegative integers are additive complements, and $B$ is an additive complement of $A$, if every sufficiently large integer is $a+b$ with $a\in A$ and $b\in B$. $S=\{1^2,2^2,\ldots\}$, so the square $0$ is not in $S$; $R_{S,B}(n)$ is the number of solutions of $n=a+b$ with $a\in S$ and $b\in B$, and $B(x)$ is the number of terms of $B$ that are at most $x$.

Theorem 1.1 (p. 412). For any additive complement $B=\{b_n\}_{n=1}^\infty$ of $S$,

$$\sum_{n=1}^{N}R_{S,B}(n)-N\ \ge\ c\,B(2\sqrt N)\log B(2\sqrt N)$$

for all sufficiently large integers $N$, where $c$ is a positive constant. In particular, for any additive complement $B$ of $S$, $\sum_{n=1}^{N}R_{S,B}(n)-N\to+\infty$ as $N\to+\infty$.

The statement does not say whether $c$ may depend on $B$; the proof on p. 417 obtains the inequality with $c=1/(2\log4)$ for every $B$, the range of $N$ depending on $B$.

Remark 1 and the example after it (pp. 412-413). The paper calls Theorem 1.1 a special case of Theorem 2.1, and shows that the analogue fails for general additive complements: with $A_0$ the finite sums of distinct $2^{2i}$ and $A_1$ the finite sums of distinct $2^{2i+1}$ ($i\ge0$), unique binary expansion makes them additive complements with $\sum_{n=1}^{N}R_{A_0,A_1}(n)-N=0$ for every nonnegative integer $N$.

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: the notation on pp. 410-411, Theorem 1.1 and Remark 1 on p. 412, the example on pp. 412-413, the proof of Theorem 1.1 on p. 417. The edition read is identified on the source card.

Read depth. Claims checked: the statement, the remark and the example were read clause by clause on the printed pages. The proof (p. 417) was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Page 417. An additive complement $B$ of $S$ is infinite and has $R_{S,B}(n)\ge1$ for all $n\ge n_0$. Theorem 2.1 applied to $D=B$ bounds the surplus over the represented $n\le N$ from below by $(1+o(1))B(2\sqrt N)\log B(2\sqrt N)/\log4$; the finitely many $n<n_0$ change $\sum_{n\le N}R_{S,B}(n)-N$ by a bounded amount, which leaves $B(2\sqrt N)\log B(2\sqrt N)/(2\log 4)$ for all large $N$.

Dependencies

Theorem 2.1 of the same paper.

Bears on

• Problem 33: every additive complement of $S$ is a set $A$ as in Problem 33, which allows $n\ge0$; the converse need not hold. The theorem says such a complement always has unboundedly many surplus representations. It bounds no counting constant and leaves both questions of Problem 33 where they stood.