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Statement

Notation (p. 411). $\mathbb N$ is the set of nonnegative integers and $S=\{1^2,2^2,\ldots\}$; the square $0$ is not in $S$. For a subsequence or subset $T$ of $\mathbb N$, $T(x)$ is the number of terms of $T$ that are at most $x$. For sequences $A$ and $B$, $R_{A,B}(n)$ is the number of solutions of $n=a+b$ with $a\in A$ and $b\in B$.

Theorem 2.1 (p. 414). Let $D=\{d_n\}_{n=1}^\infty$ be any infinite sequence of nonnegative integers. Then, for all sufficiently large $X$,

$$\sum_{\substack{n\le X\\ R_{S,D}(n)\ge1}}\bigl(R_{S,D}(n)-1\bigr) \ \ge\ \frac{1+o(1)}{\log 4}\,D(2\sqrt X)\log D(2\sqrt X).$$

No covering hypothesis is made on $D$: the left side counts, over the integers $n\le X$ that are represented at least once as a square in $S$ plus a term of $D$, the representations beyond the first.

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 p. 411, Lemma 2.1 on p. 413, Theorem 2.1 on p. 414 with its proof on pp. 414-417. The edition read is identified on the source card.

Read depth. Claims checked: the statement and its notation were read clause by clause on the printed pages. The proof (pp. 414-417) was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pages 413-417. Lemma 2.1 (p. 413): for any integer $M>1$ and any integer $n\ge1$, $x^2-y^2=2^{2M+1}n$ has at least $M$ positive integral solutions with $x<2^{2M}n$, written down explicitly. The proof chooses $M$ by (2.2) so that $M=(1+o(1))\log D(2\sqrt X)/\log 4$ (2.3), sets $K=2^{2M+1}$, and splits $D$ into its residue classes $D_i$ modulo $K$. Within a class, each term $d_l\le2\sqrt X$ above the least term $k_{i,0}$ differs from it by a multiple of $K$, so by the lemma it yields at least $M$ integers $n\le X$ represented both through $k_{i,0}$ and through $d_l$; this gives a contribution of at least $M(D_i(2\sqrt X)-1)$ per class (2.5), and summing over the $K$ classes gives $M(D(2\sqrt X)-K)$, which (2.2) turns into the stated bound.

Dependencies

Lemma 2.1 of the same paper (p. 413), an elementary factorization of $x^2-y^2$.

Bears on

• Problem 33: the theorem bounds the number of surplus representations, not the size of a complement, so on its own it gives no bound on either quantity Problem 33 asks about. It is the input to Theorem 1.1 and to the contradiction in Case 1 of the proof of Theorem 1.2.