On Problem 477 of Erdős and Graham: The Cubic Case

1. Introduction

1.1. The Problem and the Main Theorem

For subsets $A,B\subseteq \mathbb{Z}$, write $\mathbb{Z}=A\oplus B$ if every integer has exactly one representation $a+b$ with $(a,b)\in A\times B$. Erdos and Graham asked whether $f\left(\mathbb{Z}\right)$ can be a direct additive factor of $\mathbb{Z}$ for some integer-valued polynomial f of degree at least two; see [3,4]. For background on factorizations of the infinite cyclic group, see also Sekanina [8]. The quadratic case is impossible. Consequently, the cubic result below gives the least possible degree in that formulation.

The word unique is essential. Ordinary additive-complement results ask only for coverage, while a direct factorization requires both existence and uniqueness of every representation. Results on ordinary complements of positive powers, such as [2], therefore address a different problem. There are also recent non-refereed claims for higher powers, including a public manuscript treating thirteenth powers and an online claim covering degrees at least five; see [7,9]. They do not contain the cubic statement established here.

Our main theorem is as follows.

Theorem 1.

Let

$$C=\{m^3:m\in \mathbb{Z}\}.$$

There exists $A\subseteq \mathbb{Z}$ such that

$$\mathbb{Z}=A\oplus C.$$

The proof gives more than the qualitative conclusion. Put $D=C-C$ and, for a fixed noncube k, count the cube indices n for which $n^3-k$ belongs to D. The power saving in the following theorem is the analytic input needed by the final combinatorial construction.

Theorem 2.

Fix $k\in \mathbb{Z}\setminus C$. For $N\ge 1$, let

$$E_k\left(N\right)=\{n\in [N,2N)\cap \mathbb{Z}:n^3-k\in D\}.$$

For every $\delta >0$ there are constants $C_{k,\delta }>0$ and $N_{k,\delta }$ such that, for all $N\ge N_{k,\delta }$,

$$\#E_k\left(N\right)\le C_{k,\delta }{N}^{18/19+\delta }.$$

Theorem 2 is stated for distinct values of n, not for representations. Throughout the proof we are free to overcount by the auxiliary gap or by an integer triple: projecting any such collection to its n-coordinate can only decrease its cardinality.

1.2. Context and Scope

The result has two logically separate certificates. The proof printed here uses two published analytic estimates: Heath–Brown’s square sieve [5], together with the mixed character-sum estimate of Bonolis–Browning [1], for small gaps; and Heath–Brown’s fixed-level diagonal-form estimate [6] for large gaps. Their hypotheses and the substitutions used here are stated before either application.

The companion formal development follows a longer route because it does not postulate the fixed-level diagonal-form theorem as an axiom. It reconstructs the needed determinant method, common conics, binary sextics, height estimates, and factor counts in Lean. That development proves the exact statement of Theorem 1 and the independent quantitative bound $O_k\left(N^{99/100}\right)$. Machine verification establishes correctness relative to the Lean kernel and the stated foundations; it does not establish novelty, publication priority, or independent human peer review.

1.3. Proof Strategy

We now describe the proof in enough detail to display every change of scale.

1.

A finite simultaneous-escape property for a symmetric set $S\subseteq \mathbb{Z}$ is converted into a direct factorization $\mathbb{Z}=A\oplus S$ by a two-sided greedy construction.

2.

For $S=C$, escape is reduced to showing that the bad indices $E_k\left(N\right)$ have density zero for every fixed noncube k.

3.

A representation $n^3-k=y^3-x^3$ is parametrized by the positive gap $a=y-x$ and the sum $v=x+y$. Below $G_N$, the resulting square condition is estimated by the square sieve after completing the relevant Jacobi-symbol correlations. With $G_N=N^{17/19}$ this contributes $O_k\left(N^{18/19}{\log }^{4}N\right)$.

4.

Above $G_N$, the gap identity gives $\left|x\right|,\left|y\right|{\ll }_k{N}^{20/19}$. Thus every remaining representation gives a point of height $B{\ll }_k{N}^{20/19}$ on the fixed-level diagonal cubic $x^3+n^3-y^3=k$.

5.

Heath–Brown’s theorem counts these points by $O_{k,\epsilon }\left(B^{9/10+\epsilon }\right)$. Its exceptional solutions would force k to be a cube, so none occur. Since $(20/19)(9/10)=18/19$, the large-gap count is $O_{k,\epsilon }\left(N^{18/19+(20/19)\epsilon }\right)$.

6.

Combining the two ranges proves Theorem 2; finite simultaneous escape and the greedy construction then prove Theorem 1. A squarefree-kernel reorganization is included as an arithmetic cross-check of the small-gap argument.

All asymptotic statements below are for sufficiently large N, with the threshold allowed to depend on the fixed parameters displayed in the subscript. The finitely many smaller values of N are absorbed into the implied constants.

1.4. Organization and Disclosure

Section 2 fixes the conventions and the three external estimates. Section 3 proves the finite-avoidance reduction and the greedy construction. Section 4 derives the gap algebra and the root-counting estimates. Section 5 proves the small-gap bound, including all local and zero-value branches. Section 6 proves the large-gap bound and checks every hypothesis of the fixed-level theorem. Section 7 completes the density and complement arguments. The appendices give an alternative squarefree-kernel organization, an exact source-notation dictionary, and the statement and audit boundary of the companion formal certificate.

Generative models were used to explore proof strategies, translate finite algebra into Lean, and search for counterexamples to intermediate claims. Model agreement was never treated as verification. The authors remain responsible for the mathematical statements; Appendix D records the role and limits of the computational assistance.

2. Preliminaries and External Inputs

All intervals of integers are half open. The symbols $O_{k,\epsilon }(·)$ and ${\ll }_{k,\epsilon }$ allow constants depending only on the displayed subscripts. For an odd prime p, write

$${\chi }_p\left(u\right)=\left({\displaystyle \frac{u}{p}}\right),{\mathrm{e}}_p\left(u\right)=\exp (2\pi iu/p).$$

If $a\ge 1$, then $\omega \left(a\right)$ denotes its number of distinct prime factors and $d_3\left(a\right)$ its ternary divisor function.

The following three results are the only non-elementary analytic inputs in the source-level proof. We state the precise consequences used later so that the range, normalization, and exceptional sets can be checked locally.

Lemma 1

(Heath–Brown square sieve). Let $w\left(m\right)\ge 0$ be finitely supported, let $w\left(0\right)=0$, and let $\mathcal{P}$ be a set of R odd primes. Suppose $w\left(m\right)=0$ whenever $\left|m\right|\ge \exp \left(R\right)$. Then

$$\sum _{j\ge 1}w\left(j^2\right)\ll {\displaystyle \frac{1}{R}}\sum _{m}w\left(m\right)+{\displaystyle \frac{1}{R^2}}\sum _{\begin{array}{c}p,q\in \mathcal{P}\\ p\ne q\end{array}}\left|\sum _{m}w\left(m\right)\left({\displaystyle \frac{m}{pq}}\right)\right|.$$

Proof. 

This is Theorem 1 of [5], with R denoting the number of sieve primes. The support and zero-value hypotheses are verified explicitly in Section 5.    □

Lemma 2

(Mixed cubic character sum). There is an absolute constant C such that, for every odd prime p, every $b\in {\mathbb{F}}_p^{\times }$, and every $c\in {\mathbb{F}}_p$,

$$\left|\sum _{u\in {\mathbb{F}}_p}{\chi }_p(u^3+b){\mathrm{e}}_p\left(cu\right)\right|\le C\sqrt{p}.$$

If $b=0$, the same bound holds by the classical quadratic Gauss-sum formula.

Proof. 

For $b\ne 0$ this is the degree-three specialization of [1]. When $b=0$, one has ${\chi }_p\left(u^3\right)={\chi }_p\left(u\right)$; the zero frequency vanishes and every nonzero frequency is a quadratic Gauss sum of absolute value $\sqrt{p}$.    □

Theorem 3

(Heath–Brown’s diagonal fixed-level estimate). Let $d\ge 3$, let

$$F(X_1,X_2,X_3)=\pm X_1^d\pm X_2^d\pm X_3^d$$

be nonsingular, and let R be a positive integer satisfying $R{\ll }_F{B}^{3/13}$. For every $\epsilon >0$, the number of integer points $\mathbf{X}$ in the height shell

$$F\left(\mathbf{X}\right)=R,B/2<\underset{i}{\max }\left|X_i\right|\le B,$$

apart from the solutions for which one of the three displayed signed monomials already equals R, is

$$O_{F,\epsilon }\left(B^{9/10+\epsilon }{R}^{1/10}\right).$$

Proof. 

This is the final diagonal conclusion of [6]. That conclusion counts all integral points in the stated shell outside the special set; it does not impose a primitivity condition. Section 6 checks nonsingularity, the level range, the height substitution, the shell summation, and the absence of special points in the present application.    □

3. Finite Avoidance and the Greedy Construction

We first isolate the purely combinatorial part of the proof.

Definition 1.

A set $S\subseteq \mathbb{Z}$ has the finite simultaneous escape property if, for every finite set $K\subseteq \mathbb{Z}\setminus S$, there is $s\in S$ such that

$$k-s\notin S-S(k\in K).$$

Proposition 1.

If $S=-S$ and S has the finite simultaneous escape property, then there is $A\subseteq \mathbb{Z}$ such that $\mathbb{Z}=A\oplus S$.

Proof. 

Choose a sequence $z_0,z_1,\dots$ containing every integer. We construct finite sets $B_j$ such that distinct elements of $B_j$ have difference outside $S-S$. Start with $B_0=⌀$. If $z_j\in B_j+S$, put $B_{j+1}=B_j$. Otherwise, for $b\in B_j$ put $k_b=z_j-b$. Every $k_b$ lies outside S, since otherwise $z_j\in B_j+S$. By finite simultaneous escape, choose $s\in S$ such that $k_b-s\notin S-S$ for every $b\in B_j$, and put

$$b_{\mathrm{new}}=z_j-s,B_{j+1}=B_j\cup \left\{b_{\mathrm{new}}\right\}.$$

Then $z_j=b_{\mathrm{new}}+s$, and $b_{\mathrm{new}}-b=k_b-s\notin S-S$. Since $S-S$ is symmetric, the reverse difference is safe as well.

Let $A={\bigcup }_j{B}_j$. Every integer is covered. If $a+s=b+t$ with $a,b\in A$ and $s,t\in S$, then $a-b=t-s\in S-S$. The construction forces $a=b$, and then $s=t$. Thus the representation is unique.    □

For cubes, $C=-C$ and $D=-D$. The density theorem immediately supplies the escape hypothesis. 2

Corollary 1.

The set C has the finite simultaneous escape property.

Proof. 

Let $K\subseteq \mathbb{Z}\setminus C$ be finite. By 2, choose any $0<\delta <1/19$. Then

$$\#\bigcup _{k\in K}{E}_k\left(N\right)\le \sum _{k\in K}\#E_k\left(N\right)=O_{K,\delta }\left(N^{18/19+\delta }\right)=o_K\left(N\right).$$

For all sufficiently large N, some $n\in [N,2N)$ lies in none of these sets. Hence $n^3-k\notin D$ for every $k\in K$. Since $D=-D$, this is equivalent to $k-n^3\notin D$. Take $s=n^3$.    □

Theorem 1 follows from Proposition 1 and Corollary 1. It remains to prove 2.

We also include the short quadratic obstruction, in order to make the least-degree assertion independent of the formulation on the problem page. 3

Proposition 2.

If $f\in \mathbb{Q}\left[X\right]$ is integer-valued on $\mathbb{Z}$ and has degree two, then $f\left(\mathbb{Z}\right)$ has no unique additive complement in $\mathbb{Z}$.

Proof. 

There are integers $A,B,C$, with $A\ne 0$, such that

$$f\left(n\right)=A\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)+Bn+C.$$

If $2B-A\ne 0$, then

$$f\left(n\right)-f(-n)=(2B-A)n,$$

so $q\mathbb{Z}\subseteq f\left(\mathbb{Z}\right)-f\left(\mathbb{Z}\right)$ for $q=|2B-A|$. If $2B-A=0$, then

$$f(n+1)-f(n-1)=2An,$$

and the same conclusion holds with $q=2\left|A\right|$.

In either case, suppose for contradiction that $\mathbb{Z}=T\oplus f\left(\mathbb{Z}\right)$. If two distinct elements of T were congruent modulo q, their difference would belong to $q\mathbb{Z}\subseteq f\left(\mathbb{Z}\right)-f\left(\mathbb{Z}\right)$, producing two representations of one integer. Thus T has at most q elements. But a finite translate set added to the one-sided bounded quadratic set $f\left(\mathbb{Z}\right)$ is still bounded on one side, and cannot equal $\mathbb{Z}$.    □

Combining Theorem 1 and Proposition 2, degree three is the least possible degree in the Erdos–Graham problem.

4. Gap Algebra

Fix a nonzero noncube k. For sufficiently large N, every $n\in [N,2N)$ satisfies $n^3-k>0$. Suppose

$$n^3-k=y^3-x^3.$$

Then $y>x$. Put

$$a=y-x>0,v=x+y.$$

A direct calculation gives

$$4(n^3-k)=a(a^2+3v^2)$$

(1)

and

$${\displaystyle \frac{12(n^3-k)}{a}}-3a^2={\left(3v\right)}^2.$$

(2)

In particular, $a\mid n^3-k$ and

$$a^3\le 4(n^3-k){\ll }_k{N}^3,a{\ll }_{k}N.$$

(3)

We count gaps in dyadic intervals $H\le a<2H$. Let $Z_k(N,H)$ be the number of distinct $n\in [N,2N)$ admitting such a representation. We shall prove the following uniform estimate. 0

Proposition 3

(Small gaps). For fixed $k\ne 0$ and $\eta >0$,

$$Z_k(N,H){\ll }_{k,\eta }\sqrt{NH}{\log }^3\left(2N\right)$$

uniformly for $1\le H\le N^{1-\eta }$.

5. The Small-Gap Square Sieve

5.1. Cubic Roots Modulo the Gap

Define

$${\rho }_k\left(a\right)=\#\{r(moda):r^3\equiv k(moda)\}.$$

0

Lemma 3.

For fixed $k\ne 0$,

$${\rho }_k\left(a\right)\le {\left|k\right|}^2{3}^{\omega \left(a\right)}.$$

Consequently,

$$\begin{array}{cc}\hfill \sum _{H\le a<2H}{\rho }_k\left(a\right)& {\ll }_{k}H{\log }^2\left(2H\right),\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill \sum _{H\le a<2H}{\displaystyle \frac{{\rho }_k\left(a\right)}{a}}& {\ll }_k{\log }^2\left(2H\right).\hfill \end{array}$$

(5)

Proof. 

It is enough to work modulo prime powers. Fix p, put $\nu =v_p\left(k\right)$, and consider $p^e$. If $e\le \nu$, then the congruence is $r^3\equiv 0(mod{p}^e)$, and hence

$${\rho }_k\left(p^e\right)=p^{e-\lceil e/3\rceil }=p^{\lfloor 2e/3\rfloor }.$$

Suppose $e>\nu$. A root cannot vanish modulo $p^e$; writing $r=p^{s}u$ with u a unit, comparison of valuations below e gives $3s=\nu$. There are no roots if $3\nmid \nu$. If $\nu =3s$, division by $p^{\nu }$ gives a unit equation

$$u^3\equiv k/p^{\nu }(mod{p}^{e-\nu }).$$

The kernel of cubing on ${(\mathbb{Z}/p^m\mathbb{Z})}^{\times }$ has size at most three, also for $p=2,3$. Each solution modulo $p^{e-\nu }$ has exactly $p^{2\nu /3}$ lifts for u modulo $p^{e-s}$. Therefore

$${\rho }_k\left(p^e\right)\le 3p^{2\nu /3}.$$

Multiplicativity first gives the sharper real-valued estimate ${\rho }_k\left(a\right)\le {\left|k\right|}^{2/3}{3}^{\omega \left(a\right)}$; since $\left|k\right|\ge 1$, it implies the displayed integral bound ${\left|k\right|}^2{3}^{\omega \left(a\right)}$ used by the formal proof. Finally, $3^{\omega \left(a\right)}\le d_3\left(a\right)$, and the elementary mean-value estimate ${\sum }_{a\le X}{d}_3\left(a\right)\ll X{\log }^2\left(2X\right)$ gives (4). The weighted estimate follows immediately on a dyadic interval.    □

5.2. Progressions and Local Fourier Coefficients

For every root class choose $0\le r<a$ and write $n=r+at$. Put

$$q_0={\displaystyle \frac{r^3-k}{a}}\in \mathbb{Z}$$

and

$$f_{a,r}\left(t\right)=12(q_0+3r^{2}t+3ra{t}^2+a^2{t}^3)-3a^2.$$

(6)

Then $f_{a,r}\left(t\right)$ is the left side of (2). Let $I_{a,r}$ be the integer interval defined by $N\le r+at<2N$. It has length at most $N/a+1$. The total number of candidates before imposing the square condition satisfies

$$W\le \sum _{H\le a<2H}{\rho }_k\left(a\right)(N/a+1){\ll }_{k}N{\log }^2\left(2N\right),$$

(7)

where we used $H\le N$.

For a prime $p\nmid 6k$, write ${\chi }_p$ for the quadratic character and define

$$S_p\left(h\right)=\sum _{tmodp}{\chi }_p\left(f_{a,r}\left(t\right)\right){\mathrm{e}}_p\left(ht\right).$$

Lemma 4.

Uniformly in $a,r,h$ and $p\nmid 6k$,

$$|S_p\left(h\right)|\ll \sqrt{p}.$$

Proof. 

First suppose $p\nmid a$. The exact integer identity

$$a{f}_{a,r}\left(t\right)=12{(r+at)}^3-3(a^3+4k)$$

gives, modulo p, after $u=r+at$,

$$\begin{array}{cc}\hfill S_p\left(h\right)& ={\chi }_p\left(12a^{-1}\right){\mathrm{e}}_p(-hr{a}^{-1})\hfill \\ \hfill & \times \sum _{umodp}{\chi }_p(u^3-k-a^3/4){\mathrm{e}}_p\left(h{a}^{-1}u\right).\hfill \end{array}$$

The result follows from 2. If $a^3+4k\equiv 0(modp)$, the polynomial is a scalar multiple of $u^3$; equivalently, this special case reduces directly to the quadratic Gauss sum for ${\chi }_p\left(u\right)$.

Now suppose $p\mid a$. Since $r^3\equiv k(modp)$ and $p\nmid k$, we have $r\neg \equiv 0(modp)$. Moreover,

$$f_{a,r}\left(t\right)\equiv 12q_0+36r^{2}t(modp),$$

whose linear coefficient is nonzero. For a nonconstant linear polynomial $\alpha t+\beta$,

$$\sum _{tmodp}{\chi }_p(\alpha t+\beta ){\mathrm{e}}_p\left(ht\right)=\left\{\begin{array}{cc}0,\hfill & h=0,\hfill \\ {\mathrm{e}}_p(-h\beta {\alpha }^{-1}){\chi }_p\left(h{\alpha }^{-1}\right)\tau \left({\chi }_p\right),\hfill & h\ne 0.\hfill \end{array}\right.$$

This again has absolute value at most $\sqrt{p}$.    □

5.3. CRT and interval completion

Take distinct primes $p,q\in (Q,4Q]$, both coprime to $6k$, and put $M=pq$. Set

$$g\left(t\right)={\chi }_p\left(f_{a,r}\left(t\right)\right){\chi }_q\left(f_{a,r}\left(t\right)\right),{\widehat{g}}_M\left(h\right)=\sum _{bmodM}g\left(b\right){\mathrm{e}}_M\left(hb\right).$$

Let $q\overline{q}\equiv 1(modp)$ and $p\overline{p}\equiv 1(modq)$. The CRT gives the exact factorization

$${\widehat{g}}_M\left(h\right)=S_p(h\overline{q}modp)S_q(h\overline{p}modq).$$

(8)

Thus $|{\widehat{g}}_M\left(h\right)|\ll \sqrt{pq}\ll Q$ for every h. This includes all four possibilities for whether p and q divide a.

Lemma 5

(Interval completion). For every integer interval I of length L,

$$\sum _{t\in I}g\left(t\right)\ll {\displaystyle \frac{L}{Q}}+Q\log \left(2Q\right).$$

Proof. 

Split I into complete periods modulo M and one remainder. Each complete period contributes ${\widehat{g}}_M\left(0\right)=O\left(Q\right)$, giving $O(L/Q)$. On the remainder use

$$g\left(t\right)={\displaystyle \frac{1}{M}}\sum _{hmodM}{\widehat{g}}_M\left(h\right){\mathrm{e}}_M(-ht)$$

and the standard bound

$$\sum _{hmodM}\left|\sum _{j<\ell }{\mathrm{e}}_M\left(hj\right)\right|\ll M\log \left(2M\right).$$

Together with (8), this gives $O\left(Q\log \right(2M\left)\right)$, as required.    □

Since

$$f_{a,r}^{\prime }\left(t\right)=36{(r+at)}^2>0$$

on $I_{a,r}$, the additional condition $f_{a,r}\left(t\right)>0$ cuts out at most one integer interval. Summing 5 over a and its root classes gives, for the off-diagonal correlation,

$$C_{p,q}{\ll }_k\left({\displaystyle \frac{N}{Q}}+HQ\right){\log }^3\left(2N\right).$$

(9)

Indeed, the terms are bounded by

$${\displaystyle \frac{N}{Q}}\sum _{a\asymp H}{\displaystyle \frac{{\rho }_k\left(a\right)}{a}}+\left({\displaystyle \frac{1}{Q}}+Q\log \left(2Q\right)\right)\sum _{a\asymp H}{\rho }_k\left(a\right),$$

and 3 applies.

5.4. Application of the Square Sieve

Choose

$$Q=\sqrt{N/H},{\mathcal{P}}_Q=\{p\in (Q,4Q]:p\mathrm{prime},p\nmid 6k\},R=\#{\mathcal{P}}_Q.$$

For $H\le N^{1-\eta }$, we have $Q\ge N^{\eta /2}$. No prime number theorem is needed: Chebyshev’s explicit lower bound at $4Q$ and upper bound at Q give

$$\pi \left(4Q\right)\ge {\displaystyle \frac{4Q\log 2-\log (4Q+1)}{\log \left(4Q\right)}},\pi \left(Q\right)\le (\log 4+ϵ){\displaystyle \frac{Q}{\log Q}}$$

for fixed $ϵ>0$ and all sufficiently large Q. The leading constants leave a positive margin; the matching Chebyshev upper bound and deletion of finitely many primes therefore give

$$R\asymp Q/\log Q.$$

Define $w\left(m\right)$ to count triples $(a,r,t)$ above for which $f_{a,r}\left(t\right)=m>0$. It is a nonnegative, finitely supported weight with $w\left(0\right)=0$, total mass at most W, and

$$m\le 96N^3+12\left|k\right|.$$

Since $R{\gg }_{k,\eta }{N}^{\eta /2}/\log N$, the support condition $m<\exp \left(R\right)$ in 1 holds for all sufficiently large N. Furthermore,

$$\sum _{m}w\left(m\right)\left({\displaystyle \frac{m}{pq}}\right)=C_{p,q}.$$

Hence 1, (7), and (9) yield

$$\begin{array}{cc}\hfill \sum _{j\ge 1}w\left(j^2\right)& \ll {\displaystyle \frac{W}{R}}+{\displaystyle \frac{1}{R^2}}\sum _{p\ne q}\left|C_{p,q}\right|\hfill \\ \hfill & {\ll }_{k,\eta }{\displaystyle \frac{N}{Q}}{\log }^3\left(2N\right)+\left({\displaystyle \frac{N}{Q}}+HQ\right){\log }^3\left(2N\right)\hfill \\ \hfill & {\ll }_{k,\eta }\sqrt{NH}{\log }^3\left(2N\right).\hfill \end{array}$$

It remains to restore zero values. The equation $f_{a,r}\left(t\right)=0$ implies

$$4(n^3-k)=a^3.$$

For fixed a there is at most one integer n, so all zero values contribute $O\left(H\right)\le O\left(\sqrt{NH}\right)$. Every genuine representation is counted by its actual gap, root class, and progression index. This proves 3.

6. The Large-Gap Estimate

We now count the representations whose actual gap is larger than the small-gap cutoff. This part uses Theorem 3; no determinant-method assertion is left implicit.

For $0<\eta <1$, let $L_k(N,N^{1-\eta })$ be the number of distinct $n\in [N,2N)$ for which there exist $x,y\in \mathbb{Z}$ satisfying

$$n^3-k=y^3-x^3,a=y-x>N^{1-\eta }.$$

Lemma 6

(Large-gap height reduction). For fixed k and all sufficiently large N, every triple counted by $L_k(N,N^{1-\eta })$ satisfies

$$\left|x\right|,\left|y\right|,n{\ll }_k{N}^{1+\eta /2}.$$

Proof. 

Since $n^3-k>0$, the difference $y^3-x^3$ is positive and hence $a>0$. From (1),

$$3a{v}^2\le a(a^2+3v^2)=4(n^3-k)\le 4(n^3+\left|k\right|).$$

Consequently

$$\left|v\right|\le {\left({\displaystyle \frac{4(n^3+|k\left|\right)}{3a}}\right)}^{1/2}{\ll }_k{N}^{3/2}{a}^{-1/2}{\ll }_k{N}^{1+\eta /2}.$$

Equation (3) gives $a{\ll }_{k}N$, which is also $O_k\left(N^{1+\eta /2}\right)$. Finally,

$$x={\displaystyle \frac{v-a}{2}},y={\displaystyle \frac{v+a}{2}},$$

so the same bound holds for x and y; the bound for n is immediate.    □

Proposition 4

(Large gaps). Fix $0<\eta <1$. For every $\epsilon >0$,

$$L_k(N,N^{1-\eta }){\ll }_{k,\epsilon }{N}^{(1+\eta /2)(9/10+\epsilon )}.$$

Proof. 

Choose $B=c_k{N}^{1+\eta /2}$ with $c_k$ large enough for Lemma 6. We count the associated integer triples; projecting them to their n-coordinate can only decrease the count.

Suppose first that $k>0$. Then

$$F(x,n,y)=k,F(X_1,X_2,X_3)=X_1^3+X_2^3-X_3^3.$$

The projective gradient $3(X_1^2,X_2^2,-X_3^2)$ has no common nonzero zero, so F is nonsingular. The level in Theorem 3 is the fixed positive integer $R=k$. Since $B\to \infty$, the condition $R{\ll }_F{B}^{3/13}$ holds for all sufficiently large N.

The source-special solutions would satisfy one of

$$x^3=k,n^3=k,-y^3=k.$$

Each equality makes k an integer cube, the last because $-y^3={(-y)}^3$. Thus the special set is empty.

If $k<0$, rewrite the equation as

$$y^3-x^3-n^3=-k$$

and use

$$F(X_1,X_2,X_3)=X_1^3-X_2^3-X_3^3,(X_1,X_2,X_3)=(y,x,n),R=-k>0.$$

This form is nonsingular by the same gradient calculation. A special solution makes one of $y^3,-x^3,-n^3$ equal to $-k$, again forcing k to be a cube. Hence no special solution occurs in either sign case.

Partition the box $\max \left(\right|x|,|n|,|y\left|\right)\le B$ into dyadic shells $T/2<\max \left(\right|x|,|n|,|y\left|\right)\le T$. Every nonempty shell has $T\ge N$, because the second coordinate is $n\ge N$. Hence the fixed-level condition $\left|k\right|{\ll }_F{T}^{3/13}$ holds on every nonempty shell once N is sufficiently large. Since $9/10+\epsilon >0$, Theorem 3 and the geometric sum over $T=B,B/2,B/4,\dots$ give

$$L_k(N,N^{1-\eta }){\ll }_{k,\epsilon }{B}^{9/10+\epsilon }{\left|k\right|}^{1/10}{\ll }_{k,\epsilon }{N}^{(1+\eta /2)(9/10+\epsilon )}.$$

There is no primitivity reduction, and the shell sum changes only the implied constant.    □

7. Completion of the Proof

Proof 

(Proof of Theorem 2). Set $\eta =2/19$ and $G=N^{17/19}$. Cover $1\le a\le G$ by dyadic intervals $H\le a<2H$. Each lower endpoint satisfies $H\le N^{17/19}=N^{1-\eta }$, so Proposition 3 applies. There are $O\left(\log \right(2N\left)\right)$ intervals and

$$\sqrt{NG}=N^{(1+17/19)/2}=N^{18/19}.$$

Thus the representations with $a\le G$ contribute at most

$$O_k\left(N^{18/19}{\log }^4\left(2N\right)\right).$$

For $a>G$, Proposition 4 gives

$$O_{k,\epsilon }\left(N^{(20/19)(9/10+\epsilon )}\right)=O_{k,\epsilon }\left(N^{18/19+(20/19)\epsilon }\right).$$

Given $\delta >0$, choose $\epsilon <19\delta /40$. The large-gap exponent is then at most $18/19+\delta /2$, while ${\log }^4\left(2N\right){\ll }_{\delta }{N}^{\delta /2}$ for sufficiently large N. Adding the two ranges proves

$$\#E_k\left(N\right){\ll }_{k,\delta }{N}^{18/19+\delta }.$$

The finitely many smaller values of N are absorbed into the constant.    □

Proof of Theorem 1.

The density theorem implies Corollary 1: for any finite $K\subseteq \mathbb{Z}\setminus C$, a sufficiently large dyadic interval contains an n such that $n^3-k\notin C-C$ for all $k\in K$. Since $C=-C$, this is the finite simultaneous escape property for C. Proposition 1 then gives $A\subseteq \mathbb{Z}$ with $\mathbb{Z}=A\oplus C$.    □

Appendix B.1. The Square Sieve

In Theorem 1 of [5], the letter P denotes the number of sieve primes. In this paper that quantity is R, while Q denotes their size. The full translation is

Heath-Brown (1984)	Present paper
prime set $\mathcal{P}$	${\mathcal{P}}_Q=\{p\in (Q,4Q]:p\nmid 6k\}$
cardinality P	$R=\#{\mathcal{P}}_Q\asymp Q/\log Q$
weight $w\left(n\right)\ge 0$	multiplicity of positive progression values
$w\left(0\right)=0$	zero values are separated before the sieve
$\left|n\right|<\exp \left(P\right)$	$m\le 96N^3+12\left|k\right|<\exp \left(R\right)$
square count	${\sum }_{j\ge 1}w\left(j^2\right)$
Jacobi correlation	the completed sum $C_{p,q}$

The condition involving $\exp \left(P\right)$ is the reason the small-gap range is kept a fixed power below N: it makes $R\gg N^{\eta /2}/\log N$.

Appendix B.2. The Mixed Character Sum

In the branch $p\nmid a$, the local factor entering (8) is reduced by $u=r+at$ to

$$\sum _{umodp}{\chi }_p(u^3+b){\mathrm{e}}_p\left(cu\right),b=-k-a^3/4,c=h{a}^{-1}.$$

Thus Lemma 2.3 of [1] is used with odd degree 3, linear coefficient 0, constant coefficient b, and additive frequency c. The factor

$${\chi }_p\left(12a^{-1}\right){\mathrm{e}}_p(-hr{a}^{-1})$$

has modulus one. When $b=0$, the cubic is a scalar multiple of $u^3$ and the estimate is the elementary quadratic Gauss bound. The branch $p\mid a$ does not use the cited cubic lemma; it is the nonconstant-linear calculation displayed in the proof of Lemma 4.

Appendix B.3. The Fixed-Level Ternary-Form Estimate

The variables in Theorem 1 of [6] are matched as follows.

Heath-Brown (2009)	Present paper
degree	$d=3$
level N in the source	$R=\left|k\right|$ (fixed)
height	$B{\ll }_k{N}^{1+\eta /2}$, where N is our dyadic scale
range	fixed $R{\ll }_F{B}^{3/13}$ for sufficiently large N
diagonal form	$X_1^3+X_2^3-X_3^3$ or $X_1^3-X_2^3-X_3^3$
special solutions	one signed cubic monomial already equals R
height-shell estimate	$B^{9/10+\epsilon }{R}^{1/10}$

The source uses N for the level, whereas this paper uses N for the size of the cube index. The separate letter R prevents these roles from being confused. Since k is not a cube, every special solution is excluded in the application.