Erdős problem #335 — wave 5v

Date of live-page check and computation: 2026-07-26 (UTC).

Claim labels

• (a) elementary-rigorous: proved below from definitions.
• (b) rigorous-modulo-named-theorem: a cited theorem is used, with the exact source named.
• (c) plausible/structural-unverified: diagnosis or proposed direction, not a theorem.
• (d) computational-only: exhaustive finite computation, with a standalone checker.

Source/status observations are explicitly attributed and are not mathematical claims.

0. Mandatory live-page gate

I fetched the live problem page and its discussion thread through the Bright Data browser route, not by trusting the supplied YAML.

The exact current statement displayed on the live page is:

> Let \(d(A)\) denote the density of \(A\subseteq\mathbb N\). Characterise those \(A,B\subseteq\mathbb N\) with positive density such that

> \[ > d(A+B)=d(A)+d(B). > \]

Live-page gate observations:

• Status: OPEN.
• Claimed proofs: 0.
• “Currently working on this problem”: None.
• “Interested in collaborating”: None.
• The page was last edited 15 April 2026.
• There are two comments, both dated 15 April 2026:

1. Alfaiz linked Ackelsberg–Richter and said that it partially resolves the problem when \(B\) meets every residue class; the comment notes that the site was updated.

2. Ethan Ackelsberg summarized their result, explained the modular obstruction using a random density-\(1/4\) subset of the evens, and pointed to Theorem 1.4 for the precise statement.

Thus neither stop condition applies, and it is permissible to proceed.

The page also records two pieces of context:

1. Equality is produced by rotation models

\[ A=\{n>0:\{n\theta\}\in X_A\},\qquad B=\{n>0:\{n\theta\}\in X_B\}, \]

when \(\mu(X_A+X_B)=\mu(X_A)+\mu(X_B)\), and asks whether all examples arise similarly from groups.

2. It warns that unrestricted classification may be extremely complicated: a random density-\(1/4\) subset \(A\) of the even integers almost surely has \(d(A+A)=1/2\).

1. Primary-source literature check

Verified source

(b) Ethan Ackelsberg and Florian K. Richter, “An inverse theorem for sumsets of sets of positive density in the integers,” arXiv:2604.12864v1, 117 pages, submitted 14 April 2026. I downloaded and text-checked the PDF linked by the live page. Its SHA-256 at the time of this run was

c00f13c4f39029db1afd12995f2c92dd9fdf11ca0b9066359d815118abf8d1f6

(b) Theorem 1.4 of that paper assumes \(d(A)>0\), \(d(A)+d(B)<1\), and that \(B\) meets every residue class in \(\mathbb N\). Along any sequence on which the lower-bound equality is attained, it gives two alternatives: a parallel-Bohr-interval structure on \(h\mathbb N\), with the other \(h-1\) residue classes of \(B\) full up to density zero, or a residue-class/invariance alternative. This verifies the live page's summary; it does not remove the “meets every residue class” hypothesis.

(b) The same paper labels the unrestricted statement as Problem 1.5 and still calls its result a partial answer. Section 16 goes further and asks Question 16.8 even under the extra regularity assumption that every residue-class intersection of \(A\) and \(B\) has a density. This is direct primary-source evidence that the modular-obstruction regime treated below is not already claimed as solved there.

Search result and honest miss

Searches for the exact equality, the problem number, and citations of the new paper found no later primary paper claiming a complete solution as of 2026-07-26. This is a search result, not a proof of nonexistence.

The live page cites P. Erdős and R. L. Graham, Old and New Problems and Results in Combinatorial Number Theory (1980), p. 51. I verified that the bibliographic item exists, but I did not obtain an accessible primary scan of page 51. I therefore make no additional claim about its wording beyond what the live page and Ackelsberg–Richter explicitly attribute to it.

2. Complete reduction for ultimately periodic sets

This is the main rigorous progress of this run.

Definitions

Fix \(q\geq 1\) and write \(G=\mathbb Z/q\mathbb Z\). A set \(A\subseteq\mathbb N\) is ultimately \(q\)-periodic if membership of \(n\) in \(A\), for all sufficiently large \(n\), depends only on \(n\bmod q\).

For such an \(A\), define two residue supports:

Thus \(P_A\subseteq E_A\). The set \(P_A\) records permanent tail classes, while \(E_A\setminus P_A\) records classes hit only in the finite prefix. Define \(P_B,E_B\) similarly.

The finite-prefix shadow theorem

Theorem (a). Let \(A,B\subseteq\mathbb N\) be ultimately \(q\)-periodic and have positive density. Put

Then \(A+B\) is ultimately \(q\)-periodic, with permanent residue support exactly \(U\). Consequently,

In particular,

This is a necessary-and-sufficient finite characterization of every ultimately periodic pair, including arbitrary finite prefixes.

Proof (a). Choose thresholds \(N_A,N_B\) beyond which \(A,B\) are \(q\)-periodic.

First take \(u=e+p\in E_A+P_B\). Choose a fixed \(a\in A\) with \(a\equiv e\pmod q\). If \(n\equiv u\pmod q\) is sufficiently large, then \(n-a\geq N_B\), \(n-a\equiv p\pmod q\), and therefore \(n-a\in B\). Hence every sufficiently large integer in residue \(u\) lies in \(A+B\). The same argument with \(A,B\) interchanged handles \(u\in P_A+E_B\). Thus every class in \(U\) is permanent in \(A+B\).

Conversely, suppose \(n=a+b\in A+B\) and \(n\bmod q\notin U\). If \(a\geq N_A\), then \(a\bmod q\in P_A\) and \(b\bmod q\in E_B\), putting \(n\bmod q\) in \(P_A+E_B\), a contradiction. Hence \(a

An ultimately \(q\)-periodic set with \(k\) permanent classes has natural density \(k/q\). Substitution gives (2.1). \(\square\)

Realizability (a). Every quadruple of finite sets

is realized by some positive-density ultimately \(q\)-periodic pair: put one small representative in every class of \(E_A\setminus P_A\), then include all sufficiently large integers in the classes \(P_A\), and do the same for \(B\). Hence (2.1) is not merely a necessary profile constraint; it classifies all possible finite support data.

Purely periodic corollary

Corollary (a). If

with nonempty \(S,T\subseteq G\), then

Thus the pure-periodic subproblem is exactly a finite sumset problem in \(\mathbb Z/q\mathbb Z\).

The distinction between (2.1) and (2.2) is essential: finite changes to a summand do not change its density, but they can change the density of its sumset by adding a translate of the other positive-density summand.

3. An explicit deterministic infinite family

Construction (a). For every \(q\geq3\), let

Then

and, exactly,

Therefore

This is a deterministic modular-obstruction family, requiring neither randomness nor an irrational rotation. In the notation of the theorem,

The family is not claimed to settle the requested characterization; it isolates a mechanism any complete answer must retain: even a single transient element can create an entire permanent residue class in the sumset.

4. Closed small-period classification for pure periodic pairs

The following is proved without computation.

Lemma (a). In (2.2), neither \(S\) nor \(T\) can be a singleton, because \(|\{s\}+T|=|T|\), not \(1+|T|\). Hence a pure-periodic equality pair requires \(q\geq4\).

Period \(q=4\) (a). Equality forces \(|S|=|T|=2\) and \(S+T=G\). The six two-subsets of \(\mathbb Z/4\mathbb Z\) split into four adjacent pairs and two antipodal pairs. The sum is all of \(G\) exactly when one support is adjacent and the other antipodal. Hence there are

ordered labelled pairs, all with \(d(A)+d(B)=1\).

Period \(q=5\), strict case (a). Equality with density sum below \(1\) forces \(|S|=|T|=2\). After translating, write \(S=\{0,a\}\), \(T=\{0,b\}\). Their sum has four elements exactly when \(b\neq\pm a\). There are \(10\) choices for \(S\), and for each \(S\), exactly \(5\) of the \(10\) two-subsets have the other unoriented difference. Thus there are \(50\) ordered strict pairs.

Period \(q=5\), saturated case (a). The possible sizes are \((2,3)\) or \((3,2)\). More generally, if \(|S|+|T|=q\), a residue \(g\) is missing from \(S+T\) exactly when

For a fixed two-set \(S\subseteq\mathbb Z/5\mathbb Z\), the five translates in (4.1) are distinct, so \(5\) of the \(10\) three-sets work and \(5\) fail. This gives \(10\cdot5=50\) pairs of size \((2,3)\) and another \(50\) after swapping, for \(100\) saturated pairs.

These arguments give a complete pure-periodic classification through \(q=5\), not just a census.

5. Exhaustive finite computations

The companion checker is

runs/verify_erdos335_wave5v.py

Run it from the repository root with

python3 runs/verify_erdos335_wave5v.py

All counts below are ordered and residue-labelled. They are not quotient counts under translation, multiplication by units, or swapping the summands.

Purely periodic pairs

Table (d). The script exhausts every nonempty \(S,T\subseteq\mathbb Z/q\mathbb Z\) with \(|S|+|T|\leq q\), through \(q=12\), and tests (2.2).

| \(q\) | all equality pairs | density sum \(<1\) | density sum \(=1\) | no nonzero common period shift | aperiodic sum support |

|---:|---:|---:|---:|---:|---:|

| 1 | 0 | 0 | 0 | 0 | 0 |

| 2 | 0 | 0 | 0 | 0 | 0 |

| 3 | 0 | 0 | 0 | 0 | 0 |

| 4 | 16 | 0 | 16 | 16 | 0 |

| 5 | 150 | 50 | 100 | 150 | 50 |

| 6 | 1,080 | 504 | 576 | 1,080 | 432 |

| 7 | 5,292 | 2,744 | 2,548 | 5,292 | 2,744 |

| 8 | 21,200 | 10,304 | 10,896 | 21,184 | 9,472 |

| 9 | 77,436 | 33,372 | 44,064 | 77,436 | 32,562 |

| 10 | 264,550 | 89,850 | 174,700 | 264,400 | 82,600 |

| 11 | 907,016 | 224,092 | 682,924 | 907,016 | 224,092 |

| 12 | 3,152,104 | 496,584 | 2,655,520 | 3,151,008 | 451,440 |

Here “no nonzero common period shift” means

and “aperiodic sum support” means

Ultimately periodic support data

For one summand there are exactly \(3^q-2^q\) labelled pairs \((P,E)\) with \(\varnothing\neq P\subseteq E\subseteq G\): each residue is absent, transient-only, or permanent, and configurations with no permanent class are removed.

Table (d). The script exhausts every ordered pair of such support data through \(q=7\) and tests (2.1).

| \(q\) | all equality data | density sum \(<1\) | density sum \(=1\) | pure data \(E=P\) | nonpure data | permanent cores already equal | equality repaired by transients |

|---:|---:|---:|---:|---:|---:|---:|---:|

| 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

| 2 | 12 | 0 | 12 | 0 | 12 | 0 | 12 |

| 3 | 180 | 54 | 126 | 0 | 180 | 0 | 180 |

| 4 | 1,436 | 400 | 1,036 | 16 | 1,420 | 256 | 1,180 |

| 5 | 9,950 | 2,100 | 7,850 | 150 | 9,800 | 3,400 | 6,550 |

| 6 | 66,528 | 7,866 | 58,662 | 1,080 | 65,448 | 38,736 | 27,792 |

| 7 | 463,148 | 24,990 | 438,158 | 5,292 | 457,856 | 334,768 | 128,380 |

“Repaired by transients” means

but the union in (2.1) has exactly the target size. Thus the finite prefixes create equality rather than merely preserving an equality already present in the permanent tails.

6. Independent checks performed by the script

The checker uses only the Python standard library and performs all of the following on every run:

1. (d) It compares the optimized cyclic-bitset sumset routine with literal nested set addition for every pair of residue subsets through \(q=8\).

2. (a)+(d) It independently checks the saturated column without calling either sumset routine. For fixed \(S\), \(|S|+|T|=q\), the bad \(T\)'s are exactly (4.1); there are \(q/|\operatorname{Stab}(S)|\) distinct ones. This closed count agrees for every \(q\leq12\).

3. (a)+(d) For prime \(q\), it also checks the second closed formula

\[ \sum_{k=1}^{q-1}\left(\binom qk^2-q\binom qk\right). \]

4. (d) The “pure data” column of the ultimate-period census is required to reproduce the independently generated pure-period table.

5. (d) It checks the support formula using literal residue-set operations for every ultimate-period datum pair through \(q=4\).

6. (d) It builds concrete ultimately periodic representatives and directly tests late integer windows for every datum pair through \(q=4\).

7. (d) It forms all pairwise sums directly for \(A_q=q\mathbb N\cup\{1\}\), \(B_q=q\mathbb N\), for every \(3\leq q\leq12\), on a prefix of length \(4{,}000\), and compares with the claimed exact two-progressions formula.

The full default run completed successfully within the allowed few-CPU-minute budget.

7. Exact reduction showing what remains for arbitrary sets

There is an elementary residue-section identity that exposes the unresolved part. For \(r\in\{0,\ldots,q-1\}\), define

with the implicit restriction \(qk+r\geq1\). If \(C_t\) is the \(t\)-th residue section of \(A+B\), then