Erdoes #1199 - working report - Erdős Problem a Day

Erdős problem #1199: live-page gate, fresh affirmative preprint, proof audit, and exact finite cases

Date of audit: 2026-07-27 (UTC)

This report uses the requested labels:

(a) elementary-rigorous: proved here from definitions;
(b) rigorous modulo named theorem: the reduction was checked, with the

named input isolated explicitly;

(c) plausible/structural-unverified: not established;
(d) computational-only: established by the reproducible exhaustive

computation, but not promoted to a non-computational theorem.

0. Mandatory live-page check

I fetched the rendered live page, its LaTeX view, its discussion thread, and its history view through the Bright Data browser path on 2026-07-27. Direct curl was not used as authority.

The live page's verbatim LaTeX statement is:

Is it true that in any $2$-colouring of $\mathbb{N}$ there exists an infinite set $A$ such that all elements of $A+A$ are the same colour?

Source: <https://www.erdosproblems.com/latex/1199>.

The gate did not trigger:

status: OPEN;
claimed proofs: 0;
“Currently working on this problem”: None;
“Interested in collaborating”: None;
“I am working on formalising the results”: None;
likes (not work claims): Woett and Dogmachine;
the page identifies the source as [Er80, p.104];
the page says a formalised statement exists;
the page links three comments.

The page's listed mathematical context, verbatim, is:

A conjecture of Owings \cite{Ow74}. Hindman \cite{Hi79} has shown that this is false for $3$-colourings.
If we do not ask for the elements $2a$ for $a\in A$ to also be the same colour the answer is yes by Hindman's theorem (see [532]).

The three discussion comments are:

Adenwalla, 2026-04-11 09:56: “Colour $n$ red if $v_2(n)$ is

even and colour $n$ blue if $v_2(n)$ is odd. Then $n$ and $2n$ are different colours for all $n\in\mathbb N$.”

Thomas Bloom, 2026-04-11 11:38: “But $n$ and $2n$ need not both

be in $A+A$. In this colouring, for example, one could take $A$ to be the set of all integers $\equiv1\pmod4$ - then all elements of $A+A$ are blue.”

Adenwalla, 2026-04-11 11:49: “Oh yes, of course.”

The page itself warns that comments are unverified. In this instance Bloom's correction is (a): if every $a\in A$ is $1\bmod4$, then every element of $A+A$ is $2\bmod4$, and hence has $v_2=1$.

Live URLs checked:

<https://www.erdosproblems.com/1199>
<https://www.erdosproblems.com/latex/1199>
<https://www.erdosproblems.com/forum/discuss/1199>
<https://www.erdosproblems.com/history/1199>

1. Literature search: the tracker has just become stale

The original item is J. C. Owings, Jr., Problem E2494, in “Elementary Problems: E2492–E2496,” American Mathematical Monthly 81(8) (1974), 901–902. JSTOR's issue record identifies the item, authors, pages, and stable DOI <https://doi.org/10.2307/2319455>. (The v1 preprint had a typographical E2409 at one point; immutable v2 correctly says E2494.)

Neil Hindman's cited paper exists as “Partitions and sums of integers with repetition,” Journal of Combinatorial Theory, Series A 27(1) (1979), 19–32, DOI <https://doi.org/10.1016/0097-3165(79)90004-9>. The official abstract defines the relevant admissibility condition—arbitrarily long arithmetic progressions of even integers with a fixed increment in one cell—and says the resulting $B+B$ assertion is true for two cells and false for at least three. This also confirms that repetition/diagonal sums are part of the paper, rather than the restricted $b_i+b_j$, $i\ne j$, problem.

The most relevant recent partial result I found before the new preprint is Ioannis Kousek and Tristán Radić, “Infinite unrestricted sumsets of the form $B+B$ in sets with large density,” Bulletin of the London Mathematical Society 57 (2025), 48–68, <https://doi.org/10.1112/blms.13180>, preprint <https://arxiv.org/abs/2404.12201>. It proves density-threshold results and gives a syndetic three-colouring obstruction; it does not settle the arbitrary two-colouring case. The general-cardinal paper <https://arxiv.org/abs/2402.13124> likewise explicitly describes the integer/two-colour/$\aleph_0$ case as Owings's problem rather than settling it.

The decisive search hit is:

Wen Huang, Zhengxing Lian, Song Shao, Rongzhong Xiao, Leiye Xu, and Shuhao Zhang, “An affirmative answer to Owings's sumset question,” arXiv:2607.17333v2.

Primary source: <https://arxiv.org/abs/2607.17333v2>.

It was submitted on 2026-07-19 and revised on 2026-07-23, only four days before this audit. The immutable v2 TeX source has 78,997 bytes and SHA-256

d95418959f5abf77ff05d7d2fb38fc86c2c6b233d92471339f1d1e2a200afa10

The standalone verifier downloads that exact source version and checks the digest and theorem text. During this audit, the unversioned e-print endpoint served the older shift_Owings-V8.tex; pinning v2 avoided silently auditing the wrong revision.

Targeted exact-title, arXiv-ID, author, “Owings sumset,” and $B+B$-colouring searches found no independent proof, published version, retraction, or public correction as of 2026-07-27. This is a search miss, not a proof that none exists. The work is currently a very recent preprint, while the live Erdős Problems page still says OPEN and lists no proof claim.

2. Does the new theorem really have all of #1199's quantifiers?

Theorem 1.1 of arXiv:2607.17333v2 says:

Let $\mathbb N=C_1\sqcup C_2$. Then there exist $i\in\{1,2\}$ and an infinite $B\subseteq\mathbb N$ such that $B+B\subseteq C_i$.

This is an exact match, (a):

the colouring is completely arbitrary;
$B$ is infinite, not merely arbitrarily large finite;
the paper defines $B+B=\{x+y:x,y\in B\}$, so every diagonal

$2b=b+b$ is included;

there is no density, measurability, periodicity, or finite-colouring

uniformity assumption hidden in the theorem;

the single cell $C_i$ contains all of $B+B$.

Thus this is not merely the restricted pairwise-sums theorem and not a shifted $B+B+t$ result.

3. Audit of the proof of Theorem 1.1

3.1 Named inputs

The audit status is (b). It depends on:

the standard Stone–Čech semigroup facts explicitly cited in v2 from

Hindman–Strauss, Algebra in the Stone–Čech Compactification, 2nd edition (existence of nonprincipal ultrafilters, the extension of addition, ultrafilter actions, minimal left ideals, and idempotents);

Hindman's 1979 Corollary 2.10: an admissible two-cell partition of

$\mathbb N$ contains $B+B$ for an infinite $B$.

The remaining implications were checked directly below. I could verify Hindman's publication metadata and official theorem-level abstract, but Elsevier did not expose the full 1979 PDF to this environment. Therefore I deliberately retain the “modulo named theorem” label rather than pretending to have independently reproved Corollary 2.10.

There is a harmless wording looseness in v2's displayed definition of “admissible”: it writes $d\in\mathbb N$ after describing progressions of even integers, whereas the tracker and the earlier draft use an even fixed difference. The final application constructs the fixed difference $d=2$, so it satisfies the stronger original condition and is not affected.

3.2 Key identity

Assume for contradiction that a colouring $b:\mathbb N\to\{0,1\}$ has no infinite monochromatic $B+B$. Extend $b$ arbitrarily to $0$ and set

\[ c(n)=b(2n). \]

The paper records every affine sampling $c_{s,r}(t)=c(st+r)$ in a point $\mathbf c$ of a compact product shift. Let $J$ complement every bit, let $p+q$ be Stone–Čech addition, and let $Dp$ be the extension of $n\mapsto2n$. Its central identity is

(p+p)\mathbf c=J(Dp)\mathbf c \qquad(p\in\mathbb N_0^*). \tag{1}

Here is the complete combinatorial content of (1), (a) once the ultrafilter definitions are accepted. If one coordinate of (1) failed, the two uncompensated bits would be equal to some $\gamma$. The corresponding return-time set $E$ would satisfy

E\in p+p\quad\text{and}\quad E\in Dp.

Write

K=\{n:E-n\in p\},\qquad H=\{n:2n\in E\}.

Both lie in $p$. Recursively choose

v_s\in H\cap K\cap\bigcap_{i<s}(E-v_i)\cap(v_{s-1},\infty).

Nonprincipality makes the last intersection $p$-large. Then $2v_s\in E$ and $v_i+v_s\in E$, so all $v_i+v_j$, including $i=j$, lie in $E$. The affine coordinate turns this into

c(N_i+N_j+R)=\gamma,\qquad N_i=sv_i,\quad R=st+r.

Finally, $m_i=2N_i+R$ gives

\[ m_i+m_j=2(N_i+N_j+R), \]

so $b(m_i+m_j)=\gamma$ for every $i\le j$, contradicting the hypothesis. If the first $m_i$ is $0$, deleting it leaves an infinite positive sequence; v2 now states this explicitly. Thus both the off-diagonal and diagonal sums are genuinely controlled.

3.3 Minimal subsystem and the complemented copy

Let $\Omega=\omega(\mathbf c)$. Every element has the form $q\mathbf c$ for a nonprincipal ultrafilter $q$. Splitting $q$ by parity as $q=\varepsilon+Dp$, equation (1) implies $J\Omega=\Omega$.

Choose a minimal subsystem $M\subseteq\Omega$. The paper proves

M\cap JM=\varnothing. \tag{2}

The potentially delicate algebra in this step checks out, (b). If $M=JM$, the left ideal

\mathcal I_M=\{p:p\mathbf c\in M\}

contains a minimal left ideal and a nonprincipal idempotent $e$. Put $y=e\mathbf c$. Equation (1) and the affine doubling map $\Delta$ give

\Delta y=Jy,\qquad \Delta(Jy)=y.

Minimality supplies $s$ with $sy=Jy$. With $q=s+e$, the right factor $e$ makes $q$ nonprincipal, and direct use of the action law gives

(q+q)\mathbf c=y,\qquad (Dq)\mathbf c=y.

Applying (1) to this same $q$ yields $y=Jy$, impossible because coordinatewise complementation has no fixed point. This proves (2).

3.4 The thick counterexample factor and final contradiction

A clopen fundamental domain separates $M$ from $JM$. Its indicator defines a new colouring $h:\mathbb N_0\to\{0,1\}$. It retains

(p+p)h=J(Dp)h. \tag{3}

Two consequences are checked in the paper:

(a) If an infinite $B$ had $h(B+B)=\gamma$, extend $B$ and

the cofinite sets to a nonprincipal $p$. Diagonal sums give the $\gamma$-cell membership in $Dp$, and off-diagonal sums give its membership in $p+p$. The zeroth coordinate of (3) becomes $\gamma=1-\gamma$, a contradiction.

(b) The $1$-cell of $h$ is thick. The forward orbit of every

point in the minimal subsystem stays in the clopen $1$-side, and $\mathbf c$ returns arbitrarily far into every finite intersection of its inverse images. Hence $h$ has arbitrarily long consecutive blocks of $1$'s.

For every $L$, a block $[n,n+2L]$ in the $1$-cell contains

r_L,r_L+2,\ldots,r_L+2(L-1),

where $r_L$ is the least even integer at least $n$. The common difference is the same fixed value $2$ for every $L$. Thus the restricted colouring of $\mathbb N$ is admissible in Hindman's stronger original sense. Hindman's Corollary 2.10 produces an infinite monochromatic $B+B$, contradicting the first bullet.

I found no missing finiteness, uniformity, diagonal, positivity, or fixed increment step. Subject to the two named inputs, the proof is complete. This is stronger than a bare abstract check, but it is not a claim of peer review.

4. Independent exact finite computation

To produce a result independent of the infinitary preprint, define

R_\Sigma(k)=\min\left\{N: \begin{array}{l} \text{every two-colouring of }[2N]\text{ has a }k\text{-set }B\subseteq[N]\\ \text{for which }B+B\text{ is monochromatic} \end{array}\right\}.

The colour of $1$ is irrelevant. The standalone standard-library checker proves the following exact table, (d):

$k$	$R_\Sigma(k)$	avoiding colouring at $N=R_\Sigma(k)-1$	exhaustive upper-bound instance
1	1	vacuous	$B=\{1\}$
2	7	$N=6$, 15 candidate sets	$N=7$, 21 distinct constraints
3	23	$N=22$, 1540 candidate sets	$N=23$, 1771 distinct constraints

The explicit lower-bound certificates give colour $1$ to the following integers and colour $0$ to every other integer in $\{2,\ldots,2N\}$:

\begin{aligned} k=2,\ N=6:\quad& \{3,5,6,7,10,12\};\\ k=3,\ N=22:\quad& \{3,5,7,9,10,11,13,14,15,17,19,20,21,22,23,\\ &\qquad 28,30,33,34,37,38,40,42,43,44\}. \end{aligned}

Direct enumeration checks that every relevant $B+B$ contains both colours.

For the upper bounds, each distinct $E=B+B$ gives the two CNF clauses

\bigvee_{s\in E}x_s,\qquad \bigvee_{s\in E}\neg x_s.

Together they say exactly that $E$ is not monochromatic. Complement symmetry permits $x_2=0$. The included DPLL code applies sound unit propagation and then branches on both values of an unset variable. Returning UNSAT therefore exhausts every colouring under that symmetry. On this VM it used:

k=2, N=7:  nodes=7,     conflicts=4
k=3, N=23: nodes=10057, conflicts=5029

I independently cross-checked the two SAT/UNSAT answers with PySAT 1.9.dev7 using CaDiCaL 1.9.5; the deliverable does not depend on PySAT.

Finally, (a) monotonicity turns these adjacent lower and upper checks into exact thresholds: restrict an avoiding colouring at $N$ to obtain one at every smaller $N$, while a forced witness at $N$ remains available for every larger range.

This finite table does not imply the infinite theorem. It is separate, reproducible small-case information.

5. Reverification

Standalone checker:

runs/erdos1199_wave6x_verify.py

Full local and provenance check:

python runs/erdos1199_wave6x_verify.py --online

It uses only the Python standard library. The full exact $k=3$ search takes about 18–20 seconds on this VM. --quick checks both explicit certificates, the $k=2$ UNSAT result, the affine arithmetic, and the arXiv source while skipping the $k=3,N=23$ exhaustive tree.

The script also recomputes 5,157 instances of the affine/doubling and fixed-even-progression integer identities used above. Those loops are sanity checks, not a substitute for the general algebraic proof or the Stone–Čech inputs.

6. Honest status

There is no longer an honest mathematical reason to attack #1199 as if no solution were known. The authoritative tracker was still OPEN with zero claims and no worker at access time, but a matching affirmative preprint appeared eight days earlier. Its current v2 has the exact statement, all required quantifiers, and a proof whose reduction survives this audit. Conservatively, the state is “fresh claimed affirmative solution, rigorous modulo the cited standard Stone–Čech and Hindman results, not yet reflected on the tracker and not represented here as peer-reviewed.”

FOUND: arXiv:2607.17333v2 gives an exact affirmative solution to #1199; the current proof audit found no gap modulo its named Stone–Čech facts and Hindman's 1979 Corollary 2.10, and an independent exhaustive checker proves $R_\Sigma(1),R_\Sigma(2),R_\Sigma(3)=(1,7,23)$.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger

\(k\)	\(R_\Sigma(k)\)	avoiding colouring at \(N=R_\Sigma(k)-1\)	exhaustive upper-bound instance
1	1	vacuous	\(B=\{1\}\)
2	7	\(N=6\), 15 candidate sets	\(N=7\), 21 distinct constraints
3	23	\(N=22\), 1540 candidate sets	\(N=23\), 1771 distinct constraints