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:
1. 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 \(n\in\mathbb N\).”
2. 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.”
3. 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:
1. 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);
2. 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\(2v_s\in E\) 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 \(R_\Sigma(1),R_\Sigma(2),R_\Sigma(3)=(1,7,23)\).