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Erdős problems frontier

constellation seal · derived from vfr_37aec80d874a0239
id
vfr_37aec80d874a0239
license
CC-BY-4.0
findings
1,256
accepted core
6
contested
0
links
17
sources
1,234
evidence
1,256
avg conf
0.98

used by 0 · replayed by 2 producers

e1271/1271 · statement.attested · reviewer:will-blair · 2026-06-10 · null→null

Reviewable change

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verified — A frozen deterministic verifier re-checked the claim and passed.accepted

Erdős Problem #397 has status 'disproved (lean)'. Statement: Are there only finitely many solutions to $$ \prod_i \binom{2m_i}{m_i}=\prod_j \binom{2n_j}{n_j} $$ with the $m_i,n_j$ distinct? Somani, using ChatGPT, has given a negative answer. In fact, for any $a\geq 2$, if $c=8a^2+8a+1$, $\binom{2a}{a}\binom{4a+4}{2a+2}\binom{2c}{c}= \binom{2a+2}{a+1}\binom{4a}{2a}\binom{2c+2}{c+1}.$ Further families of solutions are given in the comments by SharkyKesa. This was earlier asked about in a [MathOverflow] question, in response to which Elkies also gave an alternative construction which produces solutions - at the moment it is not clear whether Elkies' argument gives infinitely many solutions (although Bloom believes that it can). This was formalized in Lean by Wu using Aristotle. Topics: number theory, binomial coefficients. Erdős prize: no. Statement is machine-verified in Lean (formal-conjectures). OEIS: N/A.

id
vpr_55faa6f9200b06c3
frontier
Erdős problems frontier
kind
finding.add
created
2026-05-30
findings
+1
state
null → b4555125

accept gate

1 of 4 on record
—
signature
reviewer:erdos-db-trust · no key registered on this bundle
✓
chain
null → b4555125
—
witness
no verifier attachment on record for this target
—
grade
in state · unreviewed

timeline

  1. 2026-05-30proposeproposed · finding.addagent — machine actor, no signing keyagent:erdos-spine-ingestvpr_55faa6f9200b06c3Candidate claim vc_14a414673be11d68 imported from artifact packet cap_61973ee16b553d57
  2. 2026-05-30acceptfinding.assertedreviewer:erdos-db-trustreviewer:erdos-db-trustnull→b4555125vev_d70e21b7740d53c9Candidate claim vc_14a414673be11d68 imported from artifact packet cap_61973ee16b553d57

proposed

reason

Candidate claim vc_14a414673be11d68 imported from artifact packet cap_61973ee16b553d57

finding type

open_question

proposed confidence

0.99

confidence basis

agent-imported candidate claim; reviewer acceptance required

provenance

proposed by

agent — machine actor, no signing keyagent:erdos-spine-ingest

actor type

agent

created at

2026-05-30

target type

finding

affected

inspect finding →

Erdős Problem #397 has status 'disproved (lean)'. Statement: Are there only finitely many solutions to $$ \prod_i \binom{2m_i}{m_i}=\prod_j \binom{2n_j}{n_j} $$ with the $m_i,n_j$ distinct? Somani, using ChatGPT, has given a negative answer. In fact, for any $a\geq 2$, if $c=8a^2+8a+1$, $\binom{2a}{a}\binom{4a+4}{2a+2}\binom{2c}{c}= \binom{2a+2}{a+1}\binom{4a}{2a}\binom{2c+2}{c+1}.$ Further families of solutions are given in the comments by SharkyKesa. This was earlier asked about in a [MathOverflow] question, in response to which Elkies also gave an alternative construction which produces solutions - at the moment it is not clear whether Elkies' argument gives infinitely many solutions (although Bloom believes that it can). This was formalized in Lean by Wu using Aristotle. Topics: number theory, binomial coefficients. Erdős prize: no. Statement is machine-verified in Lean (formal-conjectures). OEIS: N/A.

vf_4be95d2df14f26a5

Diff

Read-only frontier; diff not recomputed.

Review chain

  1. 01request

    Change request

    Erdős problems frontier receives a reviewable source, finding, caveat, replication, evaluation, or proof-affecting edit.

    open review →
  2. 02packet

    Diff packet

    The packet names affected record objects, evidence, rationale, reviewer-facing fields, and expected proof impact.

    open the campaign →
  3. 03checks

    Check output

    Schema, provenance, benchmark, contradiction, and proof checks decide whether the request is ready to read.

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  4. 04review

    Reviewer decision

    A steward accepts, rejects, caveats, revises, or retracts the request under an inspectable identity.

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  5. 05accepted

    Accepted event

    Only the accepted event mutates frontier state. Atlases, constellations, and search update from that record state.

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finding.noted · reviewer:will-blair · 1 day

renders the record as of vev_d199cb2e · 1,338 events · hub

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