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Erdős Problem #388

OPENdatabase updated 2025-08-31

No proof or disproof is known, and it isn't known whether a finite computation could settle it.

What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).

Are there only finitely many solutions of \[\prod_{1\leq i\leq k_1}(m_1+i)=\prod_{1\leq j\leq k_2}(m_2+j)\] with \(k_1,k_2>3\) and \(m_1+k_1\leq m_2\)? Here \(m_1, m_2 \geq 0\), so every factor is positive and the condition \(m_1 + k_1 \leq m_2\) makes the two blocks disjoint. Negative \(m_1\) would give trivial solutions such as \((-5)(-4)(-3)(-2) = 2 \cdot 3 \cdot 4 \cdot 5\). The source also asks to classify all solutions; see `Erdos388.erdos_388.variants.classification`.
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research open · 2 further variants / known partial results in the Lean file. The authoritative wording is on erdosproblems.com/388.
Prizenone
Topicsnumber theory
Statement in LeanFormal Conjectures
Solution in Leanno
Original pageerdosproblems.com/388

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FULL WRITE-UP BY THIS SITE · JUL 27, 2026

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Can two disjoint blocks of consecutive integers, both length >3, ever have equal products? 8·9·⋯·14 = 63·64·65·66 = 17,297,280 is the only known example. Round 1 established fixed-pair finiteness. Round 2: a real corollary narrows it further (fixed length RATIO is also finite, so any infinite family needs infinitely many distinct ratios), a 2024 paper's computer search conjectures this IS the complete classification, and a search reaching 10 million found nothing else across 1.4 billion comparisons. Round 3: a follow-up wrote out the actual proof of the fixed-unequal-length finiteness Round 2 had only cited — a clean Kulkarni–Sury case-elimination we verified by exact computation — and caught a subtle error in the naive root-comparison argument. It proves no uniform bound, so the full problem stays open; everything below the uniformity step is now settled.

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