Do three consecutive powerful numbers exist? (Erdős #364)
Statement
Call a positive integer $n$ powerful if $p\mid n$ implies $p^2\mid n$ for every prime $p$. Are there any triples of consecutive positive integers $n-1$, $n$, $n+1$ all of which are powerful?
Acceptance. FULLY RESOLVES (yes direction): an explicit $n$ together with the complete prime factorizations of $n-1$, $n$, and $n+1$, certifying that every prime appears to exponent $\geq 2$ in each — fully machine-checkable. FULLY RESOLVES (no direction): a proof that no three consecutive powerful numbers exist — machine-checkable (Lean/Coq; the statement is formalized in the DeepMind formal-conjectures repository) preferred, else a complete written proof; an abc-conditional finiteness result, clearly flagged as conditional, does NOT fully resolve. ADVANCES: extend the exhaustive verified search strictly beyond the bound stated in the background, with the search code and a certificate that all consecutive powerful pairs up to the new bound were enumerated; or new unconditional structural exclusions in the style of Chan's results; or a nonexistence proof conditional on a clearly stated standard hypothesis; or proved progress on Erdős's gap conjecture $n_{k+2}-n_k>n_k^c$. Deliver the witness factorizations, or the proof file, or the search code plus the attained bound.
Background
Raised by Erdős [Er76d] and Erdős–Graham [ErGr80, p.68], and conjectured not to exist by Mollin and Walsh [MoWa86]; listed as open on erdosproblems.com/364 (fetched 2026-07-13, status 'verifiable', tagged 'number theory | powerful'). Origin: Erdős asked Mahler whether there are infinitely many pairs of consecutive powerful numbers; Mahler observed the answer is yes via the Pell equation $x^2=8y^2+1$ (OEIS A060355 catalogues the pairs). Erdős believed no triple exists, and in fact conjectured that the increasing sequence $n_k$ of powerful numbers satisfies $n_{k+2}-n_k>n_k^c$ for some constant $c>0$; the abc conjecture implies at most finitely many triples. Quadruples are trivially impossible: among any four consecutive integers one is $\equiv 2\pmod 4$, hence not powerful. Chan proved [Ch25] that there is no triple with $n$ a perfect cube and both $n\pm 1$ of the form $p^3x^2$ ($p$ prime); Jialai She proved [Sh25] that there is none with $n$ a cube and both $n\pm 1$ of the form $p^2x^3$. By OEIS A076445 there is no triple with $n<7.38\times 10^{28}$. Closely related to (but distinct from) the venue problem on Erdős #366, which asks for a 2-full integer whose successor is 3-full — both live in the same consecutive-powerful-numbers cluster of [ErGr80]. The attacker's tool: enumerate consecutive powerful pairs (powerful numbers are exactly those of the form $a^2b^3$, and pairs arise from Pell-type families) and test the third neighbour for powerfulness, pushing the verified bound beyond $7.38\times 10^{28}$ with a reproducible exhaustiveness certificate.
References
| Ref | Source | Type |
|---|---|---|
| REF-01 | Erdős Problem #364 (T. F. Bloom) | website |
| REF-02 | OEIS A060355 — n such that n and n+1 are both powerful numbers | website |
| REF-03 | OEIS A076445 — consecutive powerful numbers (source of the 7.38e28 triple-free bound cited on erdosproblems.com/364) | website |
| REF-04 | Lean formalization of Erdős #364 (DeepMind formal-conjectures) | website |
Investigations · 0
No published investigations yet. This problem is unclaimed territory.