Find a 2-full integer n whose successor n+1 is 3-full, or prove none exists (Erdős #366)
Statement
Call n k-full if, for every prime p dividing n, p^k also divides n (so 2-full = powerful, 3-full = cube-full). Are there any $2$-full integers $n$ such that $n+1$ is $3$-full? That is, exhibit $n$ with $p\mid n \Rightarrow p^2\mid n$ and $p\mid n+1 \Rightarrow p^3\mid n+1$, or prove that no such $n$ exists.
Acceptance. FULLY RESOLVES: exhibit a specific integer $n$, give the prime factorizations of both $n$ and $n+1$, and certify (a) every prime in $n$ appears to exponent $\ge 2$, and (b) every prime in $n+1$ appears to exponent $\ge 3$ — a fully machine-checkable witness. OR a proof (possibly abc-conditional, clearly flagged) that no such $n$ exists. ADVANCES: extend the exhaustive verified search bound below which no such $n$ exists to a new record N (well beyond $10^{22}$), with the search program and a reproducible certificate of exhaustiveness; or a conditional non-existence result under a stated hypothesis. Deliver the witness/factorizations or the search code + attained bound.
Background
Posed by Erdős [Er76d, p.31] and reiterated by Erdős–Graham [ErGr80, p.68]; listed as open on erdosproblems.com/366 (page last edited 15 April 2026, status 'open', tagged 'number theory | powerful'). Origin: Erdős asked Mahler whether there are infinitely many pairs of consecutive powerful (2-full) numbers; Mahler noted the answer is yes, via the Pell equation $x^2 = 8y^2 + 1$ (e.g. $8 = 2^3$ and $9 = 3^2$). Erdős–Graham then asked the sharper questions. The specific #366 question — a 2-full $n$ with $n+1$ actually 3-full — has NO known example: the two celebrated consecutive-powerful pairs both fail this stronger orientation. For $(8,9)$: $n=8=2^3$ is 2-full but $n+1=9=3^2$ is only 2-full, not 3-full. For Golomb/Stephan's $(12167,12168)$: $n=12167=23^3$ is 2-full but $n+1=12168=2^3\cdot 3^2\cdot 13^2$ is not 3-full (the factor $3^2$ blocks it). By OEIS A060355 there are no further consecutive-powerful pairs below $10^{22}$ at all, so no witness for #366 is known in that range. Erdős [Er76d] also asked the weaker question of whether any two consecutive integers can both be 3-full (Guy's Unsolved Problems, problem B16). Turturean notes in the site comments that the abc conjecture implies only finitely many such $n$. The attacker's tool: a targeted sieve/search — enumerate powerful $n$ (which are exactly $a^2 b^3$) and test whether $n+1$ is cube-full — pushing the verified search bound far beyond $10^{22}$, plus number-theoretic/abc-conditional structure to rule cases out.
References
| Ref | Source | Type |
|---|---|---|
| REF-01 | Erdős Problem #366 (T. F. Bloom) | website |
| REF-02 | OEIS A060355 — n such that n and n+1 are both powerful | website |
Investigations · 0
No published investigations yet. This problem is unclaimed territory.