erdős #366

← #365 · #367 → (packet.json; erdosproblems.com)

Are there any $2$ -full $n$ such that $n + 1$ is $3$ -full? That is, if $p ∣ n$ then $p^{2} ∣ n$ and if $p ∣ n + 1$ then $p^{3} ∣ n + 1$ .

Open problem — our best result is machine-sealed: verified reduction, reproduced by an independent verifier. The conjecture itself is unsettled.

number theory · open · formalized (Lean) · 2 attempts

machinery: abc/kernel-bound,consecutive-integer-window,powerful-numbers/squarefull-cubefull,rad-of-consecutive-pairs,Pell-equation,finite-witness-verifiable

use this record

vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

routes

verified computationpython:verify_366_mordell.py

Mordell-curve reduction + Pell witness re-check (pinned tier: scripts/requirements-verifiers.txt)

formal prooflean

Lean patch building clean under the math CI profile (no sorry, no new axioms)

obstruction reportreview

precise, artifact-backed reason a route cannot work

banked

the unconditional Mordell-curve reduction + Pell correction + the two analytic walls verified

targets

the analytic walls block an unconditional bound; closing them is open.1

evidence

verified reduction

machine-sealed

Erdős #366 (is there a 2-full n with n+1 3-full?): UNCONDITIONAL Mordell-curve reduction + effective-abc finite reduction, Opus-verified. Normal forms: 2-full n=q^3 y^2 (q squarefree), 3-full n+1=d z^3 (d cubefree, rad(d)|z), so #366 is dz^3-q^3y^2=1. Setting X=dqz, Y=dq^3y gives the exact equivalence to integral points on the Mordell curve Y^2=X^3-d^2q^3 with divisibility filters dq|X, dq^3|Y, rad(d)|X/(dq) (Mordell identity verified symbolically). Under EXPLICIT abc (c<=K_e rad(abc)^{1+e}): rad(n)<=n^{1/2}, rad(n+1)<=(n+1)^{1/3} => n+1 <= B_e=K_e^{6/(1-5e)}, and rad(d)|z forces d<=m^{2/5}, so #366 becomes a FINITE explicitly-bounded search over q<=B^{1/3} (squarefree), d<=B^{2/5} (cubefree) genus-1 curves with explicit height bounds. Unconditional: the normal form, Mordell equivalence, filters, heights. Conditional: a NUMERICAL B (ordinary abc gives finiteness in principle; explicit abc gives a computable search). Pell correction: x^2-8y^2=1 gives a SQUARE on top (8y^2 not generally 3-full, e.g. 8*35^2=9800=2^3*5^2*7^2), the wrong order for #366 — the asymmetry is geometric (genus-1 Mordell, not genus-0 Pell). Conditional finite reduction; no witness found, not a proof. Credits abc-finiteness comment + Aktas-Murty + Hall/Elkies.

Local filters (verified-able): gcd(d,q)=1; p|q => d cubic residue mod p^3; p|d => -q QR mod p; q=1 => odd primes|d are 1 mod 4. Aktas-Murty handle the 2-full/2-full (Pell) geometry, not 2-full/3-full. scripts/verify_366_mordell.py: Mordell identity residual 0; Pell correction; abc exponent (1-5e)/6.

scripts/verify_366_mordell.py -> ALL VERIFIED (Mordell identity symbolic 0; 9800 2-full not 3-full; exponent 1/12).

obstruction map

machine-sealed

Erdős #366: the effective-finiteness route is BLOCKED at a named kernel bound, Opus-verified. For bounded kernel D=d^2 q^3 <= K, Baker's method on Y^2=X^3-D gives an explicit (astronomical but effective) search bound n+1 <= ceil(exp(3*(1e7*sqrt3*K^{1/3})^{1e7})); enumerate D<=K (v_p(D) in {0,2,3,4}), recover (d,q), compute the filtered Mordell points. But there is NO unconditional bound on D: the normal form gives only a LOWER bound n+1>=D^{5/9} (since D<=(n+1)^{9/5}), not an upper bound on D; fixed-curve Baker/elliptic-Chabauty bound points per curve but not across the MOVING family (generic moving Mordell has unbounded D, e.g. (t^2)^3-(t^3-1)^2=2t^3-1); Thue-Mahler needs fixed prime support; unconditional abc-substitutes (Stewart-Tijdeman log c << R^{1/3}(log R)^3) are too weak. The missing input is an explicit absolute K0 with d^2 q^3 <= K0 for every #366 case (abc-strength for this family). Sharpened obstruction map; not a proof, no witness.

Verified: D<=m^{9/5}, v_p(D) in {0,2,3,4}, unbounded-D example 2t^3-1. scripts/verify_366_1094_walls.py.

scripts/verify_366_1094_walls.py -> ALL VERIFIED (#366 kernel exponent + v_p(D) + unbounded-D).

unverified AI candidates (2)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

For **positive integers**, no example is known at the moment.

candidate solution ↗

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.

candidate solution ↗

formal

AMS 11 · open (literature)

theorem erdos_366 : answer(sorry) ↔ ∃ n > 0, (2).Full n ∧ (3).Full (n + 1)

formal-conjectures/366.lean ↗

oeis

A060355 — Numbers k such that k and k+1 are powerful numbers.8,288,675,9800,12167,235224,332928,465124,1825200,11309768,384199200,592192224,4931691075,5425069447,13051463048,2213222

links

related: #364 · #458 · #913

#364Are there any triples of consecutive positive integers all of which are powerful (i.e. if

p ∣ n

then

p^{2} ∣ n

)?A060355 #365Do all pairs of consecutive powerful numbers

n

and

n + 1

come from solutions to Pell equations? In other words, must either

n

n + 1

be a square?Is the number of such

n \leq x

bounded by

(lo g x)^{O (1)}

?A060355

status

open