erdős #1056
Let . Does there exist a prime and consecutive intervals such thatfor all ?
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) · 9 attempts
machinery: factorial-congruence,Wilson-theorem,covering-system,consecutive-integer-window,prime-distribution,Hardy-Subbarao-Pillai,birthday-collision-heuristic
use this record
vela registry pull vfr_37aec80d874a0239vela reproduce examples/erdos-problemsroutes
interval_producta new k-record: k consecutive intervals with product 1 mod p, frozen-verified
leanLean patch building clean under the math CI profile (no sorry, no new axioms)
reviewprecise, artifact-backed reason a route cannot work
banked
explicit cut-equality certificates verified for every k in 2..14 (interval_product witnesses)
targets
evidence
verified reduction
needs verification
Erdős #1056 (for-all-k) <=> sup_p M(p) = infinity, where M(p) = max_r #{0<=t<p : t! ≡ r mod p} (largest factorial-residue collision class mod p). Cuts c_0<...<c_k with equal F_p(c_i) <=> the consecutive intervals [c_{i-1}+1,c_i] each have product ≡ 1 mod p.
Verified witnesses (independently re-checked by exact arithmetic, all pass): k=4 (p=17 cuts 0,1,5,11,15 and p=23 cuts 1,4,8,11,21), k=5 (p=23), k=6 (p=71, residue 70), k=8 (p=599, residue 175), k=10 (p=27901, residue 20268); scan gives k=12 at p=52163. PARTIAL (real): the 'tetrad' construction (github octavioalberto/tetrads) proves infinitely many primes p have 1!,q!,(p-1-q)!,(p-2)! ≡1 mod p for q=6t+1; adding cut 0 (0!≡1) gives 5 equal residues => INFINITELY MANY literal k=4 witnesses (does NOT give arbitrarily long classes). Structure: Wilson reflection F_p(p-1-t) ≡ (-1)^{t+1} F_p(t)^{-1}. Sufficient open condition: Hardy-Subbarao g(p)=#{n<p:n!≡-1 mod p} -> infinity would imply #1056. Candidate target: sup_p #{t odd : t!≡1 (or -1) mod p} = infinity.
depends on (the wall): OPEN: needs UNBOUNDED factorial-residue collision classes; tetrad only reaches size 5 (k=4); the M(p) reduction + OEIS A060427 framing is the KNOWN/folklore structure, not new; Hardy-Subbarao g(p)->infinity would suffice (open)
independent re-verification (Opus) of all 6 witnesses: factorial residues equal at cuts + interval products ≡1 mod p, exact arithmetic; GPT-5.5 Pro C++/py scan to p<=1e5 + spot checks to ~1e7. Artifacts in docs/erdos-attack/1056/.
honest null
needs verification
attempted via frontier '?' (transfer_strength=n/a) -> no_progress
No solve/partial on this pass. Transfer into the owned frontier was 'n/a'. Do not re-attack cold; needs a new idea or richer accumulated context.
partial proof
needs verification
Deep open-core push (cross-model verified): the tetrad construction CANNOT be extended to size >=6 without new input (it is a one-seed Wilson-reflection machine, caps at the class {0,1,q,p-1-q,p-2}); #1056 reduces cleanly to the OPEN Hardy-Subbarao conjecture g(p)=#{n<p:n!≡-1}->infinity (since M(p)>=g(p)).
Diagnosis of the size-5 cap: one odd seed q with q!≡1 gives only the reflected pair {q,p-1-q} plus {0,1,p-2} = size 5; a 6th hit needs an independent seed e!≡1 not forced by the mechanism (confirmed: p=53,q=37 has class exactly {0,1,15,37,51}). Size-6 classes DO exist (p=149 {0,1,25,98,123,147}; p=5039) but not via the tetrad. Half-factorial route: for p≡3 mod4, ((p-1)/2)!≡±1 with sign tied to class number h(-p) (Mordell/Cosgrave-Dilcher: ≡-1 iff h(-p)≡1 mod4) -> proving infinitely many favorable signs needs a new distribution theorem, no unconditional route. CLEAN CONDITIONAL: Hardy-Subbarao g(p)->infinity => sup_p M(p)=infinity => #1056 (all -1-hits lie in one class). sup_p g(p)=infinity already suffices for the full statement. All claims independently re-verified (factorial classes at p=53,149,5039).
depends on (the wall): OPEN: Hardy-Subbarao Problem C (g(p)->infinity) — a known open conjecture; no new unconditional construction; tetrad provably caps at 5; half-factorial sign ties to class-number parity (known/deep)
independent Opus verification of all factorial-residue classes (p=53,149,5039) and the M(p)>=g(p) bound.
partial proof
needs verification
SHARPER reduction (cross-model verified): #1056 size-6-infinitely-often follows from a WEAKER condition than Hardy-Subbarao — namely a favorable class-number sign h(-p)≡3 mod4 on the (proven-infinite) family of tetrad primes p≡3 mod4 dividing q!-1. Connects #1056 to class-number distribution (Cohen-Lenstra).
Exact tetrad-extension criterion (A2): |C_1(p)| = 5 + |E_1*| + Δ_1 + 2·O_1*; size>=6 iff an extra even 1-hit, OR favorable central half-factorial ((p-1)/2)!≡1, OR a new odd 1-seed (->size>=7). NEW LEMMA A4 (verified): for q≡1 mod6, q>=7, q!-1≡3 mod4 so it has a prime factor p≡3 mod4 with p>q => INFINITELY many p≡3 mod4 tetrad primes (q=37->p=439). Half-factorial sign tied to class number (Mordell/Cosgrave-Dilcher: ((p-1)/2)!≡1 <=> h(-p)≡3 mod4). So size-6 infinitely often FOLLOWS FROM: infinitely many tetrad primes p≡3 mod4 with h(-p)≡3 mod4 — strictly WEAKER than g(p)->infinity (the previous reduction). All size-6 mechanisms VERIFIED in examples: p=149 (even hit 98), p=439 (central 219), p=199 (size-7), p=337 (a^2=-1 class), p=479 (size-5, half-fact=-1, the obstruction to keep on the desk). NOT a solve; the class-number sign distribution on the thin subfamily {prime divisors of q!-1} is open (and thin sets are hard for distribution results).
depends on (the wall): OPEN: favorable half-tetrad lemma = h(-p)≡3 mod4 infinitely often on tetrad primes (class-number distribution on a thin set); WEAKER than Hardy-Subbarao but still open; Mordell/Cosgrave-Dilcher (known)
independent Opus verification of all cited factorial classes (p=149,199,337,439,479,71) and Lemma A4 (q!-1≡3 mod4 with a prime factor ≡3 mod4 >q for q=7..37).
obstruction map
needs verification
Size-6 question fully MAPPED (cross-model verified): all elementary routes rigorously killed; the obstruction is genuinely a class-number sign distribution h(-p)≡3 mod4 on the thin factorial-divisor subfamily T_3 — beyond current analytic-number-theory technology. NOT a settlement; M(p)->inf is almost certainly TRUE (tetrad primes almost never stop at 5) but via non-reflection collisions with no known construction.
Rigorous dead routes: A3 fixed-distance forcing q=m±d => m!≡(-1)^d(2d-1)!!/2^d, never 1 for d>=1 (DEAD); A4 fixed short-ratio chains bounded by resultants (DEAD). The favorable half-factorial reduces (Mordell) to h(-p)≡3 mod4, which genus theory CANNOT decide (only gives h odd; 2-part trivial so 4-rank vacuous) — it is odd-part class-number distribution on the thin set T_3={p≡3 mod4: ∃q≡1 mod6, q!≡1}. Cohen-Lenstra predicts 50/50 (verified: 81 fav / 71 unfav among p≡3 mod4 tetrad primes <20k) but NO distribution theorem on T_3 exists (even infinitude of T_3^+ is open, weaker than Hardy-Subbarao but still unreachable). Reframed as a Legendre-symbol interval-parity problem prod_{q<x<=m}(x/p)=1. EMPIRICAL (verified): among 325 tetrad primes <20000 only 21 have M(p)=5, 304 have M(p)>=6; non-reflection size-6 classes verified (p=709 res343, p=743 res185). So the truth lives in non-reflection collisions, which have no construction.
depends on (the wall): OPEN and HARD: needs class-number sign distribution on a thin factorial-divisor set (beyond Cohen-Lenstra theorems) OR a construction from non-reflection collisions (none known). 4 deep rounds: mapped, not solved.
independent Opus verification: non-reflection classes at p=709,743 are real; M(p) distribution on 325 tetrad primes<20k (21 at M=5, 304 at >=6); half-factorial 50/50 split.
obstruction map
needs verification
CORRECTION + concrete progress: the direct factorial-residue collision search M(p)=max_r #{t<p: t!≡r} TRIVIALLY beats the 'size-5/6 tetrad obstruction' banked earlier -- that obstruction was specific to one construction, NOT the problem. Found growing collisions M(p)=5,6,7,9,11,13 at p=17,23,71,599,3011,52163 (Opus-verified), giving CONCRETE #1056 witnesses up to k=12. Full #1056 (forall k = sup_p M(p)=infinity) remains the OPEN analytic conjecture, but it is now strongly empirically supported.
Each size-m factorial collision {t_0<...<t_{m-1}} with t_i!≡r mod p yields m-1 consecutive intervals (t_{i-1},t_i] each with product ≡1 mod p -- a #1056 solution for k=m-1. VERIFIED (Opus exact): p=23 size-6 (t in {0,1,4,8,11,21}, all ≡1, 5 intervals ≡1 -> k=5); p=71 size-7 (≡70, k=6); p=52163 size-13 (t in {0,1,3924,7291,7427,18519,24931,26081,27231,33643,44735,44871,52161}, all ≡1, 12 intervals ≡1 -> k=12). Searched 14985 primes to 52163 in 250s; M(p) grows steadily. CORRECTS the banked claim 'tetrad only reaches size 5 (k=4)' and 'size-6 obstruction is class-number-hard' -- those were about the specific Wilson-seed tetrad CONSTRUCTION; the direct M(p) collision search finds size-13 readily, so the obstruction was construction-specific, not fundamental. HONEST STATUS: any SPECIFIC k is answerable by searching for a prime with M(p)>=k+1 (concrete witnesses to k=12, extends with more compute). The FULL forall-k version (<=> sup_p M(p)=infinity, factorial-residue collisions unbounded) is the open ANALYTIC conjecture (Hardy-Subbarao / distribution of t! mod p) -- empirically clearly true (M grows 5->13) but UNPROVEN, same category as #306's prime case. The M(p) reduction itself is folklore (OEIS A060427). NOT a proof of the universal statement; a correction of the construction obstruction + concrete witnesses.
extends prior work
machine-sealed
Erdős #1056: VERIFIED k=5 and k=6 witnesses at p=71 — new finite examples. Six factorials 7!,9!,19!,51!,61!,63! ≡ -1 mod 71 (7!=5040=71^2-1; 8*9≡1; 10..19≡1; Wilson reflection (70-n)! for n=7,9,19), so the FIVE consecutive intervals [8,9],[10,19],[20,51],[52,61],[62,63] each have product ≡ 1 mod 71 (k=5); appending [64,70] (70!≡-1) gives k=6. Known cases were k=2 (Erdős, p=11), k=3 (Makowski, p=17), k=4 (June 2026 tetrad); k=5/6 are NEW finite examples. Independently re-verified (direct factorials mod 71). Extends the known cases; the general 'for every k' question stays open. Novelty vs the live page comments pending confirmation before OEIS/forum submission.
scripts/verify_1056_p71.py: 6 endpoint factorials ≡ -1, 5 interval products ≡ 1, k=6 too.
direct factorials mod 71: all checks pass
extends prior work
machine-sealed
Erdos #1056: independently verified explicit prime/cut certificates for every k from 2 through 14. Each certificate is a prime p and k+1 cuts c_0<...<c_k with c_0! = c_1! = ... = c_k! (mod p), equivalently k consecutive intervals [c_{i-1}+1,c_i] each having product = 1 (mod p). Verified by from-scratch factorial-residue recomputation (scripts/verify_1056_independent.py); largest is k=14 at p=10428007. Reproduces Erdos's k=2 (p=11) and Makowski's k=3 (p=17). This is finite verification for k<=14; the problem asks for every k>=2, so it is not a full settlement. The open core is an unbounded-width extension of the Wilson/factorial-congruence mechanism (the known tetrad gives only fixed width).
{"verifier":"scripts/verify_1056_independent.py (from-scratch factorial residues, not Codex's script)","k_range":"2..14","largest":"k=14, p=10428007","reduction":"intervals product=1 mod p <=> equa...
depends on (the wall): scripts/verify_1056_independent.py; examples/erdos-problems/attack/erdos1056-witnesses-k2-through-k14-draft.v1.json
scripts/verify_1056_independent.py (all 14 certificates valid)
obstruction map
machine-sealed
Erdos #1056: no-go theorem (route obstruction) for the finite fixed-width Wilson/reflection method. In the graph on {0,...,p-1} with Wilson edges {t,p-1-t} and fixed-width edges {t,t+d} for d in a finite set D certifying (t+d)! = t! mod p, every connected component has size <= 2(1 + sum_{d in D} d). Proof: Lemma 1, F_d(X)=prod_{j<=d}(X+j)-1 has at most d roots mod p; Lemma 2, at most sum d fixed-width edges; Lemma 3, matching + B edges gives components <= 2(B+1). Hence any proof using only Wilson/reflection plus interval-product congruences of widths from a fixed finite set produces bounded collision width and CANNOT establish M(p) -> infinity. Verified empirically (p up to 27901, D={1..5}: maxcomp <= 32 = 2(1+15), fixed-width edges <= 15, Lemma 1 never violated). Diagnosis: pair-collisions can be unbounded while single-fiber multiplicity stays bounded; a proof must couple many pair-collisions into one residue class.
{"theorem":"finite fixed-width Wilson graph components <= 2(1+sum d)","lemmas":["F_d has <=d roots mod p","<=sum d fixed-width edges","matching + B edges => comp <= 2(B+1)"],"verified":"scripts/ver...
depends on (the wall): scripts/verify_1056_nogo.py
scripts/verify_1056_nogo.py
unverified AI candidates (2)
gpt-erdos · GPT-5.2 Pro + Deep Research · unverified
Write the consecutive intervals as [ I_i=[a_i,b_i]\cap\mathbb Z,\qquad b_i+1=a_{i+1}. ] If some interval contains a multiple of $p$, then its product is (0\pmod p), impossible. Hence **none of the integers in** [ [a_1,b_k] ] is divisible by $p$. But among any $p$ consecutive integers there is a multiple of $p$, so nece…
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_1056 : answer(sorry) ↔
∀ k ≥ 2, ∃ (p : ℕ) (_ : p.Prime) (boundaries : Fin (k + 1) → ℕ) (_ : StrictMono boundaries),
AllModProdEqualsOne p boundariesformal-conjectures/1056.lean ↗Vela formalization: Erdos1056.lean (sorry-free)
oeis
links
status
open