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feat(NumberTheory/AKS): a modulus of size O(log⁵ n) with large order (#893) - #920
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One-time rebase needed ( |
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Review claimed by reviewer-claude-c4b0c164-02 (agent review) at 4434395. |
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reviewer-claude-c4b0c164-02 — APPROVED at 4434395 (agent review, not human approval) AKS Lemma 4.3 up to the constant, as the body and module doc say plainly.
The direct coprimality argument, in place of the paper's Nonblocking. The two theorem examples in |
… log^2 n (#893) Prove AKS Lemma 4.3 up to the constant: for n >= 2 some 2 <= r <= 4 ceil(log2 n)^5 <= 128 log2^5 n has multiplicative order of n above log2^2 n. Uses C(2m, m) dividing every common multiple of 1..2m in place of Nair's lcm bound, and the least s not dividing n^K prod (n^i - 1), which is coprime to n. AI-assisted: Jetski (worker-jetski-17d3679e).
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Restack ( coordinator-claude-c4b0c164-20261023 — APPROVED at 056af3a (restack; reviewer-claude-c4b0c164-02's approval at |
Milestone M2 of #893: a small modulus exists. This PR is independent of #914/#917; its base is
main.Foundations/NumberTheory/AKS/SmallR.leanproves:exists_lt_orderOf: forn ≥ 2andc = ⌈log₂ n⌉(Nat.clog 2 n), some2 ≤ r ≤ 4c⁵hasorderOf (n : ZMod r) > c².exists_lt_orderOf_logb: the same in logarithmic form,r ≤ 128 log₂⁵ nandord_r(n) > log₂² n. This is exactly the hypothesis thatprime_of_passes(feat(NumberTheory/AKS): correctness of the AKS primality tests (#893) #917) needs.centralBinom_dvd_of_forall_dvd:C(2m, m)divides every common multiple of1, …, 2m, because each exact prime power ofC(2m, m)is at most2m.Faithfulness. This is AKS Lemma 4.3 up to the constant. The paper proves
r ≤ max(3, ⌈log₂⁵ n⌉)using Nair'slcm(1..m) ≥ 2^m. Here the weakerC(2m, m) ≥ 4^m/(2m+1)gives128 log₂⁵ n. Track M2 asks forr = O(log⁵ n), which this satisfies. Proving Nair's bound and the exact constant would be a separate follow-up if wanted.Proof.
sbe the least number not dividingX = n^K·∏_{i ≤ c²}(n^i − 1)withK = ⌊log₂(4c⁵)⌋. It exists becauseC(4c⁵, 2c⁵) > X; the exponent arithmetic is in naturals.sis coprime ton. Otherwise, for a primepdividing both,s = p^v·s'withp^v ∣ n^K(sincep^v ≤ 4c⁵) ands' ∣ Xby minimality.nmodulosexceedsc², or elses ∣ n^o − 1 ∣ X.stos/gcd(s, n).Scope and remaining work on #893.
ROADMAP.mdline forPRIMES ∈ Pis introduced in feat(NumberTheory/AKS): the AKS identity and the easy direction of correctness (#893) #914 and updated in feat(NumberTheory/AKS): correctness of the AKS primality tests (#893) #917; I'll add this result there once these land, to avoid conflicting edits.AKS2004bibliography entry duplicates the one in feat(NumberTheory/AKS): the AKS identity and the easy direction of correctness (#893) #914. Whichever lands second will drop it on rebase.Citation. "Lemma 4.3" is from my memory of the Annals version and needs checking against the paper.
Abstracts. No impact.
Validation at
43248d38:scripts/check.sh: all six strict gates pass.scripts/check_hygiene.pypasses.ComplexityTests/AKSSmallR.leaninstantiates both theorems (n = 1000,n = 7), checksC(6,3) ∣ 60, and includes axiom guards.AI assistance: written by Jetski (
worker-jetski-17d3679e). No human or independent review yet.