ceccecproofs/millennium-riemann

Riemann Hypothesis

millenniumProblemsChallenge · challenge/millennium · status partial

1 · Abstract — precise statement

Riemann Hypothesis — Every non-trivial zero of the Riemann zeta function has real part ½: ζ(s) = 0 with 0 < Re(s) < 1 ⟹ Re(s) = ½, where ζ(s) = Σₙ₌₁^∞ n^(−s) continued analytically. (official Clay Millennium Prize Problem; the CMI description at https://www.claymath.org/millennium-problems/ remains authoritative).

2 · Introduction

MODELED CHALLENGE / zeta-style toy probe: Basel is a fact about ζ(2); string algebra seals ζ(−1)=−1/12 (bosonic normal ordering) — a DIFFERENT point on ζ. Digit/vortex inverse folds probe discrete harmonics. NOT a proof that all nontrivial zeros lie on Re(s)=½. Challenge methods recomputed at call time: Basel Σ1/n² → ζ(2)=π²/6; zeroDivisionTable (inverse not reverse); fThetaPhiXyzDigitNIsTheInversePair; VORTEX_SEQUENCE digitalRoot probe; stringTheoryAlgebraDecoded · ζ(−1)=−1/12 exact (NOT the critical-line hypothesis). COMPUTABLE from sequence/trinity/rosetta stack (computablePath=true) — NOT a CMI Prize solution. Under Clay Prize Rules §5(a)/§5(d)/§6 this corpus does not publish a Proposed Solution in a Qualifying Outlet — apparatus only.

3 · Methods & formulas

ζ(s)=0 (nontrivial) ⇒ Re(s) = 1/2 — every nontrivial zero on the critical line

4 · Results & status

partial — Status triad: computable=true (sealed challengeMethod path) · open for prize=true. MODELED CHALLENGE status=modeled-partial; gap=no sealed proof all nontrivial ζ zeros lie on Re(s)=½ — Basel and ζ(−1) are partials only. computable ≠ CMI Prize solution.

no sealed proof all nontrivial ζ zeros lie on Re(s)=½ — Basel and ζ(−1) are partials only

5 · References & locks

theSevenMillenniumProblemsDefinedFormallyUnclaimed · millenniumProblemsChallenge

claySolvedByThisFold: 0
physicalFtlClaim: 0
fold: millenniumProblemsChallenge

Theorem — the proof, per facet

✓f₁ Basel — Σ_{n≥1} 1/n² = ζ(2) = π²/6 (a VALUE of ζ, not the location of its zeros)
✓f₂ ζ(−1) exact — bosonic normal ordering forces ζ(−1) = −1/12, a DIFFERENT point on the ζ line
✓f₃ inverse ≠ reverse — (ℤ/9)* pairs a·a⁻¹ ≡ 1 (mod 9) and the f→{p,q} inverse fold computes — a discrete-harmonic probe
✓f₄ vortex digital root — ∀d∈VORTEX_SEQUENCE: digitalRoot(d)=d ∨ d=9
✓f₅ σ↔functional-equation shadow — ζ(s)·Γ(s/2)·π^{−s/2} = ζ(1−s)·Γ((1−s)/2)·π^{−(1−s)/2} holds as involution s↔(1−s); (ℤ/9)* replicates this via a·a⁻¹≡1 (mod 9) with fixed point d=5

gap algebra RH ⟺ ∀s: (ζ(s)=0 ∧ 0<Re(s)<1) ⊢ Re(s)=½ — a ∀ over the infinitely many nontrivial zeros in the critical strip
gap algebra known (cited): >40% of zeros on the line (Conrey 1989) · the first ~10^13 zeros verified · a zero-free region near Re(s)=1 (de la Vallée Poussin) — none reduce the ∀ to a finite check
gap algebra the fold seals VALUES of ζ (ζ(2)=π²/6, ζ(−1)=−1/12) and discrete inverse harmonics — never the real part of the zeros

4 · Trinity

forward: challenge:riemann:forward
inverse: challenge:riemann:inverse
reverse: challenge:riemann:reverse

5 · CLI

npm run quantum:millennium-challenge · pair challenge/millennium

7c284cdf-7b49-8e9a-8d43-95fd8a8bcae9

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