# Proof attempts for three Millennium problems

**Reality Bit Engine 0.11.0 — 4 September 2026**  
**Research framework:** Anthony Michael Collado  
**Evidence:** Written restricted proofs, exact arithmetic certificates, and explicit counterexamples to proposed lemmas.  
**Millennium problems solved: zero.** No novelty or proof-assistant verification is claimed.

The targets are the Riemann Hypothesis, P versus NP, and three-dimensional Navier–Stokes existence and smoothness. They remain on the Clay Mathematics Institute's [unsolved problem list](https://www.claymath.org/millennium-problems/). This supplement develops arguments beyond finite numerical scans, states their precise scope, and records where the attempted routes fail.

| Target | Result obtained in this supplement | Missing step for the full problem |
|---|---|---|
| Riemann Hypothesis | Lagarias inequality for every integer with at most two distinct prime factors, using an analytic tail bound and exact finite base cases | A bound covering integers with arbitrarily many distinct prime factors |
| P versus NP | Exact minimum violated-clause cost and the number of minimizing assignments for every size of a specified pigeonhole CNF family | A general polynomial-time algorithm for an NP-complete problem, or a proof ruling out all such algorithms |
| Navier–Stokes | Explicit disproof of two tempting universal estimates | A sufficient global regularity estimate for all allowed data, or a breakdown construction meeting the official conditions |

The written arguments carry the infinite quantifiers. Bounded enumeration checks their implementation; enumeration alone is not the argument for arbitrary size.

## 1. Riemann route: a restricted infinite family

Define the harmonic number and sum of positive divisors by

\[
H_n=\sum_{j=1}^n\frac1j,
\qquad \sigma(n)=\sum_{d\mid n}d.
\]

The all-integer statement

\[
\sigma(n)\le H_n+e^{H_n}\log H_n\quad(n\ge1)
\tag{R}
\]

is equivalent to RH by [Lagarias's theorem](https://arxiv.org/html/math/0008177v2). Our restricted result concerns a proper subset of the integers in this quantifier.

### Theorem R

Inequality (R) holds for every positive integer with at most two distinct prime factors. Equality occurs at 1; the inequalities established below for larger integers are strict.

**Proof, large integers.** For \(n>1\) with at most two distinct prime factors,

\[
\frac{\sigma(n)}n
=\prod_{p^a\parallel n}\left(1+\frac1p+\cdots+\frac1{p^a}\right)
<\prod_{p\mid n}\frac p{p-1}\le 2\cdot\frac32=3.
\tag{R1}
\]

With only one prime factor, the product is at most 2, so the same upper bound applies.

Convexity of \(1/x\), applied separately on every interval \([j,j+1]\), gives

\[
\log n\le \sum_{j=1}^{n-1}\frac12\left(\frac1j+\frac1{j+1}\right)
=H_n-\frac12-\frac1{2n}.
\]

Consequently

\[
H_n>\log n+\frac12,
\qquad e^{H_n}>n e^{1/2}>\frac32n.
\tag{R2}
\]

We need only two coarse, rationally certified constants. The positive-term logarithm expansion gives

\[
\log2>
2\left(\frac13+\frac1{81}+\frac1{1215}\right)
=\frac{842}{1215}>\frac{69}{100}.
\]

For the exponential, the ratios after the first omitted term of the series truncated at order 12 are at most \(2/14\). Thus

\[
e^2<\sum_{j=0}^{12}\frac{2^j}{j!}
+\frac{2^{13}/13!}{1-2/14}
=\frac{134800213}{18243225}<\frac{37}{5}.
\]

If \(n\ge1024=2^{10}\), then

\[
H_n>10\log2+\frac12>\frac{37}{5}>e^2,
\qquad \log H_n>2.
\]

Combining this with (R1) and (R2),

\[
H_n+e^{H_n}\log H_n>3n>\sigma(n).
\]

This covers every allowed integer above the cutoff, with no upper limit on its size or on the exponents of its prime factors.

**Proof, finite base cases.** The accompanying certificate verifies all integers \(1\le n\le1023\), including those outside the restricted family. The equality at 1 is exact. For 2 through 1023 the lower bound on the right-hand side is strictly greater than the exact divisor sum. These are integer interval certificates, not floating-point comparisons. Combining these finite cases with the preceding analytic argument proves the stated restricted result. \(\square\)

### What the finite certificate means

The machine represents each nonnegative interval by integer endpoints \([L,U]\) at scale \(S=2^{128}\), meaning the true quantity belongs to \([L/S,U/S]\). Harmonic sums, products, logarithms, and exponentials round outward. At series budget \(m=96\), the exponential tail uses

\[
\sum_{j=m+1}^{\infty}\frac{x^j}{j!}
\le\frac{x^{m+1}/(m+1)!}{1-x/(m+2)}
\quad(0\le x<m+2).
\]

For the logarithm, with \(z=(x-1)/(x+1)\in[0,1)\), it uses

\[
\log x=2\sum_{j=0}^{m-1}\frac{z^{2j+1}}{2j+1}+R_m,
\quad
0\le R_m\le\frac{2z^{2m+1}}{(2m+1)(1-z^2)}.
\]

All endpoint comparisons are integer comparisons. The JSON retains each divisor sum and both bounds; wide integers are decimal strings to avoid truncation in JSON readers. The interval routine is the previously audited companion, `tools/residual_audit.py`. The proof depends on its stated rounding and tail bounds and its correct execution; it has not been checked by Lean, Coq, or another proof assistant.

A negative control changes \(\sigma(12)=28\) to the false value 29. The checker then reports a violation. This guards against a checker that accepts every proposed inequality.

### Where the attempt stops

For unrestricted integers, the product in (R1) is not bounded by 3. For example, adding a third prime removes that particular uniform estimate. Proving cases with arbitrarily large exponents of two primes does not cover integers with arbitrarily many prime factors.

The tempting general bound \(\sigma(n)\le nH_n\) is true, because \(\sigma(n)/n=\sum_{d\mid n}1/d\le H_n\). But using it to prove (R) would require

\[
H_n+e^{H_n}\log H_n-nH_n\ge0.
\]

The exact interval upper bound for this expression is already negative at \(n=12\). That refutes this sufficient-bound route; it does not refute (R). We therefore preserve this failed lemma as an explicit obstruction rather than adjusting its residual to zero.

## 2. SAT route: exact formulas at every size

Take \(p\ge1\) labeled pigeons and \(h\ge1\) labeled holes. Let \(x_{ij}\in\{0,1\}\). The CNF has one at-least-one-hole clause per pigeon, a negative two-literal clause for each pair of holes used by the same pigeon, and a negative two-literal clause for each pair of pigeons using the same hole. This exact encoding matters: another encoding can have a different number of violated clauses.

Let \(E(x)\) be the number of violated clauses. Put

\[
s_i=\sum_jx_{ij},\qquad t_j=\sum_ix_{ij},
\]

and let \(k=\sum_i s_i\), \(r=|\{i:s_i>0\}|\), \(q=|\{j:t_j>0\}|\). Here \(k\) counts selected pigeon–hole pairs; \(r\) and \(q\) count occupied rows and columns.

### Theorem S1: minimum error

\[
\boxed{\min_x E(x)=\max(0,p-h).}
\]

**Proof.** Direct counting gives

\[
E(x)=p-r+\sum_i\binom{s_i}{2}+\sum_j\binom{t_j}{2}.
\]

For each occupied column, use

\[
\binom t2=(t-1)+\frac{(t-1)(t-2)}2.
\]

Since \(\sum_{j:t_j>0}(t_j-1)=k-q\), we obtain the exact decomposition

\[
\boxed{
E(x)=p-h+(k-r)+\sum_i\binom{s_i}{2}+(h-q)
+\sum_{j:t_j>0}\frac{(t_j-1)(t_j-2)}2.
}
\tag{S1}
\]

Every term after \(p-h\) is nonnegative for every Boolean assignment. Hence \(E(x)\ge p-h\); also \(E(x)\ge0\).

If \(p\le h\), assigning each pigeon to a distinct hole attains zero. If \(p>h\), assigning exactly one pigeon to each hole and leaving the other \(p-h\) pigeons unassigned attains \(p-h\). These constructions attain the bound in every case. \(\square\)

### Theorem S2: number of minimizing assignments

If \(p<h\), the number is \(h!/(h-p)!\). If \(p\ge h\), it is

\[
\boxed{
N_{\min}(p,h)=
\sum_{d=0}^{\min(h,p-h)}
\binom hd\frac{p!}{(p-h-d)!\,2^d}.
}
\tag{S2}
\]

**Proof.** For \(p<h\), zero-cost assignments are exactly injections from the pigeons into the holes.

For \(p\ge h\), equality in (S1) requires every nonnegative term to vanish. Each pigeon is assigned to at most one hole, every hole is occupied, and every hole contains either one or two pigeons. Conversely, these conditions make every extra term zero and therefore attain the minimum.

Choose the \(d\) holes containing two pigeons in \(\binom hd\) ways. Assign \(h+d\) distinct pigeons to the labeled holes: there are \(p!/(p-h-d)!\) ordered choices, divided by \(2^d\) because the two pigeons in each double-occupied hole are unordered. The other pigeons are unassigned. The constraints on \(d\) are precisely \(0\le d\le\min(h,p-h)\). Summation gives (S2). \(\square\)

For the earlier four-pigeons/three-holes experiment,

\[
\min E=1,
\qquad N_{\min}=4!+3\frac{4!}{2}=24+36=60.
\]

More generally, for \(p=h+1\),

\[
N_{\min}=p!+(p-1)\frac{p!}{2}=\frac{(p+1)!}{2}.
\]

The earlier count of 60 now has an all-size explanation. It was not evidence that the contradiction could become satisfiable after a coordinate change.

### Where the attempt stops

These formulas exploit the incidence structure of a specific CNF family. They do not decide arbitrary CNF formulas. The [official P versus NP description](https://www.claymath.org/wp-content/uploads/2022/06/pvsnp.pdf) concerns general computational classes: a fast solver for this restricted family, or a lower bound for this particular encoding or solver, does not decide the question.

The independent enumeration checked all 112,938 assignments across the 21 sizes with \(1\le p\le6\), \(1\le h\le4\), \(ph\le16\). Every comparison between the direct clause evaluator, graph decomposition, and closed formulas passed. Its bounded runtime is not a claimed complexity bound for arbitrary SAT.

## 3. Navier–Stokes route: two exact obstructions

Work on the periodic domain \(\mathbb T^3=(\mathbb R/2\pi\mathbb Z)^3\), with no external forcing and \(\nu>0\). This differs from unit-period coordinates by a change of units. Write \(\langle f\rangle\) for the normalized spatial mean. Define energy and enstrophy as

\[
E=\frac12\langle|u|^2\rangle,
\qquad
Z=\frac12\langle|\nabla\times u|^2\rangle
=\frac12\langle|\nabla u|^2\rangle.
\]

The final equality follows for divergence-free periodic fields by integration by parts. The governing equation is

\[
\partial_tu+(u\cdot\nabla)u=-\nabla P+\nu\Delta u,
\qquad \nabla\cdot u=0.
\]

The [official problem statement](https://www.claymath.org/wp-content/uploads/2022/06/navierstokes.pdf) requires a global smoothness argument for every permitted initial field in an allowed setting, or a specified breakdown example. The following results disprove proposed sufficient estimates; neither is a breakdown example.

### Theorem N1: energy alone does not bound initial enstrophy

For every integer \(N\ge1\), the smooth exact solution

\[
u_N(x,y,z,t)=e^{-\nu N^2t}(\sin Nz,\cos Nz,0),\qquad P=0
\]

satisfies

\[
E(0)=\frac12,\qquad Z(0)=\frac{N^2}{2}.
\]

**Proof.** The field is divergence-free. It depends only on \(z\), while its \(z\) component is zero, so \((u_N\cdot\nabla)u_N=0\). Its Laplacian is \(-N^2u_N\), which matches its time derivative divided by viscosity. Direct differentiation also gives \(\nabla\times u_N=Nu_N\), and \(\sin^2Nz+\cos^2Nz=1\). The two mean identities follow immediately. \(\square\)

Thus no finite bound \(Z(0)\le F(E(0),\nu)\), independent of the remaining initial data, can hold for every smooth field: the right side is fixed while the left side grows without bound. Estimates depending on derivatives of the initial field or on positive time are not ruled out by this statement. Each displayed solution is globally smooth.

### Theorem N2: no fixed positive weighted energy–enstrophy sum is universally decreasing

For any fixed \(\lambda>0\) and \(\nu>0\), there is smooth divergence-free periodic initial data for which

\[
\left.\frac{d}{dt}(E+\lambda Z)\right|_{t=0}>0.
\]

**Proof.** Consider the initial field

\[
u_0=A\bigl(\sin y,\;0,\;\cos x-\cos(x+y)\bigr),\qquad A>0.
\tag{N2}
\]

It is smooth, periodic, mean zero, and divergence-free. This is initial data for a local smooth solution, not a claim that the field remains stationary. Smooth local solutions for these data are within the standard local theory discussed in the official problem description.

Set \(w=\cos x-\cos(x+y)\). Orthogonality of distinct Fourier modes gives

\[
E(0)=\frac34A^2,\qquad Z(0)=A^2,
\qquad\langle|\Delta u_0|^2\rangle=3A^2.
\]

The periodic energy identity yields

\[
E'(0)=-\nu\langle|\nabla u_0|^2\rangle=-2\nu A^2.
\]

The enstrophy identity, with the pressure term removed by integration by parts and incompressibility, is

\[
Z'=-\left\langle\sum_{i,j}\partial_j u_i\,
\partial_j\bigl((u\cdot\nabla)u_i\bigr)\right\rangle
-\nu\langle|\Delta u|^2\rangle.
\]

Only the vertical advective component contributes for (N2). Terms containing a full \(x\)-derivative integrate to zero. The remaining production term is

\[
-A^3\langle w_xw_y\cos y\rangle=\frac{A^3}{4},
\]

because \(w_x=-\sin x+\sin(x+y)\), \(w_y=\sin(x+y)\),
\(\langle\sin x\sin(x+y)\cos y\rangle=1/4\), and
\(\langle\sin^2(x+y)\cos y\rangle=0\). Consequently

\[
Z'(0)=\frac{A^3}{4}-3\nu A^2,
\]

and

\[
\boxed{
(E+\lambda Z)'(0)=\frac{\lambda A^3}{4}-(2+3\lambda)\nu A^2.
}
\]

Choosing \(A>4\nu(2+3\lambda)/\lambda\) makes this expression positive, proving the claim for every fixed positive weight. \(\square\)

For the concrete values \(A=1\), \(\nu=1/100\), \(\lambda=1\), the exact results are

\[
E'(0)=-\frac1{50},\qquad
Z'(0)=\frac{11}{50},\qquad
(E+Z)'(0)=\frac15.
\]

These values are also obtained independently by sparse Fourier algebra with rational real and imaginary coefficients. The program differentiates and multiplies the polynomials exactly and extracts their zero Fourier coefficients. No spatial mesh, time step, numerical tolerance, or floating-point cancellation is used.

### Where the attempt stops

Enstrophy growth at an instant is compatible with global regularity. The counterexample rules out monotonicity of this particular fixed-weight functional, not all possible regularity estimates, data-dependent choices of weight, or nonlinear functionals. Our restricted, \(z\)-independent initial field does not establish finite-time singularity. A valid route must control nonlinear growth with an estimate strong enough for continuation, uniformly as the numerical cutoff is removed.

## 4. What D0 can certify

For the SAT constraint, \(D_0\) can represent exact zero violation. Theorem S1 proves that some inputs have a strictly positive minimum. Their failure to reach zero is a mathematical incompatibility in the constraints, not necessarily a miscomputed factor.

More generally, let \(r\) be an equation residual and let a new coordinate map \(f\) satisfy \(f(0)=0\). If \(f\) is injective, then

\[
f(r)=0\iff r=0.
\]

**Proof.** The forward implication follows from \(f(r)=f(0)\) and injectivity; the reverse follows from \(f(0)=0\). \(\square\)

A coordinate transformation preserving distinctions cannot make a false equality true. A transformation mapping both zero and a nonzero residual to zero loses the distinction needed to certify the equation. For inequalities, satisfaction is a sign condition, so the engine keeps the signed residual as well as the nonnegative violation measure.

This gives the instruction to “freeze errors” a useful precise meaning: retain exact counterexamples with their inputs, assumptions, and rejected claim. The frozen examples prevent a later search from reusing a lemma already shown false. They do not establish that every unsolved problem has a missing coordinate that will force its residual to zero.

## 5. Reproduction and evidence boundaries

From the extracted project root, with Python 3.10 or later:

```sh
python3 tools/proof_routes.py --output results/my-proof-routes.json
python3 -m unittest discover -s tests -p 'test_proof_routes.py' -v
```

The new companion uses only the Python standard library and does not require the C++ executable. The existing C++ build instructions remain in `README.md`. The frozen result is `results/proof-routes-v1.json`; it records the source and proof-document SHA-256 values, exact base-case intervals, SAT histograms, rational fluid identities, and explicit false flags for resolution of all three Millennium problems.

The six added test methods check Gaussian rational arithmetic, Fourier orthogonality, differentiation, divergence rejection, signed amplitude scaling, known SAT counts, and preservation of nonzero contradiction cost. The executable additionally checks the complete stated SAT enumeration, all 1,023 RH base cases, the exact rational tail constants, and the fluid identities and counterexamples. Fingerprints establish file identity, not mathematical correctness.

The restricted theorems are presented with proofs and executable arithmetic support. They are not claimed to be new to mathematics. The remaining all-input proof obligations in the opening table remain unresolved. The seven-layer interpretation and theological assumptions supply none of those missing estimates by themselves.