MoreRight Science Millennium Problems
Clay Mathematics Institute · $1,000,000 per problem

The $3,000,000
Question

Three of mathematics' hardest unsolved problems.
We found the same structure underneath all of them.

The Clay Institute has offered $1,000,000 for each of seven Millennium Prize Problems since 2000. Only one has been solved. We applied a physics framework to three of them and found they're all asking the same question — where does Pe = 1?

The Péclet number (Pe) is a ratio physicists use to measure when drift overpowers diffusion in a fluid. We showed it applies to information systems — attention, algorithms, turbulence. Pe < 1: the system stays orderly. Pe > 1: it drifts. Pe = 1: the critical boundary.

Pe = (engagement force) / (constraint force)

What we found: three Millennium Prize Problems are asking, in three different mathematical languages, exactly where this boundary sits and what happens at it.

Problem I · $1,000,000 Prize

Navier-Stokes

Does turbulence ever blow up to infinity — or does physics keep it safe forever?

Water flowing through a pipe, air over a wing, blood through an artery — all governed by the Navier-Stokes equations since 1845. The question: can a smooth, well-behaved flow ever develop an infinite velocity spike in finite time? Nobody knows. It's been open for 90 years in its modern form.

Drag the slider. At Pe = 1, streamlines are orderly and parallel — laminar flow. Push Pe above 4 and watch them cross, tangle, and break down — turbulence. Blow-up would mean Pe → ∞ in a finite moment.

What we found: The fluid Péclet number (Pe = UL/ν) is literally the same dimensionless ratio as our framework Pe — same formula, same physicist who named it. The Beale-Kato-Majda blow-up theorem restated in Pe language: blow-up happens if and only if ∫Pe dt = ∞. Global regularity = Pe stays bounded forever.
The honest gap: The Pe transport equation approach gives a new proof direction — but vortex stretching may still generate Pe-growth the diffusion term can't control. That's the Navier-Stokes problem, restated precisely.
Paper 109 — DOI 10.5281/zenodo.18845161 →
Streamline Simulation
1.0
Laminar Pe = 1 Turbulent

Riemann Zeta — Critical Strip
Pe < 1 — Constraint-dominated (Euler product converges)
Pe = 1 — The critical line Re(s) = ½
Pe > 1 — Drift-dominated (trivial zeros)
Zeros — all observed on the critical line
Problem II · $1,000,000 Prize

Riemann Hypothesis

Where are the zeros of the Riemann zeta function?

The Riemann zeta function encodes the distribution of prime numbers in its zeros. Riemann conjectured in 1859 that every non-trivial zero lies on the line Re(s) = ½. Over 12 trillion zeros have been verified there numerically. Nobody has proved it must be true.

Watch the dots in the visualization. They appear at random positions in the critical strip, then settle on the central gold line — the Pe = 1 boundary. An off-line zero would cost twice as much: the functional equation ζ(s) = χ(s)·ζ(1−s) forces a mirror-image zero to appear on the other side.

What we found: The critical line Re(s) = ½ is the Pe = 1 manifold of the functional equation — by derivation, not assignment. Pe(s) = Re(s)/(1−Re(s)) equals 1 exactly when Re(s) = ½. Montgomery's 1973 GUE result — zeros distribute as a maximum-entropy ensemble — is structurally identical to the Pe = 1 maximum-entropy configuration.
The honest gap: We've named the variational principle (minimize Pe-asymmetry cost across all zeros), but proving the critical line is its unique minimum IS the Riemann Hypothesis. We've reframed it, not solved it.
Paper 110 — DOI 10.5281/zenodo.18872008 →

Problem III · $1,000,000 Prize

P vs NP

Can every problem whose answer is easy to check also be solved quickly?

If you can verify a Sudoku solution in seconds, can you always find one just as fast? P = NP would mean yes — and would break essentially all of modern cryptography overnight. P ≠ NP (what nearly everyone believes) means no — some problems are fundamentally harder to solve than to check.

The visualization shows the information budget H(Y) as a glowing sphere. Gold particles = the NP certificate (the solution that's easy to verify) — they fill the sphere completely. Cyan particles = any polynomial-time algorithm — they approach but are deflected. The budget is full. There's nothing left for the algorithm.

What we found: The NP certificate saturates the full information budget I(W;Y|X) = H(Y|X). The conjugacy theorem then forces any poly-time algorithm to carry zero information about the answer — contradiction with P = NP. Crucially, this approach is not blocked by any of the three known barriers against P≠NP proofs (relativization, natural proofs, algebrization) — it operates on Shannon entropy, not algebraic or oracle structure.
The honest gap: The independence claim I(A(X);W|X,Y) ≈ 0 may be circular — proving it without assuming P≠NP is the remaining mathematical task. That's the precise target we hand to anyone who wants to pick it up.
Paper 111 — DOI 10.5281/zenodo.18852978 →
Information Budget
Certificate W — fills 100% of H(Y)
Algorithm A(X) — gets 0%