The 11/12 Line
Start · A wall in the complex plane
A guided climb · from primes to a new proof
11⁄12

The Eleven-Twelfths Line

The Riemann zeta function is a formula that secretly encodes every prime number. Where it equals zero tells you how evenly the primes are spread out. For 167 years nobody could prove that there is a fixed safe zone, of any width at all, where it is never zero.

A new paper, The Quasi-Riemann Hypothesis (October 5, 2026), proves exactly that: no Dirichlet L-function, including Riemann’s zeta function, has a zero to the right of the line Re s = 11/12.

This page starts from what you learned in middle school (primes, fractions, exponents, angles) and climbs in turns. Each turn revisits the same ideas one level higher, until the five moves of the proof make sense.

Start at Turn 1 Jump to the proof

The paper is credited to OpenAI and is a brand-new preprint. Major claims in mathematics are accepted only after experts have checked every line, which usually takes months. This page explains the argument as the paper presents it.

The zeta function, painted. Colour shows direction, brightness shows size, and every dark pinpoint is a zero. All the zeros sit on the teal line ½. The theorem guarantees the hatched region right of the magenta line can never contain one, at any height, for zeta or any of its cousins. Hover or tap to read values.
Turn 1 · Middle-school math

Counting primes

Everything on this page grows out of one question: how many primes are there below a given number, and how accurately can we predict that count?

You need: multiplication, divisionYou’ll meet: π(x), logarithms, fractional exponents

A prime is a whole number bigger than 1 whose only divisors are 1 and itself: 2, 3, 5, 7, 11, 13, … Every whole number splits into primes in exactly one way, for example 60 = 2 × 2 × 3 × 5. Primes are the atoms that all other numbers are built from.

More than 2,200 years ago Eratosthenes found a simple way to list them. Write the numbers out, circle 2, and cross out every multiple of 2. The next number that is not crossed out (3) must be prime, so circle it and cross out its multiples. Keep going.

The sieve of Eratosthenes, 1 to 200
Interactive

You only need to sieve with primes up to 14, because 14 × 14 is just under 200: any composite number up to 200 has a prime factor of 13 or less.

Euclid proved the primes never run out. They do thin out, though. Mathematicians write π(x) for the number of primes up to x (here π is just a name, not 3.14159…). For example π(10) = 4, because 2, 3, 5 and 7 are the primes up to 10.

Around 1792, as a teenager, Carl Friedrich Gauss studied tables of primes and guessed that near a number x, about one number in every ln x is prime. Here ln is the natural logarithm, which roughly counts how many digits a number has, times 2.3. Adding up those chances gives a smooth curve called the logarithmic integral, written Li(x).

The prime count π(x) and two guesses
Computed live
π(x), the true countx / ln xLi(x), Gauss’s refined guess

Li(x) is astonishingly good. Up to a million it misses the true count of 78,498 primes by only about 130. The Prime Number Theorem (proved in 1896) says the guess is right in the long run. The deeper question is how big the error can get. To talk about that we need exponents that are fractions.

Fractional exponents

You know that x2 means x × x. A fractional exponent undoes this: x1/2 is the square root √x, because √x × √x = x. In the same way x1/12 is the number that gives x when multiplied by itself 12 times, and x11/12 is that twelfth root multiplied by itself 11 times. The bigger the exponent, the faster the quantity grows. An error of size x1/2 is tiny next to x, and an error of size x11/12 sits in between.

How big is x raised to a power?
Interactive

The last column is the important one: what fraction of x the error could be. With exponent 11/12 and x = 1024, the twelfth root of x is 100, so the error is at most about 1% of x (ignoring a slowly growing logarithm factor). Every extra digit in x shrinks that percentage further.

The error |π(x) − Li(x)| compared with √x
Computed live · up to one million
actual error√x (the size the Riemann Hypothesis predicts, up to a log factor)x11/12 (the new proven ceiling, up to constants and logs)

Both axes go up by factors of 10. The real error hugs the √x curve, just as the Riemann Hypothesis predicts. Before 2026 nobody could prove the error stays below x0.9999, let alone √x. The best proven bounds were only slightly better than “a tiny fraction of x”.

Turn 2 · Fractions that never end

A sum that knows the primes

To study the error we need a tool that “sees” every prime at once. That tool is an infinite sum.

You need: fractions, adding many numbersYou’ll meet: infinite series, ζ(s), Euler’s product

Some infinite sums settle down to a finite total. Add 1 + ½ + ¼ + ⅛ + … and you creep toward 2 without ever passing it. Others grow forever: 1 + ½ + ⅓ + ¼ + … passes any number you name, just very slowly.

In between is a whole family. Raise each denominator to a power s and add:

\[ \zeta(s) \;=\; \frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots \]

This is the zeta function (ζ is the Greek letter zeta). For s = 2 the sum is 1 + ¼ + 1/9 + 1/16 + … , and in 1734 Euler showed the total is exactly π²/6 = 1.6449…, the circle number appearing out of nowhere. Use the slider to see which powers settle down.

Adding 1/ns, one term at a time
Interactive

Why zeta knows the primes

Euler’s great discovery (1737) is that the same sum can be written as a product with one factor for each prime:

\[ \zeta(s)=\Bigl(1+\tfrac1{2^s}+\tfrac1{4^s}+\cdots\Bigr)\Bigl(1+\tfrac1{3^s}+\tfrac1{9^s}+\cdots\Bigr)\Bigl(1+\tfrac1{5^s}+\cdots\Bigr)\cdots=\prod_{p\ \text{prime}}\frac{1}{1-p^{-s}} . \]

Multiply out the brackets and pick one term from each. Choosing 1/4s from the first bracket, 1/3s from the second and 1 from all others gives 1/12s, because 12 = 4 × 3. Each whole number appears exactly once, because each number factors into primes in exactly one way. So the zeta function is unique factorization written as a formula. Whatever zeta does, the primes are doing underneath.

Building ζ(s) from primes alone
Interactive
Turn 3 · Angles and arrows

Spinning arrows

Riemann’s leap in 1859 was to feed zeta numbers that are not on the number line at all. Those numbers are best pictured as arrows.

You need: angles, the coordinate planeYou’ll meet: complex numbers, s = σ + it, zeros

A complex number is a point in the plane, or equivalently an arrow from the origin to that point. It has a length and a direction. Adding complex numbers means placing arrows head to tail. Multiplying them means multiplying the lengths and adding the angles. The number i is the arrow of length 1 pointing straight up; multiplying by i turns any arrow a quarter-turn, so i × i = −1.

We write the input of zeta as s = σ + i t. The real part σ (sigma) says how far right the point is; the height t says how far up. Here is the key fact: the term 1/ns is an arrow whose

  • length is 1/nσ, so σ controls how fast the arrows shrink, and
  • direction is the angle −t × ln n, so t controls how fast they spin.

Adding up the zeta series means laying all these arrows head to tail. When the arrows shrink fast (σ bigger than 1) the chain settles at a point, and that point is ζ(s). When σ is 1 or less the plain chain drifts off forever. Flipping the sign of every second arrow fixes that and gives a cousin called η(s), equal to (1 − 21−s) ζ(s). That chain settles for every σ above 0, and inside the strip between 0 and 1 it is zero exactly where ζ is.

The arrow chain for η(s) = 1 − 1/2s + 1/3s − 1/4s + …
Interactive

The chain draws 600 arrows; the ringed dot is where the infinite chain ends up. At a zero, thousands of spinning, shrinking arrows conspire to land exactly back at the starting point. Slide σ away from ½ at the same height and the landing point moves away from the start.

Turn 4 · Reading a colour map

The map of zeros

Riemann discovered that the zeros of zeta act like musical notes, and the primes are the music they play. Where the zeros sit decides how big the prime-counting error can be.

You need: Turn 3’s arrowsYou’ll meet: the critical strip, the Riemann Hypothesis, the explicit formula

To see zeta everywhere at once, colour every point s of the plane by the arrow ζ(s): the hue shows its direction (red pointing right, green up-left, blue down-left) and the brightness shows its length. Where ζ is zero the arrow has no length, so the point turns black and every colour meets around it. The bright white blob at s = 1 is where ζ blows up to infinity (the sum 1 + ½ + ⅓ + … grows forever).

The zeta landscape
Computed live, pixel by pixel
Hover over the map to read ζ(s).

Horizontal: the real part σ from −1 to 2. Vertical: the height t, 64 units tall. The strip between σ = 0 and σ = 1 is the critical strip. The dashed gold “old zero-free curve” is the classical guarantee σ > 1 − 1/(5.56 ln t), drawn with its proven constant. It hugs the line σ = 1 more and more tightly as you climb.

Three facts about the map, all known since the 1800s:

  • There are “trivial” zeros at −2, −4, −6, … (off this map to the left), and they are well understood.
  • Every other zero lies inside the critical strip, and the pattern is mirror-symmetric about the line σ = ½.
  • The Riemann Hypothesis (1859) says all of them lie exactly on that line. Computers have checked more than ten trillion zeros, and every one is on the line. Nobody has a proof.

Why zeros control primes

Riemann found an exact formula, made rigorous by von Mangoldt in 1895, that rebuilds a weighted prime count, written ψ(x), out of the zeros. Each zero ρ = β + iγ contributes a wave. The wave wiggles faster for higher γ, and its height grows like xβ. Add the waves from more zeros and the smooth curve snaps onto the staircase of primes.

Rebuilding the primes from zeta’s zeros
Explicit formula, live
ψ(x): jumps by ln p at each prime power pkx minus the waves from the zeros

The formula is ψ(x) = x − Σ xρ/ρ − ln 2π − ½ ln(1 − 1/x2). Now push the first zero to the right with the second slider. Its wave swells like xβ and the curve stops matching the primes. That is the whole story of this page in one picture: zeros far to the right would mean big, lopsided errors in the primes. A zero-free region is a guarantee that such waves cannot exist.

The question, stated exactly

If no zero has real part bigger than θ, then the prime-counting error is at most about xθ. The Riemann Hypothesis is the dream case θ = ½. The quasi-Riemann Hypothesis asks for much less: that there is some fixed θ below 1, the same for every height. Even that had been out of reach. The new theorem gives θ = 11/12, and it does so for zeta and every Dirichlet L-function at once.

Turn 5 · Remainders and clocks

Clocks, progressions and consequences

The theorem is not only about zeta. It covers a whole family of cousins, one for every “clock” you can count primes on. That is where most of its consequences come from.

You need: remainders (clock arithmetic)You’ll meet: Dirichlet characters, L-functions, Siegel zeros

Sort primes by their remainder when divided by 4. Apart from 2, every prime leaves remainder 1 (like 5, 13, 17) or 3 (like 3, 7, 11). In 1837 Dirichlet proved that each possible remainder gets infinitely many primes, and in the long run an equal share. Try other clock sizes:

Primes sorted by remainder
Computed live

Only remainders that share no factor with q can hold more than one prime. The dashed line is the fair share, Li(x) divided by the number of allowed remainders.

To count primes on a clock of size q, Dirichlet attached a sign or a turn to each remainder, called a character χ. For the clock of size 4: χ(remainder 1) = +1, χ(remainder 3) = −1, and χ(even) = 0. Because remainders multiply consistently (3 × 3 = 9, which is 1 on the 4-clock, and (−1) × (−1) = +1), so do the signs. Each character gives a zeta-like sum

\[ L(s,\chi)=\frac{\chi(1)}{1^s}+\frac{\chi(2)}{2^s}+\frac{\chi(3)}{3^s}+\cdots , \]

a Dirichlet L-function, with its own map of zeros. The Generalized Riemann Hypothesis says all of their zeros sit on the line ½ as well. Where the zeros of these L-functions sit controls how fairly the primes are shared among remainders.

The theorem, plainly

For every clock size q, every character χ, and every height t: L(s, χ) ≠ 0 whenever the real part of s exceeds 11/12.

The same number 11/12 works everywhere, with no exceptions. The paper proves this for an even larger family (Hecke L-functions over the number system ℤ[ω] you will meet in Turn 7) and gets the Dirichlet case for free, because each Dirichlet L-function appears as a factor of one of those.

The ghost the theorem exorcises

For a century, one possibility haunted the subject. The classical results allowed a single exception: one real-valued character χ could have a zero β extremely close to 1, called a Landau–Siegel zero. In 1935 Siegel showed such a zero cannot be too close to 1, but his proof gives no way to compute how close it may come. That “ineffective” constant spread into every theorem that used it. Below, conjure a ghost zero and watch what it would do to the primes on a clock.

What a Landau–Siegel zero would do
Thought experiment
squares on the clock (quadratic residues)non-squaresactual count of primes up to 106

A ghost zero at β adds a wave of size Li(xβ) that steals primes from the squares and hands them to the non-squares. For small clocks like these we already knew no ghost exists, and the real counts (black ticks) show none. The danger was for astronomically large clocks that no computer can check. The new theorem rules out every ghost with β above 11/12, on every clock, so Landau–Siegel zeros do not exist.

Four consequences you can test

The paper lists several consequences of a fixed zero-free half-plane. Each card below runs the real computation.

The least non-square
Number theory

For a prime p, half the remainders are perfect squares on the p-clock. Vinogradov conjectured that the smallest non-square is tiny, smaller than pε for any ε. The theorem proves it is at most a fixed power of ln p.

Primality without luck
Computer science

Miller’s 1976 test checks a number against bases 2, 3, 4, … . A composite always fails for some small base, provided zero-free regions hold. The theorem makes the test provably deterministic and polynomial-time. (Bach’s bound of 2(ln n)2 bases, used here, needs the full GRH; the 11/12 theorem gives a bound of the form (ln n)C with a larger fixed C.)

Euler’s 65 idoneal numbers
A 250-year-old list

Euler found 65 “idoneal” numbers, numbers that can not be written as ab + bc + ca with 0 < a < b < c. Weinberger showed in 1973 that at most one more could exist, and later work showed the list is complete if there are no Landau–Siegel zeros. So the theorem proves Euler’s list is complete.

Class numbers grow
Algebra

The class number h(D) measures how badly unique factorization fails in the number system built from √D. Siegel showed it grows roughly like √|D|, but with no computable constant. The theorem gives h(D) ≫ √|D| / ln ln |D| with a constant you can compute. Dots: every negative fundamental discriminant down to −6000; ringed dots have h = 1.

Turn 6 · 2,300 years in one strip

History, and why this result matters

The story of primes is a story of walls pushed slowly to the left, from σ = 1 toward ½. For a century the walls could only creep closer to 1 as you climbed higher. The new wall stands still.

Timeline
Tap a dot

    Fixed width is what is new

    Every earlier zero-free region was a sliver hugging the line σ = 1 that gets thinner the higher you go (and, for L-functions, the bigger the clock). The classical region has width about 1 / ln t; Vinogradov and Korobov (1958) improved this to about 1 / (ln t)2/3. Both widths go to zero. The new region has width 1/12 at every height and every clock.

    How far left of σ = 1 the zeros are proven absent
    Interactive
    classical, 1/(5.56 ln t)Vinogradov–Korobov with Ford’s constantnew: 1/12, forever

    The horizontal axis is the number of digits of the height, on a log scale: the far right is a height with a million digits. Even there, the old guarantees leave only a few millionths. The new one does not shrink.

    1899 · de la Vallée Poussin
    σ > 1 − c / ln t
    • shrinks with height
    • for L-functions: shrinks with clock size, plus one possible exception
    1935 · Siegel
    The exception is not too close to 1
    • constant cannot be computed
    • infects class numbers, primes in progressions
    1958 · Vinogradov–Korobov
    σ > 1 − c / (ln t)2/3(ln ln t)1/3
    • best for zeta for 68 years
    • still shrinks to nothing
    2026 · this paper
    σ > 11/12, everywhere
    • fixed width 1/12
    • every clock, every height
    • no exceptional zero
    • constants effective
    What it does not do

    It is not the Riemann Hypothesis. Zeros with real part between ½ and 11/12 are still not ruled out; the Riemann Hypothesis says there are none there except on the line itself. The paper notes that a companion preprint pushes the wall to 7/8 with the same method. And as with any new proof of this size, the expert community will need time to verify it.

    Turn 7 · The research frontier

    The proof, in five moves

    The argument never looks at a zero directly. It shows that a certain sum of ±1’s has lots of cancellation, then hides that sum in a huge crowd of similar sums, and controls the whole crowd at once using two symmetries.

    You need: Turns 1–6Each step has three levels: Picture · Mechanism · In the paper

    Toolkit before the climb

    Four ideas do almost all the work. We build each one with a picture first.

    A · Cancellation and the Möbius function

    Flip a fair coin N times, adding +1 for heads and −1 for tails. The total is rarely 0, but it is typically only about √N in size. Going from N to √N because the signs cancel is called square-root cancellation.

    The Möbius function μ(n) gives each whole number a sign: 0 if n is divisible by a square like 4 or 9, otherwise +1 or −1 according to whether n has an even or odd number of prime factors. So μ(6) = +1, μ(30) = −1, μ(12) = 0. It is not random, but it behaves like a coin. Its running total is the Mertens function M(x).

    Its link to zeta: Σ μ(n)/ns = 1/ζ(s). Littlewood showed in 1912 that the Riemann Hypothesis is equivalent to M(x) staying below x1/2+ε. More generally, if Möbius sums stay below xθ, zeta has no zeros to the right of θ.

    Möbius walk vs coin flips
    Computed live
    M(x) = μ(1) + … + μ(x)coin-flip walk±√x

    B · A number system with sixth roots of unity

    The proof works in the Eisenstein integers ℤ[ω]: numbers a + bω with whole numbers a, b, where ω is the arrow of length 1 at 120°. They form a triangular lattice. Like ordinary integers, they factor uniquely into primes.

    Its six units ±1, ±ω, ±ω² are the sixth roots of unity, the six arrows that return to 1 after six multiplications. That is why this field is the natural home for “sextic” (order-six) characters.

    The norm N(a + bω) = a² − ab + b² is the squared length. Some ordinary primes stay prime here (2, 5, 11, …), while those leaving remainder 1 on the 3-clock split: 7 = (3 + ω)(2 − ω), and each factor has norm 7. Among the six unit multiples of a number, exactly one is primary, which gives a canonical choice.

    The Eisenstein lattice
    Hover a point
    Power j
    Hover a point to see a + bω and its norm.

    C · Characters paint the lattice

    Switch the lattice to “Paint by a sextic character.” For a prime p, the sextic residue symbol (u/p)6 assigns to each lattice point u one of the six units, its colour on the clock below. It repeats with period p and respects multiplication: the colour of uv is the colour of u times the colour of v. It plays the role on ℤ[ω] that χ played on the 4-clock.

    Its powers matter. Squaring the symbol (j = 2) leaves three colours: the cubic symbol. Cubing it (j = 3) leaves only +1 and −1: the quadratic symbol, the cousin of “is this a perfect square?” Hold on to this fact. In Step 4 a sextic character turns into a quadratic one, and that rescues the proof.

    D · Gauss sums: walks that end on a circle

    Take a character χ and a uniform spinner: for each remainder x mod p, an arrow pointing at angle χ(x) plus a steady turn proportional to x. Add them head to tail. This sum is the Gauss sum. Its length is always exactly √N(p), so the walk always ends on the drawn circle. Dividing by √N(p) gives a number γ of length 1. Only its direction carries information.

    The paper uses three exact identities between these directions. For a prime p, μ(p) = −1 and α(p) = p/|p| is the direction of p itself:

    \[ \gamma_2(p)^3=\mu(p)\,\alpha(p),\qquad \gamma_1(p)\gamma_2(p)=\mu(p)\,\alpha(p)\,G(p),\qquad \gamma_1(p)\gamma_{-1}(p)=\chi_p(-1). \]

    The second identity is how Möbius signs get absorbed into Gauss sums in Step 3. The panel checks all three numerically, from scratch, for the prime you choose.

    A Gauss sum, arrow by arrow
    Computed live
    Character power j

    In 1846 Kummer computed cubic Gauss sums by hand and thought their directions were lopsided. Heath-Brown and Patterson proved in 1979 that, in the long run, the directions at primes spread out evenly. The survey button repeats Kummer’s experiment over two thousand primes. You will see the histogram lean toward direction 0. That lean is real: Patterson conjectured a small persistent bias that fades only like X−1/6, and Dunn and Radziwiłł proved it in 2024 (assuming GRH). Their cubic-theta machinery is a direct ancestor of Steps 4 and 5.

    1

    A zero would leave a fingerprint

    Reduce the theorem to showing a sum of Möbius signs has a power saving.

    Turn 4 showed that a zero ρ = β + iγ of zeta makes a wave of size xβ in the primes. The same is true for the Möbius sum: 1/ζ has a blow-up wherever ζ has a zero, and that blow-up shows up as a wave of size Dβ in the sum of μ(n) over n near D. So if we can prove the Möbius sum is always smaller than D11/12+ε, no zero can have β > 11/12. The wave would eventually burst through the ceiling.

    A hypothetical zero against the 11/12 ceiling
    Illustration

    Both axes are logarithmic, so powers of D become straight lines. Typical Möbius noise is around D1/2. A zero at β adds an oscillating wave of size Dβ. Any β above 11/12 makes a steeper line, which must eventually cross the ceiling.

    2

    Hide the sum in a crowd

    Bound a whole family of sums on average, then notice that many family members are copies of the one we care about.

    Proving that one specific sum is small is hard. Proving that a crowd of sums is small on average is often easier, because averaging lets symmetries act. So the paper builds a crowd: for each “row” u, the sum Au(D) is A₁(D) with every term recoloured by the sextic character of Toolkit C.

    Now the trick. If u is the sixth power of a prime, u = p6, every colour is a sixth root of unity raised to the sixth power, which is 1. So row p6 is a near-perfect copy of A₁. There are many such clone rows. If A₁ were large, the clones alone would blow the crowd’s average energy budget. Play with it:

    The clone-row budget
    Exponent calculator

    Each square stands for a block of rows; gold squares are clone rows p6. Bars show exponents of D: the energy the clones would carry, J × |A₁|² with J ≈ D(1+ϑ)/6 clones, against the total D2+ϑ that Step 2’s mean-square bound allows.

    3

    Flip to the dual side

    Poisson summation trades a long sum over rows for a short sum over “frequencies,” and Gauss sums appear as the exchange rate.

    Expanding |Au|² and summing over rows u produces sums of a repeating pattern χ(u) over a long, smooth range of u. Poisson summation says such a sum equals a second sum over frequencies h. A wide, gentle weight on one side becomes a narrow, sharp weight on the other, so a long sum becomes a short one. The price of the swap is a Gauss sum: the pattern’s own “spectrum”. The figure shows this exactly, for the square/non-square pattern on a clock of size m.

    Poisson summation: a long sum equals a short sum
    Exact identity, live

    Top: χ(k) e−πk²/H², about 2H terms that matter. Bottom: the dual side (√m · H/m) χ(h) e−πh²H²/m², with only about 2m/H terms that matter. The two totals agree to every digit. The factor √m comes from the Gauss sum of the square/non-square pattern.

    The scale ladder
    Exponents of D

    Columns live at size D. Original rows go up to H = D1+ϑ. After Poisson, dual rows go only up to 𝓗 = D1−ϑ. Uses the ϑ from Step 2’s slider.

    4

    The cubic theta function and a quadratic miracle

    Recognize the Gauss sums as coefficients of a symmetric object, use its symmetry, and land in a world where the strongest available sieve works.

    Some functions have a hidden symmetry that turns a long sum into a short one. The classic example is Jacobi’s theta function θ(t) = Σ e−πn²t, which satisfies θ(1/t) = √t θ(t). The figure checks it live. In 1969 Kubota built a cubic version, a function on 3-dimensional hyperbolic space, and in 1977 Patterson computed its coefficients: they are exactly the cubic Gauss sums γ₂(n) from Step 3. So the sum Bh is secretly a sum of theta coefficients, and theta’s symmetry can flip it.

    Then comes the step the paper hinges on. When you flip, the twist by the row h picks up an extra factor from the theta symmetry. The two combine as χ−1 · χ−2 = χ−3, and on the sextic clock −3 is the same as 3. A character to the power 3 is quadratic: just ±1, a “perfect square or not” test. For quadratic families there is an exceptionally strong tool, the quadratic large sieve.

    Jacobi theta: θ(1/t) = √t · θ(t)
    Exact identity, live
    The sextic clock
    Interactive
    Fourier factor χa
    Theta factor χb
    Why “quadratic” matters
    Exponent bars

    A large sieve bounds a mean square by a factor times the column energy. The target here needs factor about D¹. Quadratic: M + L. General n-th order: M + L + (ML)2/3.

    Where theta’s coefficients live
    Analogy with whole numbers
    squarefree n (b = 1): the terms we wantn × b³ with b > 1: extra cube termsfaded: coefficient is zero
    Hover a number.
    5

    Peel off the cubes, then descend

    Undo the completion, and handle what is left with a recursion that shrinks the problem by a fixed power each round.

    Step 4 controls the completed sum T, which includes extra cube terms. We want B, the squarefree part only. The cube terms could cancel against each other, so they can’t simply be thrown away. Instead, Möbius inversion writes B as a combination of copies of T at smaller scales X/N(d)³.

    Small cubes d are fine: Step 4’s bound handles them directly. Large cubes leave sums of the original kind but at a smaller column scale. Two more Poisson flips (the transfer estimate) turn those into new sums of the same shape, with both row and column scales cut by the same factor N(b)³. Each round, the row range shrinks by at least D2ϑ. After a fixed number of rounds the rows run out, and counting finishes the job.

    The descent
    Simulator, in exponents of D

    Each point is a mean square to be bounded: column scale X = Dx (across) and row scale 𝓗 = Dh (up). The proof needs every point to stay on the diagonal x − h = ϑ, and each round jumps down it by more than 2ϑ. Once x ≤ 2ϑ (the shaded corner), the large-cube terms are gone and Step 4’s bound finishes directly. “Smallest cube” is the worst case for the descent.

    The whole proof in one breath

    A zero past 11/12 would make the Möbius sum A₁ too big (Step 1). A₁ has about D(1+ϑ)/6 clones among the rows of a sextic family whose average energy is about D (Step 2), so A₁ ≤ D11/12+…. The family’s energy is controlled by Poisson summation, which swaps Möbius signs for cubic Gauss sums (Step 3). Those are coefficients of Kubota’s cubic theta function, whose symmetry turns the sextic twist into a quadratic one, where the quadratic large sieve applies (Step 4). The cube terms that theta brings along are removed by Möbius inversion and a self-similar descent that ends after about 1/(2ϑ) rounds (Step 5).

    Turn 8 · Looking back down the spiral

    What it means

    You started with Eratosthenes crossing out multiples. You end with a wall at 11/12 that holds at every height and on every clock.

    For primes

    Primes in every arithmetic progression are counted with error at most about x11/12 ln x, uniformly for all clocks q ≤ x, with constants you can compute. No hidden exception lurks for large clocks.

    For the theory

    Landau–Siegel zeros, the most famous obstruction in analytic number theory, are ruled out. Many results proved “unless a Siegel zero exists” become unconditional, and ineffective constants become effective.

    For the method

    The engine is new: hiding a Möbius sum among sextic twists, then using cubic theta functions so a quadratic large sieve can act. Combining averaging over families with metaplectic forms could reach further, as the 7/8 companion paper suggests.

    What remains open

    • The Riemann Hypothesis itself. Zeros between ½ and 11/12 are not excluded by this theorem.
    • Narrowing the gap. The 11/12 comes from 1 − ½·⅙. A different family, a better sieve, or a sharper treatment of the clone rows could lower it; the companion paper claims 7/8.
    • Verification. The paper is long and technical, and it is dated October 5, 2026. Its claims will be tested by the experts whose work it builds on.
    Glossary
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