Finite-Field Kakeya: Verification Notebook

This notebook is a self-contained mathematical and computational companion to the Finite-field Kakeya section in the paper.

The task template, including the prompt given to the agents, is available here.

The outline is as follows:

1. Problem Definitions
2. S1. A new infinite family in $d = 3$ for $p \equiv 3 \pmod 4$ . (Formal proof)
3. S2. Finite improvements and a 53-point Kakeya set in $\mathbb{F}_3^5$ .
4. S3. Structural analysis of the new infinite family. (Formal proof)
5. Higher-dimensional uniform families
6. Additional findings
7. Appendix verification

The required packages are:

NumPy

1. Problem Definitions

A finite-field Kakeya set contains a complete affine line in every direction. The problem is to construct such a set with as few points as possible; exact lower-order improvements remain meaningful even though the leading asymptotic density is known.

Let $p$ be a prime. An affine line in $\mathbb F_p^d$ is

L(w,v)=\{w+t v:t\in\mathbb F_p\},\qquad v\ne0.

Two nonzero vectors determine the same projective direction if one is a nonzero scalar multiple of the other. We use the unique representative whose first nonzero coordinate equals $1$ . There are

1+p+\cdots+p^{d-1}=\frac{p^d-1}{p-1}

such representatives. A set is Kakeya if it contains an affine line in every projective direction.

The next cell defines the shared routines used below for enumerating projective directions, constructing affine lines, and checking finite Kakeya certificates.

Show code

Code cell 3 · In [1]

from __future__ import annotations

from collections import Counter
from fractions import Fraction
from itertools import combinations, product
from math import comb, log
from pathlib import Path
import hashlib
import json

import numpy as np


ARTIFACT_DIR = Path.cwd()


def canonical_directions(p: int, d: int):
    """Unique projective representatives: first nonzero coordinate is 1."""
    for pivot in range(d):
        prefix = (0,) * pivot + (1,)
        for tail in product(range(p), repeat=d - pivot - 1):
            yield prefix + tail


def line_points(anchor, direction, p: int):
    return {
        tuple((int(a) + t * int(v)) % p for a, v in zip(anchor, direction))
        for t in range(p)
    }


def find_kakeya_witnesses(points, p: int, d: int):
    point_set = {tuple(map(int, point)) for point in points}
    witnesses = {}
    for direction in canonical_directions(p, d):
        # Every point on a contained line can serve as its anchor.
        for anchor in point_set:
            if line_points(anchor, direction, p) <= point_set:
                witnesses[direction] = anchor
                break
        else:
            raise AssertionError(f'No contained line in direction {direction}')
    return witnesses


def assert_point_certificate(array: np.ndarray, p: int, d: int, size: int):
    assert array.shape == (size, d)
    assert np.issubdtype(array.dtype, np.integer)
    assert np.all((0 <= array) & (array < p))
    assert len({tuple(map(int, row)) for row in array}) == size
    witnesses = find_kakeya_witnesses(array, p, d)
    expected = (p**d - 1) // (p - 1)
    assert len(witnesses) == expected
    return witnesses

2. S1. A new infinite family in $d = 3$ for $p \equiv 3 \pmod 4$ .

2.1 Construction and exact size

Let $Q=\{u^2:u\in\mathbb F_p\}$ , including zero, and put $S=|Q|=(p+1)/2$ . Define

B_p=\{(x,y,z):x^2+4y\in Q,\ x^2+4z\in Q\}

and, in the plane $x=0$ ,

U_p=\bigcup_{c\ne1}\{(0,t,ct+c/(c-1)):t\in\mathbb F_p\} \cup\{(0,t,t):t\in\mathbb F_p\} \cup\{(0,0,t):t\in\mathbb F_p\}.

Set $K_p=B_p\cup U_p$ .

Theorem 2.1 (coverage and exact size). For every odd prime $p$ , $K_p$ is Kakeya and

|K_p|= \begin{cases} (2p^3+7p^2-1)/8,&p\equiv1\pmod4,\\ (2p^3+7p^2+3)/8,&p\equiv3\pmod4. \end{cases}

Proof of coverage. For each graph direction $(1,a,b)$ , the line

(t,at+a^2,bt+b^2),\qquad t\in\mathbb F_p,

lies in $B_p$ , because $t^2+4(at+a^2)=(t+2a)^2$ , and similarly in the last coordinate. The displayed boundary lines cover every direction $(0,1,c)$ with $c\ne1$ , the diagonal covers $(0,1,1)$ , and the vertical line covers $(0,0,1)$ . These are all canonical directions. $\square$

Proof of the size formula. It remains to count $U_p\setminus(Q\times Q)$ in the plane $x=0$ . Write $\chi$ for the quadratic character, extended by $\chi(0)=0$ , and $\varepsilon=\chi(-1)$ .

Lemma 2.2 (finite-line selector). Put

\Delta(y,z)=(1-y-z)^2-4yz.

The finite lines with $c\ne1$ , together with the vertical line, cover exactly the pairs for which $\Delta(y,z)\in Q$ .

Proof. Eliminating $c$ from $z=cy+c/(c-1)$ gives

yc^2+(1-y-z)c+z=0.

The value $c=1$ is never a root, since substitution gives $1$ . Thus for $y\ne0$ there is an allowed root exactly when the discriminant $\Delta$ is a square. At $y=0$ , the finite intercept map $c\mapsto c/(c-1)$ covers every value except $1$ ; the vertical line restores $(0,1)$ and covers the whole row, while $\Delta(0,z)=(1-z)^2$ . $\square$

Lemma 2.3 (square-grid selector count).

T:=\#\{(y,z)\in Q^2:\Delta(y,z)\in Q\}=\frac{p^2+8p-1}{8}.

Proof. The exact indicator of $Q$ is $(1+\chi(u)+[u=0])/2$ . Hence

2T=S^2+\sum_{y,z\in Q}\chi(\Delta(y,z))+\#\{(y,z)\in Q^2:\Delta=0\}.

For $y=a^2,z=b^2$ ,

\Delta(a^2,b^2)=[1-(a+b)^2][1-(a-b)^2].

The change $(a,b)\mapsto(u,v)=(a+b,a-b)$ is bijective. The standard quadratic-character identity

\sum_{u\in\mathbb F_p}\chi(1-u^2)=-\varepsilon

therefore makes the double root-lift sum equal to $1$ . Correcting for the two axes gives

\sum_{y,z\in Q}\chi(\Delta(y,z))=(p-1)/2.

If $G(y,z)=\chi(\Delta(y,z))$ , then

\sum_{a,b}G(a^2,b^2) =4\sum_{y,z\in Q}G(y,z)-2\sum_{z\in Q}G(0,z)-2\sum_{y\in Q}G(y,0)+G(0,0).

Each axis sum is $S-1=(p-1)/2$ , and $G(0,0)=1$ .

For the zero count, the same root lift counts the union of $u=\pm1$ and $v=\pm1$ , which has $4p-4$ ordered pairs. The two axis zero counts are each $1$ , and the origin is not a zero. The identical correction formula therefore gives exactly $p$ zeros in $Q^2$ . Substitution yields the stated value of $T$ . The one-variable character identity itself follows by counting the $p-1$ solutions of $(r-u)(r+u)=-1$ and comparing them with $\sum_u(1+\chi(1-u^2))$ . $\square$

Lemma 2.4 (diagonal correction). The diagonal adds

A=\#\{y\in Q:\chi(1-4y)=-1\}=\frac{p-2+\varepsilon}{4}

new square-grid points.

Proof. On $z=y$ the selector discriminant is $1-4y$ . Moreover

\sum_{y\in Q}\chi(1-4y)=\frac{1-\varepsilon}{2},

obtained by lifting $y=a^2$ and using $\sum_a\chi(1-4a^2)=-\varepsilon$ ; $y=1/4$ is the unique zero. Solving for the number of negative values gives $A$ . $\square$

By Lemmas 2.2–2.4,

|U_p\cap Q^2|=T+A=\frac{p^2+10p-5+2\varepsilon}{8}.

Lemma 2.5 (total boundary-union count). Let $F$ be the union of the finite lines with $c\ne1$ , let $D=\{(y,y):y\in\mathbb F_p\}$ , and let $V=\{(0,z):z\in\mathbb F_p\}$ . Then

|F|=\frac{p(p-1)}2+(p-1),\qquad |D\setminus F|=\frac{p-1}2,\qquad |V\setminus(F\cup D)|=1.

Proof. On the row with first coordinate $y$ , the finite lines cover the image of

\phi_y(c)=cy+\frac{c}{c-1},\qquad c\in\mathbb F_p\setminus\{1\}.

For $y=0$ , this fractional-linear map is a bijection from $\mathbb F_p\setminus\{1\}$ onto itself, so the row has $p-1$ points. Assume $y\ne0$ . For distinct $c,d$ ,

\phi_y(c)-\phi_y(d)=(c-d)\left(y-\frac1{(c-1)(d-1)}\right).

Thus a collision occurs exactly when $(c-1)(d-1)=1/y$ . Every fiber has size at most two, because $\phi_y(c)=z$ is the quadratic from Lemma 2.2. Writing $u=c-1$ and $v=d-1$ , there are $p-1$ ordered solutions of $uv=1/y$ . The diagonal solutions satisfy $u^2=1/y$ , so there are two when $y$ is a square and none when it is a nonsquare. Hence the number of unordered distinct collisions is $(p-3)/2$ or $(p-1)/2$ , respectively, and

|\operatorname{im}\phi_y|=\frac{p+\chi(y)}2.

Summing over $y\ne0$ and using $\sum_{y\ne0}\chi(y)=0$ gives $p(p-1)/2$ off-axis points. This proves the formula for $|F|$ .

For $y\ne0$ , Lemma 2.2 says that $(y,z)\in F$ exactly when $\Delta(y,z)$ is a square. On the diagonal $z=y$ , this becomes $1-4y$ . As $y\mapsto1-4y$ is a bijection, precisely $(p-1)/2$ diagonal points have nonsquare discriminant and are new; the point at $y=0$ already belongs to $F$ . Finally the row $y=0$ of $F$ contains every point except $(0,1)$ , so the vertical line adds exactly that one point. $\square$

By Lemma 2.5,

|U_p|=\frac{p^2+2p-1}{2}.

Consequently

|U_p\setminus Q^2|=\frac{3p^2-2p+1-2\varepsilon}{8}.

Finally, every $x$ -slice of $B_p$ contains $S^2$ points, so $|B_p|=pS^2$ , and $B_p\cap\{x=0\}=Q^2$ . Adding the penalty gives Theorem 2.1. $\square$

Formal proof: Lean 4 source.

The next cell constructs $K_p$ directly, checks the displayed size formula, and verifies an explicit line in every projective direction for every odd prime through $47$ . It also defines the finite-field helpers used in later sections.

Show code

Code cell 5 · In [2]

def is_prime(n: int) -> bool:
    if n < 2:
        return False
    if n % 2 == 0:
        return n == 2
    return all(n % q for q in range(3, int(n**0.5) + 1, 2))


def quadratic_residues(p: int):
    return {x * x % p for x in range(p)}


def legendre(a: int, p: int) -> int:
    a %= p
    if a == 0:
        return 0
    return 1 if pow(a, (p - 1) // 2, p) == 1 else -1


def one_pole_construction(p: int):
    Q = quadratic_residues(p)
    body = {
        (x, y, z)
        for x, y, z in product(range(p), repeat=3)
        if (x * x + 4 * y) % p in Q and (x * x + 4 * z) % p in Q
    }
    boundary = set()
    for c in range(p):
        if c == 1:
            continue
        intercept = c * pow(c - 1, -1, p) % p
        boundary.update((0, t, (c * t + intercept) % p) for t in range(p))
    boundary.update((0, t, t) for t in range(p))
    boundary.update((0, 0, t) for t in range(p))
    return body | boundary


def one_pole_formula(p: int) -> int:
    constant = -1 if p % 4 == 1 else 3
    return (2 * p**3 + 7 * p**2 + constant) // 8


def explicit_one_pole_witnesses(p: int):
    witnesses = {}
    for a, b in product(range(p), repeat=2):
        witnesses[(1, a, b)] = (0, a * a % p, b * b % p)
    for c in range(p):
        if c != 1:
            witnesses[(0, 1, c)] = (0, 0, c * pow(c - 1, -1, p) % p)
    witnesses[(0, 1, 1)] = (0, 0, 0)
    witnesses[(0, 0, 1)] = (0, 0, 0)
    return witnesses


one_pole_rows = []
for p in [q for q in range(3, 48, 2) if is_prime(q)]:
    K = one_pole_construction(p)
    W = explicit_one_pole_witnesses(p)
    assert len(K) == one_pole_formula(p)
    assert set(W) == set(canonical_directions(p, 3))
    assert all(line_points(w, v, p) <= K for v, w in W.items())
    Q = quadratic_residues(p)
    delta_count = sum(
        ((1 - y - z) ** 2 - 4 * y * z) % p in Q for y in Q for z in Q
    )
    assert delta_count == (p * p + 8 * p - 1) // 8
    one_pole_rows.append((p, len(K), '1 mod 4' if p % 4 == 1 else '3 mod 4'))

print('Verified exact construction rows (p, size, class):')
print(one_pole_rows)

Saved output 1

Verified exact construction rows (p, size, class):
[(3, 15, '3 mod 4'), (5, 53, '1 mod 4'), (7, 129, '3 mod 4'), (11, 439, '3 mod 4'), (13, 697, '1 mod 4'), (17, 1481, '1 mod 4'), (19, 2031, '3 mod 4'), (23, 3505, '3 mod 4'), (29, 6833, '1 mod 4'), (31, 8289, '3 mod 4'), (37, 13861, '1 mod 4'), (41, 18701, '1 mod 4'), (43, 21495, '3 mod 4'), (47, 27889, '3 mod 4')]

2.2 Equivalence with the $p \equiv 1 \pmod 4$ AlphaEvolve construction

We use the set defined in AlphaEvolve’s commit-pinned kakeya2.tex. Let $g=(p-1)/4=-1/4$ in $\mathbb F_p$ and define $K^{\rm AE}=K_A\cup K_{0B}\cup K_{0C}$ by

K_A=\left\{\left(x,\frac{q_1+q_2}{2}-x^2-g,\frac{q_1-q_2}{2}\right):x\in\mathbb F_p,\ q_1,q_2\in Q\right\},

K_{0B}=\{(0,y,z):y+z^2\in Q\},\qquad K_{0C}=\{(0,y,0):y\in\mathbb F_p\}.

Theorem 2.6 (affine equivalence). If $p\equiv1\pmod4$ , then $K^{\rm AE}$ is affinely equivalent to the set $K_p$ of Theorem 2.1.

Proof. Choose $u\in\mathbb F_p^*$ with $u^2=-4$ and define

\Phi(x,y,z)=\left(ux,\frac{u^2}{4}(y+z+g),\frac{u^2}{4}(y-z+g)\right).

Its linear part has determinant $-2u\ne0$ , so it is an invertible affine map. If a point lies in $K_A$ , then its image $(X,Y,Z)$ satisfies

X^2+4Y=u^2q_1,\qquad X^2+4Z=u^2q_2.

Since $u^2$ is a nonzero square, this maps $K_A$ bijectively onto $B_p$ . On the plane $x=0$ , direct substitution, using $u^2=-4$ and $1+4g=0$ , gives

\Delta(\Phi(0,y,z))=4(y+z^2).

Therefore $\Phi(K_{0B})$ is exactly the selector set $\{(0,Y,Z):\Delta(Y,Z)\in Q\}$ . Lemma 2.2 identifies this set with the finite one-pole lines together with the vertical line. Moreover $\Phi(K_{0C})=\{(0,t,t):t\in\mathbb F_p\}$ , the diagonal line. Thus $\Phi(K^{\rm AE})=B_p\cup U_p=K_p$ . $\square$

The next cell reconstructs both sets at $p=5,13,17,29$ , applies the affine map above, and checks exact equality of the resulting point sets.

Show code

Code cell 7 · In [3]

def alphaevolve_kakeya2(p):
    assert p % 4 == 1
    Q = quadratic_residues(p)
    inv2 = pow(2, -1, p)
    g = (p - 1) // 4
    body = {
        (x, ((q1 + q2) * inv2 - x * x - g) % p,
         ((q1 - q2) * inv2) % p)
        for x, q1, q2 in product(range(p), Q, Q)
    }
    conic_boundary = {
        (0, y, z) for y, z in product(range(p), repeat=2)
        if (y + z * z) % p in Q
    }
    diagonal_source = {(0, y, 0) for y in range(p)}
    return body | conic_boundary | diagonal_source


def alphaevolve_to_one_pole(point, p):
    x, y, z = map(int, point)
    g = (p - 1) // 4
    u = next(a for a in range(1, p) if a * a % p == (-4) % p)
    inv4 = pow(4, -1, p)
    return (
        u * x % p,
        u * u * (y + z + g) * inv4 % p,
        u * u * (y - z + g) * inv4 % p,
    )


affine_equivalence_rows = []
for p in (5, 13, 17, 29):
    alpha_set = alphaevolve_kakeya2(p)
    mapped_set = {alphaevolve_to_one_pole(point, p) for point in alpha_set}
    station_set = one_pole_construction(p)
    assert mapped_set == station_set
    u = next(a for a in range(1, p) if a * a % p == (-4) % p)
    affine_equivalence_rows.append((p, u, len(mapped_set)))

print({
    'affine_equivalence_checks_(p,u,size)': affine_equivalence_rows,
})

Saved output 1

{'affine_equivalence_checks_(p,u,size)': [(5, 1, 53), (13, 3, 697), (17, 8, 1481), (29, 5, 6833)]}

Station Reference. The exact one-pole formula, corrected $y=0$ selector, and affine-equivalence calculation are based on Archive #160, message archive_160-3, and Eval #2414, with supporting checks in Evals #2399, #2406, and #2410.

Related Work. Dvir (2009) proved the finite-field Kakeya conjecture, while Bukh and Chao (2021) determined the sharp leading density and developed a general framework for completing almost-Kakeya sets using lower-dimensional Kakeya sets. The infinite family above is related to that framework; its explicitly analyzed one-pole completion yields the displayed exact size formulas. Georgiev et al. (2025) reported the AlphaEvolve families used for comparison. The $p\equiv1\pmod4$ branch is an independent rediscovery up to affine equivalence, whereas the $p\equiv3\pmod4$ branch is new and saves $(p-3)/4$ points over AlphaEvolve’s applicable infinite family.

3. S2. Finite improvements and a 53-point Kakeya set in $\mathbb{F}_3^5$ .

3.1 The 53-point construction

Theorem 3.1. If $k_5$ denotes the minimum cardinality of a Kakeya set in $\mathbb F_3^5$ , then

k_5\le53.

Proof. The point array kakeya_F3_d5_53.npy contains $53$ distinct points. The exact verification below shows that it contains an affine line in each of the $121$ projective directions of $\mathbb F_3^5$ , so it is the required Kakeya set. $\square$

The next cell authenticates the stored $53$ -point array and independently searches for a contained line in all $121$ directions. It also counts every contained affine line, providing a stronger check than replaying stored witnesses.

Show code

Code cell 10 · In [4]

DATA_HASHES = {'kakeya_F3_d3_13.npy': '8b1cac563e00cc53ecb5c0d38e23078fbbde36e6b6796f6c375769fbfa393737', 'kakeya_F3_d4_27.npy': 'dae8bf4b72750146aadad1de525be39fbfa49e78473202581363bacc82c14f2d', 'kakeya_F3_d5_53.npy': '4ddbba43b2cff772e4eb1ffe42064fb05e95a03972a2c448472acdcf803da812'}

path_53 = ARTIFACT_DIR / 'kakeya_F3_d5_53.npy'
if not path_53.is_file():
    raise FileNotFoundError(f'Required certificate is missing: {path_53}')
assert hashlib.sha256(path_53.read_bytes()).hexdigest() == DATA_HASHES[path_53.name]

K_53 = np.load(path_53, allow_pickle=False)
witnesses_53 = assert_point_certificate(K_53, p=3, d=5, size=53)

contained_line_counts = {}
point_set_53 = {tuple(map(int, row)) for row in K_53}
for direction in canonical_directions(3, 5):
    lines = {
        frozenset(line_points(anchor, direction, 3))
        for anchor in point_set_53
        if line_points(anchor, direction, 3) <= point_set_53
    }
    contained_line_counts[direction] = len(lines)

assert sum(contained_line_counts.values()) == 131
print({
    'points': len(point_set_53),
    'canonical_directions': len(witnesses_53),
    'contained_affine_lines': sum(contained_line_counts.values()),
    'minimum_lines_per_direction': min(contained_line_counts.values()),
    'maximum_lines_per_direction': max(contained_line_counts.values()),
    'sha256': DATA_HASHES[path_53.name],
})

Saved output 1

{'points': 53, 'canonical_directions': 121, 'contained_affine_lines': 131, 'minimum_lines_per_direction': 1, 'maximum_lines_per_direction': 2, 'sha256': '4ddbba43b2cff772e4eb1ffe42064fb05e95a03972a2c448472acdcf803da812'}

Station Reference. The 53 points originated in Eval #2520. The verification reconstructs witnesses independently from the $53$ points.

3.2 The complete 25-pair Station result

Table 3.1 (verified Station constructions).

Dimension $d$	Prime $p$	Number of points
3	3	13
3	5	53
3	7	128
3	11	437
3	13	697
3	19	2,030
3	23	3,504
3	29	6,833
3	31	8,288
3	37	13,861
3	41	18,701
3	43	21,495
3	47	27,889
3	53	39,677
4	3	27
4	5	161
4	7	527
4	11	2,684
4	13	4,962
4	17	13,509
4	19	20,579
5	3	53
5	5	490
5	7	2,135
5	11	16,288

Theorem 3.2. For every triple $(d,p,N)$ listed in Table 3.1, there is a Kakeya set of $N$ points in $\mathbb F_p^d$ .

Verification. For each row, the bundled certificate supplies a point set of cardinality $N$ and a witness line in every projective direction. The next cell checks the point count, distinctness, coordinate range, complete direction list, and containment of every witness line. A successful run verifies all $25$ existence claims. $\square$

The comparison with AlphaEvolve and the earlier literature is deferred to Section 7.

Show code

Code cell 12 · In [5]

STATION_MANIFEST_SHA256 = 'f19d7b03cfdd5956386f2ef4ef2aa41009783ba0b5137d8b54d0ec360b22afad'
STATION_BUNDLE_SHA256 = '105a642a701a4d26f1e360ac041504731c2d7d7bc13d1c185dc5b5013c21dfd5'

station_manifest_path = ARTIFACT_DIR / 'station_point_certificates.json'
assert hashlib.sha256(station_manifest_path.read_bytes()).hexdigest() == STATION_MANIFEST_SHA256
station_manifest = json.loads(station_manifest_path.read_text())

station_bundle_path = ARTIFACT_DIR / station_manifest['bundle']
assert hashlib.sha256(station_bundle_path.read_bytes()).hexdigest() == STATION_BUNDLE_SHA256
assert station_manifest['bundle_sha256'] == STATION_BUNDLE_SHA256

station_certificate_checks = []
with np.load(station_bundle_path, allow_pickle=False) as station_bundle:
    for record in sorted(station_manifest['records'], key=lambda row: (row['d'], row['p'])):
        d, p, size = record['d'], record['p'], record['size']
        point_array = station_bundle[record['points_key']]
        witness_array = station_bundle[record['witnesses_key']]
        directions = list(canonical_directions(p, d))

        assert point_array.shape == (size, d)
        assert witness_array.shape == (len(directions), d)
        assert np.all((0 <= point_array) & (point_array < p))
        points = {tuple(map(int, point)) for point in point_array}
        assert len(points) == size

        for direction, anchor in zip(directions, witness_array):
            assert line_points(anchor, direction, p) <= points
        station_certificate_checks.append((d, p, size, len(directions)))

assert [(d, p, size) for d, p, size, _ in station_certificate_checks] == [
    (3, 3, 13), (3, 5, 53), (3, 7, 128), (3, 11, 437),
    (3, 13, 697), (3, 19, 2030), (3, 23, 3504), (3, 29, 6833),
    (3, 31, 8288), (3, 37, 13861), (3, 41, 18701), (3, 43, 21495),
    (3, 47, 27889), (3, 53, 39677), (4, 3, 27), (4, 5, 161),
    (4, 7, 527), (4, 11, 2684), (4, 13, 4962), (4, 17, 13509),
    (4, 19, 20579), (5, 3, 53), (5, 5, 490), (5, 7, 2135),
    (5, 11, 16288),
]

print('Verified Station constructions:', len(station_certificate_checks))
print('Verified projective directions:', sum(row[3] for row in station_certificate_checks))

Saved output 1

Verified Station constructions: 25
Verified projective directions: 50066

Station Reference. The 25 Station point certificates are reconstructed from Evals #3, #12, #42, #52, #690, #730, #975, #2053, #2520, and #2529. The bundled point sets and witness lines are checked above.

Related Work. Lev (2009) recorded $k_3=13$ and $k_4\le27$ and proposed the recurrence that predicts $k_5=53$ . The previous bound $k_5\le63$ is attained by the missing-digit construction of Kopparty et al. (2011) and was also AlphaEvolve’s value. The 53-point construction improves that bound and matches Lev’s prediction.

4. S3. Structural analysis of the new infinite family.

Fix $\lambda\in\mathbb F_p^*$ and put $E_0=(\lambda Q)^2$ . Choose $A,B,r,u,v\in\mathbb F_p$ , set $D=Ar+B$ , and assume $D\ne0$ . For $c\ne r$ , define

\beta_c=\frac{Ac+B}{c-r}=A+\frac{D}{c-r},\qquad L_c=\{(y,cy+\beta_c):y\in\mathbb F_p\}.

Add $L_r(u)=\{(y,ry+u)\}$ and the vertical line $L_\infty(v)=\{(v,z)\}$ . Let $U$ be the union of these $p+1$ lines and define the boundary penalty by $\operatorname{pen}(p)=|U\setminus E_0|$ .

Theorem 4.1 (uniform full-family penalty). Let $p$ be an odd prime. For every $\lambda\in\mathbb F_p^*$ and every $A,B,r,u,v\in\mathbb F_p$ satisfying $Ar+B\ne0$ ,

\left|\operatorname{pen}(p)-\frac{3p^2}{8}\right|<5p.

Proof. Let $U_f=\bigcup_{c\ne r}L_c$ . Eliminating $c$ gives

\Delta(y,z)=(z-A-ry)^2-4Dy.

For $y\ne0$ , membership in $U_f$ is equivalent to $\Delta(y,z)\in Q$ . As a polynomial in $z$ , $\Delta$ is monic with discriminant $16Dy$ . Therefore $\sum_z\chi(\Delta)=-1$ and the number of its zeros is $1+\chi(Dy)$ , so the number of covered $z$ is $(p+\chi(Dy))/2$ . Summation over $y\ne0$ gives $p(p-1)/2$ points. At $y=0$ , the values $\beta_c=A+D/(c-r)$ run bijectively through $\mathbb F_p\setminus\{A\}$ . Hence

|U_f|=\frac{p(p-1)}2+(p-1).\tag{1}

It remains to bound $O=|U_f\cap E_0|$ . Put

Y=\chi(y/\lambda),\quad Z=\chi(z/\lambda),\quad X=\chi(\Delta),

and write $e_y,e_z,e_\Delta$ for the indicators of $y=0,z=0,\Delta=0$ . The exact square-set indicator is $(1+Y+e_y)/2$ . The selector $(1+X+e_\Delta)/2$ agrees with $U_f$ off the row $y=0$ ; on that row it additionally selects the omitted point $(0,A)$ . Thus, if $O^*$ denotes the selector overlap,

8O^*=\sum_{y,z}(1+Y+e_y)(1+Z+e_z)(1+X+e_\Delta),\qquad O=O^*-[A\in\lambda Q].\tag{2}

The part without $X$ or $e_\Delta$ is $(p+1)^2$ . We next bound the $X$ -block. The monic-quadratic identity used above gives

\sum X=\sum YX=0.

For $S_{01}=\sum\chi(z)\chi(\Delta)$ , substitute $x=z-A-ry$ . If $r\ne0$ , the inner sum over $y$ is a correlation of two linear characters. Their roots coincide precisely when

rx^2+4Dx+4DA=0.

If this quadratic has $N\in\{0,1,2\}$ roots, the correlation identity $\sum_y\chi((a y+b)(c y+d))=-\chi(ac)$ for distinct roots and $(p-1)\chi(ac)$ for equal roots gives

S_{01}=p(N-1)\chi(-4Dr).

Thus $|S_{01}|\le p$ ; when $r=0$ the inner sum is zero. Multiplication by the harmless factor $\chi(\lambda)$ proves $|\sum ZX|\le p$ .

For $S_{11}=\sum\chi(yz)\chi(\Delta)$ , restrict to $y\ne0$ and set $t=z/y$ . The inner polynomial in $y$ is

q_t(y)=(t-r)^2y^2-2(A(t-r)+2D)y+A^2,

whose discriminant is $16D(At+B)$ . If $A\ne0$ , the complete character sum is $-1$ except that it is $p-1$ at $t=-B/A$ and $0$ at $t=r$ ; subtracting the $y=0$ term and summing with weight $\chi(t)$ gives

S_{11}=p\chi(-B/A)+\chi(r).

If $A=0$ , the same calculation gives $S_{11}=\chi(r)$ . Hence $|\sum YZX|\le p+1$ . The remaining zero-coordinate $X$ -terms group as

\sum_z(1+Z+e_z)X(0,z),\qquad \sum_y(1+Y)X(y,0),

with absolute values at most $p+1$ and $p$ . The entire $X$ -block therefore has absolute value at most $4p+2$ .

Finally $\Delta=0$ has $\sum_y(1+\chi(Dy))=p$ affine solutions. The $e_\Delta$ -block is exactly four times the number of these solutions in $E_0$ , hence is at most $4p$ . From (2),

|8O^*-p^2|\le(2p+1)+(4p+2)+4p=10p+3.

Removing the possible extra point changes the numerator by at most $8$ , so, for every odd $p$ ,

|8O-p^2|\le10p+11<16p,\qquad |O-p^2/8|<2p.\tag{3}

Let $E$ be the new contribution outside $E_0\cup U_f$ from the pole and vertical lines. Since each has $p$ points, $0\le E\le2p$ . By (1) and (3),

\operatorname{pen}(p)-\frac{3p^2}{8}=\left(\frac p2-1\right)-\left(O-\frac{p^2}{8}\right)+E,

and therefore $|\operatorname{pen}(p)-3p^2/8|<(p/2-1)+2p+2p<5p$ . $\square$

Formal proof: Lean 4 source.

The next cell exhausts all admissible parameters at $p=3$ and $p=5$ . For each choice it checks the exact finite-line union count and the bound $|\operatorname{pen}(p)-3p^2/8|<5p$ proved above.

Show code

Code cell 15 · In [6]

def full_one_pole_boundary(p, lam, A, B, r, u, v):
    Q = quadratic_residues(p)
    D = (A * r + B) % p
    assert lam % p != 0 and D != 0
    finite = set()
    for c in range(p):
        if c == r:
            continue
        beta = (A * c + B) * pow(c - r, -1, p) % p
        finite.update((y, (c * y + beta) % p) for y in range(p))
    boundary = finite | {(y, (r * y + u) % p) for y in range(p)}
    boundary |= {(v, z) for z in range(p)}
    footprint = {(lam * q % p, lam * s % p) for q in Q for s in Q}
    return finite, boundary, len(boundary - footprint)


full_family_checks = 0
for p in (3, 5):
    for lam in range(1, p):
        for A, B, r, u, v in product(range(p), repeat=5):
            if (A * r + B) % p == 0:
                continue
            finite, boundary, penalty = full_one_pole_boundary(
                p, lam, A, B, r, u, v
            )
            assert len(finite) == p * (p - 1) // 2 + (p - 1)
            assert abs(penalty - Fraction(3 * p * p, 8)) < 5 * p
            full_family_checks += 1

print({'full_one_pole_parameter_checks': full_family_checks})

Saved output 1

{'full_one_pole_parameter_checks': 10324}

Station Reference. The uniform full-family penalty theorem is based on Archive #160 and the supporting checks in Evals #2399, #2406, and #2410.

Related Work. Bukh and Chao (2021) established the sharp leading density and a construction framework that determines the first two asymptotic terms. AlphaEvolve, as reported by Georgiev et al. (2025), gives particular completion families rather than a uniform bound for the full one-pole class. The theorem above proves the class-wide $3p^2/8+O(p)$ boundary penalty for every nondegenerate one-pole completion.

5. Higher-dimensional uniform families

The infinite families for $d=4$ and $d=5$ described in the paper’s limitations are presented here.

5.1 The four-dimensional prefix-sum family

Let

\mathcal Q=\{u^2:u\in\mathbb F_p\},\qquad S=|\mathcal Q|=\frac{p+1}{2}.

For $n\ge1$ , let $M_n$ be the lower-triangular matrix

(M_n)_{ij}=\begin{cases}1,&j\le i,\\0,&j>i.\end{cases}

Thus $M_n(a_1,\ldots,a_n)^T=(a_1,a_1+a_2,\ldots,a_1+\cdots+a_n)^T$ . Define

F_n(a)=-M_n^{-1} \begin{pmatrix} (M_na)_1^2\\ \vdots\\ (M_na)_n^2 \end{pmatrix}.

Set $K_1=\mathbb F_p$ . Recursively, for $n\ge1$ , define

T_{n+1}=\{(t,F_n(a)+ta):t\in\mathbb F_p,\ a\in\mathbb F_p^n\}, \qquad K_{n+1}=T_{n+1}\cup(\{0\}\times K_n). \tag{4}

Theorem 5.1. For every odd prime $p\ge5$ , the set $K_4$ defined in (4) is Kakeya in $\mathbb F_p^4$ and has size

|K_4|= \begin{cases} \dfrac18p^4+\dfrac{19}{32}p^3+\dfrac{25}{32}p^2-\dfrac3{32}p-\dfrac{13}{32},&p\equiv1\pmod4,\\[4pt] \dfrac18p^4+\dfrac{19}{32}p^3+\dfrac{25}{32}p^2-\dfrac7{32}p-\dfrac{49}{32},&p\equiv3\pmod4. \end{cases}

Proof. For each $a\in\mathbb F_p^n$ , the line

t\longmapsto(t,F_n(a)+ta)

lies in $T_{n+1}$ and has direction $(1,a)$ . By induction, $\{0\}\times K_n$ contains a line in every direction whose first coordinate is zero. Hence $K_{n+1}$ is Kakeya.

Fix $t\ne0$ . Applying $M_n$ to the transverse slice gives

M_n(F_n(a)+ta)= \bigl(-(M_na)_i^2+t(M_na)_i\bigr)_{i=1}^n.

Each coordinate ranges over a set of $S$ values, so the slice has $S^n$ points. The nonzero slices of $K_4$ therefore contain $(p-1)S^3$ points.

At $t=0$ , the top slice is

W_3=M_3^{-1}((-\mathcal Q)^3),

and the zero slice of $K_4$ is $W_3\cup K_3$ . For $\delta\ne0$ , put

N(\delta)=|\mathcal Q\cap(\mathcal Q+\delta)| =\frac{p+1+\chi(\delta)+\chi(-\delta)}4.

Partitioning $W_3\cap K_3$ by its first coordinate gives

|W_3\cap K_3| =2S^2+\sum_{x\in-\mathcal Q\setminus\{0,-4\}} N\left(-x-\frac{x^2}{4}\right)^2.

Substituting the displayed formula for $N$ and summing the quadratic characters yields

|W_3\cup K_3|= \begin{cases} \dfrac{11p^3+25p^2+5p-9}{32},&p\equiv1\pmod4,\\[4pt] \dfrac{11p^3+25p^2+p-45}{32},&p\equiv3\pmod4. \end{cases}

Adding the nonzero slices gives the stated formula. $\square$

5.2 The five-dimensional nested tangent family

Write $\chi$ for the quadratic character and retain $\mathcal Q=\{u^2:u\in\mathbb F_p\}$ and $S=(p+1)/2$ . Choose $\alpha\in\mathbb F_p^*$ with $\chi(\alpha)=\chi(-1)$ , and choose $e\in\mathcal Q$ so that

\chi(\alpha e^2+e)=-1\quad\text{if }p\equiv1\pmod4, \qquad \alpha e^2+e\ne0\quad\text{if }p\equiv3\pmod4.

Define the planar tangent completion

P_2= \bigcup_{m\in\mathbb F_p} \left\{(y,z):y=mz-\frac{(m-1)^2}{4\alpha}\right\} \;\cup\; \{(y,e):y\in\mathbb F_p\}.

For $d\ge3$ , define

A_d=\{(x_0,\ldots,x_{d-1}):x_i+x_0^2\in\mathcal Q \text{ for }1\le i<d\}.

Set

K_3=A_3\cup(\{0\}\times P_2), \qquad K_d=A_d\cup(\{0\}\times K_{d-1})\quad(d\ge4). \tag{5}

Lemma 5.2 (planar base calculation). A permitted value of $e$ exists. The set $P_2$ is Kakeya and

|P_2|=pS+S-1,

while

|P_2\cap\mathcal Q^2|=\frac{p^2+8p+7}{8}.

Consequently,

|K_3|=pS^2+|P_2\setminus\mathcal Q^2| =\frac{2p^3+7p^2+2p-11}{8}.

Proof. Let $T$ be the union of the $p$ sloped lines in $P_2$ and put $c(z)=\alpha z^2+z$ . Eliminating $m$ gives

m^2-(2+4\alpha z)m+(1+4\alpha y)=0,

whose discriminant is $16\alpha(c(z)-y)$ . Thus $(y,z)\in T$ exactly when $\alpha(c(z)-y)\in\mathcal Q$ . Every $z$ -row of $T$ has $S$ points, so $|T|=pS$ ; the horizontal line $z=e$ adds $S-1$ points. The sloped lines cover the directions $(m,1)$ and the horizontal line covers $(1,0)$ , so $P_2$ is Kakeya.

Write $\varepsilon=\chi(-1)=\chi(\alpha)$ . If $\varepsilon=-1$ , choose a nonzero square $e\ne-1/\alpha$ ; such a choice exists for $p\ge5$ , and $c(e)\ne0$ . If $\varepsilon=1$ , multiplication by $\alpha$ permutes the squares, and

\sum_{x\in\mathcal Q\setminus\{0\}}\chi(1+x)=-1.

This identity follows by expanding the square indicator and using the standard quadratic sum $\sum_x\chi(x(x+1))=-1$ . Hence some nonzero square $x$ has $\chi(1+x)=-1$ ; taking $e=x/\alpha$ gives $\chi(c(e))=-1$ .

For $c\ne0$ , expanding the same indicator gives

\#\{y\in\mathcal Q:c-y\in\mathcal Q\} =\frac{p+1+2\chi(c)}4\qquad(\varepsilon=1)

and

\#\{y\in\mathcal Q:c-y\in(\mathbb F_p\setminus\mathcal Q)\cup\{0\}\} =\frac{p+1}4\qquad(\varepsilon=-1);

at $c=0$ the corresponding count is $S$ . The polynomial $c(z)=z(\alpha z+1)$ has exactly two zeros in $\mathcal Q$ . Summing over $z\in\mathcal Q$ gives

|T\cap\mathcal Q^2|= \begin{cases} (p^2+6p+1)/8,&\varepsilon=1,\\ (p^2+6p+5)/8,&\varepsilon=-1. \end{cases}

The horizontal line adds respectively $(p+3)/4$ or $(p+1)/4$ new points of $\mathcal Q^2$ . Therefore $|P_2\cap\mathcal Q^2|=(p^2+8p+7)/8$ in both cases, and the remaining formulas follow. $\square$

Put

r=\frac{p+2+\chi(-1)}4.

For $n\ge3$ , inclusion–exclusion in (5) gives

|K_{n+1}|=pS^n+|K_n|-L_n, \qquad L_n=S^{n-1}+(S-1)r^{n-1}. \tag{6}

Theorem 5.3. For every odd prime $p\ge5$ , the set $K_5$ defined by (5) is Kakeya in $\mathbb F_p^5$ and has size

|K_5|= \begin{cases} \dfrac1{16}p^5+\dfrac{47}{128}p^4+\dfrac{25}{32}p^3+\dfrac{37}{64}p^2-\dfrac{17}{32}p-\dfrac{161}{128},&p\equiv1\pmod4,\\[4pt] \dfrac1{16}p^5+\dfrac{47}{128}p^4+\dfrac{53}{64}p^3+\dfrac{27}{32}p^2-\dfrac{25}{64}p-\dfrac{219}{128},&p\equiv3\pmod4. \end{cases}

Proof. At each lift, the new body has $pS^n$ points and supplies all graph directions; the embedded copy of $K_n$ supplies the directions in the coordinate hyperplane. Their intersection has size $L_n$ . Indeed, the overlap splits into $S^{n-1}$ points from the zero parameter and $(S-1)r^{n-1}$ points from the nonzero square parameters, where the standard quadratic-character count gives

r=|\operatorname{Sq}_p\cap(\operatorname{Sq}_p+1)| =\frac{p+2+\chi(-1)}4.

Inclusion–exclusion proves (6). Substituting the dimension-three size and expanding the cases $\chi(-1)=1$ and $\chi(-1)=-1$ through dimensions four and five gives the displayed formulas. $\square$

The next cell reconstructs both higher-dimensional families at three primes, checks their exact sizes, expands the size laws, and checks the third-coefficient comparison with AlphaEvolve.

Show code

Code cell 19 · In [7]

def poly_add(*polynomials):
    degree = max(map(len, polynomials))
    return tuple(
        sum((poly[i] if i < len(poly) else Fraction(0)) for poly in polynomials)
        for i in range(degree)
    )


def poly_scale(poly, scalar):
    scalar = Fraction(scalar)
    return tuple(scalar * coefficient for coefficient in poly)


def poly_mul(left, right):
    result = [Fraction(0)] * (len(left) + len(right) - 1)
    for i, a in enumerate(left):
        for j, b in enumerate(right):
            result[i + j] += a * b
    return tuple(result)


def poly_pow(poly, exponent):
    result = (Fraction(1),)
    for _ in range(exponent):
        result = poly_mul(result, poly)
    return result


p_poly = (Fraction(0), Fraction(1))
one_poly = (Fraction(1),)
S_poly = poly_scale(poly_add(p_poly, one_poly), Fraction(1, 2))

prefix_expected = {
    1: (Fraction(-13, 32), Fraction(-3, 32), Fraction(25, 32),
        Fraction(19, 32), Fraction(1, 8)),
    -1: (Fraction(-49, 32), Fraction(-7, 32), Fraction(25, 32),
         Fraction(19, 32), Fraction(1, 8)),
}
prefix_zero_slice = {
    1: (Fraction(-9, 32), Fraction(5, 32), Fraction(25, 32),
        Fraction(11, 32)),
    -1: (Fraction(-45, 32), Fraction(1, 32), Fraction(25, 32),
         Fraction(11, 32)),
}
prefix_body = poly_mul((-1, 1), poly_pow(S_poly, 3))
for character in (1, -1):
    assert poly_add(prefix_body, prefix_zero_slice[character]) == prefix_expected[character]


def prefix_sum_kakeya(p, dimension):
    if dimension == 1:
        return {(t,) for t in range(p)}
    n = dimension - 1
    top = set()
    for slope in product(range(p), repeat=n):
        cumulative = []
        total = 0
        for value in slope:
            total = (total + value) % p
            cumulative.append(total)
        squared = [(-value * value) % p for value in cumulative]
        anchor = tuple(
            (squared[i] - (squared[i - 1] if i else 0)) % p
            for i in range(n)
        )
        for t in range(p):
            top.add((t,) + tuple(
                (anchor[i] + t * slope[i]) % p for i in range(n)
            ))
    boundary = {(0,) + point for point in prefix_sum_kakeya(p, n)}
    return top | boundary


def poly_evaluate(poly, value):
    return sum(coefficient * value**degree
               for degree, coefficient in enumerate(poly))


prefix_construction_sizes = {}
for prime in (5, 7, 11):
    character = 1 if prime % 4 == 1 else -1
    points = prefix_sum_kakeya(prime, 4)
    expected_size = poly_evaluate(prefix_expected[character], prime)
    assert expected_size.denominator == 1
    assert len(points) == expected_size
    prefix_construction_sizes[prime] = len(points)

tangent_three = (
    Fraction(-11, 8), Fraction(1, 4), Fraction(7, 8), Fraction(1, 4)
)


def tangent_lift(size_poly, dimension, character):
    r_poly = poly_scale(poly_add(p_poly, (2 + character,)), Fraction(1, 4))
    overlap = poly_add(
        poly_pow(S_poly, dimension - 1),
        poly_mul(poly_add(S_poly, (-1,)), poly_pow(r_poly, dimension - 1)),
    )
    body = poly_mul(p_poly, poly_pow(S_poly, dimension))
    return poly_add(body, size_poly, poly_scale(overlap, -1))


tangent_five_expected = {
    1: (Fraction(-161, 128), Fraction(-17, 32), Fraction(37, 64),
        Fraction(25, 32), Fraction(47, 128), Fraction(1, 16)),
    -1: (Fraction(-219, 128), Fraction(-25, 64), Fraction(27, 32),
         Fraction(53, 64), Fraction(47, 128), Fraction(1, 16)),
}
tangent_five = {}
for character in (1, -1):
    tangent_four = tangent_lift(tangent_three, 3, character)
    tangent_five[character] = tangent_lift(tangent_four, 4, character)
    assert tangent_five[character] == tangent_five_expected[character]

def quadratic_character(value, p):
    value %= p
    if value == 0:
        return 0
    return 1 if pow(value, (p - 1) // 2, p) == 1 else -1


def tangent_kakeya(p, dimension):
    squares = quadratic_residues(p)
    character = quadratic_character(-1, p)
    alpha = next(a for a in range(1, p)
                 if quadratic_character(a, p) == character)
    if character == 1:
        e = next(x for x in sorted(squares)
                 if quadratic_character(alpha * x * x + x, p) == -1)
    else:
        e = next(x for x in sorted(squares)
                 if (alpha * x * x + x) % p != 0)

    inverse = pow(4 * alpha, -1, p)
    completion = {
        ((m * z - (m - 1) ** 2 * inverse) % p, z)
        for m in range(p)
        for z in range(p)
    }
    completion |= {(y, e) for y in range(p)}

    def leading_block(dimension):
        return {
            (x0,) + tuple((q - x0 * x0) % p for q in tail)
            for x0 in range(p)
            for tail in product(squares, repeat=dimension - 1)
        }

    points = leading_block(3) | {(0,) + point for point in completion}
    for current_dimension in range(4, dimension + 1):
        points = leading_block(current_dimension) | {
            (0,) + point for point in points
        }
    return points


tangent_construction_sizes = {}
for prime in (5, 7, 11):
    character = 1 if prime % 4 == 1 else -1
    points = tangent_kakeya(prime, 5)
    expected_size = poly_evaluate(tangent_five_expected[character], prime)
    assert expected_size.denominator == 1
    assert len(points) == expected_size
    tangent_construction_sizes[prime] = len(points)


d4_third_gap = prefix_expected[1][2] - Fraction(11, 16)
d5_third_gap = tangent_five[1][3] - Fraction(177, 256)
assert d4_third_gap == Fraction(3, 32) > 0
assert d5_third_gap == Fraction(23, 256) > 0

print('Exact coefficient gaps over AlphaEvolve:', {
    'd=4, coefficient of p^2': d4_third_gap,
    'd=5, coefficient of p^3': d5_third_gap,
})

Saved output 1

Exact coefficient gaps over AlphaEvolve: {'d=4, coefficient of p^2': Fraction(3, 32), 'd=5, coefficient of p^3': Fraction(23, 256)}

Station Reference. The nested tangent construction and its overlap identity are from Archive #50. The four-dimensional prefix-sum construction and its zero-slice decomposition are from Archive #71.

Related Work. On $p\equiv1\pmod4$ , the uniform constructions reported by Georgiev et al. (2025) have sizes

\frac18p^4+\frac{19}{32}p^3+\frac{11}{16}p^2+O(p^{3/2})

in dimension four and

\frac1{16}p^5+\frac{47}{128}p^4+\frac{177}{256}p^3+O(p^{5/2})

in dimension five. The Station formulas have the same first two coefficients but exceed these third coefficients by $3/32$ in dimension four and $23/256$ in dimension five. Since smaller Kakeya sets are better, both Station families are asymptotically worse than the corresponding AlphaEvolve families.

6. Additional findings

These are additional results that may be interesting but are not included in the spotlight.

6.1 Exact polynomial evaluation-code distance

Dvir’s polynomial method detects a small Kakeya set through a nonzero polynomial of degree at most $p-1$ that vanishes on the entire set. We determine exactly how close the classical quadratic-residue construction $C_d$ comes to admitting such a certificate: every such polynomial that is not identically zero on $C_d$ remains nonzero on at least $p+1$ points when $d=3$ , and at least $2((p+1)/2)^{d-2}$ points when $d\ge4$ , with equality in both cases. Thus the theorem gives the exact obstruction to applying Dvir’s vanishing argument to this construction, while also determining the minimum distance of the associated punctured Reed–Muller code. We prove the upper bounds with an explicit product of $p-1$ hyperplanes and the lower bounds by a weighted footprint argument, supplemented in dimension three by a shadow code and body-kernel decomposition.

The set and its evaluation code

Let $p$ be an odd prime, let

S=\frac{p+1}{2},\qquad \mathcal Q=\{u^2:u\in\mathbb F_p\},

where $0\in\mathcal Q$ , and define the quadratic-residue cone

C_d=P\sqcup B,

where

P=\{(0,x_1,\ldots,x_{d-1}):x_i\in\mathbb F_p\}

and

B=\{(x_0,\ldots,x_{d-1}):x_0\ne0,\ x_i+x_0^2\in\mathcal Q\text{ for every }i\ge1\}.

For a polynomial $f$ of total degree at most $p-1$ , its evaluation word is the function $f|_{C_d}$ . Its Hamming weight is

\operatorname{wt}_{C_d}(f)=\#\{x\in C_d:f(x)\ne0\}.

The minimum distance $\mu(C_d)$ is the least positive weight of such an evaluation word.

Theorem 6.1 (exact distance). For every odd prime $p$ ,

\mu(C_d)=2S^{d-2}\quad(d\ge4).

For $d=3$ ,

\mu(C_3)=p+1\quad(p\ge5).

The restriction $p\ge5$ in dimension three is necessary: at $p=3$ the distance is $3$ , not $4$ .

An explicit polynomial certificate

Lemma 6.2 (explicit upper certificate). For every $d\ge3$ , a polynomial of degree $p-1$ has weight exactly $2S^{d-2}$ on $C_d$ .

Proof. Take

F^*(x)=x_0x_{d-1} \prod_{c\in\mathbb F_p^*\setminus\{1/2,-1/2\}} \left(x_0+c x_{d-1}-\frac1{4c}\right).

It has $2+(p-3)=p-1$ linear factors. The first factor makes it zero on $P$ , and the second covers the part of $B$ with $x_{d-1}=0$ . At a remaining body point, one of the other factors vanishes precisely when

4x_{d-1}c^2+4x_0c-1=0.

Its discriminant is $16(x_0^2+x_{d-1})$ , a square by the definition of $B$ . If the two roots are distinct, they cannot both be the excluded pair $\{1/2,-1/2\}$ : their sum is $-x_0/x_{d-1}\ne0$ , whereas the excluded pair has sum zero. The only possible failure is therefore a double root. This requires

x_{d-1}=-x_0^2,

and the root is $c=1/(2x_0)$ . It is excluded exactly when $x_0=\pm1$ . Hence the nonzero set of $F^*$ on $C_d$ is

\{(\varepsilon,q_1-1,\ldots,q_{d-2}-1,-1): \varepsilon\in\{1,-1\},\ q_i\in\mathcal Q\},

which has $2S^{d-2}$ points. $\square$

Two elementary polynomial bounds

We use the following finite-grid form of the footprint bound.

Lemma 6.3 (Cartesian-grid footprint). Let $A_i\subseteq\mathbb F_p$ have $N_i$ elements. Suppose a nonzero polynomial $R$ , reduced so that $\deg_{X_i}R<N_i$ , has lexicographic leading monomial $X_1^{a_1}\cdots X_m^{a_m}$ . Then $R$ is nonzero at at least

\prod_{i=1}^m(N_i-a_i)

points of $A_1\times\cdots\times A_m$ .

Proof. Induct on $m$ . View $R$ as a polynomial in $X_1$ . Its leading coefficient, as a polynomial in the remaining variables, has leading exponents $a_2,\ldots,a_m$ , so by induction it is nonzero at at least $\prod_{i=2}^m(N_i-a_i)$ choices of those variables. At each such choice the resulting nonzero polynomial in $X_1$ has degree $a_1$ , hence at most $a_1$ roots in $A_1$ . $\square$

Lemma 6.4 (full-grid Reed–Muller bound). A nonzero polynomial in $r$ variables, of total degree at most $p-1$ , is nonzero at at least $p^{r-1}$ points of $\mathbb F_p^r$ .

Proof. Apply Lemma 6.3 to the full grid. If the leading exponents sum to at most $p-1$ , then

\prod_{i=1}^r(p-a_i)\ge p^{r-1}(p-\textstyle\sum_i a_i)\ge p^{r-1}.

The product inequality follows successively from $(p-a)(p-b)\ge p(p-a-b)$ . $\square$

The lower bound when $x_0$ divides the polynomial

Lemma 6.5 (body footprint). If $f=x_0h$ , $\deg f\le p-1$ , and $f|_{C_d}\ne0$ , then

\operatorname{wt}_{C_d}(f)\ge2S^{d-2}.

Proof. We have $\deg h\le p-2$ , and $f$ vanishes on $P$ . On $B$ , introduce

t=x_0,\qquad y_i=x_i+x_0^2.

This is a bijection from $B$ to

\Gamma=\mathbb F_p^*\times\mathcal Q^{d-1}.

Give $t$ weight $1$ and every $y_i$ weight $2$ . After substituting $x_i=y_i-t^2$ , the transformed polynomial $H$ has weighted degree at most $2(p-2)=4S-6$ . Reduce it by

t^{p-1}-1,\qquad \prod_{q\in\mathcal Q}(y_i-q).

These reductions preserve its values on $\Gamma$ and do not increase weighted degree. The normal form is nonzero: $f=x_0h$ vanishes on $P$ , so $f|_{C_d}\ne0$ forces $h$ , and hence $H$ , to be nonzero somewhere on $\Gamma$ . Its leading monomial in the same lexicographic order as Lemma 6.3 is

t^{a_0}y_1^{a_1}\cdots y_{d-1}^{a_{d-1}}

with

0\le a_0\le2S-3,\quad 0\le a_i\le S-1, \quad a_0+2\sum_{i=1}^{d-1}a_i\le4S-6.

Lemma 6.3 gives at least

(2S-2-a_0)\prod_{i=1}^{d-1}(S-a_i)

nonzero body values. We now minimize this product under the displayed constraints. At fixed $\sum_{i\ge1}a_i$ , moving two transverse exponents farther apart cannot increase $(S-a_i)(S-a_j)$ . Thus at a minimum all but at most one transverse exponent lie in $\{0,S-1\}$ . The degree budget permits at most one exponent equal to $S-1$ .

If none equals $S-1$ and there is no interior exponent, the product is at least $S^{d-1}\ge2S^{d-2}$ . If there is one interior exponent, maximizing it together with $a_0$ gives at least

1\cdot2\cdot S^{d-2}=2S^{d-2}.

If one exponent equals $S-1$ , write the possible other interior exponent as $g\in[0,S-2]$ . The budget gives $a_0+2g\le2S-4$ , and the product is at least

(2+2g)(S-g)S^{d-3} =\bigl(2S+2g(S-1-g)\bigr)S^{d-3} \ge2S^{d-2}.

Therefore $f$ has at least $2S^{d-2}$ nonzero values. $\square$

Proof of Theorem 6.1 for $d\ge4$ . For $d\ge4$ , the other case is immediate. If $x_0\nmid f$ , then $f(0,x_1,\ldots,x_{d-1})$ is a nonzero polynomial of degree at most $p-1$ . Lemma 6.4 gives at least $p^{d-2}$ nonzero values on $P$ , and

p^{d-2}\ge2S^{d-2}\qquad(d\ge4).

Indeed $2p/(p+1)\ge3/2$ , and its $(d-2)$ -nd power is at least $9/4>2$ . Lemmas 6.2 and 6.5 now prove Theorem 6.1 for every $d\ge4$ . $\square$

Dimension three is the sole delicate case: Lemma 6.4 gives only $p$ plane points, one short of $p+1$ . The rest of this section closes that gap.

The one-coordinate shadow code

Assume henceforth that $p=2n+1\ge5$ , so $S=n+1$ . Let $H=\mathcal Q\setminus\{0\}$ , the multiplicative group of nonzero squares.

The elementary finite-field facts used below are as follows. The squaring map on $\mathbb F_p^*$ has kernel $\{1,-1\}$ , so $|H|=n$ and $|\mathcal Q|=n+1$ . If $g$ generates the cyclic group $\mathbb F_p^*$ , then $(g^k)^n=(-1)^k$ . Thus $s^n=1$ for $s\in H$ , while $r^n=-1$ for a nonsquare $r$ . It follows in particular that $y^{n+1}=y$ for every $y\in\mathcal Q$ .

Give $\mathbb F_p[\tau,x]$ the weights

\operatorname{wdeg}(\tau)=2,\qquad\operatorname{wdeg}(x)=1,

and set

D=\{(\tau,x):\tau\in H,\ x+\tau\in\mathcal Q\}.

Define the shadow code $U\subseteq\mathbb F_p^{\mathbb F_p}$ by

U=\{x\mapsto F(0,x):F|_D=0,\ \operatorname{wdeg}F\le2n\}.

For any linear code $L$ , write $d(L)$ for the least Hamming weight of a nonzero word of $L$ .

The next lemma determines its dimension without assuming any unproved interpolation statement.

Lemma 6.6 (triangle unisolvence). The monomials

\mathcal T=\{\tau^jx^a:2j+a\le2n-1\}

form an interpolation basis on $D$ . Consequently, $\dim U=n+1=S$ .

Proof. Both $\mathcal T$ and $D$ have $n(n+1)$ elements. Put $y=x+\tau$ , so $D=H\times\mathcal Q$ . Values on the $n+1$ distinct points of $\mathcal Q$ and coefficients in the basis $1,y,\ldots,y^n$ are related by an invertible Vandermonde matrix. We may therefore work with those coefficients.

We will also use the multiplicative Fourier transform on $H$ . For a function $g:H\to\mathbb F_p$ , set

\widehat g(r)=\sum_{\tau\in H}g(\tau)\tau^{-r}, \qquad0\le r<n.

It is invertible, with

g(\tau)=\frac1n\sum_{r=0}^{n-1}\widehat g(r)\tau^r.

Indeed, after writing $H$ as a cyclic group, the finite geometric sum $\sum_{\tau\in H}\tau^m$ equals $n$ when $n\mid m$ and $0$ otherwise. Substitution in the proposed inverse leaves only the term indexed by the same residue.

Here is the algebra producing the Fourier blocks. For a column $\tau^jx^a$ ,

\tau^j(y-\tau)^a =\sum_{k=0}^a(-1)^{a-k}\binom ak y^k\tau^{j+a-k}.

The frequency- $r$ Fourier component retains exactly the terms satisfying

k\equiv j+a-r\pmod n.

Let $c$ be the residue of $j+a$ ; equivalently, a transformed entry can be nonzero only when

c\equiv j+a\equiv r+d\pmod n.

Write $j+a=c+tn$ . The degree constraint forces $t\in\{0,1\}$ . Since $j=c+tn-a$ , the two inequalities $j\ge0$ and $2j+a\le2n-1$ give

t=0:\quad0\le a\le c, \qquad t=1:\quad2c+1\le a\le n+c.

This derives the row set below. Since $a\le2n-1$ , each nonzero residue $1\le d\le n-1$ has at most the two exponents $k=d,d+n$ , and reduction by $y^{n+1}=y$ identifies $y^{d+n}$ with $y^d$ on $\mathcal Q$ . At residue zero, the exponents $k=0,n$ remain the distinct columns $1,y^n$ , because $y^n\ne1$ at $y=0$ . Multiply the monomial side by $(-1)^a$ and the coefficient side by $(-1)^d$ , then apply the determinant-one shear

C_0\longleftarrow C_0+(-1)^nC_n.

The resulting uniform block indexed by $c\in\{0,\ldots,n-1\}$ has columns $d=0,\ldots,n$ , rows

A_c=\{0,\ldots,c\}\cup\{2c+1,\ldots,n+c\},

and entries

B_c(a,d)=\binom ad+(-1)^n\binom a{d+n},

where $\binom a{2n}=0$ in the column $d=n$ .

For every affected row $a=n+r$ , the finite binomial identity

\binom{n+r}{d+n} =\sum_{k=0}^{r}\binom{n+r-k-1}{n-1}\binom{k}{d}

follows by taking the coefficient of $z^r$ in

(1-z)^{-n}\frac{z^d}{(1-z)^{d+1}} =\frac{z^d}{(1-z)^{n+d+1}}.

The convolution on the left gives the displayed sum, while the coefficient on the right is $\binom{n+r}{r-d}=\binom{n+r}{n+d}$ . The identity therefore allows the determinant-one row operation

R_{n+r}\longleftarrow R_{n+r} -(-1)^n\sum_{k=0}^{r} \binom{n+r-k-1}{n-1}R_k

to remove the second term. The required lower rows $0,\ldots,r$ all belong to $A_c$ . Thus

\det B_c =\det\left[\binom ad\right]_{a\in A_c,\,0\le d\le n} =\frac{\prod_{a<a'}(a'-a)}{\prod_{d=0}^{n}d!}.

The last equality is the Vandermonde determinant, since $\binom xd$ is a degree- $d$ polynomial with leading coefficient $1/d!$ . Every nonzero difference between nodes of $A_c\subseteq\{0,\ldots,2n-1\}$ , and every denominator factor, has magnitude less than $p=2n+1$ . Hence every block determinant is nonzero in $\mathbb F_p$ , proving unisolvence.

The space of monomials of weighted degree at most $2n$ has dimension $(n+1)^2$ , while $|D|=n(n+1)$ . Since $\mathcal T$ already evaluates bijectively, the kernel at weighted degree $2n$ has dimension $n+1$ . The slice map $F\mapsto F(0,x)$ is injective on that kernel: if the slice is zero as a function on $\mathbb F_p$ , then its degree is at most $2n=p-1$ , so one-variable root counting makes it the zero polynomial. Hence $F=\tau G$ . Because $\tau\ne0$ on $D$ , $G$ also vanishes on $D$ , but its weighted degree is at most $2n-2$ , contradicting unisolvence unless $G=0$ . Its image is $U$ , so $\dim U=n+1$ . $\square$

An explicit description of $U$

Define functions on $\mathbb F_p$

h_\ell(x)=x^\ell-\frac{2\ell}{2\ell-1}x^{n+\ell}, \qquad1\le\ell\le n,

and let

V=\operatorname{span}\{1,h_1,\ldots,h_n\}.

The denominators are nonzero because $1\le2\ell-1\le p-2$ . On $H$ , $h_\ell(s)=-s^\ell/(2\ell-1)$ . If a linear combination of the displayed generators vanishes, evaluation at $0$ first kills its constant coefficient; evaluation on the cyclic group $H$ then kills the coefficients of the $n$ distinct characters $s^\ell$ , $1\le\ell\le n$ . Thus $\dim V=n+1$ .

Lemma 6.7 (explicit lifts). Every generator of $V$ belongs to $U$ . Hence $U=V$ .

Proof. We give the lift, including the coefficient identity that makes it work. For $1\le\ell\le n$ , put

H_0(r)=(1+r)^{n+1}-(1+r)

and

G_\ell(r)=\sum_{j=0}^{\ell-1} \frac{(-1)^j(1-\ell)_j(n-\ell+1)_j} {(1-2\ell)_j\,j!}r^j,

where $(a)_j=a(a+1)\cdots(a+j-1)$ and $(a)_0=1$ . All denominators are products of nonzero integers of magnitude below $p$ .

Write $H_0G_\ell=\sum_a f_ar^a$ . We now perform the coefficient extraction explicitly. We first need the following terminating Pfaff identity: for an integer $L\ge0$ ,

\sum_{j=0}^{L}\frac{(-L)_j(b)_j}{(c)_j\,j!}z^j =(1-z)^L\sum_{j=0}^{L} \frac{(-L)_j(c-b)_j}{(c)_j\,j!} \left(\frac{z}{z-1}\right)^j.

This is a polynomial identity, so the apparent denominator on the right cancels. Indeed, the coefficient of $z^m$ on its right side is

(-1)^m\sum_{j=0}^{m} \frac{(-L)_j(c-b)_j}{(c)_j\,j!} \binom{L-j}{m-j} =\frac{(-L)_m(b)_m}{(c)_m\,m!};

after cancelling $(-L)_m/m!$ , the equality is the finite binomial identity proved in the next paragraph, with the parameters relabelled. Thus the coefficients agree term by term.

Apply this identity with $L=\ell-1$ , $b=n-\ell+1$ , $c=1-2\ell$ , and $z=-r$ . It gives

G_\ell(r) =(1+r)^{\ell-1} \sum_{j=0}^{\ell-1} \frac{(1-\ell)_j(-n-\ell)_j} {(1-2\ell)_j\,j!} \left(\frac r{1+r}\right)^j.

For $a\ge\ell+1$ , the polynomial $(1+r)G_\ell$ has degree at most $\ell$ , so it contributes nothing to $f_a$ . Consequently,

\begin{aligned} f_a=[r^a](1+r)^{n+1}G_\ell(r) &=\sum_{j=0}^{\ell-1} \frac{(1-\ell)_j(-n-\ell)_j} {(1-2\ell)_j\,j!} \binom{n+\ell-j}{a-j}\\ &=\binom{n+\ell}{a} \sum_{j=0}^{\ell-1} \frac{(1-\ell)_j(-a)_j} {(1-2\ell)_j\,j!}, \end{aligned}

where the second equality uses

\binom{n+\ell-j}{a-j} =\binom{n+\ell}{a}\frac{(-a)_j}{(-n-\ell)_j}.

The temporary denominator is legitimate: for $0\le j\le\ell-1$ , none of the factors in $(-n-\ell)_j$ is zero modulo $p=2n+1$ .

It remains to evaluate the finite sum. The required binomial identity is

\sum_{j=0}^{\ell-1} \frac{(1-\ell)_j(-a)_j}{(1-2\ell)_j\,j!} =\frac{(1-2\ell+a)_{\ell-1}}{(1-2\ell)_{\ell-1}}\,.

It gives, for $a\ge\ell+1$ ,

f_a=\binom{n+\ell}{a} \frac{(1-2\ell+a)_{\ell-1}}{(1-2\ell)_{\ell-1}}.

For an integer $L\ge0$ ,

\sum_{j=0}^{L}\frac{(-L)_j(b)_j}{(c)_j\,j!} =\frac{(c-b)_L}{(c)_L}.

It follows by induction on $L$ : split $\binom{L+1}{j}=\binom Lj+\binom L{j-1}$ , apply the induction hypothesis to the two resulting sums, and simplify

\frac{(c-b)_L}{(c)_L} -\frac bc\,\frac{(c-b)_L}{(c+1)_L} =\frac{(c-b)_{L+1}}{(c)_{L+1}}.

Taking $L=\ell-1$ , $b=-a$ , and $c=1-2\ell$ proves the displayed coefficient formula. For $\ell<a<2\ell$ , its numerator contains a zero factor, so $f_a=0$ . At the endpoints, direct extraction gives

\frac{f_{n+\ell}}{f_\ell}=-\frac{2\ell}{2\ell-1}.

Let $P=(n-\ell+1)_{\ell-1}$ and $E=(-1)^{\ell-1}(\ell-1)!\binom{n+\ell}{\ell}$ . Before normalization, $f_{n+\ell}=P/(1-2\ell)_{\ell-1}$ and $f_\ell=(E-P)/(1-2\ell)_{\ell-1}$ . Since $n=-1/2$ in $\mathbb F_p$ ,

P=\frac{(-1)^{\ell-1}3\cdot5\cdots(2\ell-1)}{2^{\ell-1}}, \qquad E=\frac{(-1)^{\ell-1}1\cdot3\cdots(2\ell-1)}{2^\ell\ell},

so $P=2\ell E$ . Here and below an empty product is $1$ . All displayed factors in $E$ are nonzero modulo $p$ , and

f_\ell=\frac{E-P}{(1-2\ell)_{\ell-1}} =\frac{E(1-2\ell)}{(1-2\ell)_{\ell-1}}\ne0.

The endpoint ratio is therefore the one claimed. Rescale $G_\ell$ , and hence every $f_a$ , so that $f_\ell=1$ .

Now define

F_\ell(\tau,x)= \sum_{a=1}^{\ell}f_a\tau^{\ell-a}x^a +\sum_{a=2\ell}^{n+\ell}f_a\tau^{n+\ell-a}x^a.

We have $f_0=0$ , because $H_0(0)=0$ , and $\deg(H_0G_\ell)\le n+\ell$ . Thus the two displayed intervals, together with the vanished gap $\ell<a<2\ell$ , account for every coefficient of $H_0G_\ell$ . The polynomial $F_\ell$ has weighted degree at most $2n$ . On $D$ , put $w=x/\tau$ . Since $\tau^n=1$ , its two displayed diagonals combine to give

F_\ell(\tau,x)=\tau^\ell(H_0G_\ell)(w).

Moreover $1+w=(x+\tau)/\tau\in\mathcal Q$ , so $(1+w)^{n+1}=1+w$ , and the right-hand side is zero. At $\tau=0$ , only its two endpoint terms remain, giving $F_\ell(0,x)=h_\ell(x)$ . Finally, $1-\tau^n$ lifts the constant function. Thus $V\subseteq U$ ; equal dimensions give $U=V$ . $\square$

The one-coordinate distance

Every word of $V$ is determined by a polynomial $A$ of degree at most $n$ : its values are $A$ on $\{0\}\cup H$ , and $A-4xA'$ on the nonsquares $N=\mathbb F_p^*\setminus H$ . Indeed, on $H$ the two terms of $h_\ell$ combine with coefficient $-1/(2\ell-1)$ , whereas on $N$ they combine with coefficient $(4\ell-1)/(2\ell-1)$ ; multiplication of the square-side coefficient by $1-4\ell$ gives the nonsquare-side coefficient.

For clarity, we also derive the matrix below. The interpolation nodes $\{0\}\cup H$ are the roots of $x(x^n-1)$ , so

A(x)=E(1-x^n)+\sum_{s\in H}u_s \frac{x(x^n-1)}{n(x-s)}.

Apply $1-4x\,d/dx$ and evaluate at $r\in N$ . Using $n=-1/2$ and $r^n=-1$ , the first term becomes $4E$ , while the $s$ -th interpolation term becomes $8r(r+s)/(r-s)^2$ . Therefore

f|_N=Tu+4E\mathbf1,\qquad T_{r,s}=\frac{8r(r+s)}{(r-s)^2},

where $E=A(0)$ , $u=(A(s))_{s\in H}$ , $r\in N$ , and $s\in H$ . Here $\mathbf1$ denotes the all-ones vector indexed by $N$ .

Lemma 6.8 (distance of the shadow code). For every prime $p\ge5$ , $d(U)=d(V)\ge3$ .

Proof. The weight is

\mathbf1_{E\ne0}+|\operatorname{supp}u| +\operatorname{wt}(Tu+4E\mathbf1).

Only inputs of weight one or two require attention.

If $u=0$ and $E\ne0$ , the last block is a nonzero constant, giving weight $n+1$ . If $E=0$ and $u$ is supported at one square $s$ , its nonsquare values are multiples of

K(t)=\frac{8t(t+1)}{(t-1)^2},\qquad t=r/s,

which vanishes only at $t=-1$ . Thus the total weight is at least $1+(n-1)=n$ , sufficient for $p\ge7$ ; for $p=5$ , $-1$ is a square and no nonsquare ratio is lost, giving weight $3$ . If $E\ne0$ and $u$ is one-sparse, the equation $K(t)=c$ is a nonzero polynomial of degree at most two and has at most two solutions. The two input coordinates plus at least $n-2$ output coordinates give total weight at least $n$ , sufficient for $p\ge7$ .

Finally, if $E=0$ and $u$ is supported at distinct $s_1,s_2$ , clearing denominators shows that its zero nonsquare coordinates are roots of the nonzero cubic

R(X)=u_1(X+s_1)(X-s_2)^2 +u_2(X+s_2)(X-s_1)^2.

It is nonzero because $R(s_1)=2u_1s_1(s_1-s_2)^2\ne0$ . It has at most three nonsquare roots, so the two input coordinates plus at least $n-3$ output coordinates give weight $n-1\ge3$ for $p\ge11$ . The two smaller primes are closed directly, without a search: with rows indexed by nonsquares and columns by nonzero squares,

T_{5}=\begin{pmatrix}3&4\\4&3\end{pmatrix} \quad\bigl(H=(1,4),\ N=(2,3)\bigr),

and

T_{7}=\begin{pmatrix}3&1&0\\1&0&3\\0&3&1\end{pmatrix} \quad\bigl(H=(1,2,4),\ N=(3,5,6)\bigr).

For $p=5$ , each nonconstant column settles the $E\ne0$ , one-sparse- $u$ case, and $\det T_5\ne0$ settles the two-sparse case. For $p=7$ , each pair of columns is independent, settling the remaining two-sparse case. These observations close precisely the cases not covered by the general root counts. Thus every nonzero word has weight at least three. $\square$

From one coordinate to the whole plane

Let $W$ be the vector space of total-degree- $\le p-1=2n$ polynomials in $x_0,x_1,x_2$ that vanish on the body $B$ , and let

\rho(f)(x_1,x_2)=f(0,x_1,x_2).

Lemma 6.9 (body-kernel decomposition).

\rho(W)=U(x_1)+U(x_2).

Proof. The lifts in Lemma 6.7, embedded separately in $x_1$ and $x_2$ , show $U(x_1)+U(x_2)\subseteq\rho(W)$ . The two copies meet exactly in the constants, so this subspace has dimension $2(n+1)-1=p$ . It remains to prove $\dim W\le p$ .

Write $\tau=x_0^2$ and decompose uniquely

f=E(\tau,x_1,x_2)+x_0O(\tau,x_1,x_2),

where $E$ has weighted degree at most $2n$ and $O$ at most $2n-1$ , with weights $(2,1,1)$ . Every $\tau\in H$ has the two square roots $\pm x_0$ ; because $p$ is odd, vanishing at both signs separates the even and odd parts. We use the following rank claim.

Weighted-grid rank claim. Evaluation on

\{(\tau,x_1,x_2):\tau\in H,\ x_i+\tau\in\mathcal Q\}

is injective through weighted degree $2n-1$ , and its kernel through weighted degree $2n$ has dimension at most $2n+1=p$ .

To prove the claim, put $z_i=(x_i+\tau)/\tau\in\mathcal Q$ and $w_i=z_i-1$ . Let $M_a$ be the coefficient vector of the remainder of $w^a$ modulo

Q_n(w)=(w+1)^{n+1}-(w+1).

Thus $M_a=e_a$ for $0\le a\le n$ , where $e_a$ is the $a$ -th standard coordinate vector. Evaluation on the $n+1$ distinct roots of $Q_n$ and the coefficient vector of the remainder modulo $Q_n$ are related by an invertible Vandermonde matrix. A monomial $\tau^jx_1^ax_2^b$ evaluates as $\tau^{j+a+b}w_1^aw_2^b$ . The invertible Fourier transform on $H$ splits the rows by the residue $e\equiv j+a+b\pmod n$ , and the row in that block is $M_a\otimes M_b$ .

For fixed $e\in\{0,\ldots,n-1\}$ , put $q=a+b$ and write $j+q=e+tn$ . Below weighted degree $2n$ , the inequalities $j\ge0$ and $2j+q\le2n-1$ give

t=0:\quad0\le q\le e, \qquad t=1:\quad2e+1\le q\le n+e.

No other $t$ is possible. Thus the totals $a+b$ arising below weight $2n$ are exactly

A_e=\{0,\ldots,e\}\cup\{2e+1,\ldots,n+e\}.

At weighted degree exactly $2n$ , the additional rows with $a,b>0$ have total $q=2e$ for $e>0$ , and $q=2n$ for $e=0$ . We add these mixed top-weight rows. Rows with $a,b\le n$ are coordinate vectors $e_a\otimes e_b$ and pivot on their matching columns; call them core rows. Every remaining row has exactly one exponent larger than $n$ . Call it horizontal when it is $(a,n+\delta)$ , grouping rows with the same low exponent $a$ into a horizontal fiber; call its transpose vertical.

We now enumerate every non-core row. When $e=0$ , only the added total $2n$ can be non-core, so $a+n+\delta=2n$ gives $\Delta_a=\{n-a\}$ for $1\le a\le n-1$ . When $e>0$ , the upper bound $q\le n+e$ gives $\delta\le e-a$ . The ordinary high totals begin at $2e+1$ , while the added mixed total is $2e$ . The latter is unavailable when $a=0$ . Therefore the possible values of $\delta$ are exactly the sets displayed next.

For $e=0$ , a horizontal fiber has $\Delta_a=\{n-a\}$ , $1\le a\le n-1$ , and we select the column $(a,n)$ . For $e>0$ , its possible $\delta$ ’s are

\Delta_0=\{\max(1,2e+1-n),\ldots,e\},

and, for $1\le a\le e-1$ ,

\Delta_a=\{\max(1,2e-n-a),\ldots,e-a\}.

After using the core pivots, a column $(a,r)$ is still unused precisely when its total $a+r$ is not one of the core-row totals. For $a=0$ , this leaves totals $e+1,\ldots,2e$ ; for $a>0$ , the mixed top total $2e$ is already used and this leaves $e+1,\ldots,2e-1$ . To keep horizontal selections disjoint from their vertical transposes, also require $r>a$ . Hence the eligible second-coordinate interval $G_a$ is

\{e+1,\ldots,\min(n,2e)\}\quad(a=0)

or

\{\max(a+1,e+1-a),\ldots,\min(n,2e-1-a)\}\quad(a>0).

A direct comparison of endpoints gives $|G_a|\ge|\Delta_a|$ . Explicitly, $|\Delta_0|=|G_0|=\min(e,n-e)$ . For $a>0$ , write $L=\max(1,2e-n-a)$ and $k=e-a-L+1=|\Delta_a|$ . If $2e-1-a\le n$ , then $L=1$ and $\max G_a-k+1=e$ ; if $2e-1-a>n$ , then $L=2e-n-a$ and the same equality follows with $\max G_a=n$ . Thus $G_a$ contains at least $k$ elements. Choose $R_a$ to be the $|\Delta_a|$ largest elements of $G_a$ , and select the columns $(a,r)$ with $r\in R_a$ . It follows that $\min R_a=e$ for $a>0$ , while $\min R_0=e+1$ . Hence every active low coordinate satisfies $a\le e-1<r$ , and also $\delta<r$ . Vertical rows use the transposed columns $(r,a)$ .

For each non-core row we selected exactly one column, and different same-orientation fibers have different fixed low coordinates. The selected horizontal and vertical sets are disjoint. When $e=0$ , every low coordinate and every $\delta=n-a$ is less than the selected terminal coordinate $n$ . When $e>0$ , every low coordinate of every active fiber is at most $e-1$ , whereas every selected terminal coordinate is at least $e$ . A horizontal row has first factor $M_a=e_a$ , so its entry on every selected vertical column is zero. The same global inequality, with the coordinates transposed, kills the other cross block after eliminating the core pivots. Together with the core pivot columns, these choices therefore form a square block-triangular minor; it remains only to prove that each diagonal arm block is nonsingular.

For each remaining arm block,

M_{n+\delta}[r] =(-1)^\delta\binom{n+\delta}{r} \binom{n+\delta-r-1}{\delta-1} =\frac{(-1)^\delta\delta\binom{n+\delta}{\delta}\binom nr} {n+\delta-r}.

To obtain this identity, first divide $w^{n+\delta}$ by $(w+1)^{n+1}$ . Replacing that divisor by $Q_n(w)$ changes the remainder only in degrees at most $\delta$ , so coefficients with $r>\delta$ are unchanged. Taylor expansion at $w=-1$ , followed by

\sum_{k=0}^{m}(-1)^k\binom Nk=(-1)^m\binom{N-1}{m},

gives the first expression; cancellation of factorials gives the second. The row scale $(-1)^\delta\delta\binom{n+\delta}{\delta}$ and column scale $\binom nr$ are nonzero: $1\le\delta\le n-1$ , $n+\delta\le p-2$ , and $0\le r\le n$ . After these nonzero scalings the block is the Cauchy matrix $1/(\delta+y)$ , $y=n-r$ . Its determinant is

\frac{\prod_{i<j}(\delta_j-\delta_i) \prod_{i<j}(y_j-y_i)}{\prod_{i,j}(\delta_i+y_j)}\,.

Here is a proof of the determinant formula. After multiplication by $\prod_{i,j}(\delta_i+y_j)$ , the determinant is a polynomial alternating separately in the $\delta_i$ ’s and in the $y_j$ ’s. It is therefore divisible by both Vandermonde products in the numerator. The degrees are equal, so the quotient is constant; comparing successive highest-degree coefficients in $\delta_m,\delta_{m-1},\ldots$ gives constant $1$ .

The determinant is nonzero: all labels are distinct and every denominator lies in $\{1,\ldots,n-1\}$ , hence is nonzero modulo $p$ . The selected minor is therefore nonsingular. This proves independence of all rows through weight $2n-1$ , even after adding the mixed top-weight rows.

The only top-weight monomials omitted from this independent augmented family have $a=0$ or $b=0$ . There are $(n+1)+(n+1)-1=2n+1=p$ of them. Thus the odd part is injective and the even kernel has dimension at most $p$ , proving the weighted-grid rank claim and $\dim W\le p$ .

We already exhibited a $p$ -dimensional subspace of $\rho(W)$ . (The intersection of the two copies of $U$ consists only of constants: a function of $x_1$ equal to a function of $x_2$ for all pairs must be constant.) Hence equality holds. $\square$

Lemma 6.10 (two-coordinate sum code). If $|X|=N$ and a linear code $L\subseteq\mathbb F^X$ over a finite field $\mathbb F$ contains the constant functions, then

d(\{a(x)+b(y):a,b\in L\})=N\,d(L).

Proof. Let $a(x)+b(y)$ be a nonzero sum-code word. If $a$ is nonconstant, every fiber $a^{-1}(t)$ has at most $N-d(L)$ elements because $a-t$ is a nonzero word of $L$ . Hence the number of zeros of $a(x)+b(y)$ is

\sum_t|a^{-1}(t)|\,|b^{-1}(-t)| \le N(N-d(L)).

If $a$ is constant, then $a+b$ is a nonzero word of $L$ , repeated $N$ times, and obeys the same lower bound. A minimum word depending only on $x$ attains $Nd(L)$ . $\square$

Lemmas 6.8–6.10 give

d(\rho(W))=p\,d(U)\ge3p.

Completion of the dimension-three proof

Proof of Theorem 6.1 for $d=3$ . Let $f$ have degree at most $p-1$ , with $x_0\nmid f$ . Its plane restriction $g=f(0,x_1,x_2)$ is nonzero, so Lemma 6.4 gives $\operatorname{wt}_P(g)\ge p$ . If $\operatorname{wt}_{C_3}(f)\le p$ , equality must hold and $f$ must vanish on all of $B$ . Thus $f\in W$ , so $g\in\rho(W)$ ; but every nonzero word in $\rho(W)$ has weight at least $3p$ , a contradiction. Hence every such $f$ has weight at least $p+1$ . Together with Lemmas 6.2 and 6.5, this proves

\mu(C_3)=p+1\qquad(p\ge5).

At $p=3$ , Lemma 6.5 gives weight at least $4$ when $x_0\mid f$ , while Lemma 6.4 gives at least $3$ nonzero plane values when $x_0\nmid f$ . The polynomial $x_1(x_1+1)$ vanishes on the body because there $x_1\in\{0,-1\}$ , and on the plane it is nonzero exactly when $x_1=1$ , at the three choices of $x_2$ . Its weight is therefore $3$ , proving $\mu(C_3)=3$ . $\square$

The next cell checks the finite-field linear algebra used in this proof. It verifies the shadow code, the body-kernel decomposition, the explicit support, and the exceptional $p=3$ case at the stated sample primes.

Show code

Code cell 23 · In [8]

def gf_rref(matrix, p):
    A = np.array(matrix, dtype=np.int64) % p
    rows, cols = A.shape
    pivots = []
    r = 0
    for c in range(cols):
        pivot = next((i for i in range(r, rows) if A[i, c] % p), None)
        if pivot is None:
            continue
        if pivot != r:
            A[[r, pivot]] = A[[pivot, r]]
        A[r] = A[r] * pow(int(A[r, c]), -1, p) % p
        for i in range(rows):
            if i != r and A[i, c] % p:
                A[i] = (A[i] - int(A[i, c]) * A[r]) % p
        pivots.append(c)
        r += 1
        if r == rows:
            break
    return A, pivots


def gf_rank(matrix, p):
    A = np.array(matrix, dtype=np.int64)
    if A.size == 0:
        return 0
    return len(gf_rref(A, p)[1])


def gf_nullspace(matrix, p):
    R, pivots = gf_rref(matrix, p)
    cols = R.shape[1]
    free = [c for c in range(cols) if c not in pivots]
    basis = []
    for c in free:
        v = np.zeros(cols, dtype=np.int64)
        v[c] = 1
        for row, pivot in enumerate(pivots):
            v[pivot] = -R[row, c] % p
        basis.append(v)
    return np.array(basis, dtype=np.int64).reshape(len(free), cols)


def row_spaces_equal(A, B, p):
    A = np.array(A, dtype=np.int64) % p
    B = np.array(B, dtype=np.int64) % p
    return gf_rank(A, p) == gf_rank(B, p) == gf_rank(np.vstack([A, B]), p)


def total_degree_exponents(variables, degree):
    return [
        exponent
        for exponent in product(range(degree + 1), repeat=variables)
        if sum(exponent) <= degree
    ]


def evaluate_monomials(points, exponents, p):
    return np.array([
        [
            np.prod([pow(int(x), int(a), p) for x, a in zip(point, exponent)],
                    dtype=np.int64) % p
            for exponent in exponents
        ]
        for point in points
    ], dtype=np.int64)


def pochhammer_mod(a, length, p):
    out = 1
    for j in range(length):
        out = out * (a + j) % p
    return out


def shadow_lift_coefficients(p, ell):
    n = (p - 1) // 2
    G = []
    for j in range(ell):
        numerator = ((-1) ** j
                     * pochhammer_mod(1 - ell, j, p)
                     * pochhammer_mod(n - ell + 1, j, p))
        denominator = pochhammer_mod(1 - 2 * ell, j, p)
        factorial = 1
        for factor in range(1, j + 1):
            factorial = factorial * factor % p
        denominator = denominator * factorial % p
        G.append(numerator * pow(int(denominator), -1, p) % p)

    H = [comb(n + 1, a) % p for a in range(n + 2)]
    H[0] = (H[0] - 1) % p
    H[1] = (H[1] - 1) % p
    raw = [0] * (len(H) + len(G) - 1)
    for i, left in enumerate(H):
        for j, right in enumerate(G):
            raw[i + j] = (raw[i + j] + left * right) % p
    scale = pow(int(raw[ell]), -1, p)
    return [(value * scale) % p for value in raw]


def shadow_code_audit(p, exhaustive=True):
    n = (p - 1) // 2
    Q0 = sorted(quadratic_residues(p))
    H = [q for q in Q0 if q]
    N = [x for x in range(1, p) if x not in H]

    D = [(tau, (q - tau) % p) for tau in H for q in Q0]
    weighted = [
        (j, a)
        for j in range(n + 1)
        for a in range(2 * n - 2 * j + 1)
    ]
    D_eval = evaluate_monomials(D, weighted, p)
    kernel = gf_nullspace(D_eval, p)
    slice_eval = evaluate_monomials([(0, x) for x in range(p)], weighted, p)
    U = kernel @ slice_eval.T % p
    triangle_columns = [
        index for index, (j, a) in enumerate(weighted) if 2 * j + a <= 2 * n - 1
    ]
    assert len(triangle_columns) == n * (n + 1)
    assert gf_rank(D_eval[:, triangle_columns], p) == n * (n + 1)
    assert gf_rank(U, p) == n + 1

    V = [np.ones(p, dtype=np.int64)]
    for ell in range(1, n + 1):
        coefficient = 2 * ell * pow(2 * ell - 1, -1, p) % p
        V.append(np.array([
            (pow(x, ell, p) - coefficient * pow(x, n + ell, p)) % p
            for x in range(p)
        ], dtype=np.int64))
    V = np.array(V, dtype=np.int64)
    assert gf_rank(V, p) == n + 1
    assert row_spaces_equal(U, V, p)

    for ell in range(1, n + 1):
        coeff = shadow_lift_coefficients(p, ell)
        assert all(coeff[a] == 0 for a in range(ell + 1, 2 * ell))
        assert coeff[n + ell] == (-2 * ell * pow(2 * ell - 1, -1, p)) % p
        for tau, x in D:
            value = 0
            for a in range(1, ell + 1):
                value += coeff[a] * pow(tau, ell - a, p) * pow(x, a, p)
            for a in range(2 * ell, n + ell + 1):
                value += coeff[a] * pow(tau, n + ell - a, p) * pow(x, a, p)
            assert value % p == 0

    minimum = None
    if exhaustive:
        minimum = p
        for coefficients in product(range(p), repeat=n + 1):
            if not any(coefficients):
                continue
            word = np.array(coefficients, dtype=np.int64) @ V % p
            minimum = min(minimum, int(np.count_nonzero(word)))
        assert minimum >= 3

    T = np.array([
        [8 * r * (r + s) * pow((r - s) % p, -2, p) % p for s in H]
        for r in N
    ], dtype=np.int64)
    if p == 5:
        assert np.array_equal(T, np.array([[3, 4], [4, 3]]))
    if p == 7:
        assert np.array_equal(T, np.array([[3, 1, 0], [1, 0, 3], [0, 3, 1]]))

    return U, minimum


def body_decomposition_audit(p, U):
    n = (p - 1) // 2
    Q0 = sorted(quadratic_residues(p))
    body = [
        (x0, (q1 - x0 * x0) % p, (q2 - x0 * x0) % p)
        for x0 in range(1, p)
        for q1, q2 in product(Q0, repeat=2)
    ]
    exponents = total_degree_exponents(3, 2 * n)
    body_eval = evaluate_monomials(body, exponents, p)
    W = gf_nullspace(body_eval, p)
    assert W.shape[0] == p

    plane = list(product(range(p), repeat=2))
    plane_eval = evaluate_monomials([(0, x1, x2) for x1, x2 in plane],
                                    exponents, p)
    rho = W @ plane_eval.T % p
    sum_code = np.array(
        [[word[x1] for x1, x2 in plane] for word in U]
        + [[word[x2] for x1, x2 in plane] for word in U],
        dtype=np.int64,
    )
    assert gf_rank(rho, p) == p
    assert gf_rank(sum_code, p) == p
    assert row_spaces_equal(rho, sum_code, p)
    return len(body), len(exponents), gf_rank(rho, p)


def certificate_support(p, d):
    Q0 = sorted(quadratic_residues(p))
    plane = [(0,) + tail for tail in product(range(p), repeat=d - 1)]
    body = [
        (x0,) + tuple((q - x0 * x0) % p for q in qs)
        for x0 in range(1, p)
        for qs in product(Q0, repeat=d - 1)
    ]
    support = set()
    excluded = {pow(2, -1, p), (-pow(2, -1, p)) % p}
    for point in plane + body:
        x0, last = point[0], point[-1]
        value = x0 * last % p
        for c in range(1, p):
            if c in excluded:
                continue
            value = value * (
                x0 + c * last - pow(4 * c % p, -1, p)
            ) % p
        if value:
            support.add(point)
    return support


audit_rows = []
for p in [5, 7]:
    U, dU = shadow_code_audit(p)
    body_size, polynomial_dimension, rho_dimension = body_decomposition_audit(p, U)
    support = certificate_support(p, 3)
    assert len(support) == p + 1
    audit_rows.append((p, dU, body_size, polynomial_dimension, rho_dimension,
                       len(support)))

for p in [11]:
    U, dU = shadow_code_audit(p, exhaustive=False)
    support = certificate_support(p, 3)
    assert len(support) == p + 1
    audit_rows.append((p, dU, 'not expanded', 'not expanded', 'not expanded',
                       len(support)))

for p in [3, 5, 7]:
    for d in [3, 4, 5]:
        S = (p + 1) // 2
        candidates = []
        for a0 in range(p - 1):
            for transverse in product(range(S), repeat=d - 1):
                if a0 + 2 * sum(transverse) <= 2 * (p - 2):
                    candidates.append(
                        (p - 1 - a0)
                        * np.prod([S - a for a in transverse], dtype=np.int64)
                    )
        assert min(candidates) == 2 * S ** (d - 2)

p3_cone = (
    [(0, x1, x2) for x1, x2 in product(range(3), repeat=2)]
    + [
        (x0, (q1 - x0 * x0) % 3, (q2 - x0 * x0) % 3)
        for x0 in [1, 2]
        for q1, q2 in product(sorted(quadratic_residues(3)), repeat=2)
    ]
)
assert sum((x1 * (x1 + 1)) % 3 != 0 for _, x1, _ in p3_cone) == 3

print('Polynomial-distance audits (p, d(U), |B|, polynomial dimension, '
      'dim rho(W), certificate weight):')
print(audit_rows)

Saved output 1

Polynomial-distance audits (p, d(U), |B|, polynomial dimension, dim rho(W), certificate weight):
[(5, 3, 36, 35, 5, 6), (7, 3, 96, 84, 7, 8), (11, None, 'not expanded', 'not expanded', 'not expanded', 12)]

Station Reference. The explicit certificate and body-footprint proof originate in Archives #124 and #126. The dimension-three proof is reconstructed from Archives #127, #128, #130, #132, #142, #145, and #147. These identifiers record the provenance of the proof above.

Related Work. The Cartesian-grid footprint bound and the full-grid Schwartz–Zippel bound are classical; see Geil and Høholdt (2000) and Dvir (2009). Here these bounds are combined with an explicit upper certificate and a separate dimension-three analysis to obtain equality. The exact distance formula for this quadratic-residue support appears to be new.

6.2 Exact inversion-map shear profile

We determine the image size of every shear $x\mapsto x^{-1}+tx$ and use it to compute the normalized second moment exactly. For $p\equiv1\pmod4$ , all nonzero shears have the same image size; for $p\equiv3\pmod4$ , their image sizes differ by one according to the quadratic character of $t$ . The proof reduces the image count to a quadratic discriminant calculation, revealing the precise role of the secant geometry of the inversion conic in this one-pole completion.

Let $f(0)=0$ and $f(x)=x^{-1}$ for $x\ne0$ . Put $P_t=|\operatorname{im}(f+t\,\mathrm{id})|$ , $S=(p+1)/2$ , $\delta_t=P_t-S$ , $E=\sum_t\delta_t$ , and, when $E>0$ ,

\rho(f)=\frac{\sum_t\delta_t^2}{2(S-1)E}.

Theorem 6.11.

\rho(x^{-1})= \begin{cases} (p+3)/(2(p-1)),&p\equiv3\pmod4,\\ 1/2,&p\equiv1\pmod4. \end{cases}

Proof. At $t=0$ , inversion is a bijection, so $P_0=p$ and $\delta_0=S-1$ . For $t\ne0$ and $x\ne0$ , a value $v$ is attained precisely when

tx^2-vx+1=0

has a solution. Its discriminant is $v^2-4t$ . The identity

\sum_v\chi(v^2-4t)=-1

follows by counting the $p-1$ solutions of $(v-u)(v+u)=4t$ . Therefore the image from nonzero $x$ has size $(p+\chi(t))/2$ . The separately supplied value $g_t(0)=0$ is already present exactly when $x^2=-1/t$ is soluble.

If $\chi(-1)=-1$ , square $t$ gives $\delta_t=1$ and nonsquare $t$ gives $\delta_t=-1$ . Hence $E=S-1$ and

\sum_t\delta_t^2=(S-1)^2+(p-1)=\frac{(p-1)(p+3)}4.

If $\chi(-1)=1$ , every nonzero $t$ has $\delta_t=0$ , so $E=S-1$ and $\sum_t\delta_t^2=(S-1)^2$ . Substitution gives the two formulas. $\square$

The next cell evaluates the shear profile directly at every odd prime from $5$ through $53$ and checks the two exact formulas according to $p\bmod4$ .

Show code

Code cell 26 · In [9]

def inversion_rho(p: int):
    inv = [0] + [pow(x, -1, p) for x in range(1, p)]
    S = (p + 1) // 2
    deltas = []
    for t in range(p):
        P_t = len({(inv[x] + t * x) % p for x in range(p)})
        deltas.append(P_t - S)
    E = sum(deltas)
    return Fraction(sum(delta * delta for delta in deltas), 2 * (S - 1) * E), deltas


inversion_rows = []
for p in [q for q in range(5, 54, 2) if is_prime(q)]:
    rho, deltas = inversion_rho(p)
    expected = Fraction(p + 3, 2 * (p - 1)) if p % 4 == 3 else Fraction(1, 2)
    assert rho == expected
    inversion_rows.append((p, str(rho), deltas[0], sorted(set(deltas[1:]))))
print('Inversion checks (p, rho, delta_0, nonzero-slope delta values):')
print(inversion_rows)

Saved output 1

Inversion checks (p, rho, delta_0, nonzero-slope delta values):
[(5, '1/2', 2, [0]), (7, '5/6', 3, [-1, 1]), (11, '7/10', 5, [-1, 1]), (13, '1/2', 6, [0]), (17, '1/2', 8, [0]), (19, '11/18', 9, [-1, 1]), (23, '13/22', 11, [-1, 1]), (29, '1/2', 14, [0]), (31, '17/30', 15, [-1, 1]), (37, '1/2', 18, [0]), (41, '1/2', 20, [0]), (43, '23/42', 21, [-1, 1]), (47, '25/46', 23, [-1, 1]), (53, '1/2', 26, [0])]

Station Reference. The inversion-map theorem and its statistic are based on Archive #72 and Eval #967.

Related Work. The maps $x\mapsto tx+x^{-1}$ are scalar multiples of the classical $x+a/x$ maps associated with Dickson polynomials; see Chou, Gómez-Calderón, and Mullen (1988). We combine their classical value-set structure with the shear statistic $\rho$ to derive the displayed closed formula.

7. Appendix verification

This section reconstructs the 25-entry literature comparison from explicit constructions and witness lines.

Write $\operatorname{Sq}_p=\{u^2:u\in\mathbb F_p\}$ and $S=(p+1)/2$ .

7.1 Saraf–Sudan quadratic-residue construction

This is the quadratic-residue construction of Mockenhaupt and Tao (2004), in the form recorded by Saraf and Sudan (2008).

Theorem 7.1. For odd prime $p$ , define

D_d=\{(t,y_1,\ldots,y_{d-1}):y_i+t^2\in\operatorname{Sq}_p\text{ for every }i\},\qquad H=\{0\}\times\mathbb F_p^{d-1}.

Then $C_d=D_d\cup H$ is Kakeya and

|C_d|=(p-1)S^{d-1}+p^{d-1}.

Proof. The hyperplane $H$ contains a line in every direction whose first coordinate is zero. For direction $(1,a_1,\ldots,a_{d-1})$ , put $b_i=a_i/2$ ; the line

t\longmapsto(t,a_1t+b_1^2,\ldots,a_{d-1}t+b_{d-1}^2)

lies in $D_d$ because $a_it+b_i^2+t^2=(t+b_i)^2$ . There are $pS^{d-1}$ points in $D_d$ , $p^{d-1}$ in $H$ , and $S^{d-1}$ in their intersection. Inclusion–exclusion gives the formula. Thus the often-quoted sum $pS^{d-1}+p^{d-1}$ is a valid upper bound but is not the exact size of the displayed union. $\square$

7.2 Bukh–Chao recursion with full translations

The recursion is due to Bukh and Chao (2021).

Let

G_d=\{(t,y_1,\ldots,y_{d-1}):y_i+t^2/4\in\operatorname{Sq}_p\text{ for every }i\}.

Starting with $R_1=\mathbb F_p$ , recursively set

R_d=G_d\cup\left((c_d,h_d)+(\{0\}\times R_{d-1})\right),\qquad(c_d,h_d)\in\mathbb F_p\times\mathbb F_p^{d-1}. \tag{7}

Every translation is allowed. The certificate uses two explicitly recorded special cases: complete transverse histories with every $c_d=0$ , and slice-placement histories with every $h_d=0$ . Together they improve twelve cells of the earlier column.

Theorem 7.2. Let $d\ge2$ . For every sequence $((c_j,h_j))_{j=2}^d$ with $c_j\in\mathbb F_p$ and $h_j\in\mathbb F_p^{j-1}$ , the set $R_d$ defined by (7) is Kakeya.

Proof. A graph direction $(1,a_1,\ldots,a_{d-1})$ has the line with anchor $(0,a_1^2,\ldots,a_{d-1}^2)$ in $G_d$ , since

a_i^2+a_it+t^2/4=(a_i+t/2)^2.

By induction, a witness $w+tv$ in $R_{d-1}$ becomes $(c_d,h_d+w)+t(0,v)$ in the translated slice. These are all remaining directions. Translation changes only the overlap, not this proof. $\square$

The code reads the complete parameters for every changed cell from the local certificate. Unchanged cells are regenerated by the earlier deterministic transverse recursion. In every case it materializes the point set and checks the displayed witnesses directly.

7.3 The KLSS image-stratum construction

This construction is due to Kopparty et al. (2011).

For $f(x)=x^2$ , put $I(t)=\{x^2+tx:x\in\mathbb F_p\}$ . Define

L_d=\bigcup_{j=0}^{d-1}\{(x_1,\ldots,x_j,t,0,\ldots,0):x_i\in I(t)\}.

Theorem 7.3. The set $L_d$ is Kakeya.

Proof. Let $v=(v_1,\ldots,v_d)$ be nonzero and let $j$ be the last index for which $v_j\ne0$ . Along the line with $j$ th coordinate $t=sv_j$ , choose, for $i<j$ ,

x_i(s)=\left(\frac{v_i}{v_j}\right)^2+s v_i=\left(\frac{v_i}{v_j}\right)^2+t\frac{v_i}{v_j}\in I(t).

All later coordinates are zero, so this line lies in the $j$ th stratum. $\square$

KLSS state the convenient upper bound $p\sum_{j<d}S^j$ . The displayed strata can overlap, so the code also reports the smaller cardinality of their actual union.

7.4 The KLSS missing-digit construction

This construction is also from Kopparty et al. (2011).

Theorem 7.4. For every prime power $q$ and $d\ge1$ ,

M_d=(\mathbb F_q^\times)^d\cup\{0,1\}^d

is Kakeya and has $(q-1)^d+2^d-1$ points.

Proof. For direction $v=(v_i)$ , take the anchor $b_i=1$ when $v_i=0$ and $b_i=0$ otherwise. At parameter zero, $b\in\{0,1\}^d$ . At every nonzero parameter $t$ , every coordinate of $b+tv$ is either $1$ or the nonzero value $tv_i$ , so $b+tv\in(\mathbb F_q^\times)^d$ . Finally, the two parts intersect only at $(1,\ldots,1)$ , proving the size formula. $\square$

7.5 Sharp planar sets, products, and characteristic three

Proposition 7.5. In dimension two, Theorem 7.1 has size $p(p+1)/2+(p-1)/2$ , the sharp value proved by Blokhuis and Mazzocca (2008). If $A\subseteq\mathbb F_p^a$ and $B\subseteq\mathbb F_p^b$ are Kakeya, then $A\times B$ is Kakeya of size $|A||B|$ .

Proof. Substitution of $d=2$ into Theorem 7.1 gives the planar size. For a direction $(u,v)$ , synchronize witness-line parameters in the two factors; if one component is zero, hold an arbitrary point of that factor constant. This gives a line in $A\times B$ , whose size is exactly the product. $\square$

The supplied $p=3$ arrays of sizes $13$ , $27$ , and $53$ are independently checked below. Only $13$ is used as an exact minimum; $27$ and $53$ are upper bounds.

7.6 Complete 25-cell comparison

The next cell reconstructs every named literature family, uses the 25 Station certificate checks from S2, and checks every AlphaEvolve integer against the commit-pinned provenance map. It then performs the baseline comparison and computes the 14-improvement, 11-tie, 0-loss tally.

Show code

Code cell 29 · In [10]

def assert_explicit_witnesses(points, witnesses, p: int, d: int):
    point_set = {tuple(map(int, point)) for point in points}
    directions = set(canonical_directions(p, d))
    assert set(witnesses) == directions
    for direction, anchor in witnesses.items():
        assert line_points(anchor, direction, p) <= point_set
    return len(directions)


def classical_qr_cone_with_witnesses(p: int, d: int):
    Q = quadratic_residues(p)
    plane = {(0,) + tail for tail in product(range(p), repeat=d - 1)}
    body = {
        (t,) + tail
        for t in range(p)
        for tail in product(range(p), repeat=d - 1)
        if all((y + t * t) % p in Q for y in tail)
    }
    points = plane | body
    inv2 = pow(2, -1, p)
    witnesses = {}
    for direction in canonical_directions(p, d):
        if direction[0] == 0:
            witnesses[direction] = (0,) * d
        else:
            witnesses[direction] = (0,) + tuple(
                (a * inv2) ** 2 % p for a in direction[1:]
            )
    return points, witnesses


def quadratic_graph_block(p: int, d: int):
    slices = {
        t: {(a * a + t * a) % p for a in range(p)}
        for t in range(p)
    }
    points = {
        (t,) + tail
        for t in range(p)
        for tail in product(slices[t], repeat=d - 1)
    }
    witnesses = {
        (1,) + slopes: (0,) + tuple(a * a % p for a in slopes)
        for slopes in product(range(p), repeat=d - 1)
    }
    return points, witnesses


def transverse_extension(p, child, child_witnesses, shift):
    d = len(next(iter(child))) + 1
    graph, witnesses = quadratic_graph_block(p, d)
    embedded = {
        (0,) + tuple((x + h) % p for x, h in zip(point, shift))
        for point in child
    }
    witnesses.update({
        (0,) + direction:
        (0,) + tuple((x + h) % p for x, h in zip(anchor, shift))
        for direction, anchor in child_witnesses.items()
    })
    return graph | embedded, witnesses


def transverse_history_set(p: int, d: int, history):
    points = {(t,) for t in range(p)}
    witnesses = {(1,): (0,)}
    points, witnesses = transverse_extension(p, points, witnesses, (0,))
    assert len(history) == d - 2
    for n, shift in enumerate(history, start=3):
        assert len(shift) == n - 1
        points, witnesses = transverse_extension(
            p, points, witnesses, tuple(shift)
        )
    return points, witnesses


def slice_placement_set(p: int, d: int, placements):
    if d == 1:
        assert len(placements) == 0
        return {(t,) for t in range(p)}, {(1,): (0,)}
    assert len(placements) == d - 1
    c = int(placements[0])
    child, child_witnesses = slice_placement_set(p, d - 1, placements[1:])
    graph, witnesses = quadratic_graph_block(p, d)
    embedded = {(c,) + point for point in child}
    witnesses.update({
        (0,) + direction: (c,) + anchor
        for direction, anchor in child_witnesses.items()
    })
    return graph | embedded, witnesses


def greedy_transverse_set(p: int, d: int):
    if d == 1:
        return {(t,) for t in range(p)}, {(1,): (0,)}, []
    child, child_witnesses, history = greedy_transverse_set(p, d - 1)
    Q_grid = list(product(sorted(quadratic_residues(p)), repeat=d - 1))
    overlap_counts = Counter(
        tuple((s - x) % p for s, x in zip(square_point, child_point))
        for child_point in child
        for square_point in Q_grid
    )
    maximum = max(overlap_counts.values())
    shift = min(h for h, value in overlap_counts.items() if value == maximum)
    points, witnesses = transverse_extension(p, child, child_witnesses, shift)
    return points, witnesses, history + [shift]


def klss_image_strata(p: int, d: int):
    image = {
        t: {(x * x + t * x) % p for x in range(p)}
        for t in range(p)
    }
    points = set()
    for j in range(d):
        for t in range(p):
            for prefix in product(image[t], repeat=j):
                points.add(prefix + (t,) + (0,) * (d - j - 1))
    witnesses = {}
    for direction in canonical_directions(p, d):
        j = max(i for i, value in enumerate(direction) if value)
        inv_last = pow(int(direction[j]), -1, p)
        anchor = tuple(
            (int(direction[i]) * inv_last) ** 2 % p for i in range(j)
        ) + (0,) * (d - j)
        witnesses[direction] = anchor
    return points, witnesses


def missing_digit_set(p: int, d: int):
    nonzero = set(product(range(1, p), repeat=d))
    binary = set(product((0, 1), repeat=d))
    points = nonzero | binary
    witnesses = {
        direction: tuple(1 if value == 0 else 0 for value in direction)
        for direction in canonical_directions(p, d)
    }
    return points, witnesses


CERTIFICATE_HASHES = {
    'benchmark_25_cells.json':
        '286014dfc5d7899dd83b0d7ab102b0e5807236a72e34d1f4b7fe64bf049d78f6',
    'literature_nested_shift_improvements.json':
        '89311a8f76b855878a5a1f88a79c6ff34a1b4e87cb2636afadc5a70df09154c7',
    'station_point_certificates.json':
        'f19d7b03cfdd5956386f2ef4ef2aa41009783ba0b5137d8b54d0ec360b22afad',
    'alphaevolve_provenance.json':
        'eb140c6f3a164e2c3ac6489f8dd932d1757cd8f671f466dbbfef5122bf31dc0a',
}
certificate_data = {}
for filename, expected_hash in CERTIFICATE_HASHES.items():
    path = ARTIFACT_DIR / filename
    if not path.is_file():
        raise FileNotFoundError(f'Required certificate is missing: {path}')
    assert hashlib.sha256(path.read_bytes()).hexdigest() == expected_hash
    certificate_data[filename] = json.loads(path.read_text())

assert station_manifest['benchmark_sha256'] == CERTIFICATE_HASHES['benchmark_25_cells.json']
benchmark_rows = certificate_data['benchmark_25_cells.json']['rows']
assert len(benchmark_rows) == 25
benchmark = {
    (d, p): {'literature': literature, 'ae': ae, 'station': station}
    for d, p, literature, ae, station in benchmark_rows
}
improvement_rows = certificate_data[
    'literature_nested_shift_improvements.json'
]['improvements']
improvements = {(row['d'], row['p']): row for row in improvement_rows}
assert len(improvements) == 12



alphaevolve_provenance = certificate_data['alphaevolve_provenance.json']
alphaevolve_values = {(row['d'], row['p']): row['size']
                      for row in alphaevolve_provenance['records']}
assert len(alphaevolve_values) == 25
assert alphaevolve_values == {(d, p): ae for d, p, _, ae, _ in benchmark_rows}
assert alphaevolve_provenance['official_notebook'].startswith('https://github.com/google-deepmind/')


small_arrays = {}
for filename, p, d, size in [
    ('kakeya_F3_d3_13.npy', 3, 3, 13),
    ('kakeya_F3_d4_27.npy', 3, 4, 27),
    ('kakeya_F3_d5_53.npy', 3, 5, 53),
]:
    path = ARTIFACT_DIR / filename
    if not path.is_file():
        raise FileNotFoundError(f'Required certificate is missing: {path}')
    assert hashlib.sha256(path.read_bytes()).hexdigest() == DATA_HASHES[filename]
    array = np.load(path, allow_pickle=False)
    assert_point_certificate(array, p, d, size)
    small_arrays[d] = {tuple(map(int, row)) for row in array}


def selected_literature_set(d: int, p: int):
    if p == 3 and d in (3, 4):
        points = small_arrays[d]
        return points, find_kakeya_witnesses(points, p, d), 'finite p=3 certificate'
    if p == 3 and d == 5:
        points, witnesses = missing_digit_set(p, d)
        return points, witnesses, 'missing-digit construction'
    if (d, p) in improvements:
        row = improvements[(d, p)]
        if row['method'] == 'transverse_shift_history':
            points, witnesses = transverse_history_set(p, d, row['parameters'])
        elif row['method'] == 'slice_placement_history':
            points, witnesses = slice_placement_set(p, d, row['parameters'])
        else:
            raise AssertionError(f"Unknown method: {row['method']}")
        return points, witnesses, row['method']
    points, witnesses, _ = greedy_transverse_set(p, d)
    return points, witnesses, 'greedy transverse recursion (unchanged cell)'


literature_checks = []
for (d, p), reported in benchmark.items():
    points, witnesses, method = selected_literature_set(d, p)
    direction_count = assert_explicit_witnesses(points, witnesses, p, d)
    assert len(points) == reported['literature'], (
        d, p, len(points), reported['literature']
    )
    literature_checks.append((d, p, len(points), direction_count, method))


family_checks = []
for d, p, _, _, _ in benchmark_rows:
    classical, classical_witnesses = classical_qr_cone_with_witnesses(p, d)
    assert len(classical) == (p - 1) * ((p + 1) // 2) ** (d - 1) + p ** (d - 1)
    assert_explicit_witnesses(classical, classical_witnesses, p, d)

    strata, strata_witnesses = klss_image_strata(p, d)
    assert len(strata) <= p * sum(((p + 1) // 2) ** j for j in range(d))
    assert_explicit_witnesses(strata, strata_witnesses, p, d)

    missing, missing_witnesses = missing_digit_set(p, d)
    assert len(missing) == (p - 1) ** d + 2 ** d - 1
    assert_explicit_witnesses(missing, missing_witnesses, p, d)
    family_checks.append((d, p, len(classical), len(strata), len(missing)))


for p in sorted({p for _, p, _, _, _ in benchmark_rows}):
    planar, planar_witnesses = classical_qr_cone_with_witnesses(p, 2)
    assert len(planar) == p * (p + 1) // 2 + (p - 1) // 2
    assert_explicit_witnesses(planar, planar_witnesses, p, 2)

product_3_5 = {a + b for a in small_arrays[3] for b in small_arrays[5]}
assert len(product_3_5) == 13 * 53
assert len(find_kakeya_witnesses(product_3_5, 3, 8)) == (3**8 - 1) // 2


before_after = [
    (row['d'], row['p'], row['before'], row['after'])
    for row in improvement_rows
]
assert all(benchmark[(d, p)]['literature'] == after for d, p, _, after in before_after)
verdicts = []
for row in benchmark.values():
    prior = min(row['literature'], row['ae'])
    verdicts.append(prior - row['station'])
assert (
    sum(value > 0 for value in verdicts),
    sum(value == 0 for value in verdicts),
    sum(value < 0 for value in verdicts),
) == (14, 11, 0)

print('Verified selected literature sets (d, p, points, directions, method):')
for row in literature_checks:
    print(row)
print('Before/after literature-reference corrections:')
for row in before_after:
    print(row)
print('Verified all named uniform families at all 25 cells.')
print('Verified all 25 Station point sets and every supplied witness line.')
print('Matched all 25 AlphaEvolve integers to the commit-pinned official notebook map.')
print('Verified finite p=3 sizes:', {d: len(points) for d, points in small_arrays.items()})
print('Verified |K_3 x K_5| in dimension 8:', len(product_3_5))
print('Benchmark tally: 14 strictly better, 11 ties, 0 worse.')

Saved output 1

Verified selected literature sets (d, p, points, directions, method):
(3, 3, 13, 13, 'finite p=3 certificate')
(3, 5, 53, 31, 'greedy transverse recursion (unchanged cell)')
(3, 7, 129, 57, 'greedy transverse recursion (unchanged cell)')
(3, 11, 440, 133, 'greedy transverse recursion (unchanged cell)')
(3, 13, 698, 183, 'slice_placement_history')
(3, 19, 2034, 381, 'greedy transverse recursion (unchanged cell)')
(3, 23, 3509, 553, 'greedy transverse recursion (unchanged cell)')
(3, 29, 6837, 871, 'greedy transverse recursion (unchanged cell)')
(3, 31, 8295, 993, 'greedy transverse recursion (unchanged cell)')
(3, 37, 13866, 1407, 'slice_placement_history')
(3, 41, 18708, 1723, 'slice_placement_history')
(3, 43, 21504, 1893, 'greedy transverse recursion (unchanged cell)')
(3, 47, 27899, 2257, 'greedy transverse recursion (unchanged cell)')
(3, 53, 39686, 2863, 'slice_placement_history')
(4, 3, 27, 40, 'finite p=3 certificate')
(4, 5, 163, 156, 'transverse_shift_history')
(4, 7, 528, 400, 'transverse_shift_history')
(4, 11, 2689, 1464, 'greedy transverse recursion (unchanged cell)')
(4, 13, 4972, 2380, 'transverse_shift_history')
(4, 17, 13521, 5220, 'slice_placement_history')
(4, 19, 20586, 7240, 'transverse_shift_history')
(5, 3, 63, 121, 'missing-digit construction')
(5, 5, 497, 781, 'transverse_shift_history')
(5, 7, 2142, 2801, 'transverse_shift_history')
(5, 11, 16307, 16105, 'transverse_shift_history')
Before/after literature-reference corrections:
(3, 13, 699, 698)
(3, 37, 13867, 13866)
(3, 41, 18709, 18708)
(3, 53, 39687, 39686)
(4, 5, 164, 163)
(4, 7, 529, 528)
(4, 13, 4973, 4972)
(4, 17, 13524, 13521)
(4, 19, 20593, 20586)
(5, 5, 503, 497)
(5, 7, 2145, 2142)
(5, 11, 16348, 16307)
Verified all named uniform families at all 25 cells.
Verified all 25 Station point sets and every supplied witness line.
Matched all 25 AlphaEvolve integers to the commit-pinned official notebook map.
Verified finite p=3 sizes: {3: 13, 4: 27, 5: 53}
Verified |K_3 x K_5| in dimension 8: 689
Benchmark tally: 14 strictly better, 11 ties, 0 worse.

Result and table provenance

For the AlphaEvolve column, alphaevolve_provenance.json maps each integer to the displayed constructor in the official commit-pinned notebook. The integers are obtained by executing the indicated constructor, deduplicating its returned points, and counting them; the notebook’s sub-unit random score jitter is not part of the cardinality. All 25 integers were independently replayed from that source.

1. Problem Definitions

2. S1. A new infinite family in d=3d = 3d=3 for p≡3(mod4)p \equiv 3 \pmod 4p≡3(mod4).

2.1 Construction and exact size

2.2 Equivalence with the p≡1(mod4)p \equiv 1 \pmod 4p≡1(mod4) AlphaEvolve construction

3. S2. Finite improvements and a 53-point Kakeya set in F35\mathbb{F}_3^5F35​.

3.1 The 53-point construction

3.2 The complete 25-pair Station result

4. S3. Structural analysis of the new infinite family.

5. Higher-dimensional uniform families

5.1 The four-dimensional prefix-sum family

5.2 The five-dimensional nested tangent family

6. Additional findings

6.1 Exact polynomial evaluation-code distance

The set and its evaluation code

An explicit polynomial certificate

Two elementary polynomial bounds

The lower bound when x0x_0x0​ divides the polynomial

The one-coordinate shadow code

An explicit description of UUU

The one-coordinate distance

From one coordinate to the whole plane

Completion of the dimension-three proof

6.2 Exact inversion-map shear profile

7. Appendix verification

7.1 Saraf–Sudan quadratic-residue construction

7.2 Bukh–Chao recursion with full translations

7.3 The KLSS image-stratum construction

7.4 The KLSS missing-digit construction

7.5 Sharp planar sets, products, and characteristic three

7.6 Complete 25-cell comparison

Result and table provenance

2. S1. A new infinite family in $d = 3$ for $p \equiv 3 \pmod 4$ .

2.2 Equivalence with the $p \equiv 1 \pmod 4$ AlphaEvolve construction

3. S2. Finite improvements and a 53-point Kakeya set in $\mathbb{F}_3^5$ .

The lower bound when $x_0$ divides the polynomial

An explicit description of $U$