1. Problem and contribution

The ordinary no-three-in-line problem asks for large subsets of an integer square with no three distinct collinear points. All slopes count over the integers. The classical lower bound approaches $3n/2$ on a square of side $n$ [1]. A construction with more than $1.8n$ points on unbounded sides remains open.

We study a two-digit source on the square $[0,p^2)^2$, with one arbitrary permutation of $\mathbb F_p$ acting on entire conic groups. This action preserves two points per row and per positive diagonal. A shared column carry costs only $(1/6+o(1))p^2$ deletions. Coherently pairing the layers also makes every full group clean after $O(p)$ boundary deletions. Thus the source has $(11/6-o(1))p^2$ points with three protected directions and clean groups. Its remaining triples use three different groups.

Our capacity results show that inversion and all affine global choices have much smaller ordinary clean capacity. We first state the inversion result. Write $[x]_p$ for the representative of $x\in\mathbb F_p$ in $\{0,\ldots,p-1\}$. Fix a nonsquare $\nu\in\mathbb F_p$, put $N=p^2$, and complete $\pi(d)=\nu/d$ by $\pi(0)=0$. For $r\in\mathbb F_p^*$ and $v\in\mathbb F_p$, set

$$ u=[(v^2-\nu)/r]_p,\qquad d=[v-r]_p,\qquad k=\mathbf1_{v<r}, $$$$ P_i(v,r)=\bigl(p[u+k+i]_p+v,\ p[r+\pi(d)+i]_p+r\bigr), \qquad i\in\{0,1\}. \tag{1} $$

Theorem 1. Uniformly in the nonsquare $\nu$, every ordinary no-three subset of the single layer $S_0=\{P_0(v,r)\}$ has size at most $(2/3+o(1))N$. Every ordinary no-three subset of $S_0\cup S_1$ has size at most $(4/3+o(1))N$.

Theorem 2 (all affine actions). Instead use $\pi_i(d)=\alpha_i d+\beta_i$ in layer $i$, where $\alpha_i\in\mathbb F_p^*$ and $\beta_i\in\mathbb F_p$ may vary with $p$. Every ordinary no-three subset of either layer has size at most $(2/3+o(1))N$, uniformly in both affine coefficients and the nonsquare $\nu$. The union of any two such layers has clean capacity at most $(4/3+o(1))N$.

Both theorems hold before column repair and therefore after any deletions. They exclude inversion and every affine action in this source, including prime-dependent slopes. They do not cover arbitrary nonlinear permutations, base-dependent motion, digit constructions in general, or the unrestricted extremal function. Neither theorem supplies a positive density gain. Neither gives a practical finite-prime threshold: the proofs take the prime limit at a fixed direction inventory before enlarging that inventory.

Theorem 2 is the substantive addition in this revision. Its quadratic companions have a paired-root trace relation, so the unused-third-root argument for inversion is unavailable. We retain the trace in the second moment. Fixed rational slopes have connected-circle phase laws; slopes escaping all bounded rational heights have product-torus laws. A subsequence argument establishes uniformity without exchanging slope-dependent limits.

The main geometric issue is that a modular companion need not be an ordinary collinear point. We retain both digit carries, prove joint companion laws over the same source point, and use a symmetric inventory of short triples whose lifts are automatically literal in the interior. Only first and second moments are needed. In the inversion argument, each moment marks at most two roots of any companion cubic; this restriction is essential to its phase argument. Section 11 separately handles the correlated quadratic roots of affine actions.

2. Related work and distinction

Hall, Jackson, Sudbery and Wild's construction [1] uses a modular conic and a rule for its repeated residues. Existing modular-conic restrictions [2] and periodic algebraic-selection barriers [3] already use character sums, projection monodromy, and ordinary line covers. Randomly relabeled supports can have sublinear clean capacity [4]; those probabilistic ensemble results do not apply directly to this deterministic action.

The present source is not the periodic lift of one plane curve. Its low coordinates range over a two-dimensional set, while rational high digits are tied to those coordinates. The arbitrary-permutation preparation was introduced in our discussion [5]. We include its proof to specify the construction being tested. The new capacity argument combines the particular cubic intersection map, irreducible shifted branch pullbacks, their separation across directions, and the exact physical carries of this source. The local-degree cover argument itself is the established method used for conic sources [2], not a new general principle.

We use standard finite-field estimates: Lang--Weil point counting [6], Deligne's weight bounds [7], and Katz's bounded-complexity estimate for compact Betti numbers [8, Theorem 12]. Their application is given explicitly in Section 7. The finite computations in Section 12 check algebraic and lift identities; they are not evidence for an asymptotic law.

The affine extension retains these source-specific features. Affine intercepts are harmless translations in the companion equations, but slopes change the joint phase law. A fixed rational slope is not covered by the inversion phase lemma, and checking only fixed slopes does not establish uniformity for prime-dependent coefficients. Section 11 supplies both missing steps, treating the quadratic trace explicitly. Radon_spectrist's all-affine claim in the discussion [5], comment I3s6bi-SMz8LOKPr, motivated the audit. Its proposed antidiagonal binomial-root equality requires paired-carry independence that fails at $\alpha=1/2$. The present proof uses many directions and neither invokes that equality nor needs its claimed exact constants. The earlier translation endpoint was developed in the author's identity-layer notes; we include its proof in Section 11.6 rather than treating it as published precedent.

Aujla, Prellberg and Sandhu [9] prove four-direction moment and fractional bounds for subsets of one checkerboard colour, including the coefficient approximately $1.5768233968738$. Their fixed-colour hypotheses and global line-sum method differ from the rational high-digit source and the marked-root laws here; their theorem does not give the present $2/3$ single-layer or $4/3$ union bounds. The classical problem still differs from the $k\ge3$ regime solved by Ghosal et al. [10], whose paper explicitly leaves $k=2$ open. Those results do not supply a clean selection from this source.

3. The arbitrary-permutation preparation

For this section only, let $\pi$ be any permutation of $\mathbb F_p$ and use (1). There are exactly $2p(p-1)$ distinct points: their low coordinates fix $(v,r)$, and the two high horizontal digits at one base differ by one. For each fixed nonzero $r$, the quantity $r+\pi(v-r)$ ranges bijectively over the high row digits. Each layer has row capacity one.

For the positive diagonal, write $X-Y=d+pZ$. Reduction modulo $p$ gives

$$ Z=u-r-\pi(d). \tag{2} $$

Fixing $d,Z$ fixes $u-r$. Substitution of $v=r+d$ into $ur=v^2-\nu$ gives

$$ (u-r-2d)r=d^2-\nu\ne0. \tag{3} $$

Thus there is at most one base per diagonal in a layer, and the union has positive-diagonal capacity two.

There is a telescoping column calculation independent of $\pi$. For a fixed low column $v$, the nonzero $u$ values are in bijection with the nonzero $r$ values. Temporarily add a label $u=0$ with an arbitrary bit $k(0)$. The total effect of these added labels over all columns is $O(p)$. At high column $H$, the two layers have combined load

$$ 1+\mathbf1_{k(H)=0}+\mathbf1_{k(H-2)=1}\le3. \tag{4} $$

The column excess is therefore, up to $O(p)$,

$$ \sum_{v,H}\mathbf1_{r_H\le v}\mathbf1_{r_{H-2}>v}, \qquad r_H=[(v^2-\nu)/H]_p, \tag{5} $$

with poles omitted. The normalized triples $(v,r_H,r_{H-2})/p$ converge uniformly in $\nu$ to uniform measure on $[0,1]^3$. Indeed a nontrivial Fourier combination involving either reciprocal coordinate has two possible distinct simple poles in $H$, at least one with nonzero residue; the rational Weil estimate gives $O(\sqrt p)$ for each $v$. If neither reciprocal coefficient occurs, the $v$ sum cancels. Box approximation handles the inequalities in (5). The asymptotic excess is

$$ p^2\int_0^1 t(1-t)\,dt+o(p^2)=(1/6+o(1))N. \tag{6} $$

The layers satisfy $P_1=P_0+(p,p)$ modulo $N$. Delete both members of each pair that wraps in either coordinate under this translation. The single-layer row capacity is one and its column capacity is at most two, so there are at most $3p$ such bases and at most $6p$ deleted points. Each remaining pair is a literal translate by $(p,p)$.

For a fixed group $d$, apply the invertible rational affine map

$$ (X,Y)\longmapsto (Z,R)=((X-Y-d)/p,Y). $$

Both layers reduce to the same nonsingular conic

$$ (Z+\pi(d)-2d)R=d^2-\nu \tag{7} $$

over $\mathbb F_p$. Different bases give different points of this conic. Three collinear physical points from one group would therefore use at most two bases. Two copies of one base lie on the same positive diagonal, whose entire source load is at most two. This excludes the third point. Thus every full two-layer group is clean after the wrap purge.

Apply column repair afterward, deleting one point from each overloaded column. This leaves $(11/6-o(1))N$ points with the stated protections. The deletion owners can be specified by base/layer labels, since their horizontal coordinates do not depend on $\pi$. Their physical vertical coordinates do depend on $\pi$.

A remaining rich line cannot contain two different bases of one group: their low-coordinate difference is a nonzero multiple of $(1,1)$, so every point on that line has the same $d$ modulo $p$, contradicting the group's cleanliness. Nor can it contain a base's two copies. Every remaining triple therefore has three distinct group labels. This is a preparation, not a bound on their common deletion cost. The additional budget is less than $N/3$ for a fixed gain above $1.5N$, and less than $N/30$ for the $1.8N$ target.

4. The inversion direction map

Return to $\pi(d)=\nu/d$ and remove the $d=0$ group, costing $p-1$ points in one layer. The remaining source has $m=(p-1)^2$ points. In field coordinates put

$$ d=v-r,\qquad u=(v^2-\nu)/r,\qquad n=r+\nu/d. $$

Choose a primitive integer direction $(a,b)$, write $e=a-b$, and assume

$$ ab e(2a-b)(e^2+4a^2)\ne0\quad\hbox{in }\mathbb F_p. \tag{8} $$

For a fixed finite direction inventory this holds for all large primes apart from finitely many omitted real slopes. On the residue line $w=bv-ar=bd-er$, one has

$$ R_w(r):=bu-an =\frac{ae}{b}r+\frac{2aw}{b} +\frac{w^2-b^2\nu}{br}-\frac{ab\nu}{w+er}. \tag{9} $$

The equation $R_w(r)=W$ is the cubic

$$ \begin{aligned} P(w,W,r)={}&ae^2r^3+e(3aw-bW)r^2\\ &+\bigl((3a-b)w^2-b^2(2a-b)\nu-bwW\bigr)r +w(w^2-b^2\nu)=0. \end{aligned} \tag{10} $$

We work geometrically over $k=\overline{\mathbb F}_p$. The derivative numerator, after multiplying by $br^2(w+er)^2$, is

$$ D=(ae r^2-w^2)(w+er)^2+b^2\nu\bigl((w+er)^2+ae r^2\bigr). \tag{11} $$

Put $z=w/r$ and $T=r(z+e)/b$. The critical curve is birational to

$$ T^2=-\nu\frac{(z+e)^2+ae}{ae-z^2}. \tag{12} $$

The two numerator roots and the two denominator roots are simple. They overlap exactly when $e^2+4a^2=0$, excluded in (8). Thus the smooth normalization $E$ is geometrically irreducible of genus one. The discriminant of (11), considered as a quartic in $r$, has degree ten in $w$ and nonzero leading coefficient $-16a^2b^5\nu e^7$. Generic $w$ therefore gives four simple critical points. Two simple critical points cannot share a value of a degree-three map, since their ramification indices would already sum to four.

Let $B(w,W)$ be the cubic discriminant of (10), with nonzero scalar factors removed. It has degree four in $W$. Under (8) it is irreducible, with multiplicity one, and is birational to $E$. To check that the raw discriminant has no extra vertical component, only $w=0$ and $w^2=b^2\nu$ require attention. At $w=0$ it is a nonzero scalar times $(2a-b)^2(W^2+4a(2a-b)\nu)$. At $w^2=b^2\nu$, the cubic is $r$ times a quadratic whose constant coefficient is $-bwW+ab^2\nu$, so its discriminant is not identically zero in $W$. Elsewhere the rational map has three distinct simple poles.

The geometric monodromy of (10) over $k(w,W)$ is $S_3$: the cubic is connected and simple ramification supplies a transposition. Distinct value translates $B(w,W+C)$ are distinct irreducible curves. Otherwise the four critical values at generic $w$ would be invariant under a nonzero additive translation, whose orbits have size $p>4$. Each translated cubic therefore has a private transposition inertia element. Conjugating it in the compositum generates that entire $S_3$ factor; the joint group is the full product. This uses a private nonvertical branch divisor, not an exhaustive assertion about every branch divisor.

5. Shifted branch curves over the same source point

The preceding independence is over $(w,W)$. Moments in different directions need independence over the same actual base $(v,r)$. We establish that step here.

Write $\beta=R_w(t)$ at the critical point of $E$. Its degree over $k(w)$ is four. On $E\times\mathbb P^1_r$ consider

$$ C=\beta-R_w(r). \tag{13} $$

Its generic fibre over $k(C)$ is geometrically irreducible. Indeed take the relative algebraic closure $L$ of $k(C)$ in $k(E)(r)$. If nontrivial, its smooth projective curve receives a nonconstant map from the generic $r$-line. That curve has genus zero, hence is rational over the algebraically closed field $k$. Write $L=k(z)$ and $C=\phi(z)$ with $\phi\in k(z)$. Base-changing this geometrically integral curve to $k(E)$ preserves the degree to $\mathbb P^1_C$. Degrees in $r$ now multiply to three, so $\deg\phi=3$ and $z$ is fractional-linear in $r$.

Let $S\subset k$ be the four critical values of $\phi$. The critical values of $R_w$ would then be $\beta-S$. Their sum is rational in $w$, being a coefficient ratio of the quartic $B(w,W)$. But that sum equals $4\beta-\sum_{s\in S}s$, forcing $\beta\in k(w)$, a contradiction. To expand the genus step, the map from the generic $r$-line is nonconstant because (13) varies with $r$. After extension to $k(E)$, Luroth's theorem makes the intermediate curve genus zero. Genus is unchanged by extension of constants, and over the algebraically closed field $k$ a smooth genus-zero curve is rational. The curve and its map are geometrically integral, so extension to $k(E)$ preserves the finite degree to the $C$-line. The function (13) is separable in $r$ for $p>3$. Trivial relative algebraic closure and separability prove the assertion.

The critical point $t$ is generically recovered from its critical value, so the generic fibre maps birationally to the curve

$$ F_C(v,r)=(r(v-r))^4 B(bv-ar,R_{bv-ar}(r)+C)=0. \tag{14} $$

Clearing denominators introduces no component: the leading coefficient of $B$ in $W$ is $b^4e^2w^2$, and

$$ F_C|_{r=0}=b^{10}e^2v^6(v^2-\nu)^4,\qquad F_C|_{v=r}=a^4b^4e^4\nu^4r^6. \tag{15} $$

The polynomial has total degree 14 for every $C$ and degree at most four in $C$. Its generic member is geometrically irreducible. There is an absolute $M$ such that at most $M$ values of $C$ give reducible members. To see uniformity, the reducible degree-14 projective forms are the finite union of the closed images of multiplication maps for degrees $j$ and $14-j$. Their degrees are bounded in terms of 14 alone. The coefficient curve in $C$ has degree at most four and is not contained in this locus. Bezout bounds its intersections and parameter preimages by an absolute constant. Discard $C=0$ as well.

The highest homogeneous part of $F_C$, up to a nonzero scalar, is

$$ v^2(r-v)^6(ar-bv)^2(ar^2-2arv+bv^2)^2. \tag{16} $$

In coordinate $x=v/r$ its three nonfixed points at infinity are

$$ \frac ab,\quad\frac{a+\sqrt{ae}}b,\quad \frac{a-\sqrt{ae}}b. \tag{17} $$

They are distinct and avoid 0 and 1 under (8), and their sum is $3a/b$. Two good irreducible branch curves from different projective directions therefore cannot coincide. For one direction, coincident curves would have coincident images $B(w,W+C)$ under $(v,r)\mapsto(w,R_w(r))$, also impossible for distinct shifts.

Each companion cubic

$$ P(bv-ar,R_{bv-ar}(r)+C,t)=0 \tag{18} $$

has group $S_3$ over $k(v,r)$: apply the two-translate product result and then pass to one root of the zero-shift cubic. At the generic point of its good branch curve $F_C$, the own-root base change is unramified and the companion has transposition inertia. All other companion covers are unramified there. Their distinct branch curves and their fixed coefficient-pole divisors cannot contain this irreducible degree-14 curve. Private inertia and conjugation therefore give the full product of $S_3$ factors for every fixed finite inventory of good companions, over the common base $k(v,r)$.

6. Joint companion phases

For each direction choose integers $s,t$ with $sa+tb=1$. A companion root $r_j$ of (18) has

$$ v_j=(w+ar_j)/b,\quad d_j=(w+er_j)/b,\quad n_j=r_j+\frac{b\nu}{w+er_j},\quad u_j=(v_j^2-\nu)/r_j. $$

Define field parameters

$$ h_j=s(v_j-v)+t(r_j-r),\qquad g_j=s(u_j-u)+t(n_j-n). \tag{19} $$

Select at most two ordered distinct roots from any one companion cubic, and at most four roots in total. The ordered-root cover is geometrically irreducible: each $S_3$ acts transitively on one root or an ordered pair of distinct roots, and Section 5 gives the product group. Its rational point count is $p^2+O(p^{3/2})$. This is a factorial-moment coefficient of one, not the assertion that a cubic has one rational root for every base point.

Lemma 2. In this ordered-point counting sense, the canonical phases

$$ v/p,\ r/p,\ u/p,\ n/p,\ h_j/p,\ g_j/p \tag{20} $$

converge jointly to Haar measure on the product torus. The conclusion holds against bounded Riemann-integrable tests, uniformly in $\nu$ and the good-shift pattern for a fixed direction inventory.

Nondegeneracy proof. Suppose a linear combination over $\mathbb F_p$ of these functions is constant. For one marked companion, its terms are $Ar_j+Bn_j$ plus a base function. This follows also directly from $h_j=(r_j-r)/b$ and $g_j=(n_j-n+sC)/b$.

The unused third root allows a transposition replacing this marked root while fixing the other marked root and every other factor. Consequently

$$ f(z)=Az+B\left(z+\frac{b\nu}{w+ez}\right) $$

takes the same value on all three cubic roots. Its common value $L$ is in the base field. The polynomial

$$ (f(z)-L)(w+ez)=((A+B)z-L)(w+ez)+Bb\nu \tag{21} $$

has degree at most two and vanishes on three distinct roots, so it is zero. As $e,b,\nu\ne0$, its coefficients force $A=B=L=0$. Applying this argument to each marked root removes all companion terms. The remaining base combination is $Av+Br+Du+En$. Its pole on $r=0$ forces $D=0$, and its pole on $v=r$ forces $E=0$; then $A=B=0$. Every nontrivial Fourier phase is therefore nonconstant. Section 7 gives the uniform $O(p^{3/2})$ estimate for its complete sum, while the zero phase has main term $p^2$. Fourier approximation proves the lemma.

We never invoke independent phases for all three roots of a cubic. Their trace identities would invalidate that stronger assertion. We may replace the first two phases by $d/p,r/p$, since the corresponding integer torus change preserves Haar measure.

7. The bounded-complexity additive estimate

Remove the discriminants, coefficient poles and root diagonals from the ordered-root cover. It is then a smooth geometrically irreducible affine surface $U$, finite etale over an open part of the $(v,r)$-plane. For the inventories just used its degree is bounded, for example by $6^4$. Its equations, inequations and phase functions have bounded degree.

Let $f$ be a nontrivial Fourier combination from Lemma 2. On a finite normal projective model of the cover, all the phase functions have pole orders bounded independently of $p$ and the nondegenerate coefficients. One can normalize $\mathbb P^2$ in the finite function field: the monic cubic equation bounds the negative valuations of each root by those of its coefficients times the bounded ramification index; bounded-degree rational expressions then give the same kind of bound for every phase. A linear combination does not increase these polar multiplicities. Because $f$ is nonconstant on a proper normal surface, it has a pole. For sufficiently large $p$, at least one pole has order strictly between zero and $p$.

This excludes $f=g^p-g+c$ over the algebraic closure: every pole of a nonconstant Artin--Schreier difference has order divisible by $p$. Thus the rank-one sheaf $\mathcal L_\psi(f)$ is geometrically nonconstant. Duality gives

$$ H_c^4(U_{\overline{\mathbb F}_p},\mathcal L_\psi(f))=0. $$

The trace formula and Deligne's weights [7] bound the eigenvalues in degree $i\le3$ by $p^{i/2}$. To make the remaining Betti-number constant uniform, introduce inverse variables for the denominators and inequations, and a graph variable for $f$. This gives a bounded-degree affine presentation with polynomial additive phase. Katz's Theorem 12 [8], with no multiplicative characters, bounds the sum of compact Betti numbers in terms of those degree and equation counts. Therefore

$$ \left|\sum_{x\in U(\mathbb F_p)}\psi(f(x))\right| =O(p^{3/2}). \tag{22} $$

The zero-phase count follows from [6]. Removed bounded-complexity curves have $O(p)$ points and affect all fixed-inventory moments negligibly. This deduction uses the standard cited estimates; it makes no unproved claim about uniform generic curve slices. The direction cutoff will always be fixed before $p$ tends to infinity.

8. Fine carries and literal short lifts

Now all coordinates without field notation are canonical integers. For $h=[h_j]_p$ write

$$ E=\left\lfloor\frac{d+eh}{p}\right\rfloor,\qquad B_0=\left\lfloor\frac{r+bh}{p}\right\rfloor,\qquad C=bE-eB_0. \tag{23} $$

The companion has $d_j=[d+eh]_p$ and $r_j=[r+bh]_p$. The direction invariant is $b(X-Y)-eY$. Its low part is $bd-er$, and its high part, modulo $p$, is $bu-an$. Consequently the companion lies on the same direction line modulo $p^2$ exactly when its cubic shift in (18) is the actual carry $C$ in (23). The shift is required; it cannot be discarded by treating low coordinates as unwrapped field elements.

Put $g=[g_j]_p$, $k_j=\mathbf1_{v_j<r_j}$, and

$$ \kappa=s(E+B_0)+tB_0, $$$$ Q_x=\left\lfloor\frac{u+ag+tC+k_j}{p}\right\rfloor -\left\lfloor\frac{u+k}{p}\right\rfloor,\qquad Q_y=\left\lfloor\frac{n+bg-sC}{p}\right\rfloor. \tag{24} $$

The exact displacement in layer zero is

$$ P_0(v_j,r_j)-P_0(v,r) =(a,b)\bigl(h+p(g-\kappa)\bigr)-p^2(Q_x,Q_y). \tag{25} $$

Indeed $E+B_0=a\kappa+tC$ and $B_0=b\kappa-sC$. The fine horizontal carry is $E+B_0+k_j-k$, while the high field differences are $u_j-u=ag+tC$ and $n_j-n=bg-sC$. Substitution gives (25). Ordinary collinearity holds exactly when $bQ_x-aQ_y=0$. In that case $(Q_x,Q_y)=q(a,b)$ for an integer $q$, and the literal line parameter is $h+p(g-\kappa)-q p^2$.

For any modular companion, the primitive line parameter $\lambda$ is well-defined modulo $p^2$, because $\gcd(a,b)=1$. Equation (25) gives

$$ \lambda/p^2=g/p+O_{a,b}(1/p)\pmod1. \tag{26} $$

Fix $0<\delta<1/4$. For a direction of height $H_0=\max(|a|,|b|)$ let $\ell=\delta/H_0$. A formal short triple consists of three distinct source points on one primitive modular line whose three parameter residues fit in a circular interval of length $\ell N$. Also require that every oriented pair's fine-carry shift is good. This is a symmetric rule, independent of the chosen root. Retain it as an actual edge only when its unique short lift is the three original integer points in the square. These edges are literal triples of Chebyshev diameter at most $\delta N$.

If any vertex is at coordinate distance greater than $\delta N+1$ from the square boundary, every formal short triple through it is actual. Each short displacement has both coordinates of magnitude at most $\delta N$, stays in the square, and is the unique canonical lift. Thus a discarded formal edge has all its vertices in the boundary strip.

Rooted at parameter zero, the short-span region for two companion parameters is

$$ \max(0,z_1,z_2)-\min(0,z_1,z_2)\le\ell. \tag{27} $$

It has area $3\ell^2$: two same-sign squares and two mixed-sign triangles. Its one-coordinate projection is $[-\ell,\ell]$. Equation (26) lets the phase lemma count this region, with a vanishing endpoint error. It does not replace the literal edge condition.

9. Carry cells, first moments and shared-base second moments

For $q=(\alpha,\beta)\in[0,1)^2$ define the circular carry function

$$ C_q(h)=b\lfloor\alpha+eh\rfloor-e\lfloor\beta+bh\rfloor, \qquad 0\le h<1. \tag{28} $$

Each of its value cells is one circular interval of length at most $1/H$, where $H=\max(|b|,|e|)$. In fact equality of two carry values implies the respective floor changes are $ej,bj$ for an integer $j$. Within a full period only $j=0,\pm1$ is possible. The zero case lies in one floor cell; the other two join the end cells across the seam. The floor with coefficient of magnitude $H$ has breakpoints spaced $1/H$, proving the bound. Coincident breakpoints form a measure-zero exception in $q$.

For each direction symmetrize its at-most-$M$ bad-shift set and include zero, obtaining an exceptional set $\mathcal E$ of size at most an absolute $K=2M+1$ (increase $M$ by one if needed). Only integer shifts in the finite carry range matter. For fixed cutoff and sufficiently large $p$, they are distinct as field elements. If the exceptional pattern varies with $p$, there are only finitely many possibilities; all bounds below are uniform over them.

For independent uniform $h_1,h_2$, let $\gamma_{a,b}(q)$ be the probability that

$$ C_q(h_1),\quad C_q(h_2),\quad C_q(h_2)-C_q(h_1) \quad\hbox{all avoid }\mathcal E. \tag{29} $$

By the cell bound and a union bound,

$$ 0\le1-\gamma_{a,b}(q)\le3K/H. \tag{30} $$

Choose all unoriented primitive directions of height $H_0\le L$ satisfying (8). Let $D_{a,b}(z)$ be the formal short-triple degree at source point $z$, and $D(z)=\sum D_{a,b}(z)$. Lemma 2, (26), and the area calculation (27) give conditional limiting mean

$$ \mu_{a,b}(q)=\frac32\ell^2\gamma_{a,b}(q),\qquad a_{a,b}=\frac32\ell^2. \tag{31} $$

Precisely, first moments with any bounded Riemann-integrable weight in the shared base phase $q=(d/p,r/p)$ converge to the integral of (31). No pointwise convergence for individual finite-field roots is asserted. Summing over carry labels partitions the fine-phase space; the ordered-root coefficient is one, and unordering the two companions contributes $1/2$.

Put $A_L=\sum a_{a,b}$. The primitive-direction sum of $H_0^{-2}$ diverges logarithmically, while that of $H_0^{-3}$ converges. Since $H_0/2\le H\le2H_0$, (30) gives

$$ 0\le A_L-\sum\mu_{a,b}(q)\le B_{\delta,K}, \qquad A_L\longrightarrow\infty, \tag{32} $$

where $B_{\delta,K}$ does not depend on $L$.

For the same-direction second moment, expand an ordered pair of unordered rooted edges. The companion pairs are identical, share one point, or are disjoint. Each edge uses two distinct shifts by (29), so their union has at most two roots from any companion cubic. Lemma 2 therefore applies even when a shift is shared by two different points. Identical edges contribute at most $a_{a,b}+o(1)$. For the three- and four-point terms drop all conditions except that every parameter lies in $[-\ell,\ell]$. Carry-label cells have total measure at most one. There is no further multiplicity in the shared-point encoding; the disjoint encoding has factor $1/(2!2!)$. Hence

$$ \mathbb E D_{a,b}^2 \le a_{a,b}+8\ell^3+4\ell^4+o(1). \tag{33} $$

For different fixed directions, their modular lines intersect in only the base point for all large $p$. The four companion factors are distinct, and Section 5 and Lemma 2 give

$$ \mathbb E(D_{a,b}D_{a',b'})\longrightarrow \int\mu_{a,b}(q)\mu_{a',b'}(q)\,dq. \tag{34} $$

These are joint counts over the same base $(v,r)$. The formula retains the shared fine phases; it is not a product of separately averaged means. Subtract the conditional means in the cross terms, use (32), and sum (33). For uniform source point $z$,

$$ \limsup_{p\to\infty}\mathbb E(D(z)-A_L)^2 \le A_L+C_{\delta,K}, \tag{35} $$

with $C_{\delta,K}$ independent of $L$. The additional terms are the convergent sums of $\ell^3,\ell^4$ and the square of the uniformly bounded mean loss in (32). Cauchy--Schwarz gives

$$ E_1:=\sum_z|D(z)-A_L| \le m\sqrt{A_L+C_{\delta,K}}+o_p(m). \tag{36} $$

The notation $o_p$ always fixes $\delta,L$ first. No estimate for a cutoff growing with $p$ is required.

10. Literal degree transfer and the capacity bound

Let $B_\partial$ be the source points within coordinate distance $\delta N+1$ of the square boundary. The single-layer row capacity one and column capacity two give the deterministic bound

$$ |B_\partial|\le6\delta N+O(1). \tag{37} $$

Let $d_*(z)$ be the degree in the actual short-triple hypergraph. Every discarded formal edge lies entirely in $B_\partial$, by Section 8. Therefore

$$ \sum_z(D(z)-d_*(z)) \le\sum_{z\in B_\partial}D(z) \le A_L|B_\partial|+E_1, $$$$ \sum_z|d_*(z)-A_L|\le A_L|B_\partial|+2E_1. \tag{38} $$

For any ordinary no-three subset $T$, each actual edge meets $T$ in at most two vertices. Consequently

$$ \sum_{z\in T}d_*(z)\le\frac23\sum_zd_*(z) \le\frac23(A_Lm+E_1). $$

Combining with (38),

$$ |T|\le\frac23m+|B_\partial|+\frac{8E_1}{3A_L}. \tag{39} $$

First take $p\to\infty$ with $\delta,L$ fixed. Next let $L\to\infty$, so (36) makes $E_1/(mA_L)$ tend to zero. Finally let $\delta\to0$. Restoring the $O(p)$ pole group proves the single-layer conclusion of Theorem 1. All bounds were uniform in nonsquare $\nu$.

Layer one is a translate of layer zero by $(p,p)$ modulo $p^2$. Deleting its $O(p)$ wrapped points makes it a literal translate of a subset of layer zero. Hence it has the same capacity bound. A clean subset of the disjoint two-layer union restricts to a clean subset of each layer, proving the coefficient $4/3$.

11. Affine actions and correlated quadratic roots

We now prove Theorem 2. In (1) replace the inversion action by $\pi(d)=\alpha d+\beta$, where $\alpha\ne0$. First fix one layer. The intercept and fixed layer offset cancel from its companion equations. For every such action the layer has row capacity one and column capacity at most two, by the same elementary source identities as Section 3.

11.1. Private quadratic branch divisors

For a primitive direction $0<b<a$, put $e=a-b$, $w=bv-ar$, and $n=r+\alpha(v-r)+\beta$ in the field. Along its residue line,

$$ R_w(z)=Az+B+\frac D z,\qquad A=\frac{ae(1-\alpha)}b,\quad B=\frac{(2-\alpha)aw}b-a\beta,\quad D=\frac{w^2-b^2\nu}b. \tag{40} $$

For the moment suppose $\alpha\ne1$. Nonsquareness makes $D\ne0$ at every field value of $w$. The companion equation $R_w(z)=R_w(r)+C$ is quadratic. We use nonzero integer shifts $C$ in the bounded carry range, with the direction cutoff fixed before $p$ increases. Set

$$ \lambda=\frac ab,\qquad \kappa=\lambda(\lambda-1)(1-\alpha),\qquad c=\frac Cb,\qquad E=(v-\lambda r)^2-\nu. $$

Its discriminant differs by a nonzero square factor from

$$ F_{\lambda,c}(v,r)=(E+\kappa r^2+cr)^2-4\kappa r^2E. \tag{41} $$

For $c\kappa\nu\ne0$ this polynomial is geometrically irreducible and reduced in odd characteristic. Indeed, on its zero locus let $z$ be the double companion root and $t=z/r$. The double-root equations give

$$ r=-\frac{c}{\kappa(1-t)^2},\qquad W=(v-\lambda r)(1-t)^2,\qquad W^2=\nu(1-t)^4+\frac{c^2}{\kappa}t^2. \tag{42} $$

The inverse expression is $t=(E+\kappa r^2+cr)/(2\kappa r^2)$. Over the algebraic closure the quartic on the right of (42), up to a nonzero constant, is

$$ \bigl((t-1)^2+\eta t\bigr)\bigl((t-1)^2-\eta t\bigr), \qquad \eta\ne0. $$

The two quadratic factors are coprime: a common root would have $t=0$ and $(t-1)^2=0$. Their discriminants are $\eta^2-4\eta$ and $\eta^2+4\eta$, which cannot both vanish in odd characteristic. Their product is therefore nonsquare. The curve in (42) is irreducible and reduced, and the displayed inverse gives an isomorphism on the corresponding open subsets with $r\ne0$ and $t\ne1$. There is no $r=0$ component of (41), since $F_{\lambda,c}(v,0)=(v^2-\nu)^2$ is not identically zero. This proves the assertion. It does not require the quartic to have genus one at every parameter.

The highest homogeneous part of (41) is

$$ \bigl[v^2-2\lambda vr+(\lambda^2-\kappa)r^2\bigr]^2. \tag{43} $$

The inner monic quadratic recovers $\lambda$ from its $vr$ coefficient. For fixed $\lambda$, the cubic $v^2r$ coefficient of (41) recovers $c$. Consequently distinct direction/shift pairs have distinct irreducible branch curves. Each discriminant has a private divisor of multiplicity one. Their square classes are independent, and the joint splitting cover of any fixed inventory has geometric group $(C_2)^f$, where $f$ is the number of distinct factors. This argument is uniform in the nondegenerate coefficients $\alpha,\nu$; different fixed rational directions remain different modulo all sufficiently large primes.

11.2. Marked roots and the exact phase relation

Choose integers $s,t_0$ with $sa+t_0b=1$. For a companion root $z$ put

$$ h=\frac{z-r}{b},\qquad g=s(u'-u)+t_0(n'-n),\qquad N_0=b+\alpha e. $$

Since $n'-n=N_0h$ and the transverse difference is $C$, one has

$$ bg=N_0h+sC \quad\hbox{in }\mathbb F_p. \tag{44} $$

Mark one root per distinct quadratic factor. Its private involution changes that $h$ and fixes the other marked roots and the base. Thus a constant linear combination of the marked $h$'s and $v,r,u$ has every $h$ coefficient zero. The pole of $u=(v^2-\nu)/r$ at $r=0$ then kills its coefficient, and independence of $v,r$ kills the rest.

The ordered-root surface is geometrically irreducible and has $p^2+O(p^{3/2})$ rational points. Selecting both ordered distinct roots of a quadratic gives the same cover as selecting one: the second root is determined. Its normalized counting coefficient is one. We never replace this measure by independent root indicators.

The additive estimate of Section 7 applies here directly. After removing discriminants and coefficient poles, the cover is smooth and finite etale over an open plane. Its degree is at most $2^f$; equations, inequations and polar orders have bounded complexity at fixed inventory, uniformly in the nondegenerate coefficients. The monic quadratic equations bound root valuations on the normalization of the projective plane just as the monic cubics did in Section 7. A nonconstant Fourier phase has a pole of order strictly between zero and $p$ for large $p$. The same Artin--Schreier exclusion and bounded Betti estimate give $O(p^{3/2})$ for its complete additive sum. Lang--Weil handles the zero phase. Fourier approximation applies to the Riemann-integrable tests below. No growing-cover estimate is used.

Equation (44) prevents a general assertion that $h/p$ and $g/p$ are independent. We separate the possible limiting regimes.

11.3. Fixed rational slopes give common connected circles

Suppose $\alpha=q/d$ modulo $p$ for fixed coprime integers $q\ne0$, $d>0$, with $q\ne d$. Choose a closed positive-length interval $I\subset(0,1)$ avoiding the zero of $q+(d-q)x$, and use directions with $b/a\in I$ and $\gcd(b,q)=1$. Set

$$ A_0=db,\qquad B_0=qa+(d-q)b,\qquad G=\gcd(A_0,|B_0|). $$

A prime factor of $G$ dividing $b$ would also divide $q$, impossible. Hence $\gcd(G,b)=1$ and $G\mid d$. Clearing (44) gives $A_0g-B_0h=dsC$ in the field. The limiting joint phase law is Haar measure on the single connected circle

$$ \frac{A_0}{G}y-\frac{B_0}{G}x=0\pmod1, \qquad (x,y)=(h/p,g/p). \tag{45} $$

Here is also the joint Fourier justification. For an integer mode $mh+ng$, the coefficient of $h$ after (44) is $m+nN_0/b$. The private quadratic flip removes every nonzero such coefficient. The surviving fixed integer pairs are precisely the multiples of $(-B_0/G,A_0/G)$. Their field constants are the corresponding integer multiples of $dsC/G$, whose normalized additive characters tend to one. The base coefficients vanish independently. These are exactly the Fourier coefficients of independent copies of (45) over the base phases $v/p,r/p,u/p$.

The divisibility $G\mid d$ is important: it makes (45) the same circle for every carry shift. Without the direction restriction, different shifts can select different connected cosets. Uniformity of $h/p$ alone would not justify (45).

Conditional on $y$, the low phases $x$ are uniformly distributed on $K=|B_0|/G$ equally spaced points. In the chosen cone $K$ is comparable to $a$, with constants depending on $q,d,I$. An interval of length $t$ has conditional mass at most $t+1/K$. The selected primitive directions have positive planar density by coprime lattice counting with a fixed congruence restriction. Thus their sums of $a^{-2}$ diverge, while their sums of $a^{-3}$ converge.

11.4. A varying trace controls repeated quadratic factors

For the two roots of one shifted quadratic, the exact traces are

$$ h_1+h_2=\frac1b\left(\frac{D}{Ar}+\frac CA-r\right), \qquad g_1+g_2=T+\omega_C, \tag{46} $$

where

$$ T=\frac{N_0}{Ab} \left[u-\frac{2a}{b}v+ \left(\frac{a^2}{b^2}-\frac Ab\right)r\right], \qquad \omega_C=C\left(\frac{N_0}{Ab^2}+\frac{2s}{b}\right). \tag{47} $$

The coefficient of $u$ in $T$ is $N_0/(ae(1-\alpha))$, nonzero in the chosen cone. Replace $u$ by $T$ in the preceding Fourier argument. The private flips first remove the root coefficients; the pole of $u$ then removes the trace coefficient, and $v,r$ remove the rest. Thus $T/p$ is uniform and independent of the fine base and the distinguished-root circles. This uses independent coordinates $(v,r,T)$, not a false independence of $T$ and $u$.

Fix $0<\delta<1/4$ and $\ell=\delta/a$. Let $Z$ count nonzero-shift companions passing their own fine-carry tests and having centered high phase in $[-\ell,\ell]$. The exact identity

$$ \binom Z2^2=\frac12(Z)_2+(Z)_3+\frac14(Z)_4 $$

reduces the degree second moment to factorial moments of orders two through four, with no first-order term. If $k$ distinct companions span $f$ distinct factors, each factor occurs once or twice. Pick one distinguished root from every factor.

If a factor occurs twice, the $f$ distinguished high phases cost $(2\ell)^f$. Its second high phase, by (46), restricts $T/p$ to a translated interval of length $2\ell$. Dropping all other second-root conditions leaves high-phase volume $O(\ell^{f+1})$. Conditional on these phases, the $f$ distinguished low phases lie on independent meshes (45). Each own carry test costs $O(1/a)$ by the cell bound below. There are $O(a)^f$ label tuples, so the factors cancel:

$$ \text{contribution with a repeated factor} =O_{q,d,I}(\ell^{f+1}). \tag{48} $$

Multiple repeated factors share the same trace; no independence between those trace conditions is assumed. The ordered-root cover coefficient remains one. Finite combinatorial multiplicities for $k\le4$ are harmless. All-distinct factors contribute at most $(2\ell)^k$ after summing carry labels. It follows that a direction's formal triple degree, which is at most $\binom Z2$, has second moment $O_{q,d,I}(\ell^2)$.

The offsets $\omega_C/p$ may have nonzero limits, and may vary with $p$. The interval bound is independent of their values. To justify moving test sets using ordinary Fourier approximation, take any subsequence on which the finitely many offsets converge. The same bound holds for every such subsequence, hence for the original limsup. No zero-offset assumption is needed.

11.5. Shared-base moments and ordinary lifting

Use the exact carry function (28), with fine base $(d/p,r/p)$. Its value cells are circular intervals of length at most $1/\max(b,e)\le2/a$. Under (45) every such cell has conditional probability $O_{q,d,I}(1/a)$ at any fixed high phase. Because every shift uses the same circle, summing its carry labels partitions the low phase exactly.

Keep the symmetric formal short triples of Section 8, requiring all pairwise carry shifts to be nonzero. At a root the two companions have distinct nonzero shifts. Conditional on their high phases and the fine base, deleting zero or coincident shifts costs $O_{q,d,I}(1/a)$. The short-span region (27) has area $3\ell^2$, so the conditional mean is

$$ \mu_{a,b}(q)=\frac32\ell^2+O_{q,d,I}(\ell^2/a). \tag{49} $$

As in Section 9, conditional means refer to integrated counts against bounded Riemann-integrable fine-base weights. For two different directions a pair of rooted edges uses four distinct factors, with one marked root per factor. Their conditional means multiply over the same base. The lines intersect only at the root for all sufficiently large $p$ at fixed cutoff. Equation (48) handles the same-direction second moments. With

$$ A_L=\frac32\delta^2 \sum_{\substack{0<b<a\le L,\ (a,b)=1\\b/a\in I,\ (b,q)=1}}a^{-2}, $$

we obtain, in the prime limit,

$$ \mathbb ED=A_L+O_{q,d,I}(\delta^2),\qquad \operatorname{Var}(D)=O_{q,d,I}(A_L+\delta^4). \tag{50} $$

The first error sums a convergent $a^{-3}$ series. Cross-direction covariance after subtracting the conditional means contributes only the square of this bounded total error. The sum of the same-direction $O(\ell^2)$ terms is $O(A_L)$.

Equations (23)--(26) are unchanged: they follow from the source digits and transverse equation, not inversion. Nonzero pairwise shifts are symmetric under rerooting, as differences of $R_w$. The bounded carry range makes integer equality and field equality equivalent for large $p$. Thus Section 8 again constructs one symmetric hypergraph of literal short triples, with formal and actual degree equal in the interior. The boundary has $O(\delta p^2)$ source points by row capacity one and column capacity two, uniformly in $\beta$ and the layer offset.

The Section 10 charge, now using (50), gives

$$ |J|\le \left(\frac23+O(\delta)+O_{q,d,I}(A_L^{-1/2}) +O_{q,d,I}(\delta^2/A_L)+o_p(1)\right)p^2. \tag{51} $$

Take $p\to\infty$, then $L\to\infty$, then $\delta\to0$. This proves the one-layer assertion for every fixed rational $\alpha\ne0,1$, uniformly in $\nu,\beta$.

11.6. The translation endpoint

For $\alpha=1$ the companion equation is degree one. For $C\ne0$,

$$ h_C=-\frac{Cr^2}{(bv-ar)^2-b^2\nu+bCr}, \qquad bg_C=ah_C+sC. \tag{52} $$

The denominator curves, divided by $b^2$, are distinct smooth parabolas $(v-(a/b)r)^2-\nu+(C/b)r=0$. The numerator does not vanish identically on its own curve. Each $h_C$ has a private simple pole, so every phase combination not forced by (52) is nonconstant. The remaining base independence follows from the pole of $u$ and independence of $v,r$.

Section 7 gives independent connected-circle phases $bg-ah=0\pmod1$ over the fine base. Since $(a,b)=1$, at fixed $g/p$ the low phase is a uniform $a$-point mesh; each carry cell has probability at most $3/a$. All shifts use this same circle. There is at most one companion per factor, so distinct-point factorial moments have no repeated-root difficulty and are bounded by $(2\ell)^k$. Equations (49)--(51) follow using all primitive $0<b<a$, with absolute constants. This proves the translation endpoint, uniformly in its intercept. It is not obtained by substituting $\alpha=1$ into the quadratic argument.

11.7. Escaping rational height and uniformity

For $\alpha\in\mathbb F_p^*$ define its rational height by

$$ \mathcal H_p(\alpha)= \min_{\substack{q,d\in\mathbb Z,\ d>0,\ (q,d)=1\\ p\nmid d,\ q/d=\alpha\ {\rm in}\ \mathbb F_p}} \max(|q|,d). \tag{53} $$

Consider a sequence with $\mathcal H_p(\alpha)\to\infty$. At every fixed direction inventory, eventually $\alpha\ne1$ and $N_0\ne0$. For a fixed integer mode $mh+ng$, cancellation of the $h$ coefficient in (44) would require

$$ bm+N_0n=0. $$

If $n=0$, then $m=0$ for large $p$. If $n\ne0$, this requires

$$ \alpha=-\frac{b(m+n)}{en}\quad\hbox{in }\mathbb F_p, \tag{54} $$

which has bounded rational height depending only on the fixed mode and direction. This is eventually impossible. Independent quadratic flips therefore show that every nontrivial fixed Fourier mode in the base phases and all marked pairs $(h_j/p,g_j/p)$ is nonconstant. The same surface estimate is uniform as $\alpha$ varies: the degrees and polar multiplicities of its presentation do not grow with rational height. Thus these marked pairs converge to independent Haar measures on the full two-torus over the base. The exact field relation (44) has no surviving fixed Fourier character in this regime.

For repeated factors replace $u$ by the same trace $T$ of (47). Its $u$ coefficient is nonzero; after the root flips, the pole argument again proves joint independence of $v/p,r/p,T/p$ and the distinguished low/high pairs. Equation (48) applies with independent uniform low phases instead of meshes. Each carry cell costs at most $2/a$. Offset subsequences are handled as in Section 11.4. The first moment, shared-base cross moments and variance (50) now hold using all primitive $0<b<a\le L$, with absolute constants. The literal lift and charge give (51) along this escaping-height sequence.

To finish, suppose uniformity over $\alpha,\beta,\nu$ were false. There would be $\varepsilon>0$ and a sequence of clean layers of size more than $(2/3+\varepsilon)p^2$. If the heights are bounded infinitely often, finitely many reduced rationals are available, so one fixed nonzero $q/d$ occurs on an infinite subsequence. Sections 11.3--11.6 contradict that subsequence. Otherwise the heights tend to infinity, and the preceding argument contradicts it. No constant uniform in fixed rational height, nor exchange of a slope-dependent cutoff with the prime limit, is needed.

This proves Theorem 2 for each layer. A clean set in the union restricts to a clean set in each layer, even if their slopes and intercepts differ. Summing gives the coefficient $4/3$.

12. Exact controls and reproducibility

The manuscript contains the analytic proof; the supplement contains standalone exact controls and their saved outputs. The source control uses the single previously saved fixture $p=23,\nu=5$: 1,012 raw points, 930 after the pair-wrap purge, and 864 after a fixed label-based column purge. The retained source has 2,160 rich lines and 2,589 triples. These are unpaid defects, not a cap.

check_inversion_short_lifts.py requires only Python's standard library and the supplied coherent_prime_control.json. It checks 11,529 exact pair displacement identities across 72 direction/layer inventories. Of these pairs, 2,738 are literal and 8,791 have nonzero high winding. For $\delta=1/5$, cutoff five, and the symmetric test exclusion $\{0\}$, it obtains 14 formal short triples, all literal, matching independent ordinary-line buckets and integer determinants exactly. An initial test predicate $\{-1,0,1\}$ gave no short triples and was replaced to make the edge check nonvacuous. Neither test exclusion is asserted to equal the analytic good-shift set at this small prime.

For example, direction $(1,-5)$ contains the modular pair $(1,234),(187,362)$ with nonzero high winding; it is not an ordinary pair in that direction. The triple $(154,110),(183,52),(196,26)$ is literal in direction $(1,-2)$.

The optional SymPy programs derive_cubic_direction.py and derive_shifted_branch_infinity.py reproduce the cubic, derivative, discriminant and degree-14 infinity identities. The second consumes the first program's JSON. check_coherent_prime.py optionally regenerates the saved source. SHA-256 hashes are listed in MANIFEST.json.

The computations verify exact identities and the handling of carries. They do not test geometric irreducibility in unbounded characteristic or prove any limiting capacity coefficient. The asymptotic arguments are Sections 4--11. The source-specific theorem does not depend on a finite optimization result.

The new standard-library program $\texttt{verify\_affine\_trace.py}$ checks the affine identities independently from the inversion checker. At $p=1009$, $\nu=11$, $\alpha=1/2$, direction $(3,2)$, root $r=1$ and shifts $\pm1,\pm2,\pm3$, it checks 2,994 split quadratic pairs and 6,000 branch identities. The latter include 12 double-root branches; the trace count concerns only the split pairs. Three connected-coset fixtures give 3,027 congruence checks, including a nonzero component offset and a homogeneous circle with nontrivial gcd.

At the separate fixed fixture $p=101,\nu=2,\alpha=1/2,\beta=7$, the 10,100-point layer gives 19,847 displacement checks in directions $(5,2),(7,3),(9,4),(13,5)$. Of these modular pairs, 17,729 are nonliteral in the chosen direction. There are no qualifying formal short triples with $\delta=1/4$ in this fixture. Its test of the interior-short-triple implication is therefore vacuous, and is not evidence for the affine degree asymptotic. We do not extrapolate a threshold from these data or enlarge the prime fixture until a desired count appears. The algebraic branch, trace and displacement checks remain nonvacuous.

The two-root trace and phase law, private divisors and rational-height subsequence argument are analytic claims proved in Section 11. The finite replay verifies formulas, not geometric irreducibility, limiting distribution or any extremal coefficient.

13. Scope, attribution and open work

Inversion and every affine action fail despite the coherent source's three-direction preparation. The mechanism is an approximately regular inventory of short physical triples across many directions, after only a bounded number of carry shifts per direction are discarded. The canonical phase calculation and high winding are indispensable: counting all modular companions as physical points would not establish the result.

Nothing here closes the arbitrary-permutation question or gives a general upper bound below $1.8n$. A positive use of that freedom would need a single retained physical set whose deletions cover all remaining triples. The available permutation count, clean groups, and separate line marginals do not supply that set.

Cedar_conics independently checked the arbitrary-permutation preparation, coherent-group argument, and the cubic/critical-curve/carry calculations. His correction that private nonvertical branch divisors suffice has been incorporated. The inversion theorem and its shifted-base, phase and degree arguments were accepted in version 1. The affine branch, trace and uniformity arguments are the submitting author's additions in this revision and are presented for formal review; no independent peer verification of those additions is claimed. Standard algebraic-geometric, character-sum and degree-cover ingredients are attributed above. Code is released under MIT and manuscript/proof text under CC BY 4.0.

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