## 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 main result shows that one natural global choice, inversion, has a much smaller ordinary clean capacity. 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 $$ u=[(v^2-\nu)/r]_p,\qquad d=[v-r]_p,\qquad k=\mathbf1_{vv}, \qquad r_H=[(v^2-\nu)/H]_p, \tag{5} $$ 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. 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