# Free Möbius actions, sharp coordinate capacity and oblique obstructions ## 1. Problem and contribution The no-three-in-line problem asks for large subsets of an integer square with no three collinear points, in all Euclidean directions. Classical constructions approach density $3/2$ relative to the side length [1]; the present community target is an unbounded family of density greater than $9/5$. We study a restricted construction mechanism: selecting points from a fixed union of periodic rational graphs. An obstruction for such a support is not an upper bound for unrestricted grid sets. Local row/column covers test a support before attempting a construction. A difficulty arises when successively explored polynomial fibers are treated as independent. Relations among affine output maps can invalidate this independence at greater depth. We replace the affine action by an explicit free Möbius action and prove that its translates of an entire branch set are disjoint. This establishes independence at every prescribed finite depth and justifies taking a limit of finite cover bounds. Set $$ A(v)=\frac{13v+8}{8v+5},\qquad B(v)=\frac{93v-872}{8v-75}=10+A(v-10). \tag{1} $$ For a polynomial $Q$ and prime $p$, let $\mathcal P_p(Q)\subseteq\mathbb F_p^2$ be the union $$ y=Q(x),\qquad y=A(Q(x)),\qquad y=B(Q(x)), \tag{2} $$ with undefined rational values omitted. Its periodic lift consists of integer points whose residues lie in this support. We use three applications: $$ Q_3(X)=X^3-3X,\qquad Q_4(X)=X^4-X,\qquad Q_5(X)=X^5-X. $$ **Theorem 1.** For $j\in\{3,5\}$, every sufficiently large prime $p$, and every no-three selection $S$ from the lift of $\mathcal P_p(Q_j)$ in every translated aligned square of integer side $N\ge p$, $$ \frac{|S|}{N}\le b_j+O\!\left(p^{-1/2}\log p\right), \qquad b_3=\frac{357849}{200000}=1.789245,\quad b_5=\frac{71991}{40000}=1.799775. \tag{3} $$ The constants are independent of $N$ and the square position. Both coefficients are below $9/5$. Let $s_3,s_5\in(0,1)$ be the unique roots of $$ 3s_3^4-6s_3^2+6s_3-2=0,\qquad 25s_5^4-90s_5^2+120s_5-44=0. \tag{4} $$ Define $$ \begin{aligned} c_3&=2-2s_3^3+3s_3^4-2s_3^6=1.789191831376919\ldots,\\ c_5&=2-2s_5^3+\frac94s_5^4-\frac56s_5^6 =1.798584319949732\ldots. \end{aligned} \tag{5} $$ Uniformly over the same windows and selections, $$ \frac{|S|}{N}\le c_j+o(1)\qquad(p\longrightarrow\infty). \tag{6} $$ We do not claim the error rate in (3) at the limiting constants (5). All three limiting capacities are identified exactly for the row/column relaxation on complete windows, as proved below. The fixed-depth constants may be enormous. **Theorem 2.** For each $j=3,4,5$, let $G_{j,p}$ be the residue support graph of (2). Its maximum matching number satisfies $$ \frac{2\nu(G_{j,p})}{p}\longrightarrow c_j. $$ Here $c_3,c_5$ are given by (5), while $$ 2s_4^4-16s_4^2+24s_4-9=0,\qquad 00. \tag{10} $$ Every finite point in (10) has degree four over $\mathbb Q$, by the irreducibility of $X^4-5$, taking reciprocals, and rational rescaling. **Lemma 3.** For any of the three branch sets $E_j$, $j=3,4,5$ (with $E_5:=E$ from (10)), if $g,h\in\Gamma$ are distinct, then their inverse translates by $g,h$ are disjoint. **Proof.** The cubic and quartic cases are proved in Sections 3.2–3.3. For the quintic, a nonidentity rational Möbius map cannot fix a degree-four point: its fixed-point equation has degree at most two. Section 3.1 excludes fixing infinity for a nonidentity element of $\Gamma$. Nor can a positive-determinant rational Möbius map send distinct points of $E$ to each other. A rational point and a degree-four point cannot be interchanged. Real maps preserve the real axis and, for positive determinant, each of the upper and lower half-planes. The only distinct pair not excluded this way is $\beta,-\beta$. If $(a\beta+b)/(c\beta+d)=-\beta$, then $$ c\beta^2+(a+d)\beta+b=0. $$ Degree four forces $c=b=0$ and $a=-d$, contradicting positive determinant. The reverse direction is identical. Hence no nonidentity element of $\Gamma$ maps a point of $E$ to a point of $E$. Apply this to $hg^{-1}$. $\square$ For each finite list of words, Lemma 3 survives modulo every sufficiently large prime. Indeed all maps have rational coefficients, and the finitely many branch-set nonintersection conditions can be expressed by nonzero resultants after clearing denominators. Excluding their prime divisors and finitely many bad-reduction primes suffices. This assertion is made separately for each fixed depth; it does not claim a free action of an infinite group on a finite projective line. ## 4. Independent fibers at every fixed depth The incidence relation on a variable source value is $$ v\longmapsto\{v,A(v),B(v)\}. \tag{11} $$ At a row $y$, the three inverse arguments are $y,A^{-1}(y),B^{-1}(y)$. Formal nonbacktracking walks correspond to reduced walks in the graph with two vertices and three parallel edges. Its fundamental group is free of rank two, with generators represented by $A,B$. Therefore (11) has a formal tree neighborhood at every fixed depth: a nontrivial reduced walk cannot give an identity of rational source functions. For $Q=Q_j$, $j=3,4,5$, the splitting field of $Q(X)-T$ has geometric Galois group $S_j$. The finite inertia generators are transpositions, since its finite critical values are distinct and simple. They generate a transitive group, and a transitive group generated by transpositions is the full symmetric group. The same argument applies in sufficiently large characteristic. For a local source argument $g(y)$, the splitting field of $Q(X)-g(y)$ has branch set $g^{-1}(E_j)$. Lemma 3 makes these sets pairwise disjoint for distinct local arguments. These Galois closures are jointly linearly disjoint over the algebraic closure of the constants. Inductively, the intersection of a new closure with the previous compositum can ramify only in the intersection of their branch sets, which is empty. There is no nontrivial connected unramified cover of the projective line. Consequently the geometric group is a direct product of $S_j$ groups. The arithmetic group embeds in the same permutation product and contains this full geometric group; it is therefore the same product, with no additional constant-field dependence. It follows that the root counts of distinct local fibers are independent, with the fixed-point distribution of a uniform permutation in $S_j$. Write $z=\Pr(R=0)$, $o=\Pr(R=1)$, $d=\Pr(R=2)$ for the fiber's number $R$ of distinct rational roots. Then $$ (z,o,d)=\begin{cases} (1/3,1/2,0),&j=3,\\ (3/8,1/3,1/4),&j=4,\\ (11/30,3/8,1/6),&j=5. \end{cases} \tag{12} $$ These are exact Frobenius probability masses. Actual finite-field counts have square-root errors and finitely many branch exceptions. There is no congruence restriction on the primes. The argument establishes more than a shallow numerical coincidence: at each fixed depth the incidence tree and its independent probability law have a characteristic-zero proof and persist outside a finite set of characteristics. This is the reason the later limiting upper bound is justified. ## 5. A finite grouped cover A source group $v$ comprises the columns $x$ satisfying $Q(x)=v$. All its columns have the same available target neighbors in (11), with undefined rational values omitted. Initially cover every row adjacent to a group with $R\ge3$, and every column in a group with $R=1$ or $2$. This covers every support edge. In one synchronous round, remove each currently covered group with at most $R$ uncovered target neighbors, and cover those neighbors. Removing $R$ columns and adding at most $R$ rows preserves the cover and does not increase its size. This is an exact statement about the finite support graph, including branch fibers, pole values, and coincident neighbors. For the probability calculation, exceptional values are discarded and the generic tree of Section 4 is used. Let $r_{t-1}$ be a directed row's survival probability when its parent source group is omitted. Initially $$ r_0=(z+o+d)^2. $$ Facing an uncovered parent row, a singleton must have at least one of its other two rows uncovered, and a doubleton must have both uncovered. Thus, with $r=r_{t-1}$, define $$ \begin{aligned} Z_t&=z+o(2r-r^2)+dr^2, & r_t&=Z_t^2,\\ U_t&=Z_t^3, & S_t&=o(3r^2-2r^3)+2dr^3. \end{aligned} \tag{13} $$ The two other groups of a directed row pass independently, giving $r_t$. A full row has three tests, giving its uncovered mass $U_t$. A full singleton column remains when at least two of its three rows remain available; a doubleton group remains only when all three do. The latter contributes two actual columns, explaining the factor two in $S_t$. Earlier-survival factors are redundant by monotonicity: if enough neighboring rows remain at the current round, the group could not already have been removed. This is also a direct way to derive (13) from the synchronous finite cover, without assuming a stationary recursion. The final line-cover density is $1-U_t+S_t$. Every physical horizontal or vertical line contains at most two points of a no-three set, so the leading coefficient is $$ c_t=2(1-U_t+S_t). \tag{14} $$ Only row and column consequences of no-three-in-line are used. ## 6. Rational certificate and the limit of the bounds ### 6.1 Eight rounds The supplied standard-library programs evaluate (13) with exact rational outward rounding. The denominator is $10^{20}$ for the cubic and $10^{12}$ for the quintic. This avoids enormous exact numerators while retaining rigorous bounds. For $$ Z(r)=z+2or+(d-o)r^2,\qquad S(r)=o(3r^2-2r^3)+2dr^3, $$ both functions are nondecreasing on $[0,1]$, since $$ Z'(r)=2o(1-r)+2dr\ge0,\qquad S'(r)=6or(1-r)+6dr^2\ge0. $$ An enclosure $l\le r\le h$ therefore gives the coefficient upper bound $$ 2\bigl(1-Z(l)^3+S(h)\bigr). \tag{15} $$ At round eight, exact rational comparison gives $$ c_{3,8}<1.789243830339<\frac{357849}{200000},\qquad c_{5,8}<1.799774687636<\frac{71991}{40000}. $$ For the quintic, the input enclosure is $$ \frac{228155200313}{500000000000}\le r_7 \le\frac{456310400629}{1000000000000}. $$ For orientation, the outward upper bounds, displayed rounded upward, are: | Round | Cubic | Quintic | |---:|---:|---:| | 1 | 1.803372013 | 1.862696860 | | 2 | 1.794549541 | 1.834659160 | | 4 | 1.790209591 | 1.810349166 | | 8 | 1.789243831 | 1.799774688 | The proof uses the exact comparisons in the supplied certificates, not the displayed decimals. ### 6.2 Taking the limit in the correct order The map $r\mapsto Z(r)^2$ is increasing on $[0,1]$. Since $r_1\le r_0$, the iteration decreases to a fixed point. Writing its square root as $s$, equation $s=Z(s^2)$ gives the corresponding equation in (4). The derivatives of the two polynomials have respective positive minima $$ 6-\frac8{\sqrt3}>0,\qquad 120(1-\sqrt{3/5})>0 $$ on $[0,1]$. Both polynomials are negative at zero and positive at one. Each therefore has a unique root in this interval. This paragraph concerns $j=3,5$; the quartic capacity is treated in Section 8. Hence $r_t\to s_j^2$, and (14) tends to the corresponding expression in (5). For any $\varepsilon>0$, first choose a fixed $t$ with $c_tX_1+X_2\}},\qquad X=\mathbf1_{\{Y_1+Y_2=0\}}. \tag{A1} $$ Putting $r=\Pr(X=1)$ and $s=\Pr(Y=0)$ yields $$ r=s^2,\qquad s=z+o(2r-r^2)+dr^2=Z(r). \tag{A2} $$ The objective in the cited theorem becomes $$ H(s)=\mathbb E\min\{R,\operatorname{Bin}(3,s^2)\} +\Pr\{\operatorname{Bin}(3,1-s)\ge2\}. \tag{A3} $$ To verify its second term directly, let $K$ be the number of incoming $Y$ messages equal to one at a row. The theorem's three clipped residual-capacity terms have sum three when $K=0$, sum one when $K=1$, and sum zero when $K\ge2$. Its positive-part correction is therefore exactly $\mathbf1_{\{K\ge2\}}$. The first term in (A3) is $$ 3(1-z)s^2-3os^4+(o-d)s^6, $$ since capacities at least three never truncate a binomial variable with three trials. Consequently $$ H(s)=1-3zs^2+2s^3-3os^4+(o-d)s^6. \tag{A4} $$ Its difference from the cover density is the exact polynomial identity $$ H(s)-\bigl(1-s^3+S(s^2)\bigr) =3s^2\bigl(s-Z(s^2)\bigr). \tag{A5} $$ Thus the allocation objective agrees with the cover at every scalar fixed point. We still must address the infimum over all solutions; uniqueness does so here. For $j=3,4,5$, the equations $s=Z(s^2)$ are respectively $$ 3s^4-6s^2+6s-2=0,\quad 2s^4-16s^2+24s-9=0,\quad 25s^4-90s^2+120s-44=0. $$ Each polynomial is negative at zero and positive at one and increases strictly on $[0,1]$. The cubic and quintic derivative bounds were proved in Section 6. For the middle polynomial, $$ \frac{d}{ds}(2s^4-16s^2+24s-9) =8(s-1)(s^2+s-3)>0\quad(0\le s<1). $$ Every Bernoulli recursive solution therefore has the same parameter $s_j$, and hence the same objective. There is a Bernoulli solution for this parameter by (A1)–(A2); message values at unused edge capacities can be defined by the same clipping rule without affecting the objective. The infimum in [7, Theorem 1] equals $H(s_j)$. We conclude $$ \frac{2\nu(G_{j,p})}{p}\longrightarrow 2\bigl(1-s_j^3+S(s_j^2)\bigr)=c_j. \tag{A6} $$ Exact rational root isolation gives $1.802335973 For every finite projection value and each component, at least one root of that projection fiber is unramified in its test cover. The exclusion of critical images makes every finite test-branch fiber have $N_i$ distinct root blocks. Its inertia is a translation with identity on each unramified block. Critical inertia remains a labelled transposition confined to one component. Finite inertia normally generates the composed geometric group by the tame one-branch-point argument. Its conjugates preserve the stated forms and proper-support property. The graph of critical transpositions is connected, since the projection group is generated by its finite transpositions. The group lemma applies. Thus all the selected-label conclusions, including mixed patterns and the actual degree-one arithmetic cosets, remain valid under this weaker hypothesis. Several test-branch points may share one projection image, even within the same component. The exceptional configuration now to rule out is a **whole projection fiber contained in the test branch set**, rather than an arbitrary pair collision. This cover statement assumes sufficiently large characteristic for all covers to be tame; for fixed cover complexity this excludes finitely many characteristics. As in the original theorem, its constants may depend on fixed exploration depth. A separation condition cannot be dropped entirely. Pure $S_N$ together with the diagonal translation by the nonidentity element of $C_2$ is transitive on labelled blocks but has only a diagonal $C_2$ kernel. For $N\ge3$, omitting one block does not give independent selected labels. The supplement enumerates this counterexample as well as nonabelian and mixed proper-support examples. ### 10.3. An exact finite-round inequality Take $D=32$. The standard-library certificate evaluates (9.1)--(9.3) with outward-rounded rational endpoints of denominator $10^{40}$. For a cavity interval $[r_-,r_+]$, it uses the rigorous lower bound $$ \rho_D\ge3o(1-r_+)^2(z+or_-^2)^2. $$ The gain $\rho^2/2-\rho^3/12$ is increasing for $0\le\rho\le1$. The certificate therefore proves, in rational arithmetic, $$ C_{32}-\frac{\rho_{32}^2}{2}+\frac{\rho_{32}^3}{12} <\frac{179999}{100000}<\frac95. $$ Its computed upper endpoint is approximately $1.799989601834655$. For perspective, the limiting cavity value gives $\rho_\infty=0.0689185618781408\ldots$ and the limiting bound $1.7999883682254115\ldots$. These decimals only describe the exact certificate. The proof uses the fixed 32-round rule and the rational inequality, not a limit taken simultaneously with the prime. The script is `mobius_single_diagonal_exposure_certificate.py`, importing the existing standard-library cavity certificate `mobius_quintic_tree_certificate.py`. The independent implementation control `mobius_certified_exposure_control.py` checks the six-message event against the actual parallel coordinate cover, rather than using the probability formula. Every certified row has exactly one private column in 24 combinations of prime and depth; the same check passes at p=100019 and p=1000003 at depth 32. At the latter prime the certified density is 0.0689017933, compared with the limiting finite-depth prediction 0.0689004614. These controls check the event implementation only. Both authors independently reproduced the 32-round rational endpoint and audited the forced event, mixed selected-label application and physical-fiber inequality. ## 11. The original quartic scale Genericity alone does not decide the fixed scale $m=1$. We verify the weaker hypotheses of Lemma 6 directly for $Q(u)=u^4-u$. Set $g_0(v)=v$, $g_1(v)=A(v)$ and $g_2(v)=B(v)$, with $A,B$ as in (1), and $F_j(u)=g_j(Q(u))-u$. Let $E$ be the three finite critical values of $Q$ together with infinity, and put $$ \mathcal B=\{u\in\overline{\mathbb Q}:Q(u)\in\Gamma E\}. $$ Every finite test branch at every fixed exploration depth belongs to this set. We prove that its images avoid all finite projection-critical values and that it contains no whole finite projection fiber. These statements concern the entire characteristic-zero orbit, so every fixed-depth cover inherits them after reduction outside finitely many primes. ### 11.1. Critical values avoid every local test branch **Lemma 7.** The three projections are Morse, their finite critical-value sets are pairwise disjoint, and $F_j(u)$ is never a finite critical value of any $F_i$ when $u\in\mathcal B$ and $F_j(u)$ is finite. **Proof.** Let $a^3=1/2$. The finite critical values of $F_0$ are $-3a/2$ and its conjugates. The finite branch values of $Q$ are $\beta=-3a^2/4$ and its conjugates, satisfying $\beta^3=-27/256$. They have the same normal closure $L=\mathbb Q(a,\zeta_3)$, whose Galois group is $S_3$. The critical-value resultants for $F_1,F_2$ are irreducible of degree eight. The exact source `quartic_scale_one_irreducibility.py` computes both resultants and certifies irreducible reductions modulo $7$ and $13$, respectively. Their degree equals the number of zeros of the derivative numerator, so irreducibility in characteristic zero also proves simple critical points with distinct critical values. The two octics are distinct, and neither shares a root with the irreducible cubic for $F_0$. This proves the first two claims. If $Q(u)$ belongs to a finite branch orbit, its value generates a cubic field, since a rational Möbius map is invertible. Thus $[\mathbb Q(u):\mathbb Q]=3d$ for an integer $1\le d\le4$. If $Q(u)$ belongs to the infinity orbit, it is rational and this degree is at most four. None of these degrees is divisible by eight. Since $F_j(u)\in\mathbb Q(u)$, it cannot equal a critical value of $F_1$ or $F_2$. Suppose $F_j(u)=t=-3a/2$. Every orbit value $v=Q(u)$ belongs to $L$, so $u=g_j(v)-t$ also belongs to $L$. For $j=1,2$, the quintic fiber polynomial is irreducible over $\mathbb Q(a)$. Indeed, at the prime ideal above $5$ with $a\mapsto2$, both reduce after making monic to $$ X^5+X^4+4X^2+4X+4. $$ The source checks $\gcd(f,X^5-X)=1$ and $X^{5^5}\equiv X\pmod f$, proving irreducibility. A root cannot lie in $L$, which has degree two over $\mathbb Q(a)$. For $j=0$ the fiber factors as $$ F_0(u)-t=(u-a)^2(u^2+2au+3a^2). $$ Its quadratic roots are $a(-1\pm\sqrt{-2})$. Neither lies in $L$, whose unique quadratic subfield is $\mathbb Q(\sqrt{-3})$. Thus $u=a$ and $v=Q(a)=-a/2$. This is not rational. If $v=M(\beta')$ for a finite branch value $\beta'$, then $\mathbb Q(\beta')=\mathbb Q(a)$, forcing $\beta'=-3a^2/4$ among the three branches. The unique rational Möbius map sending this value to $v$ is $z\mapsto3/(16z)$. Uniqueness follows because equality of two Möbius maps at a cubic irrational gives a polynomial relation of degree at most two. Its matrix determinant is $-48$, whereas every element of $\Gamma$ has positive determinant. This is impossible. All cases are excluded. The certificate was reproduced independently; both octic and quintic finite-field irreducibility tests passed. The argument above is uniform over the infinite branch orbit and uses no finite word cutoff. **Lemma 8.** No finite fiber of any $F_j$ is contained entirely in $\mathcal B$. The next three subsections give a proof using the valuation at $2$ and real-root counts. Section 12 gives an independent proof using reduction at $5$. ### 11.2. The branch set and its two types Let Gamma be generated by the determinant-one integer matrices of g1 and g2. Their diagonal entries are odd and off-diagonal entries divisible by 8. These properties persist under products and inverses. The finite branch values of Q are the three roots beta of $\beta^3=-27/256$. Put $$ L=\mathbb Q(\sqrt[3]{2},\zeta_3),\qquad \mathcal B=\{u\in\overline{\mathbb Q}:Q(u)\in\Gamma\{\infty,\beta_1,\beta_2,\beta_3\}\}. $$ Every branch point of every fixed-depth exposure test cover lies in this set, or at infinity. Pulling back a Q-splitting cover can only branch above its finite critical values or infinity; taking composita and subcovers introduces no additional branch points. Branch points that are poles of a projection map to infinity and need no finite separation. Fix an extension of the 2-adic valuation to the algebraic closure, normalized by v(2)=1. For any beta and any matrix M in Gamma, the denominator of M(beta) is a unit: its terms have valuations at least 1/3 and 0. Its numerator has valuation -8/3. Consequently $$ v(M(\beta))=-8/3,\qquad v(u)=-2/3\quad\text{if }Q(u)=M(\beta). $$ Since each gj M also belongs to Gamma, $$ v(F_j(u))=-8/3 \tag{11.1} $$ at every such beta-type branch point. All these Q-values belong to L. The completion of L at 2 has ramification index 3: the pure cubic extension is totally ramified of degree 3, while adjoining zeta3 is unramified of degree 2. Thus values of nonzero elements of L belong to $(1/3)\mathbb Z$. The other, infinity-type, branch points have rational Q-value q=M(infinity). If Fj is finite there, a=gj(q) is rational. Such a point cannot have v(Fj(u))=-8/3. To prove this without an assumption about cancellation, suppose t=a-u has that valuation. If v(a)>-8/3 (including a=0), then v(u)=-8/3 and v(Q(u))=-32/3, contradicting rationality of q. Otherwise v(a)=-l for an integer l>=3. In $$ Q(a-t)-q=(a^4-a-q)+(1-4a^3)t+6a^2t^2-4at^3+t^4, $$ the four nonconstant terms have valuations $$ -3l-2/3,\quad -2l-13/3,\quad -l-6,\quad -32/3. $$ For l=3 the last is uniquely smallest. For l>=4 the first is uniquely smallest. That minimum is nonintegral and so cannot cancel the rational constant term. This proves the contradiction. In particular the two branch types never have the same projected image. ### 11.3. A beta-type image cannot contain a whole fiber Consider a projection value t with v(t)=-8/3. For j=1,2 write gj=(av+b)/(cv+d). Its degree-five fiber polynomial is $$ cu^5+(ct-a)u^4-cu^2+(d-ct+a)u+(dt-b). $$ Here c=8, both a,d are odd, a+d=18, and v(b)=3. Its nonzero coefficient valuations, ordered by exponent, are $$ (0,-8/3),\ (1,1/3),\ (2,3),\ (4,0),\ (5,3). $$ The lower Newton polygon has slopes 2/3 of length four and 3 of length one. Thus one root has valuation -3. It cannot be beta-type, whose roots have valuation -2/3, and cannot be infinity-type by Section 11.2. At least this root is outside the test branch set. For j=0, suppose one root u is a branch point. It is beta-type and v(u)=-2/3. The nonzero differences e from u to the other roots of $u^4-2u=t$ satisfy $$ e^3+4ue^2+6u^2e+(4u^3-2)=0. $$ The coefficient valuations from constant to cubic are 0,-1/3,4/3,0. Its lower polygon has slopes -1/3 of length one and 1/6 of length two. Two other roots therefore differ from u by valuation -1/6. They cannot be branch points: by Section 11.2 any branch root u' in this fiber is beta-type, and equality of F0 values gives $$ u'-u=Q(u')-Q(u)\in L, $$ whose valuation must be in $(1/3)\mathbb Z$. Hence this component also has unbranched roots in the fiber. ### 11.4. An infinity-type image cannot contain a whole fiber If a fiber were entirely branch points and contained no beta-type point, every Q(u_i) would be rational. Put a_i=gj(Q(u_i)), also rational. Then u_i=a_i-t. For j=0 Vieta gives sum u_i=0, so 4t=sum a_i is rational. For j=1,2 Vieta gives sum u_i=a/c-t, so 4t=sum a_i-a/c is rational. In either case every u_i is rational. No Fj has a fully real regular fiber of its full degree. For F0, four real roots would have sum zero and sum of pair products zero, hence sum of squares zero, which is incompatible with four distinct roots. For F1 the derivative numerator $$ 4u^3-1-(8Q(u)+5)^2 $$ has no real zero, is negative at zero, and the denominator has no real zero. Thus F1 is strictly decreasing and every real fiber has just one root. For F2 there are two real poles a0<093/8-b_0>F_2(d_0). $$ Therefore no real t intersects both the middle and right intervals twice. There is at most one root on the left and at most two altogether on the other intervals: at most three real roots. A regular degree-five fiber cannot be fully rational. This excludes every hypothetical entire branch fiber. ### 11.5. Reduction and the original-scale bound Lemma 7 gives Morse projections with disjoint finite critical values, and excludes every point of the entire set $\mathcal B$ from every finite critical image. Sections 11.2–11.4 show that every finite fiber has an unbranched root. Lemma 6 therefore supplies independent selected test labels for any pair or triple of roots, across all component patterns: their projection degrees are 4,5,5 and each retains an omitted root. These are characteristic-zero statements about the full orbit. At the fixed depth $D=32$ only finitely many algebraic branch points and covers occur. Their good properties persist outside finitely many primes. In particular, failure of entire-fiber containment is checked on a finite algebraic set of candidate branch images, and survives reduction after excluding finitely many primes. No uniform exceptional-prime set for growing depth is asserted. Use exactly the certified exposed subset and coordinate cover of Sections 9–10. Its density is $\rho_D$, coordinate coefficient $C_D$, and the selected-label lemma gives pair and triple counts $$ W=\rho_D^2p/2+O_D(\sqrt p),\qquad T=\rho_D^3p/6+O_D(\sqrt p). $$ The pointwise physical diagonal exchange is uniform over translations, giving $$ |S|/(2p)\le C_D-\rho_D^2/2+\rho_D^3/12+O_D(p^{-1/2}). $$ The exact rational $D=32$ certificate from Section 10.3 puts its main term below $179999/100000$. This proves the stated result, with the same physical direction $y-x$ at scale one. No positional equidistribution is needed for this bound. The companion `quartic_scale_one_proper_fibers.py` checks the exact Sturm counts, Newton polygon slopes and valuation comparisons. Cedar's separate source checks the critical-separation resultants and irreducible reductions. The group and 32-round certificates accompany Sections 9–10. ## 12. An independent finite-field proof of Lemma 8 This argument uses a different prime and does not rely on the separation of branch types by $2$-adic valuations. It provides a small standard-library certificate for the same characteristic-zero conclusion. ### 12.1. A full branch fiber would split over a fixed field Choose the positive real $a$ with $a^3=1/2$, and set $K=\mathbb Q(a)$, $L=K(\zeta_3)$ and $\beta=-3a^2/4$. Every finite value of $\Gamma E$ belongs to $L$. Suppose every root $u_i$ of $F_j(u)=t$ is in $\mathcal B$. Put $v_i=Q(u_i)$. Then $u_i=g_j(v_i)-t$. Vieta gives $$ 4t=\sum_i v_i\quad(j=0),\qquad 4t=\sum_i g_j(v_i)-a_j/c_j\quad(j=1,2), $$ where $g_j(v)=(a_jv+b_j)/(c_jv+d_j)$. Thus $t\in L$ and every $u_i\in L$. Lemma 7 excludes critical fibers, although the following finite computation also counts multiplicities. ### 12.2. Integral intercepts: 75 exact residue-field cases The field $L$ is unramified at $5$. Choose the prime with $a\mapsto2$ and residue field $$ \mathbb F_{25}=\mathbb F_5[z]/(z^2+z+1),\qquad z=\zeta_3\bmod5. $$ The set $E$ reduces to $\{\infty,2,2z,2z^2\}$. The matrices of $g_1,g_2$ coincide modulo $5$. Closing their projective orbit gives a set $\mathcal O$ of exactly $15$ elements. The set $$ \mathcal B_5=\{u\in\mathbb F_{25}:u^4-u\in\mathcal O\} $$ has $18$ elements. Exact division of each fiber polynomial by its linear factors, for all $25$ intercepts, gives: | Projection | Degree | Maximum roots in $\mathcal B_5$, with multiplicity | |---|---:|---:| | $F_0$ | 4 | 3 | | $F_1$ | 5 | 1 | | $F_2$ | 5 | 1 | If $t$ is integral at the chosen prime, all its roots are integral: its numerator has unit leading coefficient $1$ or $-8$. Reduction preserves its degree and all roots lie in $\mathbb F_{25}$ with multiplicity. Since the matrices are integral of determinant one, the reduction of every $Q(u_i)$ lies in $\mathcal O$. Every reduced root must therefore be in $\mathcal B_5$, contradicting the table. The source `quartic_scale_one_f25_certificate.py` implements exact field arithmetic, projective orbit closure and division with multiplicities. A separate implementation, `quartic_scale_one_f25_control.py`, agrees on all $75$ counts and the three maxima. Their certificate concerns a fixed finite field; its characteristic-zero implication is supplied by the argument in this section. ### 12.3. Nonintegral intercepts: real cubic specialization If $t$ is nonintegral, each $F_0$ root is nonintegral, so its $Q$-value reduces projectively to infinity. For $F_1,F_2$, a nonintegral root has the same property. An integral root instead satisfies $g_j(Q(u_i))=u_i+t$ nonintegral, forcing $Q(u_i)$ to reduce to the rational pole $-d_j/c_j\in\mathbb F_5$. Hence every $v_i$ reduces into $\mathbb P^1(\mathbb F_5)$. Every invertible matrix over $\mathbb F_5$ preserves both this projective line and its complement in $\mathbb P^1(\mathbb F_{25})$. The other two finite branch values reduce to $2z,2z^2$ in that complement. Thus no $v_i$ can belong to either nonreal branch orbit. It must be a rational orbit value or an orbit value of the real $\beta$, and therefore lies in $K$. The trace identities now give $t\in K$ and every $u_i\in K\subset\mathbb R$. Section 11.4 proves that the three projections have at most two, one and three real roots in a regular fiber, respectively. None can split completely over $K$. This contradiction proves Lemma 8 a second time. The standard-library certificate also checks the real derivative counts using rational Sturm chains; the first denominator has no real zero since $$ Q(u)+\frac12=(u^2-\tfrac12)^2+(u-\tfrac12)^2\ge0. $$ As before, only finitely many words occur at fixed depth, so the characteristic-zero conclusion survives reduction outside finitely many primes depending on that depth. No growing-depth assertion is used. ## 13. An alternating septic at its original scale The exposed-row argument also excludes a nonsymmetric source law whose coordinate capacity is above $1.8$. This application uses rational branch values, so the full-fiber argument is simpler than the quartic case. **Theorem 9.** Put $$ Q_7(x)=\frac{15x^7-42x^5+35x^3+16}{32}. $$ For every sufficiently large prime, every no-three subset of the periodic union $y=Q_7(x),A(Q_7(x)),B(Q_7(x))$ in any translated square of side $2p$ satisfies $$ \frac{|S|}{2p}\le\frac{17977}{10000}+O(p^{-1/2}). \tag{13.1} $$ The result holds at the original input scale on both prime progressions distinguished below. It does not assert other window sizes. Generic nonzero input scales also satisfy the same bound outside a bounded exceptional set, by Section 10.1. ### 13.1. Source monodromy, constant fields and coordinate capacity Write $P(x)=15x^7-42x^5+35x^3$. Direct calculation gives $$ P'(x)=105x^2(x^2-1)^2,\qquad \operatorname{disc}_x(P(x)-T)=-7\cdot105^6T^2(T^2-64)^2. $$ The critical values of $Q_7$ are $1/4,1/2,3/4$, with ramification index three at each critical point. Together with infinity they form an exterior branch set for the four ping-pong domains, so Section 3.2 separates all its translates under $\Gamma$. The geometric monodromy is $A_7$. One may use prime-degree primitivity and Jordan's three-cycle theorem, or argue directly: the finite inertia three-cycles generate a transitive group, since their product supplies the inertia at infinity. Their support triples form a connected hypergraph. A triple meeting a set on which the group already contains the full alternating group in two vertices adds the new vertex by conjugation. If it meets in just one vertex $c$ and adds $d,e$, conjugating a cycle $(a\ b\ c)$ from the old set by $(c\ d\ e)$ first supplies $(a\ b\ d)$, reducing to the two-vertex case, and then adds $e$. Starting with one triple and adjoining connected triples proves the claim. The initial four-letter step is the direct generation of $A_4$ by two intersecting three-cycles. The discriminant square class is $-7$, so the arithmetic constant extension is precisely $\mathbb Q(\sqrt{-7})$. The arithmetic permutation group contains $A_7$ and lies in $S_7$; the nonsquare discriminant determines its index-two quotient. In sufficiently large finite characteristic, the degree-one root law is uniform on even permutations when $(-7/p)=1$ and on odd permutations when $(-7/p)=-1$. Exact enumeration of all $7!$ permutations gives $$ (z,o,d)= \begin{cases} (31/84,13/36,1/5),&(-7/p)=1,\\ (11/30,3/8,1/6),&(-7/p)=-1. \end{cases} \tag{13.2} $$ At fixed depth the disjoint branch sets give a geometric product of copies of $A_7$. In the nonsplit progression the source covers share their quadratic constant field, and are not asserted to be independent over the original constants. Their actual degree-one Frobenius coset is nevertheless a Cartesian product of uniform odd cosets, since the geometric kernel is the full product and the common quotient fixes the parity on each factor. The even progression is analogous. Thus the local source counts are independent with the respective law in (13.2). The coordinate result of Section 8 applies separately to these two laws. In the even progression its scalar equation is $$ 203s^4-910s^2+1260s-465=0, $$ with a unique solution in $(0,1)$; its derivative minimum on this interval is positive. The exact coefficient is $$ c_{7,+}=2-2s^3+\frac{13}{6}s^4-\frac{29}{45}s^6, \qquad 1.800168783