# Permutations destroying weighted relations at all sufficiently large cyclic orders ## 1. Matched relations and the main construction Fix an integer $Q\ge2$. For $0K$ be an odd prime for which every nonzero integer $a$ with $|a|\le K$ is a square modulo $p$. Such primes include every prime $p>K$ in the progression $$ p\equiv1\pmod {D_K},\qquad D_K=8\prod_{\substack{\ell\le K\\ \ell\text{ an odd prime}}}\ell. \tag{3} $$ Choose any nonsquare $\lambda\in\mathbb F_p^\times$. For $z=x+py$, with $0\le x,yQ$, each of $r,q,q-r$ is a unit modulo $p$ and modulo $p^2$. The proof nevertheless keeps the original integer coefficients: the carry bounds depend on their sizes, so dividing a common factor is unnecessary and would change the error window. ## 2. Relation to earlier work Hegarty studied permutations avoiding arithmetic patterns [1]. His Proposition 2.3(ii) constructs arithmetic-progression-destroying permutations on finite integer intervals, and his Lemma 3.5 gives a subgroup-and-quotient lifting principle. Such permutations can serve as finite factors in recursive constructions. Hegarty–Martinsson prove eventual existence of arithmetic-progression-destroying permutations of cyclic groups and permutations destroying almost arithmetic progressions [2, arXiv Theorem 1.1; published Theorem 2]. Their bounded-error theorem concerns the midpoint coefficients $(1,-2,1)$ and permits errors on both sides. Section 7 adapts their block-interleaving architecture to a finite weighted family, using the CRT master of Theorem 6 and a weighted interval construction established there. Their Question 4.1 concerns destroying a conjunction of linear equations; our finite-family condition destroys each equation separately. Their Question 4.2 in the arXiv version (Question 9 in the published version) proposes a criterion for one linear equation at every sufficiently large cyclic order. Corollary 8 proves its sufficiency direction for invariant equations in three variables with nonzero coefficients. It makes no assertion about the converse or the invariant cases with at least four variables. Theorem 3 retains the separate conclusion of bounded modifications of inversion at all sufficiently large prime orders. Geneson proves a related interval result: when both spacing parameters are odd, an even/odd recursion avoids ordered generalized three-term progressions on every integer interval [3, Proposition 8]. This restricted subsequence-avoidance theorem does not treat arbitrary weights, cyclic destruction, or bounded input and output errors. Elkies–Swaminathan use inversion and a bounded sequence of transpositions to destroy midpoint progressions in elementary odd-primary groups [4, Lemma 3 and Sections 2.2–2.3]. Their method is the direct source for Section 4: swap inversion at 0 and 1, use nonsquare discriminants to protect untouched inverse points, and control interactions with earlier exceptional points. We give a common discriminant factorization and a compatible character prescription for a finite weighted family. Sawhney–Stoner settle the cyclic midpoint problem apart from orders $2,3,5,7$ [5, Theorem 5]. Their Theorem 9 states the character-sum estimate used in our finite-parameter count. Jungić–Sahasrabudhe characterize breakable minimal linear forms on countably infinite abelian groups [6, Theorem 1]. In their terminology, an invariant ternary form with three nonzero coefficients is minimal. Their criterion requires $G/\Omega_{|\beta|}$ to be infinite for each nonempty proper coefficient subsum $\beta$, where $\Omega_m$ is the subgroup annihilated by $m$. The countably infinite hypothesis is part of that theorem. Their finite-prime result counts permutations destroying four-term progressions, with asymptotic proportion $e^{-1}$ [6, Section 3]. Neither statement asserts the finite weighted theorem proved here. These comparisons concern the stated scopes of the cited results. The quadratic prime-square construction, the finite-family inversion construction, and the weighted CRT and interval arguments are proved here. The final all-orders assembly explicitly adapts the interleaving method of Hegarty–Martinsson. We make no priority claim beyond this comparison. ## 3. Quadratic shears and digit products ### 3.1. The prime condition and bijectivity If $p$ satisfies (3), then $-1$ and $2$ are squares modulo $p$. For every odd prime $\ell\le K$, quadratic reciprocity and $p\equiv1\pmod\ell$ give $(\ell/p)=1$. Factoring any positive integer at most $K$, and then accounting for its sign, shows that every nonzero integer of absolute value at most $K$ is a square modulo $p$. Dirichlet's theorem gives infinitely many primes in (3), since the residue 1 is coprime to $D_K$. The two shears in (4) are invertible. Given $u,v\in\{0,\ldots,p-1\}$, recover $$ x=[v-\lambda u^2]_p,\qquad y=[u-\lambda x^2]_p. \tag{5} $$ Thus (4) defines a permutation of all $p^2$ residues, including zero. ### 3.2. Carry control and destruction of the weighted relations Fix $0K$ satisfying (3), then the least integer $1\le\lambdaL$, a permutation $\alpha$ witnesses $\mathcal D(Q,L;N)$, and a permutation $\beta$ witnesses $\mathcal D(Q,0;m)$. Then $$ F(Na+b)=N\beta(a)+\alpha(b),\qquad 0\le a2$ and $\phi$ witnesses $\mathcal D(Q,Q;n)$. Define $$ S_\phi=\{(x,[\phi(x)+e]_n):0\le x0$. This rules out a positive-linear cap along the fixed-prime depth sequence. It supplies no uniform bound when the prime grows, as in the depth-one family above. ## 4. Bounded inversion patches at every sufficiently large prime For an integer $Q\ge2$, put $$ \mathcal W_Q=\{(a,b):0P_Q$, there is a permutation $\phi$ of $\mathbb F_p$ satisfying (12) for every $(a,b)\in\mathcal W_Q$, with exceptional support of size at most $s_*$. It can be constructed by swapping the outputs of $\iota_p$ at 0 and 1, followed by at most $6M$ transpositions of outputs at previously untouched nonzero inputs. The choice of all transpositions can be made deterministically. For $p>Q$, the exact conclusion also holds for all $0\max\{P_Q,G\}$, a single permutation with at most $12|\mathcal W_Q|+2$ exceptional inputs destroys each matched form $\sum_i c_i x_i$. *Proof.* A nonzero invariant integer triple has two coefficients of one sign and one of the other. After changing its common sign, let the two positive coefficients divided by $g_c$ be $b_c-a_c$ and $a_c$. The negative coefficient is then $-b_c$, and the definition of the gcd gives $\gcd(a_c,b_c)=1$. Since $p>G$, the scalar $\pm g_c$ is invertible in the field. Coordinate permutation and division by this scalar reduce both matched equations to a member of $\mathcal W_Q$. Apply Theorem 3. $\square$ ### 4.1. Initial permutation and its complete exception list Throughout this section $p>P_Q$. In particular, $p$ is odd and all coefficients of (11) are nonzero in $\mathbb F_p$. If two coordinates of a solution of $\Delta_{a,b}=0$ coincide, the third coincides with them, since the coefficient of the remaining coordinate is nonzero. Every forbidden nonconstant tuple therefore has three distinct inputs. Start with $$ f_0(0)=1,\qquad f_0(1)=0,\qquad f_0(x)=x^{-1}\quad(x\notin\{0,1\}). \tag{16} $$ For any input relation, write $(x_0,x_1,x_2)=(x,x+ad,x+bd)$. If all three inputs are nonzero, direct calculation gives $$ \frac{b-a}{x}-\frac b{x+ad}+\frac a{x+bd} =\frac{ab(b-a)d^2}{x(x+ad)(x+bd)}. \tag{17} $$ For a nonconstant tuple $d\ne0$, so the right-hand side is nonzero. Equivalently, three distinct inverse graph points cannot be collinear on the conic $xy=1$. Every bad tuple for (16) must therefore contain 0 or 1. Fix one coefficient triple and assign its coefficients $A,B,C$ to three distinct coordinate positions $i,j,\ell$. Thus $A+B+C=0$. If the tuple contains 0 but not 1, put $x_i=0$. The two equations are $Bx_j+Cx_\ell=0$ and $A+B/x_j+C/x_\ell=0$. Their unique possible solution is $$ x_j=1-C/B,\qquad x_\ell=1-B/C, \tag{18} $$ when $B\ne C$. If $B=C$, there is no solution. The two displayed values are distinct and avoid 0 and 1 because $A,B,C,B-C\ne0$. If the tuple contains 1 but not 0, put $x_i=1$. Solving $A+Bx_j+Cx_\ell=0$ and $B/x_j+C/x_\ell=0$ gives $$ x_j=\frac B{B-C},\qquad x_\ell=\frac C{C-B}, \tag{19} $$ again only when $B\ne C$. These values are the reciprocals of the entries in (18), and are distinct and outside $\{0,1\}$. Finally, suppose both 0 and 1 occur, at positions with coefficients $A$ and $B$. The third input is $-B/C$. Preservation of the output relation would imply $AB=C^2$, or $$ A^2+AB+B^2\equiv0\pmod p. \tag{20} $$ For the integer coefficients under consideration, $03Q^2$ in (14). Consequently (18)–(19), with one choice for each family and each distinguished position containing 0 or 1, give the complete list $\mathcal B_0$ of bad ordered tuples. There are at most $6M$ records, each recording the family as well as the ordered tuple. Let $U$ be the union of their coordinates other than 0 and 1. Then $$ U\subseteq \{1-C/B,\ (1-C/B)^{-1}: B\ne C \text{ are coefficient positions of one family}\}, \qquad |U|\le12M. \tag{21} $$ Here and below all collections are reduced to sets in the field. Coincidence between rational descriptions can only decrease their sizes. In particular, $U\cap\{0,1\}=\varnothing$. ### 4.2. One modified point and two inverse points Suppose a graph point has input $u$, output $v$, and coefficient $A$. Give two other nonzero inputs $X,Y$, whose outputs are their inverses, coefficients $B,C$. Eliminating $Y=-(Au+BX)/C$ from $$ Au+BX+CY=0,\qquad Av+B/X+C/Y=0 $$ gives $$ BvX^2+(Auv+C-B)X+Bu=0. \tag{22} $$ For every genuine tuple both cleared denominators are nonzero, so every such tuple supplies a root of (22). Write $z=uv$. The discriminant factors as $$ \begin{aligned} D&=(Az+C-B)^2-4B^2z\\ &=(z-1)\bigl(A^2z-(B-C)^2\bigr) =A^2(z-1)(z-\rho), \qquad \rho=\left(\frac{B-C}{A}\right)^2. \end{aligned} \tag{23} $$ In particular, $\rho\ne1$: equality would imply $(B-C)^2=(B+C)^2$, and hence $4BC=0$. Now swap the inverse outputs at two distinct untouched nonzero inputs $u,y$, and write $y=ut$. Their new input-output products are $1/t$ and $t$, respectively. Let $\chi$ be the quadratic character of $\mathbb F_p$, with $\chi(0)=0$. If $\rho\ne0$, then it is a square, and $$ (1/t-1)(1/t-\rho) =\frac{\rho}{t^2}(t-1)(t-\rho^{-1}). $$ Thus the conditions $$ \chi((t-1)(t-\rho)) =\chi((t-1)(t-\rho^{-1}))=-1 \tag{24} $$ make the discriminant nonsquare at both new points. When $\rho=0$, the corresponding sufficient conditions are $$ \chi(t(t-1))=\chi(1-t)=-1. \tag{25} $$ The leading coefficient in (22) is nonzero for both new points, because their outputs are nonzero. A nonsquare discriminant therefore excludes every tuple consisting of that new point and two untouched inverse points. Let $R$ be the set containing 0 and every nonzero $\rho$ and $\rho^{-1}$ obtained from all families and all positions of $A$. Interchanging $B,C$ leaves $\rho$ unchanged. Hence $$ 1\notin R,\qquad k_0:=1+|R|\le6M+2=k. $$ Put $\epsilon=\chi(-1)$. All requirements (24)–(25) follow from the single prescription $$ \chi(t-1)=-\epsilon,\qquad \chi(t-r)=\epsilon\quad(r\in R). \tag{26} $$ Indeed, each product in (24) has character $(-\epsilon)\epsilon=-1$; the two expressions in (25) have characters $\epsilon(-\epsilon)$ and $\epsilon(-\epsilon)$ as well. Every root other than 1 receives the same prescribed sign. Distinct rational roots may coalesce in the field without creating conflicting requirements. No root can coalesce with 1, by the $4BC\ne0$ calculation. Also, $p>2Q$ ensures that a nonzero integer difference $B-C$ cannot vanish modulo $p$. These observations account for all root collisions, including the case $B=C$. ### 4.3. Counting parameters with the prescribed signs We use the standard quadratic-character estimate in the following form: if $h\in\mathbb F_p[T]$ is a product of $d\ge1$ distinct linear factors, then $$ \left|\sum_{t\in\mathbb F_p}\chi(h(t))\right| \le(d-1)\sqrt p+1. \tag{27} $$ This follows from the Weil bound; [5, Theorem 9] states a form at least as strong as (27). The case $d=1$ also follows immediately by summing the character over the field. Let $T=R\cup\{1\}$, let the signs $\delta_r$ be those in (26), and let $N_p$ count parameters satisfying (26). Outside $T$, the product $$ I(t)=2^{-k_0}\prod_{r\in T}(1+\delta_r\chi(t-r)) $$ is exactly the indicator of (26). At each of the at most $k_0$ roots, it lies in $[0,1]$. Expanding the product, every nonempty subset $J\subseteq T$ contributes the character sum of a product of distinct linear factors, to which (27) applies. Since $\sum_{J\subseteq T}|J|=k_0 2^{k_0-1}$, the expansion gives $$ \begin{aligned} N_p &\ge\frac p{2^{k_0}}-\frac{k_0}{2}\sqrt p-k_0-1\\ &\ge\frac p{2^k}-\frac k2\sqrt p-k-1. \end{aligned} \tag{28} $$ The additive 1 bounds the normalized sum of the additive errors in (27), and the term $k_0$ removes all root contributions. In particular, no independence assumption about the character values is needed. ### 4.4. Complete finite exclusions for patch interactions Let $f$ be the current permutation, agreeing with inversion outside a set $S$ containing 0 and 1. Write $s=|S|$. Choose $u\in U\setminus S$. We will swap the values at $u$ and a fresh $y=ut$, where $t$ satisfies (26). Before the swap both outputs are their inverse values. Define a forbidden set $E$ of possible inputs $y$, initially $$ E=S\cup U\cup\{u\}. $$ For every coefficient family, consider all six assignments of its coordinate positions to the ordered roles $(A,B,C)$. Equal coefficient values cause no problem; repetitions among the resulting exclusions may simply be removed. Make the additions below. An inverse is used only when its displayed denominator is nonzero. **A tuple containing $u$ and an old patched point.** For each $z\in S$, assign coefficients $A,B$ to $u,z$. The input equation fixes the third input as $$ w=-(Au+Bz)/C. $$ Add $w$ to $E$; this covers the possibility $w=y$. For a tuple with three distinct inputs and $y\ne w$, the third output is the unchanged value $f(w)$, whether or not $w\in S$. Put $D=Bf(z)+Cf(w)$. If $D\ne0$, add $$ y=-A/D \tag{29} $$ to $E$, since the output equation would be $A/y+D=0$. If $D=0$, that equation has no solution. At most two values are added for each assignment and $z$. A repeated input is a constant tuple by Section 4.1 and needs no exclusion. **A tuple containing $y$ and an old patched point.** Assign coefficients $A,B$ to $y,z$, with $z\in S$. The third input is $w(y)=-(Ay+Bz)/C$. For every $w\in S\cup\{u\}$, add $$ y=-(Bz+Cw)/A \tag{30} $$ to $E$. This covers every case where the third output is also changed, including $w=0$ and $w=u$. Otherwise, in a tuple with distinct inputs, the third output is $1/w(y)$. Set $H=A/u+Bf(z)$. If $H\ne0$, the remaining output equation forces $w(y)=-C/H$, so add $$ y=(C^2/H-Bz)/A. \tag{31} $$ If $H=0$, the remaining term $C/w(y)$ cannot vanish. These rules add at most $s+2$ values per assignment and $z$. The case $w(y)=y$ is a repeated input and would force $y=z$, already forbidden by $y\notin S$. **A tuple containing both new points and no old patched point.** Assign coefficients $A,B$ to $u,y$. The third input is nonzero and untouched. Substituting its value from the input equation into the output equation gives $$ (Au+By)^2=C^2uy. $$ Writing $y=ut$ and using $A+B+C=0$, this becomes $$ (t-1)(B^2t-A^2)=0. \tag{32} $$ The value $t=1$ is already forbidden. For each coefficient assignment, add $u(A/B)^2$ to $E$. If the third point is in $S$, it was covered by the preceding cases. **A tuple containing exactly one new point and no old patched point.** Its other two outputs are inverse values. Equations (22)–(26) exclude it by a nonsquare discriminant. These cases exhaust all potentially new bad tuples: a tuple avoiding both $u$ and $y$ has unchanged outputs. They include tuples with multiple old patched points and old outputs equal to zero. The exclusions involving $S\cup\{u\}$ cover every place where an inverse formula for the third point would be invalid. The number of forbidden values is bounded by $$ \begin{aligned} |E| &\le s+|U|+1+12Ms+6Ms(s+2)+6M\\ &=6Ms(s+4)+6M+s+|U|+1. \end{aligned} \tag{33} $$ When $s\le s_*$ and $|U|\le12M$, this is at most $B_*$. By (14), $\sqrt p>k2^k$, and (28) therefore gives $$ N_p>\frac p{2^{k+1}}-k-1>B_*. \tag{34} $$ Multiplication by nonzero $u$ is a bijection, so there is a parameter satisfying (26) for which $ut\notin E$. The resulting swap creates no bad tuple in any coefficient family. ### 4.5. Termination Maintain the exact set $\mathcal B$ of surviving records from $\mathcal B_0$. Each initial bad tuple contains one of 0 and 1 and two inputs in $U$. If $\mathcal B$ is nonempty, select one record and one of its coordinates outside $\{0,1\}$ as $u$. Inductively, both such coordinates remain outside $S$: every earlier chosen $u$ destroyed all surviving records that contained it, and every auxiliary $y$ was chosen outside $U$. Choose $y$ by Section 4.4. In any surviving bad tuple containing $u$, only that coordinate changes output, because $y$ is not a coordinate of any initial bad tuple. If its coefficient is $A$, the change in output discrepancy is $$ A(1/y-1/u)\ne0. \tag{35} $$ Thus every surviving bad record containing $u$ is destroyed. Every other surviving record is unchanged, and Section 4.4 prevents new bad tuples. Deleting precisely the records containing $u$ therefore maintains the exact set $\mathcal B$. Each step removes at least one of the at most $6M$ records and uses a previously unused point of $U$. There are at most $\min\{6M,|U|\}$ steps, each adding two new inputs to $S$. The support never exceeds $2+12M=s_*$, which also justifies the bound used in (34) at every step. When the process terminates, $\mathcal B$ is empty and every family is destroyed. Each operation is a transposition of outputs of a permutation, so the final map remains a permutation. This proves Theorem 3. $\square$ ### 4.6. Deterministic construction For definiteness, order field elements by their representatives in $\{0,\ldots,p-1\}$, order families by $(b,a)$, and order bad records by their family and ordered tuple. Store only the exceptional values of the current permutation. Off the stored set its value is inversion. ```text Input: Q and a prime p satisfying (14). Form the families (b-a,-b,a) and initialize f as in (16), S={0,1}. Build Bad from (18) and (19), recording the family and ordered tuple. Remove duplicate records, and form U from their coordinates outside {0,1}. Form the distinct field root set R used in (26). While Bad is nonempty: Take the first record in the fixed order. Let u be its least coordinate outside S. Construct E using (29)--(32), together with S union U union {u}. Enumerate t=0,1,...,p-1 until (26) holds and y=u*t is outside E. Swap f(u) and f(y). Add u and y to S. Delete exactly those records of Bad containing u. Output: the dictionary of f on S, with the inverse rule elsewhere. ``` In constructing $E$, the value $w$ preceding (29) is added as well as the displayed values (29)–(32). Formula (32) contributes the input $u(A/B)^2$. Euler's criterion computes each character value by modular exponentiation. Inequality (34) proves that each parameter enumeration succeeds, and Section 4.5 proves termination of the outer loop. No exhaustive search over permutations is required. The finite sets and the number of swaps are bounded in terms of $Q$; this argument does not assert a bound polynomial in $\log p$ for the parameter enumeration. For implementation, the parameters satisfying (26) may be cached between steps. Their character requirements depend only on the family and the field, while the exclusion set depends on the current patch. Rescanning a cache in increasing order and extending it when needed implements the same first-admissible-parameter rule. ### 4.7. The $Q=4$ specialization The five coefficient triples are $$ (1,-2,1),\quad(2,-3,1),\quad(1,-3,2),\quad (3,-4,1),\quad(1,-4,3). \tag{36} $$ Equations (18)–(19) give four ordered cases for the first family and six for each of the other four, hence 28 cases in total. Their noninitial coordinates belong to the following 25 rational values: $$ \begin{gathered} -2,-1,-\tfrac12,\tfrac15,\tfrac14,\tfrac13,\tfrac25, \tfrac37,\tfrac12,\tfrac47,\tfrac35,\tfrac23,\tfrac34, \tfrac45,\\ \tfrac54,\tfrac43,\tfrac32,\tfrac53,\tfrac74, 2,\tfrac73,\tfrac52,3,4,5. \end{gathered} \tag{37} $$ Reduction modulo the field can only decrease the size of this set. The distinct character roots before reduction are $$ R\cup\{1\} =\{0,1,9,1/9,4,1/4,25,1/25,25/9,9/25,49,1/49\}. \tag{38} $$ Thus at most 25 swaps are needed after initialization, giving at most 52 exceptional inputs. Substituting $M=5$, $s=52$, and $|U|=25$ in (33) gives $$ B=87{,}468. $$ Use $k=12$ in (28). The two sufficient character-count thresholds are $$ k^2 4^k=2{,}415{,}919{,}104, \qquad 2^{k+1}(B+k+1)=716{,}644{,}352. \tag{39} $$ The first also exceeds $3Q^2=48$. Consequently every prime $p>2{,}415{,}919{,}104$ admits a permutation destroying all five families with support at most 52. ## 5. Two certified fixtures ### 5.1. A bounded-error fixture at order $73^2$ The first fixture fixes $Q=3$, $L=1$, $p=73$, and $\lambda=5$, giving $n=5329$. The prime satisfies $p\equiv1\pmod {24}$, the progression in (3) for $K=4$. Trial division proves primality, and direct character checks show that $\pm1,\pm2,\pm3,\pm4$ are squares and that 5 is the least positive nonsquare. Formula (4) specifies the full permutation. The independent checker reads only the saved permutation table and verifies that it is a bijection. It does not use the shear formula or the carry proof. For each $(r,q)\in\{(1,2),(1,3),(2,3)\}$ and every ordered pair $(z_0,z_2)$, it forms the unique completion $$ z_1=[q^{-1}((q-r)z_0+rz_2)]_n. $$ Because $q$ is a unit modulo $n$, this enumerates every raw ordered input relation, including relations whose representatives wrap around the modulus. It skips only constant triples and directly computes each output discrepancy modulo $n$. The complete check covers $3n^2=85{,}194{,}723$ ordered pairs, of which $3n=15{,}987$ give constant triples. All $3n(n-1)=85{,}178{,}736$ nonconstant relations have discrepancies outside $\{-1,0,1\}$ modulo $n$. The minimum cyclic absolute discrepancy is 4 for $(1,2)$ and 3 for each of $(1,3),(2,3)$. These minima describe this fixed fixture, not an improvement of the general tolerance stated in Theorem 1. The checker stores the complete discrepancy histogram for each family and an attaining triple for each minimum. The saved run used approximately 0.00218 CPU seconds for generation and 0.555 CPU seconds for the checker process, excluding compilation. The audit status is PASS. The theorem does not depend on this computation. ### 5.2. An exact fixture at $p=4{,}294{,}967{,}291$ The accompanying implementation fixes $$ Q=4,\qquad p=4{,}294{,}967{,}291. \tag{40} $$ It proves primality by trial division by 2 and every odd integer through $\lfloor\sqrt p\rfloor=65{,}535$. Starting with the 28 initial bad tuples, it performs 14 repair swaps and produces the 30 exceptional values listed in Appendix A. The sparse dictionary and the inversion rule specify the full permutation without materializing $p$ entries. The recorded generation used 0.195271213 CPU seconds. It tested 13,770 distinct candidate parameters with 27,566 character evaluations; the five distinct parameters passing the full character prescription were $$ 5275,\quad8183,\quad11192,\quad11638,\quad13769. $$ An admissible character parameter may be used in several swaps, provided its resulting input avoids that step's exclusion set. These timings describe this fixture and are not asymptotic performance guarantees. The general existence proof is independent of the computation. ### 5.3. Independent complete algebraic validation The independent validator reads the final exceptional-value dictionary and does not import the generator or use its swap trace. It first checks primality and verifies that the listed outputs are distinct and form exactly the set $\{\iota_p(x):x\in S\}$. This proves that replacing those inversion values gives a permutation of the whole field. It then covers every possible nonconstant bad tuple by the number of its inputs in $S$. 1. If no input is patched, equation (17) excludes the tuple. 2. If exactly one input is patched, choose its family, position, and input $u\in S$, and put $v=\phi(u)$. Solve the polynomial (22) exactly over the field. The validator handles quadratic, double-root, linear, and constant cases, filters out roots whose other two inputs meet $S$, and checks both original equations on every remaining candidate. Clearing denominators therefore introduces no false certificate of validity. A polynomial identically zero would have all coefficients zero; since $B\ne0$, this would force $u=v=0$. The fixture has $\phi(0)=1$, so that case cannot occur. 3. If at least two inputs are patched, choose every ordered pair of distinct elements of $S$ and every assignment to two coefficient positions. The input equation uniquely fixes the third input, because its coefficient is nonzero. Check the output equation directly. This includes triples with three patched inputs. Repeated-input nonconstant tuples are impossible by Section 4.1, so this exhausts all field tuples. In the final fixture there are $5\cdot3\cdot30=450$ distinguished-point polynomials and $5\cdot6\cdot30\cdot29=26{,}100$ ordered patched-pair completions. The polynomial classifications are: | Type | Count | | --- | ---: | | Quadratic with nonsquare discriminant | 420 | | Quadratic with two roots | 14 | | Linear | 14 | | Quadratic with a double root | 1 | | Nonzero constant | 1 | No candidate with exactly one patched input survives the filters, and no violating tuple survives the pair-completion checks. As a control, the same validator applied to (16) recovers all 28 initial bad ordered tuples. The recorded independent audit passes and used 0.079355596 CPU seconds. ## 6. Every nonzero discrepancy survives a bounded inversion patch For a permutation $\phi$ and an exact input relation $\Delta_{a,b}(x_0,x_1,x_2)=0$, call $\Delta_{a,b}(\phi(x_0),\phi(x_1),\phi(x_2))$ its output discrepancy. Theorem 3 avoids the zero discrepancy on nonconstant tuples. The following proposition gives an exact obstruction to avoiding any additional field discrepancy with a bounded patch of inversion. **Proposition 5 (positive-error limitation).** Suppose $p>Q$ is prime, $S\subseteq\mathbb F_p$, and $\phi$ is a permutation satisfying $\phi(x)=\iota_p(x)$ for every $x\notin S$. For every $(a,b)\in\mathcal W_Q$ and every $\eta\in\mathbb F_p^\times$, there are at least $p-3-6|S|$ distinct ordered triples of three distinct inputs outside $S$ such that $$ \Delta_{a,b}(x_0,x_1,x_2)=0, \qquad \Delta_{a,b}(\phi(x_0),\phi(x_1),\phi(x_2))=\eta. \tag{41} $$ In particular, each nonzero output discrepancy occurs whenever $$ p>6|S|+3. \tag{42} $$ *Proof.* Put $C=ab(b-a)\ne0$. For $t\in\mathbb F_p\setminus\{0,-a,-b\}$, define $$ x_0=\frac C{\eta(t+a)(t+b)},\quad x_1=\frac C{\eta t(t+b)},\quad x_2=\frac C{\eta t(t+a)}. \tag{43} $$ Writing $d=C/[\eta t(t+a)(t+b)]$, we have $x_1-x_0=ad$ and $x_2-x_0=bd$. Thus the input relation holds and the three entries are distinct. Direct expansion gives $$ \begin{aligned} \frac{b-a}{x_0}-\frac b{x_1}+\frac a{x_2} &=\frac\eta C\bigl((b-a)(t+a)(t+b)-bt(t+b)+at(t+a)\bigr)\\ &=\eta. \end{aligned} \tag{44} $$ For each fixed nonzero $s\in S$ and each coordinate $i$, the equation $x_i=s$ is a nonzero quadratic equation in $t$, hence removes at most two parameters. None of the $x_i$ is zero, so $s=0$ removes none. In addition to the three poles, at most $6|S|$ parameters are excluded. At least $p-3-6|S|$ parameters remain. They give distinct ordered triples, because the triple recovers $d=(x_1-x_0)/a$ and then $t=x_0/d$. Their outputs are their inverses, so (44) proves the stated count and (41); condition (42) makes the count positive. $\square$ For clarity, give a residue $r\in\mathbb F_p$ the cyclic distance $\|r\|_p=\min_{n\equiv r\pmod p}|n|$. A positive integer tolerance $L\ge1$ would require the output discrepancy of every nonconstant exact input relation to avoid $\{r:\|r\|_p\le L\}$. Proposition 5 rules this out for a bounded inversion patch once (42) holds, already by taking $\eta=1$. More generally it rules out avoidance of any prescribed nonzero discrepancy. The proposition concerns the discrepancy in the field and makes no assertion about a chosen integer lift having that same small integer discrepancy. The constructions supply distinct ingredients. Theorem 1 gives a fixed positive output window on selected prime-square orders and their digit powers. Theorem 3 gives exact destruction throughout all sufficiently large prime orders with bounded exceptional support. Proposition 5 limits the tolerance obtainable from that bounded-patch form. In the next section, a CRT construction upgrades the first ingredient to simultaneous input and output tolerance; a weighted interval construction and arithmetic-progression interleaving then give every sufficiently large cyclic order. These permutation results make no no-three-in-line density claim. ## 7. Simultaneous errors at every sufficiently large cyclic order The CRT construction, the weighted interval bootstrap, and the quantitative extension in this section are due to lattice_lift_research. The final interleaving follows the architecture of Hegarty–Martinsson [2, Section 2], with a common master for all the weighted relations and an interval construction proved below for those same weights. Write $\mathcal D_2(Q,E,L;n)$ if there is a permutation $F$ of $\mathbb Z/n\mathbb Z$ such that, for every $02H^2,\qquad \log N\le\Gamma Q(Q+E+L). \tag{7.3} $$ Such masters also exist at unbounded selected orders. For symmetric windows $E=L=R$, denote the order selected in (7.3) by $N_R$; if $2R\ge Q$, it satisfies $N_R>8R^2$. *Proof.* Two applications of Bertrand's postulate give primes $$ H<\ell_+<2H<\ell_-<4H. \tag{7.4} $$ For $i\in\{+,-\}$, define $$ s_i(a)=[a^2]_{\ell_i}\quad(0\le a<\ell_i),\qquad D_i=Q(\ell_i-1),\qquad K=Q\ell_-. \tag{7.5} $$ Put $$ P_K=8\prod_{\substack{v\le K\\v\text{ an odd prime}}}v, $$ and select $p_+$ as the least prime congruent to $1$ modulo $P_K$, and $p_-$ as the least prime congruent to $1$ modulo $p_+P_K$. Every prime at most $K$ divides $P_K$. Hence $p_+>K$: otherwise reduction of its defining congruence modulo $p_+$ gives $0\equiv1\pmod{p_+}$. Also $p_-=1+jp_+P_K$ for an integer $j\ge1$, so $p_->p_+>K$. The four primes $\ell_+,\ell_-,p_+,p_-$ are consequently pairwise distinct. Both high primes are $1$ modulo $P_K$. For tolerance $D_i$, the sufficient progression modulus in Theorem 1 divides $P_K$, because $D_i+Q=Q\ell_i\le K$. The same theorem therefore gives a permutation $f_i$ witnessing $\mathcal D(Q,D_i;m_i)$, where $m_i=p_i^2$. In particular $m_i>D_i$. The factor $8$ in $P_K$ includes the sign and the prime $2$ in the quadratic-residue condition. The four moduli $\ell_+,m_+,\ell_-,m_-$ are pairwise coprime. In their CRT coordinates, define $$ \begin{aligned} F(a_+,b_+,a_-,b_-) =\bigl(&a_+,\ f_+(b_+)+s_+(a_+),\\ &-a_-,\ f_-(b_-)+s_-(a_-)\bigr), \end{aligned} \tag{7.6} $$ where each output is reduced in its own coordinate modulus, and each $a_i$ has its canonical representative in $[0,\ell_i-1]$. This is a permutation of order $N=\ell_+m_+\ell_-m_-$: the low outputs recover $a_+$ and $a_-$, after which the high outputs recover $b_+$ and $b_-$ using the inverses of $f_+$ and $f_-$. Fix one literal weight pair and suppose the relations in (7.1) hold. The two low coordinates give $$ h-e\equiv0\pmod{\ell_+},\qquad h+e\equiv0\pmod{\ell_-}. $$ Both left sides have absolute value at most $E+L<\min\{\ell_+,\ell_-\}$. Thus $h=e$ and $h=-e$ as integer equalities, implying $e=h=0$. This step uses no division by $2$ modulo the cyclic order. For each sign $i\in\{+,-\}$, the remaining relations are now $$ \Delta b_i\equiv0\pmod{m_i},\qquad \Delta f_i(b_i)\equiv-\Delta s_i(a_i)\pmod{m_i}. \tag{7.7} $$ Here $a_i$ and $b_i$ each denote the triple in that coordinate. By (7.2), $|\Delta s_i(a_i)|\le q(\ell_i-1)\le D_i$, so the one-sided property of $f_i$ forces its three $b_i$ coordinates to agree. Their image discrepancy is then zero, and $\Delta s_i(a_i)\equiv0\pmod{m_i}$. Since $m_i>D_i$, this is the integer equality $\Delta s_i(a_i)=0$. The input relation also gives $\Delta a_i\equiv0\pmod{\ell_i}$. Use the polynomial identity $$ \begin{aligned} q\Delta(a_i^2) &-\bigl((q-r)a_{i,0}+qa_{i,1}+ra_{i,2}\bigr)\Delta a_i\\ &=r(q-r)(a_{i,2}-a_{i,0})^2. \end{aligned} \tag{7.8} $$ Its reduction modulo $\ell_i$, together with $s_i(a)\equiv a^2\pmod{\ell_i}$, forces $a_{i,2}=a_{i,0}$ in that field. Indeed $r$ and $q-r$ are nonzero modulo $\ell_i>Q$. The input relation then gives $q(a_{i,0}-a_{i,1})=0$, and $q$ is also a unit, so all three canonical $a_i$ coordinates agree. This argument applies independently to both signs. All four coordinate triples are constant, and CRT proves equality of the original inputs. The same permutation handles every weight pair, including triples with repeated coordinates. It remains to bound the least-prime choices. We use the effective form of Linnik's theorem stated in [9, Section 1.3, equation (1.5), and Theorem 1.1]: there are absolute effectively computable constants $A,\lambda\ge1$ such that the least prime in any reduced residue class modulo an integer $h\ge2$ is at most $Ah^\lambda$. Both progressions above have residue $1$ and therefore satisfy the coprimality hypothesis. Applying the theorem twice gives $$ \begin{aligned} \log p_+&\le\log A+\lambda\log P_K,\\ \log p_-&\le\log A+\lambda(\log p_++\log P_K)\\ &\le(1+\lambda)\log A+(\lambda^2+\lambda)\log P_K. \end{aligned} \tag{7.9} $$ The second application uses the modulus $p_+P_K$; it does not require a bound for the second prime in the first progression. The least primes themselves have already met every lower cutoff needed for the construction. For completeness, the product of primes in $(u,2u]$, for integer $u\ge1$, divides $\binom{2u}{u}\le4^u$. If $2^{j-1}\ell_+\ell_->2H^2$. When $E=L=R$ and $2R\ge Q$, we have $H=2R$, giving $N_R>8R^2$. The indicated prime searches and the searches for the nonsquares in Theorem 1 are effective. To obtain unbounded selected orders, keep the two guards and $p_+$ fixed and let $p_-$ range through the infinitely many primes $1\pmod{p_+P_K}$ supplied by Dirichlet's theorem. The proof of the permutation property remains valid; the size estimate (7.11) concerns the least-prime choices. $\square$ ### 7.2. Consecutive orders and integer intervals Call a positive integer $n$ **good** if there is a permutation $\sigma$ of $\{0,\ldots,n-1\}$ such that, for every original pair $04Q+2$ has a $(3Q,3Q)$ master, then both $B$ and $d=B-1$ have $(2Q,2Q)$ masters. *Proof.* Translate the output of the order-$B$ master $\pi$ to arrange $\pi(B-1)=B-1$. Every discrepancy is unchanged modulo $B$, since its coefficients sum to zero. Its restriction $\tau$ to $\{0,\ldots,B-2\}$ is a permutation of that set. Suppose $\Delta x\equiv e\pmod d$ and $\Delta\tau(x)\equiv h\pmod d$, where $|e|,|h|\le2Q$. With canonical representatives, write $$ \Delta x=vd+e,\qquad \Delta\tau(x)=wd+h. $$ By (7.2), $$ |v|d\le q(B-2)+2Q\le QB. $$ Since $d>Q$, it follows that $|v|72Q^2>4Q+2$ permits Lemma 7.1. Both lengths $d,B$ are good. We prove by induction on $j\ge1$ that every integer in $$ [d^j,B^j] \tag{7.18} $$ is good. The assertion at $j=1$ is exactly the two initial lengths. Applying Lemma 7.2 to the entire preceding band gives the intervals $$ [d^{j+1},dB^j]\qquad\text{and}\qquad[Bd^j,B^{j+1}]. $$ They overlap, because $Bd^j\le dB^j$ for $j\ge1$. Their union is the next band, proving the induction, including all integer lengths between the endpoints. The binomial theorem gives $(1+1/d)^d\ge2$, so $$ \left(\frac Bd\right)^J\ge2^H\ge d. $$ Consequently $B^j\ge d^{j+1}$ for every $j\ge J$. Successive bands (7.18) from that point onward overlap, and their upper endpoints tend to infinity. Their union therefore contains every integer at least $d^J=T$. Theorem 6 gives $$ \log B\le\Gamma Q(Q+6Q)=7\Gamma Q^2, $$ using $Q\ge2$. Since $d\ge2$, both $H\le d$ and $\log d\le d$ hold. Hence $$ \log T=dH\log d\le d^38R^2$ implies $$ M>2Q(t+2)+1,\qquad M\ge(t+1)(t+2),\qquad M>t+1. \tag{7.24} $$ Indeed $R>2Q(t+2)+1$ and $R>t+2$. These inequalities follow from the chosen master itself; no enlargement of $M$ in terms of $T$ is required. We construct the desired permutation for every $n\ge\mathcal N$. Write $n=kM+u$, $0\le ub$, and we may use $b$ groups of size $t+2$ and $a-b$ groups of size $t+1$. A group of $g$ chunks containing $H_0$ long chunks is a consecutive interval of length $gk+H_0$. If its first point is $c$, it is partitioned by the $g$ arithmetic progressions $$ \bigl\{c+h+gj:0\le j\log2$. Condition (7.21) consequently ensures $n\ge\mathcal N$. The finite prime selections of Theorem 6, the recursive interval construction, and the assignments within the two length classes make the whole construction effective. The constant is expressed in terms of any fixed effective Linnik constants; the bound asserts neither an optimized numerical threshold nor efficient computation at that threshold. $\square$ ### 7.4. Invariant ternary forms **Corollary 8 (finite invariant families at all large cyclic orders).** Let $\mathcal F$ be a nonempty finite family of integer coefficient triples $c=(c_0,c_1,c_2)$ with $c_i\ne0$ for every $i$ and $c_0+c_1+c_2=0$. Put $$ Q_{\mathcal F}=\max_{c\in\mathcal F}\sum_{i:c_i>0}c_i. \tag{7.32} $$ For any integers $E,L\ge0$, put $t=\max\{E,L\}$. If $$ \log n\ge\exp(CQ_{\mathcal F}^2) +CQ_{\mathcal F}^2(t+1), \tag{7.33} $$ there is one permutation $F$ of $\mathbb Z/n\mathbb Z$ for which, for every $c\in\mathcal F$, $$ \sum_{i=0}^2c_ix_i\equiv e\pmod n,\qquad \sum_{i=0}^2c_iF(x_i)\equiv h\pmod n,\qquad |e|\le E,\quad |h|\le L \quad\Longrightarrow\quad x_0=x_1=x_2. \tag{7.34} $$ *Proof.* Each coefficient triple has two entries of one sign and one of the other. Choose a common sign and a coordinate order so that the two positive coefficients are $q-r$ and $r$, and the negative one is $-q$. These are the original coefficient magnitudes, with no common factor removed. Since their sum is zero, $q$ is the positive coefficient mass of the original triple, regardless of the sign choice. Thus $0