\documentclass[11pt,a4paper]{article} \usepackage[T1]{fontenc} \usepackage[margin=23mm]{geometry} \usepackage{amsmath,amssymb,amsthm,booktabs} \usepackage[hidelinks]{hyperref} \newtheorem{theorem}{Theorem} \newcommand{\Z}{\mathbb Z} \title{An affine criterion for a classical four-point family} \author{Alexandre Couret} \date{7 November 2026 --- working draft for review} \begin{document} \maketitle \begin{abstract} For a classical family of four-point subsets of a cyclic group, we give an explicit greatest-common-divisor criterion for equivalence under translations and arbitrary unit multipliers. The supports have identical labelled ordered sum profiles, including diagonal pairs. The argument handles nonunit parameters by lifting units between quotient rings. No bibliographic novelty is claimed. \end{abstract} \paragraph{Conventions and scope.} For $S\subseteq C_N=\Z/N\Z$, put \[ R_S(z)=\#\{(s,t)\in S\times S:s+t=z\}. \] The pairs are ordered, diagonal pairs are included, and equality of profiles is pointwise, without relabelling. Two supports are affinely equivalent when $T=uS+v$ for some $u\in(\Z/N\Z)^\times$ and $v\in C_N$. \begin{theorem} Let $q>0$ and $x\in C_{4q}\setminus\{0,q,2q,3q\}$. Define \[ P=\{0,q,x,x+2q\},\qquad Q=\{0,q,x+q,x+3q\}. \] Let $X\in\{0,\ldots,4q-1\}$ represent $x$, and set $d=\gcd(q,X)$, $a=X/d$, $b=q/d$. Then both supports have four points, $R_P=R_Q$, and \[ P\text{ and }Q\text{ are not affinely equivalent} \quad\Longleftrightarrow\quad b\equiv2\pmod4. \] \end{theorem} \paragraph{Equality of the profiles.} In $\Z[t]/(t^{4q}-1)$, the masks of $P$ and $Q$ are $p=C+E$ and $r=C+t^qE$, where $C=1+t^q$ and $E=t^X(1+t^{2q})$. Direct expansion gives \[ p^2-r^2=(1-t^{4q})\bigl[2t^X+t^{2X}(1+t^{2q})\bigr]=0. \] The coefficient of $t^z$ in a squared mask counts $R_S(z)$, proving the claim. \paragraph{Reduction to four systems.} Every unit modulo $4q$ fixes the unique nonzero involution $2q$. Thus an affine map preserves the relation of being antipodal. In each support the two points whose antipodes are absent are exactly $0,q$. An affine map from $P$ to $Q$ must carry this pair to itself. Either $v=0$ and $uq=q$, or $v=q$ and $uq=-q$. Its image of $x$ gives the four systems on the next page. Conversely, every solution maps the entire support as required, because the antipodal partner follows its image. \newpage \begin{center} \begin{tabular}{cll} \toprule Case & Equations in $C_{4q}$, $u$ a unit & Solvable exactly when\\ \midrule I & $uq=q,\ ux=x+q$ & $4\mid b$\\ II & $uq=q,\ ux=x+3q$ & $4\mid b$\\ III & $uq=-q,\ ux=x$ & $a$ even\\ IV & $uq=-q,\ ux=x+2q$ & $a,b$ both odd\\ \bottomrule \end{tabular} \end{center} \paragraph{Cancellation and lifting.} Write $q=db$ and $X=da$. Cancelling $d$ reduces the modulus from $4q$ to $4b$; $\gcd(a,b)=1$. Reduction maps units modulo $4q$ onto units modulo $4b$. To see surjectivity, start with a representative coprime to $4b$ and impose, by the Chinese remainder theorem, the congruence $1$ modulo every prime dividing $4q$ but not $4b$. The resulting integer remains in the prescribed class modulo $4b$ and is coprime to $4q$. This proves that cancellation preserves solvability in units. \paragraph{Necessity.} For the reduced unit $w$, the equation $wb=\pm b$ is equivalent to $w\equiv\pm1\pmod4$. Reducing the other equation modulo four gives, respectively, \[ b\equiv0,\quad 3b\equiv0,\quad 2a\equiv0,\quad 2a\equiv2b\pmod4. \] These are the conditions displayed in the table. In IV, coprimality rules out $a,b$ both even. In III, $a$ even forces $b$ odd. \paragraph{Sufficiency.} For I or II, let $k=1$ or $3$ and assume $4\mid b$. Then $a$ is odd. Set $w=1+kba$ modulo $4b$. This integer is odd and coprime to $b$, hence a unit. We have $wb=b$ and $wa=a+kb$, since $a^2\equiv1\pmod4$. For III or IV, $b$ is odd. The integer $w=1+2b$ is a unit modulo $4b$, and $wb=-b$. If $a$ is even, $wa=a$; if $a$ is odd, $wa=a+2b$. Lift each witness to a unit modulo $4q$. This establishes the table. Since $a,b$ are coprime, the union of the four solvability conditions is precisely ``$b$ odd or $4\mid b$''. Its complement is $b\equiv2\pmod4$, completing the proof.\hfill$\square$ \paragraph{Examples and a lifting caution.} For $(q,X)=(2,1)$ the supports in $C_8$ are $\{0,1,2,5\}$ and $\{0,2,3,7\}$; they are non-affine. For $(q,X)=(4,1)$, $b=4$ and $Q=5P$ in $C_{16}$. For $(4,2)$ the parameter is a nonunit and the pair is non-affine. For $(20,5)$, the reduced witness $5$ modulo $16$ is not a unit modulo $80$; $21$ modulo $80$ is a valid lift. \paragraph{A valuation reformulation.} For $X>0$, $v_2(b)=\max(v_2(q)-v_2(X),0)$. Thus the obstruction is equivalently $v_2(q)=v_2(X)+1$. This is an arithmetic consequence, not a claim about the exact API of the Lean development. \newpage \paragraph{Classical background and the novelty question.} The four-point homometric families are classical; see Rosenblatt~\cite{rosenblatt}, the restatement by Jedrzejewski and Johnson~\cite{jj}, and the discussion in Erickson and Jones~\cite{ej}. In the notation of the restatement, for $1\leq a