\documentclass[11pt]{amsart} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{textcomp} \usepackage{amsmath,amssymb,amsthm} \usepackage[margin=1.1in]{geometry} \usepackage{booktabs,longtable,microtype,array,pdflscape} \usepackage[hidelinks,unicode]{hyperref} \makeatletter\g@addto@macro\UrlBreaks{\do\_\do/}\makeatother \hypersetup{pdftitle={The Bannai-Munemasa-Venkov lattice method for strongly regular graphs: a uniform discriminant-form condition and new nonexistence results},pdfauthor={Anton Kova\v{l}}} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{computation}[theorem]{Computation} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathbb{F}} \newcommand{\srg}{\mathrm{srg}} \newcommand{\ip}[2]{\langle #1,#2\rangle} \newcommand{\one}{\mathbf 1} \newcommand{\half}{\tfrac12} \newcommand{\fn}[1]{\nolinkurl{#1}} \newcommand{\ybar}{\bar y} \newcommand{\ph}{\operatorname{ph}} \newcommand{\NKILL}{70} \newcolumntype{L}[1]{>{\raggedright\arraybackslash}p{#1}} \setlength{\LTcapwidth}{\textwidth} \emergencystretch=2em \raggedbottom \title[The BMV lattice method for strongly regular graphs]{The Bannai--Munemasa--Venkov lattice method for strongly regular graphs: a uniform discriminant-form condition and new nonexistence results} \author{Anton Kova\v{l}} \address{AI Lead, V\'UB banka, Bratislava, Slovakia} \email{akoval@pobox.sk} \date{Version 1.1, October 2026} \subjclass[2020]{Primary 05E30; Secondary 11E08, 11H06, 11E12} \keywords{strongly regular graph, feasibility condition, even lattice, discriminant form, Milgram's formula, maximal overlattice, tight spherical design} \begin{document} \begin{abstract} Bannai, Munemasa and Venkov, and later Nebe and Venkov, excluded tight spherical designs --- among them strongly regular graphs of a special kind --- by studying the lattice spanned by the design: they bound its discriminant group, pass to an even (over)lattice, and compare Gauss sums by the Milgram--Braun formula. We extend this method to arbitrary strongly regular graphs with integral eigenvalues. For an integral representation $G=b(A-\rho I)+cJ$ that is positive semidefinite with even diagonal, an elementary identity for dual vectors ties every class of the discriminant form of a maximal even overlattice to the class of a rescaled sum of the vertex vectors. This gives a necessary condition on the parameters $(v,k,\lambda,\mu)$ alone: a finite quadratic module with a short list of properties must exist. Its simplest case applies the textbook congruence $\det\equiv(-1)^{n/2}\pmod 4$ for even lattices of odd determinant to the centred representation. A sweep of Brouwer's table ($v\le1300$) finds \NKILL{} complementary pairs of parameter sets listed as open that fail the condition, among them $(96,35,10,14)$, the triangle-free set $(176,25,0,4)$ and $(169,42,5,12)$; the grade of evidence is given for each (hand proof, or exact computation by at least two independent programs). No existing parameter set of the table fails in any representation tested, real graphs satisfy every condition with their actual discriminant forms, and the method gives short independent second proofs of the known nonexistence of $\srg(75,32,10,16)$ and $\srg(289,54,1,12)$. The simplest case also yields an infinite family: if $q$ is a power of a prime $p\equiv1\pmod4$ and $m$ is odd, there is no strongly regular graph with the parameters $\bigl(q^2,\,m(q+1),\,-q+m^2+3m,\,m(m+1)\bigr)$ of the negative-Latin-square type $\mathrm{NL}_m(q)$. This family, the parity case and the nonexistence of $\srg(169,42,5,12)$ are formalized in Lean~4 with Mathlib. \end{abstract} \maketitle \setcounter{tocdepth}{1} \tableofcontents %====================================================================================================================== \section{Introduction}\label{sec:intro} A strongly regular graph $\srg(v,k,\lambda,\mu)$ is a $k$-regular graph on $v$ vertices, neither complete nor empty, in which two adjacent vertices have exactly $\lambda$ common neighbours and two non-adjacent vertices exactly $\mu$. Its adjacency matrix $A$ satisfies $A^2=kI+\lambda A+\mu(J-I-A)$; apart from $k$ (on $\one$) it has two eigenvalues $r>s$, the roots of $x^2-(\lambda-\mu)x-(k-\mu)$, with multiplicities $f$ and $g$. The standard feasibility conditions \cite{BvL84,BvM} are the integrality of $f$ and $g$, the Krein conditions, the absolute bound, the claw bound, the conditions for $\mu\in\{1,2\}$, and, for conference graphs, the condition that $v$ be a sum of two squares; a few parameter sets have ad hoc proofs. None of these is a condition on integral quadratic forms. Brouwer's online table \cite{BrouwerTable} lists every parameter set with $v\le1300$ that survives the known conditions, with status ``$+$'' (a graph exists), ``$!$'' (classified), ``$-$'' (known not to exist) or ``?'' (open). \subsection*{The method and its origin} Lattices have accompanied strongly regular graphs since Cameron, Goethals, Seidel and Shult \cite{CGSS} showed that smallest eigenvalue $-2$ forces a root lattice; Venkov \cite{Venkov} connected spherical designs and lattices, and Koolen, Rehman and Yang \cite{KRY} studied the integrability of srg lattices for $s\ge-3$. The arithmetic of the discriminant group appears in the language of $p$-ranks and Smith normal forms \cite{BvE}, \cite[\S9.4]{BvM}: only primes dividing $r-s$ are interesting, which is the statement that the vertex lattice of a centred representation has discriminant group supported on the primes of $r-s$ and of the scaling. The method used here is that of Bannai, Munemasa and Venkov \cite{BMV}, who proved the nonexistence of certain tight spherical $4$-, $5$- and $7$-designs by studying the lattice generated by the design: the design identities bound its determinant, an even sublattice or overlattice is formed, and the Milgram--Braun formula excludes the remaining cases (for one dimension after gluing to a unimodular lattice and comparing theta series). Nebe and Venkov \cite{NV} refined it with Sylow-wise anisotropy and Gauss sums of the Sylow subgroups, excluding infinitely many dimensions. Tight $4$-designs \emph{are} strongly regular graphs, as \cite{BMV} note; the case of dimension $46$ there is $\srg(1127,486,165,243)$, which Brouwer's table marks ``$-$'' with a reference to Makhnev. Akiyama \cite{Akiyama} recently used a maximal integral overlattice with anisotropic discriminant form, followed by gluing and theta series, to exclude $\srg(266,45,0,9)$. With related lattice methods we recently excluded the parameter sets $(69,20,7,5)$ \cite{K69}, $(99,42,21,15)$ \cite{K99}, $(105,52,21,30)$ \cite{K105}, $(154,72,26,40)$ \cite{K154}, $(162,69,36,24)$ \cite{K162}, $(288,105,52,30)$ \cite{K288} and $(405,132,63,33)$ \cite{K405}. Several of these preprints use the same chain --- vertex lattice, maximal even overlattice, Milgram's formula, gluing, theta series --- for individual parameter sets. \subsection*{Contribution} What we have not found in the literature, and claim here subject to the caveats of Section~\ref{sec:novelty}, is: \begin{enumerate} \item the dual-vector identity and the class $\ybar$ for an arbitrary representation $G=b(A-\rho I)+cJ$ of an arbitrary strongly regular graph with integral eigenvalues (Lemmas~\ref{lem:dual}--\ref{lem:y}), turning the first, genus-level step of the method into a uniform necessary condition on the parameters (Theorem~\ref{thm:main}, Proposition~\ref{prop:weak}); \item its simplest special case for the centred representation (Corollary~\ref{cor:parity}), an application of the textbook congruence for even lattices of odd determinant \cite{ConwaySloane,Elkies}, and its consequence that no strongly regular graph has the parameters of $\mathrm{NL}_m(q)$ with $q$ a power of a prime $\equiv1\pmod4$ and $m$ odd (Corollary~\ref{cor:nl}), which contains $\srg(289,54,1,12)=\mathrm{NL}_3(17)$; \item a systematic sweep of Brouwer's table, with the resulting nonexistence of the open parameter sets of Table~\ref{tab:kills}, each marked with its grade of evidence, and the controls of Section~\ref{sec:controls}. \end{enumerate} The condition uses no theta series, no gluing and no symmetry assumption. It is strictly weaker than the full method: it does not exclude $\srg(1127,486,165,243)$, for which \cite{BMV} needed more (Section~\ref{sec:limits}). \subsection*{Organisation} Section~\ref{sec:setup} sets up the lattice, Section~\ref{sec:main} proves the condition, Section~\ref{sec:parity} the parity case, Section~\ref{sec:hand} gives hand proofs for $(169,42,5,12)$, $(96,35,10,14)$ and $(176,25,0,4)$ and short proofs for six further open sets and for two known cases, Section~\ref{sec:table} lists every open parameter set found to fail, Section~\ref{sec:controls} describes implementations and controls, Section~\ref{sec:limits} the limits --- in particular, the genus of the vertex lattice is \emph{not} determined by the parameters (Remark~\ref{rem:w3q43}) --- and Section~\ref{sec:novelty} the literature check. \subsection*{Verification} Besides the derivation, a blind adversarial audit and an independent review were carried out (see the acknowledgements); each row of Table~\ref{tab:kills} is marked with the grade of evidence it has. The parity corollary (Corollary~\ref{cor:parity}), the nonexistence of $\srg(169,42,5,12)$ (Section~\ref{sec:hand}) and Corollary~\ref{cor:nl} are formalized in Lean~4 with Mathlib \cite{Lean} (theorems \texttt{Parity.srg\_parity\_paper}, \texttt{Parity.no\_srg\_169\_42\_5\_12} and \texttt{Parity.no\_srg\_NL}). The two nonexistence theorems start from Mathlib's definition \texttt{SimpleGraph.IsSRGWith} with no further hypotheses (eigenvalues, Gram identity and lattice determinant are all derived), and use only the standard axioms (\texttt{propext}, \texttt{Classical.choice}, \texttt{Quot.sound}); no \texttt{sorry} or \texttt{native\_decide}. In the formal proof the rank $n$ is defined by $n\ell=v\nu$ (the trace of $G$); its identification with the multiplicity of $\theta$ is true but not formalized, and not needed for any conclusion. \subsection*{Conventions} A lattice is a free $\Z$-module $L\subset\R^d$ of full rank with the standard inner product $\ip{\cdot}{\cdot}$; $x^2=\ip{x}{x}$; $L$ is \emph{integral} if $\ip{L}{L}\subseteq\Z$ and \emph{even} if moreover $x^2\in2\Z$ for all $x\in L$. $L^\#=\{x\in\R^d:\ip{x}{L}\subseteq\Z\}$. For even $L$, the discriminant module $D(L)=L^\#/L$ carries the bilinear form $b(\bar x,\bar y)=\ip{x}{y}\bmod\Z$ and the quadratic form $q(\bar x)=x^2\bmod2\Z$. The Gauss sum of a finite quadratic module $(D,q)$ is $\gamma(D)=\sum_{x\in D}e^{\pi iq(x)}$; when $|\gamma(D)|=\sqrt{|D|}$ we write $\gamma(D)=\sqrt{|D|}\,e^{2\pi i\ph(D)/8}$ and call $\ph(D)\in\Z/8$ its \emph{phase} (the Brown invariant). Representations are written $(\rho,b,c)$ for $G=b(A-\rho I)+cJ$. %====================================================================================================================== \section{Setup}\label{sec:setup} \subsection{Representations and the vertex lattice} \begin{definition}\label{def:rep} Let $\Gamma$ be an $\srg(v,k,\lambda,\mu)$ with integral restricted eigenvalues $r>s$, multiplicities $f$ (of $r$) and $g$ (of $s$). A \emph{representation} is a triple $(\rho,b,c)$ of integers with $\rho\in\{r,s\}$ and $b\ne0$; let $\theta$ be the other restricted eigenvalue and $n$ its multiplicity, and put \[ e=b(\theta-\rho),\qquad T=b(k-\rho)+cv,\qquad \nu=c-b\rho . \] It is \emph{admissible} if $e>0$, $T\ge0$ and $\nu$ is even. It is \emph{centred} if $T=0$, and \emph{primitive} if $b=\pm1$ and $T>0$. \end{definition} The matrix $G=b(A-\rho I)+cJ$ acts as $T$ on $\one$, as $e$ on the $\theta$-eigenspace $E_\theta$ and as $0$ on $E_\rho$. For an admissible representation it is therefore positive semidefinite of rank $d=n+1$ if $T>0$ and $d=n$ if $T=0$; its diagonal entries equal $\nu$ and its off-diagonal entries are $c+b$ (adjacent) or $c$ (non-adjacent). Write $G=V^{\mathsf T}V$ with $V$ a real $d\times v$ matrix whose columns $v_1,\dots,v_v$ span $\R^d$, and let $L$ be the lattice they generate (it is discrete: a basis of $\R^d$ among the $v_i$ generates a full-rank integral sublattice $L_0$, and all $v_i$ lie in $L_0^\#$). $L$ is even because $\nu$ is even. Put \[ \Sigma=\sum_iv_i=V\one,\qquad \ip{\Sigma}{v_i}=T,\qquad \Sigma^2=vT . \] Every integral overlattice of $L$ lies between $L$ and $L^\#$, which contains $L$ with finite index; hence maximal even overlattices exist. We fix one, $M$; thus $L\subseteq M\subseteq M^\#\subseteq L^\#$. \subsection{The dual-vector identity} Put $\kappa=(T-e)/(vT)$ if $T>0$ and $B=(k-\theta)/v$. \begin{lemma}[Dual-vector identity]\label{lem:dual} Let $(\rho,b,c)$ be admissible, $x\in L^\#$, $z_i=\ip{x}{v_i}$ and $\sigma=\sum_iz_i=\ip{x}{\Sigma}$. Then $z\in\Z^v$ and \begin{enumerate} \item if $T>0$: $VV^{\mathsf T}=eI+\kappa\,\Sigma\Sigma^{\mathsf T}$, hence $\sum_iz_i^2=e\,x^2+\kappa\sigma^2$ and $e\,x=\sum_iz_iv_i-\kappa\sigma\,\Sigma$; \item if $T=0$: $\Sigma=0$, $VV^{\mathsf T}=eI$, hence $\sum_iz_i^2=e\,x^2$, $\sigma=0$ and $e\,x=\sum_iz_iv_i\in L$; \item $Az=\theta z+B\sigma\one$; in particular $B\sigma\in\Z$; \item $\sum_iz_i^2\equiv\sigma\pmod 2$. \end{enumerate} \end{lemma} \begin{proof} $z_i\in\Z$ since $v_i\in L$. The matrices $VV^{\mathsf T}$ and $V^{\mathsf T}V=G$ have the same nonzero eigenvalues. If $T>0$, $VV^{\mathsf T}\Sigma=VG\one=T\Sigma$ and $VV^{\mathsf T}$ acts as $e$ on $\Sigma^\perp$ (the image of $E_\theta$), so $VV^{\mathsf T}=eI+(T-e)\Sigma\Sigma^{\mathsf T}/\Sigma^2$, which is (1) since $\Sigma^2=vT$. Then $Vz=VV^{\mathsf T}x$ and $z^{\mathsf T}z=x^{\mathsf T}VV^{\mathsf T}x$ give the two identities. If $T=0$, $\Sigma^2=0$, and $G$ has the single nonzero eigenvalue $e$, so $VV^{\mathsf T}=eI$. For (3), $z=V^{\mathsf T}x$ lies in the column space of $G$, which is $\R\one\oplus E_\theta$ (or $E_\theta$ if $T=0$); writing $z=(\sigma/v)\one+z_\theta$ gives $Az=(k\sigma/v)\one+\theta z_\theta=\theta z+B\sigma\one$, and $Az-\theta z$ is an integer vector. (4) holds because $z_i^2\equiv z_i\pmod2$. \end{proof} For tight designs the identities (1) and (4) specialise to the moment identities used in \cite[(7)--(9)]{BMV}. \subsection{The class \texorpdfstring{$\ybar$}{y}} Assume $T>0$. Let $H$ be the set of integers $h$ with \begin{equation}\label{eq:H} Bh\in\Z,\qquad \kappa h^2\in\Z,\qquad \kappa h^2\equiv h\pmod 2 . \end{equation} \begin{lemma}\label{lem:subgroup} $H$ is a subgroup of $\Z$. \end{lemma} \begin{proof} (Supplied by the Codex review \cite{CodexC31}.) Clearly $0\in H$ and $H=-H$, since $\kappa h^2-(-h)=(\kappa h^2-h)+2h$. Let $h_1,h_2\in H$. The first condition is additive. If the denominator of $\kappa$ divides $h_1^2$ and $h_2^2$ then, prime by prime, it divides $h_1h_2$, so $\kappa h_1h_2\in\Z$ and $\kappa(h_1-h_2)^2\in\Z$. Finally \[ \kappa(h_1-h_2)^2-(h_1-h_2)=(\kappa h_1^2-h_1)+(\kappa h_2^2-h_2)-2\kappa h_1h_2+2h_2\in2\Z. \qedhere \] \end{proof} Let $\gamma>0$ be the generator of $H$; $H\ne0$ because $T\in H$ (proof of Lemma~\ref{lem:y}(1) below, applied to $w=v_i$). Equivalently, $\gamma=\gcd\{h\in H:h\mid T\}$, which is how the implementations compute it. Put \[ y=\Sigma/\gamma,\qquad Y=y^2=vT/\gamma^2,\qquad C=\kappa\gamma^2,\qquad r_y=\min\{j\ge1: j^2Y\in2\Z\}. \] Since $\gamma\in H$, $C$ is an integer and $C\equiv\gamma\pmod2$. \begin{lemma}\label{lem:y} Let $(\rho,b,c)$ be admissible with $T>0$. \begin{enumerate} \item $\{\ip{w}{\Sigma}:w\in M\}=h_0\Z$ with $h_0\in H$ and $h_0\mid T$; hence $\gamma\mid h_0$ and $y\in M^\#$. \item The class $\ybar\in D(M)$ has order exactly $r_y$, and $q(\ybar)\equiv Y\pmod 2$. \item $r_y\mid C$; consequently $e\,D(M)\subseteq\langle\ybar\rangle$ and the exponent of $D(M)$ divides $e\,r_y$. \end{enumerate} \end{lemma} \begin{proof} (1) Let $w\in M$ and $\sigma=\ip{w}{\Sigma}$. Since $M\subseteq L^\#$, Lemma~\ref{lem:dual} applies: $B\sigma\in\Z$; and since $w^2$ is an even integer and $e\in\Z$, $\kappa\sigma^2=\sum z_i^2-e\,w^2\in\Z$ and $\kappa\sigma^2\equiv\sum z_i^2\equiv\sigma\pmod2$. So every value of $\sigma$ on $M$ lies in $H$ (in particular $T=\ip{v_i}{\Sigma}\in H$), and so does the generator $h_0$; hence $\gamma\mid h_0$, and $h_0\mid T$. Therefore $\ip{w}{y}=\sigma/\gamma\in\Z$ for all $w\in M$. (2) If $jy\in M$ then $j^2Y=(jy)^2\in2\Z$. Conversely, if $j^2Y\in2\Z$, then $M+\Z jy$ is integral (since $y\in M^\#$ and $(jy)^2\in\Z$) and even (since $(w+tjy)^2=w^2+2tj\ip{w}{y}+t^2j^2Y$), so $jy\in M$ by maximality. (3) The proof that $r_y\mid C$ is the repair of a gap found by the Codex review \cite{CodexC31}; the original derivation asserted (3) without it. First, $CY=\kappa vT=T-e\in\Z$. Let $p$ be a prime and $\pi=v_p(Y)$. Unpacking the definition of $r_y$ ($j^2Y\in2\Z$ means $2v_p(j)\ge-\pi$ for odd $p$ and $2v_2(j)\ge1-\pi$), we have $v_p(r_y)=\max(0,\lceil-\pi/2\rceil)$ for odd $p$ and $v_2(r_y)=\max(0,\lceil(1-\pi)/2\rceil)$. If $\pi\ge1$ ($p=2$) or $\pi\ge0$ ($p$ odd), $p\nmid r_y$. If $p$ is odd and $\pi<0$, then $v_p(r_y)=\lceil-\pi/2\rceil \le-\pi\le v_p(C)$, the last step because $CY\in\Z$. If $p=2$ and $\pi<0$, then $v_2(r_y)=\lceil(1-\pi)/2\rceil\le-\pi\le v_2(C)$. There remains $p=2$, $\pi=0$, where $v_2(r_y)=1$ and we must show that $C$ is even. Suppose $C$ is odd. Then $\gamma$ is odd ($C\equiv\gamma$), and since $Y=vT/\gamma^2$ is a $2$-adic unit, $v$ and $T$ are odd; from $T-e=CY$, $e$ is even. But the trace of $G$ is $v\nu=T+ne$, and $v\nu$ is even because $\nu$ is even, so $ne=v\nu-T$ is odd and $e$ is odd, a contradiction. Hence $r_y\mid C$. Now let $x\in M^\#\subseteq L^\#$ and $t=\ip{x}{y}$. Since $r_yy\in M$, $r_yt\in\Z$. By Lemma~\ref{lem:dual}(1), $e\,x=\sum_iz_iv_i-\kappa\gamma t\cdot\gamma y=\sum_iz_iv_i-(C/r_y)(r_yt)\,y$, with $\sum z_iv_i\in L\subseteq M$ and $(C/r_y)(r_yt)\in\Z$. Thus $e\bar x\in\langle\ybar\rangle$, and $e\,r_y\bar x=0$. \end{proof} \subsection{Finite quadratic modules} We use the following standard facts; see \cite{Nikulin,ConwaySloane,MirandaMorrison,Scharlau}. \begin{enumerate} \item[(F1)] (\emph{Anisotropy.}) If $M$ is a maximal even overlattice of $L$ then $D(M)$ is anisotropic: $q(\bar x)=0$ only for $\bar x=0$ (an isotropic class $\bar x\neq0$ gives the even overlattice $M+\Z x$). \item[(F2)] (\emph{Anisotropic modules.}) An anisotropic finite quadratic module is the orthogonal sum of its $p$-parts. For odd $p$ the $p$-part is elementary abelian of rank at most $2$. The $2$-part has order at most $8$ and is one of $0$, $\Z/2$, $(\Z/2)^2$, $(\Z/2)^3$, $\Z/4$, $\Z/4\oplus\Z/2$ with an anisotropic quadratic form. (For $p=2$ the implementations V1, V3, V4 of Section~\ref{sec:impl} enumerate these by brute force and obtain the same list; for odd $p$ they use the classification, V3 additionally filtering for anisotropy.) \item[(F3)] (\emph{Gauss sums.}) For odd $p$ and $q(x)=2ax^2/p$ on $\Z/p$, $\gamma=\left(\frac ap\right)\varepsilon_p \sqrt p$ with $\varepsilon_p=1$ for $p\equiv1$ and $\varepsilon_p=i$ for $p\equiv3\pmod4$. Hence the phase of any module of odd order $p^j$ with $p\equiv1\pmod4$ lies in $\{0,4\}$, and of a cyclic module of order $p\equiv3\pmod4$ in $\{2,6\}$. On $\Z/2$, $q=\half$ has phase $1$ and $q=\frac32$ phase $7$. An anisotropic $2$-elementary module on which $q$ takes only integral values (\emph{even type}) is $0$ or the plane $\mathsf v$ with $q\equiv1$ on its three nonzero elements, of phase $0$ resp.\ $4$. \item[(F4)] (\emph{Milgram's formula}.) If $M$ is an even positive definite lattice of rank $d$, then $\gamma(D(M))=\sqrt{|D(M)|}\,e^{2\pi id/8}$, i.e.\ $\ph(D(M))\equiv d\pmod 8$ \cite[Appendix~4]{MH}. \item[(F5)] (\emph{Even lattices of odd determinant}.) If $L$ is even and $\det L$ is odd, then the rank $n$ of $L$ is even and $\det L\equiv(-1)^{n/2}\pmod4$. \end{enumerate} \begin{proof}[Proof of (F5)] Work over the $2$-adic integers $\Z_2$, where $\det L$ is a unit. The Gram matrix modulo $2$ is symmetric with zero diagonal and nonsingular, hence alternating and nondegenerate over $\F_2$; so $n$ is even, and for a basis vector $e_1$ there is $e_2$ with $\ip{e_1}{e_2}$ odd. The plane $P=\Z_2e_1+\Z_2e_2$ has Gram matrix $\left(\begin{smallmatrix}2\alpha&\beta\\\beta&2\delta\end{smallmatrix}\right)$ with $\beta$ odd and determinant $4\alpha\delta-\beta^2\equiv-1\pmod4$, a unit; so $L\otimes\Z_2=P\perp P^\perp$ with $P^\perp$ again even and unimodular. By induction $L\otimes\Z_2$ is an orthogonal sum of $n/2$ planes of determinant $\equiv-1\pmod 4$, and $\det L$ differs from the product of their determinants by the square of a $2$-adic unit, which is $\equiv1\pmod8$. \end{proof} %====================================================================================================================== \section{The obstruction}\label{sec:main} \begin{theorem}\label{thm:main} Let $\Gamma$ be an $\srg(v,k,\lambda,\mu)$ with integral restricted eigenvalues and let $(\rho,b,c)$ be an admissible representation with $T>0$, of rank $d=n+1$. Then there is a finite quadratic module $(D,q)$, namely $D=D(M)$ for a maximal even overlattice $M$ of the vertex lattice, and an element $\ybar\in D$ such that \begin{enumerate} \item[(i)] $D$ is anisotropic; \item[(ii)] $\ph(D)\equiv d\pmod 8$; \item[(iii)] $\ybar$ has order $r_y$ and $q(\ybar)\equiv Y\pmod2$; \item[(iv)] $eD\subseteq\langle\ybar\rangle$, in particular $\exp D\mid e\,r_y$; \item[(v)] \textup{(C)} for every $\bar x\in D$ there is a rational $t\equiv b(\bar x,\ybar)\pmod1$ such that $\gamma t\in\Z$, $B\gamma t\in\Z$, and, for any rational representative $Q$ of $q(\bar x)$, \[ eQ+Ct^2\in\Z\qquad\text{and}\qquad eQ+Ct^2\equiv\gamma t\pmod 2 . \] \end{enumerate} Consequently, if for some such representation no pair $(D,\ybar)$ satisfies \textup{(i)--(v)}, then no $\srg(v,k,\lambda,\mu)$ exists. \end{theorem} The proof below does not use the size of $b$; the implementation V1 and the hand proofs use only primitive representations ($b=\pm1$), and rescaled ones enter only through Proposition~\ref{prop:weak}. \begin{proof} (i) is (F1), (ii) is (F4), (iii) is Lemma~\ref{lem:y}(2) and (iv) is Lemma~\ref{lem:y}(3). For (v), choose a lift $x\in M^\#$ of $\bar x$ and put $t=\ip{x}{y}$; then $t\equiv b(\bar x,\ybar)\pmod1$ and $\sigma=\ip{x}{\Sigma}=\gamma t$ is an integer (Lemma~\ref{lem:dual}), with $B\sigma\in\Z$ by Lemma~\ref{lem:dual}(3). By Lemma~\ref{lem:dual}(1),(4), $e\,x^2+Ct^2=e\,x^2+\kappa\sigma^2=\sum_iz_i^2$ is an integer congruent to $\sigma=\gamma t$ modulo $2$. Finally $x^2=Q+2j$ for some integer $j$, and $e\cdot2j\in2\Z$, so $x^2$ may be replaced by any representative $Q$. \end{proof} \begin{remark}\label{rem:oneRep} One representation with no admissible pair suffices: (i)--(v) are necessary conditions for each representation separately. Different representations of the same graph give different lattices, and a parameter set can pass the test in one representation and fail it in another; the implementations of Section~\ref{sec:impl} test several. \end{remark} \begin{remark} In (v) the choice of lift matters only through $t$ modulo the values $\ip{w}{y}$, $w\in M$, which are integers; the implementations test all $t\equiv b(\bar x,\ybar)\pmod 1$ in a full period of the conditions, which is finite. \end{remark} \subsection{A weaker form}\label{sec:weak} The following form drops the class $\ybar$ but applies to every admissible representation, including rescaled ($|b|>1$) and centred ones; it is what the implementation V3 of Section~\ref{sec:impl} tests. For an admissible representation let $S\subseteq\Q/2\Z$ be the set of values \[ \frac{Z-\kappa\sigma^2}{e}\ \ (T>0),\qquad \frac Ze\ \ (T=0,\ \sigma=0), \] over all integers $Z,\sigma$ with $Z\equiv\sigma\pmod2$ and $B\sigma\in\Z$. \begin{proposition}\label{prop:weak} Let $(\rho,b,c)$ be admissible of rank $d$. Then $D=D(M)$ is anisotropic, $\ph(D)\equiv d\pmod 8$, and $q(\bar x)\in S$ for every $\bar x\in D$. \end{proposition} \begin{proof} Anisotropy and the phase are (F1), (F4). For $x\in M^\#\subseteq L^\#$, Lemma~\ref{lem:dual} gives $x^2=(\sum z_i^2-\kappa\sigma^2)/e$ (resp.\ $\sum z_i^2/e$) with $Z=\sum z_i^2\equiv\sigma$ and $B\sigma\in\Z$. \end{proof} \begin{lemma}[Odd representations]\label{lem:odd} Let $(\rho,b,c)$ satisfy $e>0$ and $T\ge0$, but with $\nu$ odd; let $L$ and $d$ be as above, let $L_0=\{x\in L:x^2\in2\Z\}$, and let $M$ be a maximal even overlattice of $L_0$. \begin{enumerate} \item[(a)] $L_0$ is an even sublattice of index $2$ in $L$, of rank $d$, and $L_0\subseteq M\subseteq M^\#\subseteq L_0^\#$. \item[(b)] For $x\in L_0^\#$ and $z_i=\ip{x}{v_i}$, either $z\in\Z^v$ (and then $x\in L^\#$) or $z\in(\half+\Z)^v$. \item[(c)] In both cases the identities of Lemma~\ref{lem:dual}(1)--(3) hold: $\sum_iz_i^2=e\,x^2+\kappa\sigma^2$ (resp.\ $=e\,x^2$ with $\sigma=0$ if $T=0$) and $Az=\theta z+B\sigma\one$. In the half-integral case $2\sigma\in\Z$, $2\sigma\equiv v\pmod2$, $2B\sigma\in\Z$ with $2B\sigma\equiv k-\theta\pmod2$, and $4\sum_iz_i^2\equiv v\pmod 8$. \item[(d)] Consequently $D(M)$ is anisotropic, $\ph(D(M))\equiv d\pmod8$, and $q(\bar x)\in S\cup S_{1/2}$ for every $\bar x$, where $S$ is as in Proposition~\ref{prop:weak} and $S_{1/2}$ is the set of values $(Z-\kappa\sigma^2)/e$ (resp.\ $Z/e$ with $\sigma=0$ if $T=0$) over rationals $Z,\sigma$ with $2\sigma\in\Z$, $2\sigma\equiv v\pmod2$, $2B\sigma\in\Z$, $2B\sigma\equiv k-\theta\pmod2$, $4Z\in\Z$, $4Z\equiv v\pmod8$. \item[(e)] If $T=0$ then $v\nu$ is even; so a centred odd representation has $v$ even. \end{enumerate} \end{lemma} \begin{proof} (a) Since $(x+y)^2=x^2+y^2+2\ip xy$, the map $x\mapsto x^2\bmod2$ is a homomorphism $L\to\Z/2$; it is onto because $v_i^2=\nu$ is odd, so its kernel $L_0$ has index $2$. A subgroup of finite index in a lattice spans the same real space (here $2L\subseteq L_0\subseteq L$), so $L_0$ is again a lattice of rank $d$. $L_0$ is even by definition, and every integral overlattice of $L_0$ lies in $L_0^\#$. (b) $v_i-v_j\in L_0$ (its norm is $2\nu-2G_{ij}$) and $2v_i\in L_0$, so $z_i-z_j\in\Z$ and $2z_i\in\Z$. Hence all $z_i$ lie in one coset of $\Z$ in $\half\Z$: either every $z_i$ is an integer or every $z_i$ is in $\half+\Z$; no mixed case occurs. If $z$ is integral, $x$ pairs integrally with the generators $v_i$ of $L$, so $x\in L^\#$. (c) The proof of Lemma~\ref{lem:dual}(1)--(3) uses only $x\in\R^d$ and $z=V^{\mathsf T}x$. In the half-integral case put $w_i=2z_i$, all odd. Then $2\sigma=\sum_iw_i\equiv v$; $2(Az-\theta z)_i=\sum_{j\sim i}w_j-\theta w_i\equiv k-\theta\pmod 2$ (a sum of $k$ odd numbers), and this equals $2B\sigma$; and $4\sum_iz_i^2=\sum_iw_i^2\equiv v\pmod8$ since odd squares are $\equiv1\pmod8$. (d) Anisotropy and the phase are (F1), (F4) for $M$, of rank $d$. For $x\in M^\#\subseteq L_0^\#$ apply (b), (c) and solve the identity for $x^2$; the integral case gives $S$ as in Proposition~\ref{prop:weak}. (e) $\one^{\mathsf T}G\one=vT=0$, and $\one^{\mathsf T}G\one=v\nu+2\sum_{i0$, $T=0$, and $\nu$ is even. The vertex lattice $L$ is even of rank $n$, and by Lemma~\ref{lem:dual}(2) $\ell L^\#\subseteq L$, so $|L^\#/L|=\det L$ divides $\ell^n$. Apply (F5) with $\Delta=\det L$; here $\Delta>0$ because the Gram matrix of a basis of $L$ is positive definite. (The sign matters: a negative determinant $-13^j\equiv-1\pmod4$ would not contradict (F5) at $n=126$.) The same argument applies verbatim to any centred representation, i.e.\ to any integer $b\ne0$ with $b(\theta-\rho)>0$, $c=-b(k-\rho)/v\in\Z$ and $\nu=c-b\rho$ even, with $\ell=b(\theta-\rho)$. \end{proof} No overlattice, no Gauss sum and no computation beyond the parameters is needed. The congruence (F5) is textbook (\cite[Ch.~15]{ConwaySloane}, \cite[Prop.~1]{Elkies}); its unimodular special case (an even unimodular lattice has rank divisible by $8$) is the step by which \cite[Thm.~4.3]{BMV} and \cite[Thm.~3.3]{NV} exclude tight designs. The corollary is the case of Proposition~\ref{prop:weak} (centred, $T=0$) in which the lattice $L$ itself already has odd determinant; it is the case in which Milgram's formula is applied to $L$ itself: when all primes of $\ell$ are $\equiv1\pmod4$ the Gauss sum of $D(L)$ is real and Milgram gives exactly $n\equiv0\pmod4$; for primes $\equiv3\pmod4$ Milgram contains more information than (F5), which we do not use here. \begin{corollary}[Negative-Latin-square parameters]\label{cor:nl} Let $q=p^e$ with $p\equiv1\pmod4$ prime, and let $m$ be odd. There is no strongly regular graph with the parameters of $\mathrm{NL}_m(q)$, \[ \bigl(q^2,\ m(q+1),\ -q+m^2+3m,\ m(m+1)\bigr). \] \end{corollary} \begin{proof} The restricted eigenvalues are $r=m$ and $s=m-q$ (their sum is $\lambda-\mu=2m-q$, their product $-(k-\mu)=-m(q-m)$), and the multiplicity of $s$ is $g=m(q+1)$ (from $k+fr+gs=0$ and $f+g=q^2-1$). Take the centred representation with $\rho=r=m$, $\theta=s=m-q$, $b=-q$, $c=m$, that is \[ G=q(mI-A)+mJ . \] Then $T=-q(k-m)+mq^2=0$, $e=\ell=b(\theta-\rho)=q^2$, whose only prime is $p\equiv1\pmod4$, and $\nu=c-b\rho=m(q+1)$ is even. By Corollary~\ref{cor:parity} (in the form for an arbitrary centred representation stated in its proof), $n=g=m(q+1)\equiv0\pmod4$. But $q\equiv1\pmod4$ gives $q+1\equiv2\pmod4$, and $m$ is odd, so $n\equiv2\pmod4$, a contradiction. No passage to the complement and no condition on $p\mid m$ or $p\mid m+1$ is needed. \end{proof} The range $1\le m\le q-2$ matters only for the parameter set to be feasible at all (for other $m$ the parameters are not those of a graph); it plays no role in the proof. Bondarenko and Radchenko's $(289,54,1,12)$ is $\mathrm{NL}_3(17)$, and $14$ of the $21$ pairs of Remark~\ref{rem:paritycover} are instances (all those with $v\in\{169,289,625,841\}$; \fn{out/milgram_paper/nl_check.txt}). The corollary is consistent with the known graphs: $\mathrm{Paley}(q^2)$ has the parameters of $\mathrm{NL}_{(q-1)/2}(q)$ with $(q-1)/2$ even, and Latin-square parameters $\mathrm{LS}_m(q)$ have $n=m(q-1)\equiv0\pmod4$. \begin{remark}[Which rows it covers]\label{rem:paritycover} The hypotheses force $a$ and $|\theta-\rho|$ to be odd, but not $v$: for $T(9)=\srg(36,14,7,4)$ and $\rho=-2$ one has $a=9$, $G=9(A+2I)-4J$, $\nu=14$, $\ell=63$, and (F5) holds ($\det L=3^{14}$, rank $8$, computed by the referee). All exclusions below have odd $v$. In Brouwer's table ($v\le1300$, snapshot of 8 October 2026; script \fn{out/milgram_paper/parity_scan.py}, output \fn{parity_scan.txt}) the corollary fails for exactly $20$ open complementary pairs and for one pair marked ``$-$'', and for no row marked ``$+$'' or ``$!$'' ($1096$ such rows with integral eigenvalues, counted per orientation). The killed open pairs are: \begin{itemize}\sloppy \item $v=p^2$ with $r-s=p$, $p\in\{13,17,29\}$, $\ell=p^2$: $(169,42,5,12)$, $(169,70,27,30)$, $(289,90,23,30)$, $(289,126,53,56)$, and the five pairs $(841,k,\lambda,\mu)$ with $k\in\{150,210,270,330,390\}$; \item $\ell\in\{5^3,5^4\}$: $(375,154,53,70)$, $(375,170,85,70)$, $(875,184,33,40)$, and $(625,k,\lambda,\mu)$ with $k\in\{130,182,234,286\}$; \item $\ell=5\cdot13$: $(235,52,9,12)$; $\ell=13^2$: $(507,44,1,4)$; $\ell=5^2\cdot17$: $(391,140,39,56)$; $\ell=13\cdot29$: $(871,348,137,140)$. \end{itemize} The pair marked ``$-$'' is $(289,54,1,12)$, whose nonexistence is a theorem of Bondarenko and Radchenko \cite{BR13}: with $\rho=s=-14$ one gets $a=17$, $G=17(A+14I)-4J$, $\nu=234$, $\ell=17^2$ and $n=f=234\equiv2\pmod4$. Bondarenko and Radchenko proved it by a different route (Euclidean representation, vertex-transitivity, an $\F_3$-structure on the vertex set, and a Diophantine equation), so the three lines above are an independent second proof; they are also a consistency check of the corollary. In the first family ($v=p^2$, $r-s=p$, $p\equiv1\pmod 4$) the corollary demands $n\equiv0\pmod4$; Latin-square-type parameters $\mathrm{LS}_m(p)$, with $n=m(p-1)$, satisfy this automatically, which is consistent with their existence. \end{remark} %====================================================================================================================== \section{Worked proofs}\label{sec:hand} \subsection{\texorpdfstring{$\srg(169,42,5,12)$}{srg(169,42,5,12)}} The eigenvalues are $r=3$, $s=-10$ (roots of $x^2+7x-30$), with $f=126$, $g=42$. Take $\rho=s=-10$: $k-\rho=52$, $\gcd(169,52)=13$, $a=13$, \[ G=13(A+10I)-4J . \] On $\one$ it acts by $13\cdot52-4\cdot169=0$, on $E_r$ by $13\cdot13=169$, on $E_s$ by $0$; its diagonal is $\nu=130-4=126$, even. So the vertex lattice is even of rank $n=f=126$ with $169L^\#\subseteq L$, hence $\det L=13^j$ for some $j$, and $13^j\equiv1\not\equiv(-1)^{63}\pmod4$, contradicting (F5). \emph{No $\srg(169,42,5,12)$ exists.} (The other centred representation, $G=13(3I-A)+3J$ of rank $g=42\equiv2\pmod 4$, gives the same contradiction. The original derivation used instead the representation $A'+4I$ of the complement $\srg(169,126,95,90)$, of rank $43$, with $\gamma=130$, $Y=\frac{13}{10}$, $r_y=10$, $C=90$, where the phases of the $2$-, $5$- and $13$-parts force $\ph(D)\in\{1,5\}\ne3$; Codex's implementation confirms that representation independently.) \subsection{\texorpdfstring{$\srg(96,35,10,14)$}{srg(96,35,10,14)}} This row carries the comment ``$\mathrm{pg}(5,6,2)$?'' in Brouwer's table. Pass to the complement $\srg(96,60,38,36)$, with eigenvalues $r=6$, $s=-4$ (roots of $x^2-2x-24$), $f=32$, $g=63$, and take the primitive representation $G=A+4I$, i.e.\ $(\rho,b,c)=(-4,1,0)$: $e=10$, $T=64$, $\nu=4$, $d=33$. Directly $G^2=10G+36J$. Then \[ \kappa=\frac{54}{96\cdot64}=\frac{9}{1024},\qquad B=\frac{54}{96}=\frac9{16}. \] For $h\in H$: $\frac9{16}h\in\Z$ forces $16\mid h$; $\frac{9h^2}{1024}\in\Z$ forces $32\mid h$; for $h=32u$, $\frac{9h^2}{1024}=9u^2\equiv h\equiv0\pmod 2$ forces $u$ even. The only divisor of $64$ in $H$ is $64$, so $\gamma=64$, \[ y=\Sigma/64,\qquad Y=\frac{96\cdot64}{64^2}=\frac32,\qquad r_y=2,\qquad C=\frac{9\cdot64^2}{1024}=36 . \] \emph{Exponent.} For $x\in M^\#$ with $t=\ip{x}{y}\in\half\Z$, Lemma~\ref{lem:dual} gives $10x=\sum z_iv_i-36t\,y=\sum z_iv_i-18(2t)y$, and $2y\in M$ (Lemma~\ref{lem:y}(2)), so $10D=0$: $D=D_2\oplus D_5$ with both parts elementary abelian, and $\ybar\in D_2$. \emph{The $2$-part.} $b(\ybar,\ybar)=Y\bmod1=\half$ is a unit, so $D_2=\langle\ybar\rangle\perp\ybar^\perp$. For $\bar x\in\ybar^\perp\cap D_2$ we have $t\in\Z$, and (C) reads $10Q+36t^2\equiv64t\equiv0\pmod2$, so $10Q\in2\Z$, i.e.\ $Q\in\frac15\Z$. Since $2\bar x=0$, $Q\in\half\Z$; hence $Q\in\Z$, so $\ybar^\perp\cap D_2$ is anisotropic of even type, of phase $0$ or $4$ by (F3). $\langle\ybar\rangle$ with $q=\frac32$ has phase $7$. So $\ph(D_2)\in\{3,7\}$. \emph{The $5$-part} has phase $0$ or $4$ by (F3). Hence $\ph(D)\in\{3,7\}$, whereas $d=33\equiv1\pmod8$: (ii) fails. \emph{No $\srg(96,35,10,14)$ exists.} The same conclusion holds in every representation of either orientation that the implementation V1 tested (Table~\ref{tab:kills}); the blind auditor's weaker decider admits $4$ modules, among them $\langle\half\rangle$, in this representation (it does not use $\ybar$; \fn{out/milgram_paper/rerun_v3.tsv}, line ``$(96,60,38,36)$ R a=1 c=0'') and excludes the row through the odd representation $A'+4I+J=3I-A+2J$ instead (Lemma~\ref{lem:odd}). \subsection{\texorpdfstring{$\srg(176,25,0,4)$}{srg(176,25,0,4)}} Pass to the complement $\srg(176,150,128,126)$, eigenvalues $r=6$, $s=-4$, $f=55$, $g=120$, and take $G=A+4I$: $e=10$, $T=154$, $\nu=4$, $d=56$; $G^2=10G+126J$. Here $\kappa=\frac{144}{176\cdot154}=\frac9{1694}$ with $1694=2\cdot7\cdot11^2$, and $B=\frac{144}{176}=\frac9{11}$. For $h\in H$: $11\mid h$ and $1694\mid9h^2$, so $154\mid h$; $h=154$ satisfies the parity condition ($\kappa\cdot154^2=126$). Thus \[ \gamma=154,\qquad Y=\frac{176}{154}=\frac87,\qquad r_y=7,\qquad C=126 . \] By Lemma~\ref{lem:y}(3), $10D\subseteq\langle\ybar\rangle\cong\Z/7$ and $\exp D\mid70$, so $D=D_2\oplus D_5\oplus D_7$. Multiplication by $10$ is bijective on $D_7$, so $D_7=10D_7\subseteq\langle\ybar\rangle$, i.e.\ $D_7=\langle\ybar\rangle$ with $q(\ybar)=\frac87=2\cdot\frac47$; as $\left(\frac47\right)=1$ and $7\equiv3\pmod4$, its phase is $2$. On $D_2$ (elementary, orthogonal to $\ybar$) $t\in\Z$ and (C) gives $10Q\equiv154t\equiv0\pmod 2$, so as before $Q\in\Z$ and $\ph(D_2)\in\{0,4\}$; $\ph(D_5)\in\{0,4\}$. Hence $\ph(D)\in\{2,6\}$, whereas $d=56\equiv0$. \emph{No $\srg(176,25,0,4)$ exists.} \subsection{Six further sets}\label{sec:four} The same computation, with the constants checked by three implementations (V1, V4, and V5 of Section~\ref{sec:impl}; Table~\ref{tab:kills}): \begin{itemize} \item $\srg(616,205,90,57)$, $r=37$, $s=-4$, $G=A+4I$: $e=41$, $T=209$, $d=56$, $\kappa=\frac3{2299}$, $\gamma=209$, $Y=\frac{56}{19}$, $r_y=19$, $C=57$. $\exp D\mid41\cdot19$; $41$ is invertible on $D_{19}$, so $D_{19}=\langle\ybar\rangle$ with $q(\ybar)\equiv\frac{18}{19}=2\cdot\frac9{19}$ of phase $2$; $\ph(D_{41})\in\{0,4\}$. So $\ph(D)\in\{2,6\}\ne0$. \item $\srg(300,26,4,2)$, $r=6$, $s=-4$, $G=A+4I$: $e=10$, $T=30$, $d=118$, $\kappa=\frac1{450}$, $B=\frac1{15}$, $\gamma=30$, $Y=10$, $r_y=1$ ($y\in M$), $C=2$. Then $10D=0$; on $D_2$, (C) gives $10Q+2t^2\equiv30t\pmod2$ with $t\in\Z$, so $Q\in\Z$ and $\ph(D_2)\in\{0,4\}$; $\ph(D_5)\in\{0,4\}$; $d\equiv6$. \item $\srg(430,39,8,3)$, $r=9$, $s=-4$, $G=A+4I$: $e=13$, $T=43$, $d=130$, $\kappa=\frac3{1849}$, $\gamma=43$, $Y=10$, $r_y=1$. Then $13D=0$, $13\equiv1\pmod 4$, $\ph(D)\in\{0,4\}$, whereas $d\equiv2$. \item $\srg(690,318,121,168)$, via the complement $\srg(690,371,220,175)$ with $r=49$, $s=-4$, $G=A+4I$: $e=53$, $T=375$, $d=46$, $\kappa=\frac7{5625}$, $B=\frac7{15}$, $\gamma=75$, $Y=46$, $r_y=1$. Then $53D=0$, $53\equiv1\pmod4$, $\ph(D)\in\{0,4\}$, whereas $d\equiv6$. \item $\srg(508,247,126,114)$, $r=19$, $s=-7$, $f=127$, $G=A+7I+J$, i.e.\ $(\rho,b,c)=(-7,1,1)$: $e=26$, $T=762$, $\nu=8$, $d=128$, $\kappa=\frac{92}{48387}$ ($48387=3\cdot127^2$), $B=\frac{57}{127}$. For $h\in H$: $381\mid h$, and $h=381$ fails the parity condition ($\kappa h^2=276$), $h=762$ passes, so $\gamma=762$, $Y=\frac23$, $r_y=3$, $C=1104$. $\exp D\mid e\,r_y=2\cdot3\cdot13$; $26$ is invertible on the elementary group $D_3$, so $D_3=26D_3\subseteq\langle\ybar\rangle$, i.e.\ $D_3=\langle\ybar\rangle$ with $q(\ybar)\equiv\frac23$ of phase $2$; $\ph(D_{13})\in\{0,4\}$; on the elementary $D_2\perp\ybar$, $t\in\Z$ and (C) gives $26Q+1104t^2\equiv762t$, i.e.\ $26Q\in2\Z$, so $Q\in\Z$ and $\ph(D_2)\in\{0,4\}$. So $\ph(D)\in\{2,6\}$, whereas $d\equiv0$. \item $\srg(1072,391,150,138)$, $r=23$, $s=-11$, $f=335$, $G=A+11I+J$, i.e.\ $(\rho,b,c)=(-11,1,1)$: $e=34$, $T=1474$, $\nu=12$, $d=336$, $\kappa=\frac{45}{49379}$ ($49379=11\cdot67^2$), $B=\frac{23}{67}$, $\gamma=737$ ($737\mid h$ for $h\in H$, and $\kappa\cdot737^2=495$ is odd like $737$), $Y=\frac{32}{11}$, $r_y=11$, $C=495$. $\exp D\mid 2\cdot11\cdot17$; as before $D_{11}=\langle\ybar\rangle$ with $q(\ybar)\equiv\frac{32}{11}\equiv2\cdot\frac5{11}$, $\left(\frac5{11}\right)=1$, phase $2$; $\ph(D_{17})\in\{0,4\}$; on $D_2$, $34Q+495t^2\equiv737t$ with $t\in\Z$ gives $34Q\in2\Z$, $Q\in\Z$, $\ph(D_2)\in\{0,4\}$. So $\ph(D)\in\{2,6\}$, whereas $d\equiv0$. \end{itemize} The last two rows are not excluded by V3 (Section~\ref{sec:impl}), which does not use $\ybar$: there the trivial module ($M$ unimodular, possible since $d\equiv0\pmod 8$) and several others pass Proposition~\ref{prop:weak}; it is condition (iii) --- an element of order $r_y=3$ resp.\ $11$ must exist --- together with (iv), which pins the $r_y$-part to $\langle\ybar\rangle$, that excludes them. The programs V1, V4 and V5 also exclude the second primitive representation $(19,-1,1)$ resp.\ $(23,-1,1)$ (Table~\ref{tab:kills}); that case is not written out here. \subsection{Two known cases}\label{sec:known} \emph{$\srg(75,32,10,16)$} (Azarija--Marc \cite{AzarijaMarc}: star complements and an exhaustive computer search). Here $r=2$, $s=-8$, $f=56$, $g=18$. Take $(\rho,b,c)=(2,-1,2)$, $G=2I-A+2J$: $e=10$, $T=-30+150=120$, $\nu=4$, $d=g+1=19$, $\kappa=\frac{110}{75\cdot120}=\frac{11}{900}$, $B=\frac{40}{75}=\frac8{15}$. For $h\in H$: $15\mid h$ and $900\mid11h^2$ give $30\mid h$; $h=30$ fails the parity condition ($\kappa\cdot30^2=11$ odd, $30$ even) and $h=60$ satisfies it ($\kappa\cdot60^2=44$). So $\gamma=60$, $Y=\frac{75\cdot120}{3600}=\frac52$, $r_y=2$, $C=44$. As for $(96,35,10,14)$: $10x\equiv-22(2t)y\pmod M$ with $2y\in M$, so $10D=0$; $\ybar$ splits off $D_2$ with $q(\ybar)=\frac52\equiv\half$, phase $1$; on $\ybar^\perp\cap D_2$, (C) gives $10Q+44t^2\equiv60t\equiv0$, so $Q\in\Z$ and the phase is $0$ or $4$; $D_5$ has phase $0$ or $4$. Hence $\ph(D)\in\{1,5\}$, whereas $d=19\equiv3\pmod8$. This is a computer-free second proof. (V1 excludes the row in all six representations it tried, V3 in the centred representation $5(2I-A)+2J$.) \emph{$\srg(289,54,1,12)$} (Bondarenko--Radchenko \cite{BR13}): Remark~\ref{rem:paritycover}. %====================================================================================================================== \section{All rows excluded}\label{sec:table} {\sloppy Table~\ref{tab:kills} lists every open complementary pair of Brouwer's table that failed the condition in at least one representation, by any implementation of Section~\ref{sec:impl}. Data cut-off: 9 October 2026, 18:00 CEST; table snapshot of 8 October 2026 ($v\le1300$). The V1 scan is \emph{complete}: all $1402$ open complementary pairs with integral eigenvalues, both sides ($2804$ representations), with the $227$ representations that timed out at $90$\,s re-run at $900$\,s (\fn{out/milgram_final/coverage.txt}). It excludes 70 pairs; $1318$ pairs admit a module on both sides, and $14$ pairs admit one on one side while the other side is still undecided after $900$\,s (listed in Section~\ref{sec:limits}). No other implementation excludes a pair that V1 does not. Of the 70 pairs, 28 are proved by hand in this paper (Section~\ref{sec:hand} or Corollary~\ref{cor:parity} --- the parity rows need only the arithmetic shown in the last column), and 42 rest on exhaustive finite computations by at least two independent programs, namely $(105,40,15,15)$, $(112,36,10,12)$, $(120,34,8,10)$, $(120,35,10,10)$, $(133,32,6,8)$, $(209,52,15,12)$, $(352,26,0,2)$, $(391,182,93,77)$, $(441,190,89,76)$, $(456,130,24,42)$, $(456,195,74,90)$, $(494,85,12,15)$, $(507,138,49,33)$, $(507,240,106,120)$, $(508,234,100,114)$, $(560,247,102,114)$, $(630,68,1,8)$, $(664,170,24,50)$, $(690,260,94,100)$, $(715,272,102,104)$, $(721,220,69,66)$, $(732,238,92,70)$, $(741,260,91,91)$, $(750,210,55,60)$, $(750,214,63,60)$, $(760,252,82,84)$, $(780,246,76,78)$, $(780,247,78,78)$, $(780,369,158,189)$, $(836,375,190,150)$, $(875,304,78,120)$, $(931,180,29,36)$, $(1026,451,212,187)$, $(1140,391,126,138)$, $(1148,481,210,195)$, $(1183,192,26,32)$, $(1190,492,202,204)$, $(1190,493,204,204)$, $(1236,195,34,30)$, $(1248,203,22,35)$, $(1276,204,14,36)$, $(1288,195,26,30)$. No pair rests on one program only. Every (row, representation) pair that V1 excludes ($151$ in all) was re-decided by V4 and by V5: $151$ agreements, no disagreement, no timeout; so every C2 row is excluded by V1, V4 and V5 alike, and some also by V3. Codex's own table run (review C35, \fn{out/codex/c35/table_results.json}) ran V4 on both orientations of the $48$ rows of the previous version of this table: $81$ representations without admissible module, $15$ timeouts, none admissible. Every pair in the table has status ``?'' in the snapshot of 8 October 2026; the snapshot of 9 October held in the repository covers only $v\le750$, and agrees there. A ``C'' grade means: an exact enumeration of all finite quadratic modules allowed by (F2) on the relevant primes found none satisfying the conditions; there is no certificate beyond rerunning the enumeration. Timeouts are never counted as exclusions; where an implementation timed out on some representation of a row, the row is listed on the strength of the representations that were decided. \par} \begin{landscape} {\footnotesize\setlength{\tabcolsep}{3.5pt} \begin{longtable}{@{}rrrr rr ccccc cc L{5.2cm}L{4.6cm}@{}} \caption{Open complementary pairs of Brouwer's table ($v\le1300$) that fail the condition, in Brouwer's layout: the first line of a pair is $(v,k,\lambda,\mu)$ with $2k0$ and even $L$: $y\in M^\#$, $\gamma\mid\ip{M}{\Sigma}$ and $\operatorname{ord}\ybar=r_y$ in every record, $0$ violations (\fn{scripts/audit_milgram_fable_check_y.py}, rerun, output \fn{out/milgram_paper/check_y_rerun.txt}; also reproduced by the referee). Independently, the derivation lane checked $112$ (graph, representation) pairs on $28$ graphs against every condition of Theorem~\ref{thm:main}, including (iv) and (C) on every class: $0$ violations (\fn{out/s4lat/realcheck.log}). \item \emph{Every existing row.} V1 on all $557$ complementary pairs of the table with status ``$+$'' or ``$!$'' and integral eigenvalues ($v\le1300$, both sides): $1114$ representations, $1043$ decided, $71$ timeouts at $90$\,s, $0$ exclusions (\fn{out/s4lat/allopen_known.tsv}). V3 on all $458$ existing rows with $v\le400$ (each orientation counted): $4034$ representations, $0$ exclusions (\fn{brouwer_scan_existing.tsv}). V4 on $112$ existing rows: $222$ representations admissible, $2$ timeouts, $0$ exclusions (\fn{out/codex/c31/controls.json}). P on all $1096$ existing rows (each orientation counted) with $v\le1300$: $0$ exclusions. V5 (full mode, all of (i)--(v)) on the same $1114$ representations as V1: $1114$ admissible, $0$ timeouts at $300$\,s, $0$ exclusions (\fn{out/milgram_final/ycheck_known_full.tsv}). \item \emph{A recent construction.} $\srg(85,30,11,10)$ (Hancock, status ``$+$'' since 2026, smallest eigenvalue $-4$) and its complement: V1 finds exactly one admissible module in each of the $12$ primitive representations tried, V3 accepts all $16$ standard representations (\fn{out/milgram_paper/controls_known.txt}, \fn{out/audit_milgram_fable/control_85_30.txt}); the audit reports that V4 accepted the $4$ representations it was run on (no log of that run is archived). \item \emph{Known nonexistence.} $\srg(75,32,10,16)$ fails in all $12$ primitive representations V1 tried (both orientations) and in the centred representation for V3; $\srg(289,54,1,12)$ fails in $5$ of $5$ for V1, $3$ of $8$ for V3, and fails P. The known nonexistent rows $(49,16,3,6)$, $(57,14,1,4)$, $(95,40,12,20)$, $(96,38,10,18)$ and $(1127,486,165,243)$ are \emph{not} excluded: V3 accepts every standard representation of both orientations of the first four (\fn{out/milgram_paper/control_unreproduced_v3.txt}); for the last see Section~\ref{sec:novelty}. The condition is not a complete test. \end{enumerate} %====================================================================================================================== \section{Limits}\label{sec:limits} \begin{remark}[The genus is not a parameter invariant]\label{rem:w3q43} The test asks whether \emph{some} finite quadratic module satisfies (i)--(v); it never assumes that the parameters determine $D(M)$. They do not. The generalized quadrangles $W(3)$ and $Q(4,3)$ are both $\srg(40,12,2,4)$; in the same representation $G=A+4I$ (rank $25$) the blind audit computed $D(M)\cong\langle\half\rangle$ ($\det M=2$) for $W(3)$ and $D(M)$ of order $72$, with invariants $\Z/6\oplus\Z/6\oplus\Z/2$, for $Q(4,3)$ (\fn{out/audit_milgram_fable/truth/W_3.jsonl}, \fn{Q4_3.jsonl}). Pairs related by Seidel or Godsil--McKay switching (e.g.\ $T(8)$ and the Chang graphs) do not show this, because switching preserves the rational class of the eigenspace. Any argument that pins $|D(M)|$, the Hasse invariants or the genus from the parameters alone is therefore unsound; the implementations only test existence. \end{remark} \begin{itemize} \item \emph{Representations.} Only finitely many representations are tested per row. A row that passes in every tested representation is not thereby shown to pass the condition; conversely a single failing representation suffices (Remark~\ref{rem:oneRep}). \item \sloppy\emph{Scan coverage.} The V1 scan of the open rows is complete (Section~\ref{sec:table}), but $14$ open pairs are undecided on one side after $900$\,s (the other side admits a module): $(381,114,29,36)$, $(676,150,23,36)$, $(736,350,162,170)$, $(775,288,98,112)$, $(868,408,182,200)$, $(931,390,173,156)$, $(946,243,60,63)$, $(981,340,115,119)$, $(993,310,89,100)$, $(993,352,131,121)$, $(1036,405,138,171)$, $(1055,186,17,36)$, $(1156,567,258,297)$, $(1208,459,150,189)$. Only the smallest admissible $c$ is tested on each side, and V3's scan of open rows covers only $v\le750$ (it had reached $v=639$ at the cut-off). Further exclusions in other representations are therefore possible. \item \emph{Strength.} The test is genus-level. Where admissible modules exist (e.g.\ $\srg(88,27,6,9)$, $\srg(99,14,1,2)$, $\srg(162,21,0,3)$, each with a unique admissible module in its smallest representation) it says nothing; it also misses $\srg(1127,486,165,243)$, which the full method of \cite{BMV} excludes (Section~\ref{sec:novelty}); and the theta-series refinements of \cite{K69,K105,K154} were found too weak at smallest eigenvalue $-4$ (derivation notes, \fn{docs/s4lat_2026_10_08.md}, \S\S3--6). \item \emph{Computer-assisted rows.} The C2 rows of Table~\ref{tab:kills} rest on exhaustive finite enumerations repeated by at least two independent programs; no row of the current table rests on one program only, but no certificate other than rerunning the enumeration exists. \end{itemize} %====================================================================================================================== \section{Novelty}\label{sec:novelty} This section summarises a literature search made on 9 October 2026 (\fn{docs/novelty_milgram_obstruction_2026_10_09.md}). It is a web and library search; it did not include MathSciNet citation searches forward from \cite{BMV,NV} or the Russian-language literature (in particular the Makhnev school). \emph{The method is not new.} Taking the lattice spanned by a Euclidean representation of a two-distance set, bounding its discriminant group by the design identities, passing to an even sublattice or overlattice, requiring anisotropy and comparing Gauss sums via the Milgram--Braun formula, with $2$-adic congruences for characteristic vectors, is the method of Bannai, Munemasa and Venkov \cite{BMV} (who credit Martinet for one case) and of Nebe and Venkov \cite{NV}, developed for tight spherical designs; tight $4$-designs are strongly regular graphs. Akiyama \cite{Akiyama} used the step ``maximal integral overlattice $\Rightarrow$ anisotropic discriminant form, killed by an explicit exponent'' for $\srg(266,45,0,9)$ (an unrefereed preprint). The finite-quadratic-form facts are classical \cite{Nikulin,ConwaySloane,MH}; the parity congruence (F5) is textbook \cite{ConwaySloane,Elkies}. \emph{What appears to be new.} (a) The dual-vector identity with the class $\ybar$ for an arbitrary integral representation $b(A-\rho I)+cJ$ (centred and non-centred, rescaled, and odd via the even sublattice) of an arbitrary strongly regular graph with integral eigenvalues, and the resulting uniform condition on the parameters; (b) the application of (F5) to the centred representation (Corollary~\ref{cor:parity}) and the resulting exclusion of all $\mathrm{NL}_m(q)$ parameter sets with $q$ a power of a prime $p\equiv1\pmod 4$ and $m$ odd (Corollary~\ref{cor:nl}); (c) the sweep of Brouwer's table and the exclusions of Table~\ref{tab:kills}. The feasibility conditions collected by Brouwer and van Lint \cite{BvL84} and by Brouwer and Van Maldeghem \cite{BvM} contain no lattice, genus or Gauss-sum condition (the only quadratic-form condition is the sum-of-two-squares condition for conference graphs), Brouwer's reference list contains none of \cite{BMV,NV}, and every open row of Table~\ref{tab:kills} carries a bare ``?'' (or a comment about an associated partial geometry) with no feasibility reason. Searches for $(176,25,0,4)$, $(169,42,5,12)$ and $\mathrm{pg}(5,6,2)$ found no exclusion. \emph{Reproductions.} Both known cases of Section~\ref{sec:known} were proved by non-lattice methods --- $\srg(75,32,10,16)$ by star complements and computer search \cite{AzarijaMarc}, $\srg(289,54,1,12)$ by a Euclidean representation, an $\F_3$-structure and a Diophantine equation \cite{BR13} --- so the proofs here are independent second proofs by a different method, not restatements. Conversely, $\srg(1127,486,165,243)$, excluded in the table with a reference to Makhnev and, as the tight $4$-design in dimension $46$, also by \cite{BMV}, is \emph{not} excluded by our condition: V1 finds $16$ admissible modules in the representation $A+4I$ of the complement $(1127,640,396,320)$ (rank $47$) and V3 accepts all $16$ standard representations of both orientations (\fn{out/milgram_paper/control_1127_v3.txt}); \cite{BMV} needed gluing and theta series for it. \emph{Further checks.} A specialist in lattices and spherical designs may know whether the general dual-vector identity and the $\ybar$ refinement appear elsewhere; a citation-forward search on \cite{BMV,NV} and a search of the Russian-language literature would settle this. The rows of Table~\ref{tab:kills} and the two reproductions of Section~\ref{sec:known} can be checked against Brouwer's table and by rerunning the deciders from the data archive. %====================================================================================================================== \section*{Acknowledgements and use of AI systems}\label{sec:ai} This work was done with substantial assistance from AI systems. They are not authors; the author is responsible for all content. Claude (Anthropic), working in the author's repository under the author's direction, produced the derivation, the hand proofs of Section~\ref{sec:hand} (in their original form), the implementations V1 and the table scans, and drafted this text, including Corollary~\ref{cor:parity} in the general form stated here and its scan. A separate Claude session on a different model family (Fable) carried out a blind adversarial audit: it re-derived the obstruction from first principles before reading the derivation, wrote the decider V3, computed actual discriminant forms of $254$ real graphs, and found the $W(3)$/$Q(4,3)$ example of Remark~\ref{rem:w3q43}. Codex (OpenAI's coding agent) independently reviewed the uniform lemma and the two main hand proofs, found the missing step in Lemma~\ref{lem:y}(3) and supplied its repair, and wrote V4. A further Fable session made the literature check of Section~\ref{sec:novelty}. The author coordinated the sessions. No step rests on the judgement of an AI system: every claim is either proved in the text or is an exact finite computation that can be rerun. \section*{Data availability} \sloppy The Lean~4 formalization of Section~\ref{sec:parity} is at \url{https://github.com/tonykoval/srg-parity-lean}. The scripts, scan outputs and notes cited by path are deposited with this preprint on Zenodo in the data archive \texttt{milgram\_obstruction\_data.zip} (repository-relative paths as cited; the archive's README lists its contents and how to rerun the deciders). The table of Section~\ref{sec:table} is regenerated by \fn{out/milgram_paper/make_table.py} from the scan outputs (\fn{out/s4lat/allopen_open_*.tsv}, \fn{out/milgram_final/}) together with \fn{compile_kills.py}, \fn{rerun_rows.py}, \fn{v3_parity_rows.py} and \fn{parity_scan.py}; the controls are listed in Section~\ref{sec:controls}. \begin{thebibliography}{99} \bibitem{AzarijaMarc} J.~Azarija and T.~Marc, There is no $(75,32,10,16)$ strongly regular graph, \emph{Linear Algebra Appl.} \textbf{557} (2018), 62--83; arXiv:1509.05933. \bibitem{Akiyama} K.~Akiyama, Nonexistence of a strongly regular graph with parameters $(266,45,0,9)$: a certificate-free Lean proof, preprint, 2026, arXiv:2609.08319. \bibitem{BMV} E.~Bannai, A.~Munemasa and B.~Venkov, The nonexistence of certain tight spherical designs, \emph{Algebra i Analiz} \textbf{16}:4 (2004), 1--23; English transl.\ \emph{St. Petersburg Math. J.} \textbf{16} (2005), 609--625. \bibitem{BvE} A.~E. 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