\documentclass[11pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{amsmath,amssymb,amsthm} \usepackage{booktabs} \usepackage[margin=28mm]{geometry} \usepackage{xurl} \usepackage[hidelinks]{hyperref} \usepackage{microtype} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem*{conjHZ}{Conjecture 5.10 of Hou and Zhao} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} \newcommand{\F}{\mathbb{F}} \newcommand{\Tr}{\operatorname{Tr}} \hypersetup{pdftitle={The Welch function is not third-order sum-free},pdfauthor={Deep Bhattacharjee}} \title{The Welch function is not third-order sum-free} \author{Deep Bhattacharjee\thanks{Corresponding author. Email: \href{mailto:d.bhattacharjee@erl-forschung.de}{d.bhattacharjee@erl-forschung.de}, \href{mailto:itsdeep@live.com}{itsdeep@live.com}, \href{mailto:egspl@proton.me}{egspl@proton.me}. ORCID: \href{https://orcid.org/0000-0003-0466-750X}{0000-0003-0466-750X}.}\\[2pt] \normalsize Formerly, Electro-Gravitational Space Propulsion Laboratory (EGSPL),\\ \normalsize Bhubaneswar, Odisha 751030, India\\[1pt] \normalsize Formerly, University of Calcutta, Senate House, 87/1 College Street,\\ \normalsize Kolkata 700073, West Bengal, India} \date{26 October 2026} \begin{document} \maketitle \begin{abstract} \itshape Let $n=2m+1$ be odd and let $\F_{2^n}$ be the field with $2^n$ elements. A function on $\F_{2^n}$ is third-order sum-free when its values over any eight elements forming a three-dimensional affine subspace never add up to zero. The Welch function is the power map $x\mapsto x^{2^m+3}$; it is one of the classical almost perfect nonlinear functions, and Hou and Zhao conjectured that for $n\ge7$ it is not third-order sum-free. We prove this. The question comes down to a linearised polynomial $g(x,y)$ in $y$ whose roots form a vector space $Z(x)$ over $\F_2$ containing $1$ and $x$, and the Welch function fails to be third-order sum-free exactly when $Z(x)$ has dimension three for some $x$ outside $\F_2$. We show that the dimension is always $2$ or $3$, and that it is $3$ if and only if two explicit trace conditions hold: with $u=x^{2^m}$, both $\Tr(1/(x+u))$ and $\Tr(xu/(x+u))$ equal $1$. The proof applies the field automorphism $z\mapsto z^{2^{m+1}}$, whose square is the Frobenius map, to turn $g$ into an additive polynomial of degree eight whose roots always lie in $\F_{2^n}$. Counting the elements that satisfy both trace conditions gives exactly $(2^n+1-3K_n)/4$ of them, where $K_n$ is the binary Kloosterman sum of $\F_{2^n}$, and Weil's bound $|K_n|\le 2^{n/2+1}$ makes this count positive for every $n\ge7$. The Welch function sums to zero over exactly $(2^n+1-3K_n)(2^n-1)/168$ three-dimensional linear subspaces, and the count $2+(2^n+1-3K_n)/4$ agrees with the numbers that Hou and Zhao computed for $n\le15$. Every statement is proved by hand. \end{abstract} \medskip \noindent\textbf{Keywords.} sum-free function; Welch function; almost perfect nonlinear function; Kloosterman sum. \smallskip \noindent\textbf{MSC 2020.} 11T06; 11T23; 94A60; 06E30. \medskip \begin{quote} \small\sloppy \noindent\textbf{Note.} The results of this paper were first made public on 9 October 2026, as release \texttt{v1.0.0} of the repository \url{https://github.com/creelie/eConjectureWelchSumFree}; later versions (\texttt{v1.1.0} of 22 October and \texttt{v1.2.0} of 26 October) are archived on Zenodo as \href{https://doi.org/10.5281/zenodo.23553010}{10.5281/zenodo.23553010} and \href{https://doi.org/10.5281/zenodo.23662272}{10.5281/zenodo.23662272} (concept DOI \href{https://doi.org/10.5281/zenodo.23553009}{10.5281/zenodo.23553009}). An earlier and incomplete version of the paper appeared on Hexagon on 22 October 2026 as the preprint \texttt{hexagon:2610.00472}, available at \url{https://hexagonmath.org/2610.00472}. That version cited the conjecture and the corollary of Hou and Zhao as Conjecture~4.10 and Corollary~4.9, numbers that do not match their paper, listed the programs that check its statements in the text, and had a short abstract. This version is complete and self-contained: the references follow the arXiv paper of Hou and Zhao (Conjecture~5.10, Corollary~5.9, Section~5.3), every reference has been checked at its source, the abstract states what is proved, and every statement is proved by hand, so that no statement rests on a computer. The source of this revision is in release \texttt{v2.0.0} of the same repository, archived as \href{https://doi.org/10.5281/zenodo.23663217}{10.5281/zenodo.23663217}, with programs that recompute Table~\ref{tab} and check the criterion on small fields. \end{quote} \section{Introduction}\label{sec:intro} A function $f\colon\F_{2^n}\to\F_{2^n}$ is \emph{$k$th order sum-free} if $\sum_{a\in A}f(a)\ne0$ for every $k$-dimensional affine subspace $A$ of $\F_{2^n}$ over $\F_2$. Carlet introduced this notion~\cite{Carlet-JC,Carlet-DCC}; for $k=2$ it is the APN property. Throughout, $n=2m+1$ is odd, $q=2^n$, $F=\F_q$ and $\Tr\colon F\to\F_2$ is the absolute trace. The \emph{Welch function} is \[ W_n(X)=X^{2^m+3}. \] It is APN~\cite{Dobbertin}, and decimating a binary $m$-sequence by $2^m+3$ gives a three-valued cross-correlation, as Welch conjectured~\cite{CCD}. For $m\ge2$ it has algebraic degree $3$, so the first open question about its higher-order sum-freedom is the third order. For $m=2$ the Welch function is $X^7=X^{2^3-1}$, which is third-order sum-free by a result of Carlet~\cite[Proposition~1]{Carlet-JC}. Hou and Zhao~\cite{HZ} reduced the question to the rank of a Dickson matrix, computed the answer for $5\le n\le15$, and proposed the following statement~\cite[Conjecture~5.10]{HZ}. \begin{conjHZ} For $n=2m+1\ge7$, the Welch function on $\F_{2^n}$ is not third-order sum-free. \end{conjHZ} We prove this conjecture. The reduction of Hou and Zhao goes through the polynomial \begin{equation}\label{eq:g} g(X,Y)=Y^{2^m}(X^2+X)+Y^2(X^{2^m}+X)+Y(X^{2^m}+X^2). \end{equation} For $x\in F$ let $Z(x)=\{y\in F: g(x,y)=0\}$. Since $g(x,Y)$ is an additive polynomial in $Y$, $Z(x)$ is a vector space over $\F_2$, and it contains $1$ and $x$. By Lemma~\ref{lem:reduction} below, $W_n$ fails to be third-order sum-free exactly when $\dim_{\F_2}Z(x)\ge3$ for some $x\in F\setminus\F_2$. Our main result decides this dimension for every $x$. \begin{theorem}\label{thm:criterion} Let $n=2m+1\ge3$ and $x\in F\setminus\F_2$, and put $u=x^{2^m}$. Then $\dim_{\F_2}Z(x)\in\{2,3\}$, and $\dim_{\F_2}Z(x)=3$ if and only if \[ \Tr\Bigl(\frac{1}{x+u}\Bigr)=1\qquad\text{and}\qquad\Tr\Bigl(\frac{xu}{x+u}\Bigr)=1 . \] \end{theorem} Let \[ K_n=\sum_{z\in F^*}(-1)^{\Tr(z+z^{-1})} \] be the Kloosterman sum of $F$. The conditions of Theorem~\ref{thm:criterion} turn the count of the elements $x$ with $\dim Z(x)=3$ into Kloosterman sums. \begin{theorem}\label{thm:count} Let $n=2m+1\ge3$, and let $N_n$ be the number of $x\in F\setminus\F_2$ with $\dim_{\F_2}Z(x)=3$. Then \[ N_n=\frac{2^n+1-3K_n}{4}. \] \end{theorem} By Weil's bound~\cite{Weil} (see also~\cite[Chapter~5]{LN}), $|K_n|\le2\cdot2^{n/2}$. Hence \[ N_n\ge\frac{2^n+1-6\cdot2^{n/2}}{4}=\frac{(2^{n/2}-3)^2-8}{4}>0\qquad\text{for }n\ge7 . \] \begin{corollary}\label{cor:conj} The conjecture of Hou and Zhao is true. More precisely, for $n=2m+1\ge5$ the Welch function sums to zero over exactly \[ \frac{N_n\,(2^n-1)}{42} \] three-dimensional linear subspaces of $F$, and over $2^{n-3}$ times as many three-dimensional affine subspaces. This number is positive for every $n\ge7$ and is zero for $n=5$. \end{corollary} By a theorem of Carlitz~\cite{Carlitz} (see also~\cite[Chapter~5]{LN}), $K_n=-(\omega^n+\bar\omega^n)$ with $\omega=(-1+\sqrt{-7})/2$, because $K_1=1$. So $K_0=-2$, $K_1=1$ and $K_n=-K_{n-1}-2K_{n-2}$, which gives Table~\ref{tab}. Hou and Zhao~\cite[Corollary~5.9]{HZ} showed that $\dim Z(x)=3$ for some $x\in F\setminus\F_2$ if and only if the polynomial $\det D^*(X)$ built from a Dickson matrix has more than two roots in $F$; the number of its roots in $F$ is $2+N_n$. For $5\le n\le15$ these are the values Hou and Zhao computed~\cite[Table~1]{HZ}. \begin{table}[ht] \caption{The Kloosterman sum $K_n$, the count $N_n$ of Theorem~\ref{thm:count} and the number $2+N_n$ of roots of $\det D^*(X)$ in $\F_{2^n}$.}\label{tab} \centering \begin{tabular}{rrrr} \toprule $n$ & $K_n$ & $N_n$ & $2+N_n$\\ \midrule 5 & 11 & 0 & 2\\ 7 & $-13$ & 42 & 44\\ 9 & $-5$ & 132 & 134\\ 11 & 67 & 462 & 464\\ 13 & $-181$ & 2184 & 2186\\ 15 & 275 & 7986 & 7988\\ 17 & $-101$ & 32844 & 32846\\ 19 & $-797$ & 131670 & 131672\\ 21 & 2795 & 522192 & 522194\\ \bottomrule \end{tabular} \end{table} The proof of Theorem~\ref{thm:criterion} uses the automorphism $\rho(z)=z^{2^{m+1}}$ of $F$, whose square is the Frobenius map $z\mapsto z^2$ and whose inverse is $z\mapsto z^{2^m}$. Applying $\rho$ to the equation $g(x,y)=0$ and eliminating $y^{2^m}$ produces an additive polynomial of degree $8$ in $y$. Its roots form a three-dimensional space $V$ that always lies in $F$ and contains $Z(x)$, and a linear map built from $g(x,\cdot)$ moves $V$ into $Z(x)$. Whether $Z(x)=V$ is then decided by two elements of $\F_2$, and both turn out to be traces of explicit rational functions of $x$ and $u$, because an element of $\F_2$ equals its own trace when $n$ is odd. The counting of Section~\ref{sec:count} is a finite count attached to the action of the Galois group on $F$, of the same kind as the counts in the author's earlier work~\cite{Bha22a,Bha22b,BB26}. Section~\ref{sec:prelim} collects facts about the trace and the reduction of Hou and Zhao. Section~\ref{sec:fixed} studies a fixed $x$, Section~\ref{sec:criterion} proves Theorem~\ref{thm:criterion}, and Section~\ref{sec:count} proves Theorem~\ref{thm:count} and Corollary~\ref{cor:conj}. \section{Preliminaries}\label{sec:prelim} \begin{lemma}\label{lem:frob} Let $\rho(z)=z^{2^{m+1}}$ for $z\in F$. \begin{enumerate} \item[(a)] $\rho\bigl(z^{2^m}\bigr)=z=\rho(z)^{2^m}$ and $\rho(\rho(z))=z^2$ for every $z\in F$. \item[(b)] The elements of $F$ fixed by $z\mapsto z^{2^m}$, and those fixed by $\rho$, are $0$ and $1$. \end{enumerate} \end{lemma} \begin{proof} (a) holds because $z^{2^n}=z$, $2^m\cdot2^{m+1}=2^n$ and $2^{m+1}\cdot2^{m+1}=2\cdot2^n$. For (b), the fixed field of $z\mapsto z^{2^k}$ is $\F_{2^{\gcd(k,n)}}$, and $\gcd(m,2m+1)=\gcd(m+1,2m+1)=1$. \end{proof} Let $T_0=\{w\in F:\Tr(w)=0\}$, put $\wp(z)=z^2+z$, and for $z\in F$ let \[ T_m(z)=\sum_{i=0}^{m-1}z^{2^i}. \] \begin{lemma}\label{lem:trace} \begin{enumerate} \item[(a)] $\Tr\bigl(z^{2^k}\bigr)=\Tr(z)$ for all $z\in F$ and $k\ge0$; in particular $\Tr(\rho(z))=\Tr(z)$. \item[(b)] $\Tr(1)=1$, so $\Tr(\varepsilon)=\varepsilon$ for $\varepsilon\in\F_2$. \item[(c)] Each of the maps $z\mapsto\wp(z)$, $z\mapsto z^{2^m}+z$ and $z\mapsto\rho(z)+z$ is $\F_2$-linear with kernel $\F_2$ and image $T_0$. \item[(d)] $\wp(T_m(z))=T_m(\wp(z))=z^{2^m}+z$ for all $z\in F$. \end{enumerate} \end{lemma} \begin{proof} (a) is clear and (b) holds because $n$ is odd. In (c), the kernels are $\F_2$ by Lemma~\ref{lem:frob}(b), so each image has $q/2$ elements; each image lies in $T_0$ by (a), and $|T_0|=q/2$. For (d), both sides telescope: $\wp(T_m(z))=\sum_{i=1}^{m}z^{2^i}+\sum_{i=0}^{m-1}z^{2^i}=z^{2^m}+z$, and $T_m(z^2+z)$ is the same sum. \end{proof} The next lemma is the reduction of Hou and Zhao~\cite[Section~5.3]{HZ}; we include the short proof. \begin{lemma}\label{lem:reduction} \begin{enumerate} \item[(a)] For $x,y\in F$, \[ \sum_{a_1,a_2,a_3\in\F_2}W_n(a_1x+a_2y+a_3)=g(x,y). \] \item[(b)] Let $A=c+A_0$ be a three-dimensional affine subspace of $F$ with linear part $A_0$. Then $\sum_{a\in A}W_n(a)=\sum_{a\in A_0}W_n(a)$, and $\sum_{a\in A_0}W_n(a)=b^{2^m+3}\sum_{a\in b^{-1}A_0}W_n(a)$ for every $b\in A_0\setminus\{0\}$. \item[(c)] $W_n$ is not third-order sum-free if and only if $\dim_{\F_2}Z(x)\ge3$ for some $x\in F\setminus\F_2$. \end{enumerate} \end{lemma} \begin{proof} (a) Since $a_i^{2^k}=a_i$ for $a_i\in\F_2$, \[ W_n(a_1x+a_2y+a_3)=(a_1x^{2^m}+a_2y^{2^m}+a_3)(a_1x^2+a_2y^2+a_3)(a_1x+a_2y+a_3). \] Expanding, we get a sum of terms $a_ia_ja_k\,\alpha\beta\gamma$ with $\alpha\in\{x^{2^m},y^{2^m},1\}$, $\beta\in\{x^2,y^2,1\}$ and $\gamma\in\{x,y,1\}$. Summed over $(a_1,a_2,a_3)\in\F_2^3$, a monomial in the $a_i$ gives $0$ unless all three of $a_1,a_2,a_3$ occur in it, because $\sum_{a\in\F_2}1=0$. So only the six terms in which $i,j,k$ are distinct survive, each with sum $1$, and they add up to $g(x,y)$. (b) Write the elements of $A$ as $c+a$ with $a\in A_0$ and expand $W_n(c+a)=(c^{2^m}+a^{2^m})(c^2+a^2)(c+a)$. Apart from $a^{2^m}a^2a$, every one of the eight products is a constant times a product of at most two $\F_2$-linear functions of $a$. In coordinates $a=a_1e_1+a_2e_2+a_3e_3$ with respect to a basis of $A_0$, such a product is a combination of monomials in at most two of the $a_i$, and its sum over $A_0$ vanishes as in (a). The second statement holds because $W_n$ is a power function. (c) By (b), $W_n$ is not third-order sum-free if and only if it sums to zero over a three-dimensional linear subspace that contains $1$. Such a subspace is $\langle 1,x,y\rangle$ with $1,x,y$ linearly independent over $\F_2$, and by (a) the sum over it is $g(x,y)$. Since $Z(x)$ always contains $\langle1,x\rangle$, a suitable $y$ exists if and only if $\dim Z(x)\ge3$. \end{proof} \section{A fixed element \texorpdfstring{$x$}{x}}\label{sec:fixed} In this section and the next, $x\in F\setminus\F_2$ is fixed and \[ u=x^{2^m},\qquad t=x^2+x,\qquad d=x+u,\qquad v=u^2+u . \] By Lemma~\ref{lem:frob}, $\rho(x)=u^2$ and $\rho(u)=x$, so \begin{equation}\label{eq:rho} \rho(t)=v^2,\quad \rho(d)=d+v=x+u^2,\quad \rho(d+v)=d^2,\quad \rho(x^2+u)=u^4+x,\quad t^{2^m}=v . \end{equation} The elements $t$, $u^2+u=v$, $d$ and $d+v=x+\rho(x)$ are nonzero by Lemma~\ref{lem:frob}(b), and \begin{equation}\label{eq:t} t=d^2+d+v . \end{equation} Write $L(y)=g(x,y)$, so that \[ L(y)=t\,y^{2^m}+d\,y^2+(x^2+u)\,y , \] an $\F_2$-linear map of $F$ with kernel $Z(x)$. One checks directly that $L(1)=L(x)=0$. \begin{lemma}\label{lem:Tm} $L(y)=t\,T_m(\wp(y))+d\,\wp(y)$ for every $y\in F$. \end{lemma} \begin{proof} By Lemma~\ref{lem:trace}(d), $t\,T_m(\wp(y))+d\,\wp(y)=t(y^{2^m}+y)+d(y^2+y)=t\,y^{2^m}+d\,y^2+(t+d)\,y$, and $t+d=x^2+u$. \end{proof} Let \[ Y(y)=\frac{d\,y^2+(x^2+u)\,y}{t},\qquad P(y)=v^2\,y+(d+v)\,Y(y)^4+(u^4+x)\,Y(y)^2 , \] and define $\Phi\colon F\to F$ by $\Phi(y)=\rho\bigl(L(y)/t\bigr)$. Then $Y$, $P$ and $\Phi$ are $\F_2$-linear, $L(y)=t\bigl(y^{2^m}+Y(y)\bigr)$, and $\Phi(y)=0$ if and only if $L(y)=0$. \begin{lemma}\label{lem:key} $P(y)=\rho\bigl(L(\Phi(y))\bigr)$ for every $y\in F$. \end{lemma} \begin{proof} Let $\ell=L(y)/t$, so that $y^{2^m}=Y(y)+\ell$ and, by Lemma~\ref{lem:frob}(a), $\rho(y)=\bigl(y^{2^m}\bigr)^2=(Y(y)+\ell)^2$. Applying $\rho$ to $L(y)$ and using~\eqref{eq:rho}, \[ \rho(L(y))=v^2y+(d+v)\rho(y)^2+(u^4+x)\rho(y)=P(y)+(d+v)\ell^4+(u^4+x)\ell^2 . \] On the other hand $\rho(L(y))=\rho(t\ell)=v^2\rho(\ell)$. Hence $P(y)=v^2\rho(\ell)+(d+v)\ell^4+(u^4+x)\ell^2$. Now put $\ell'=\Phi(y)=\rho(\ell)$. By Lemma~\ref{lem:frob}(a), $\ell'^{\,2^m}=\ell$ and $\rho(\ell')=\ell^2$, so \[ \rho(L(\ell'))=\rho\bigl(t\ell+d\ell'^2+(x^2+u)\ell'\bigr)=v^2\ell'+(d+v)\ell^4+(u^4+x)\ell^2=P(y). \qedhere \] \end{proof} Let $\Pi(y)=y(y+1)(y+x)(y+x+1)=\wp(y)\bigl(\wp(y)+t\bigr)=y^4+(t+1)y^2+ty$. It is an additive polynomial whose roots are the elements of $\langle1,x\rangle$. Put \[ \lambda=\frac{(d+v)\,d^4}{t^4},\qquad c=\frac{t^3v^2}{d^4(d+v)},\qquad a=\frac{c}{t^2}=\frac{t\,v^2}{d^4(d+v)} . \] \begin{lemma}\label{lem:factor} As polynomials in $y$ over $F$, $P(y)=\lambda\,\Pi(y)\bigl(\Pi(y)+c\bigr)$. The roots of $P$ in an algebraic closure of $F$ form a three-dimensional $\F_2$-space $V$ that contains $\langle1,x\rangle$, and $V\setminus\langle1,x\rangle=\{y:\Pi(y)=c\}$. \end{lemma} \begin{proof} $P$ is an additive polynomial. Its coefficient of $y^8$ is $\lambda\ne0$ and its coefficient of $y$ is $v^2\ne0$, since $Y(y)^4$ and $Y(y)^2$ contain only $y^8$, $y^4$ and $y^2$. So $P$ has $8$ distinct roots, and they form a three-dimensional $\F_2$-space $V$. Since $Y(1)=1$ and $Y(x)=u$, \[ P(1)=v^2+(d+v)+(u^4+x)=0,\qquad P(x)=v^2x+(d+v)u^4+(u^4+x)u^2=0 , \] so $\langle1,x\rangle\subseteq V$. Choose $y_0\in V\setminus\langle1,x\rangle$. Then $V=\langle1,x\rangle\cup(y_0+\langle1,x\rangle)$ and, since $\Pi$ is additive, \[ P(y)=\lambda\,\Pi(y)\,\Pi(y+y_0)=\lambda\,\Pi(y)\bigl(\Pi(y)+\Pi(y_0)\bigr). \] The coefficient of $y$ on the right is $\lambda\,t\,\Pi(y_0)$, because $\Pi(y)^2$ has only even powers of $y$. Comparing with $v^2$ gives $\Pi(y_0)=v^2/(\lambda t)=c$. Finally $c\ne0$, so the roots of $\Pi(y)+c$ are exactly the four elements of $V$ outside $\langle1,x\rangle$. \end{proof} \begin{lemma}\label{lem:inF} $\Tr(a)=0$. If $z\in F$ satisfies $\wp(z)=a$, then $\Tr(tz)=0$. Consequently $V\subseteq F$. \end{lemma} \begin{proof} From~\eqref{eq:t} one checks the identity \begin{equation}\label{eq:a} a=\wp\Bigl(\frac{v}{d^2}\Bigr)+\frac1d+\frac1{d+v}; \end{equation} multiplied by $d^4(d+v)$ it reads $tv^2=v^2(d+v)+vd^2(d+v)+d^3(d+v)+d^4$, and the right side equals $v^2(d^2+d+v)$. By~\eqref{eq:rho}, $1/(d+v)=\rho(1/d)$, so $\Tr(1/(d+v))=\Tr(1/d)$ and $\Tr(a)=0$. By Lemma~\ref{lem:trace}(c) there are exactly two elements $z\in F$ with $\wp(z)=a$, and they differ by $1$; since $\Tr(t)=\Tr(\wp(x))=0$, the value $\Tr(tz)$ does not depend on the choice. Let $h=1/d$. Since $\Tr\bigl(h+\Tr(h)\bigr)=0$ by Lemma~\ref{lem:trace}(b), Lemma~\ref{lem:trace}(c) gives $z_2\in F$ with $\rho(z_2)+z_2=h+\Tr(h)$. By Lemma~\ref{lem:frob}(a), \[ \wp(z_2)=\rho(\rho(z_2))+z_2=\rho\bigl(\rho(z_2)+z_2\bigr)+\bigl(\rho(z_2)+z_2\bigr)=\rho(h)+h=\frac1{d+v}+\frac1d , \] so by~\eqref{eq:a} we may take $z=v/d^2+z_2$. First, by~\eqref{eq:t}, \[ \frac{tv}{d^2}=v+\frac vd+\frac{v^2}{d^2}=\wp(u)+\wp\Bigl(\frac vd\Bigr), \] which has trace $0$. Second, let $e=d+v=x+\rho(x)$. Then $\rho(e)=d^2$ by~\eqref{eq:rho}, and $\rho(e)+e=x^2+x=t$. Using Lemma~\ref{lem:trace}(a) and Lemma~\ref{lem:frob}(a) to move $z\mapsto z^{2^m}$ and $\rho$ through the trace, \begin{align*} \Tr(tz_2)&=\Tr(\rho(e)z_2)+\Tr(ez_2)=\Tr\bigl(e\,z_2^{2^m}\bigr)+\Tr(ez_2)=\Tr\bigl(e\,(\rho(z_2)+z_2)^{2^m}\bigr)\\ &=\Tr\bigl(\rho(e)\,(\rho(z_2)+z_2)\bigr)=\Tr\bigl(d^2(h+\Tr(h))\bigr)=\Tr(d)+\Tr(h)\Tr(d^2)=0, \end{align*} since $\Tr(d)=\Tr(x+x^{2^m})=0$. Hence $\Tr(tz)=0$. Now Lemma~\ref{lem:trace}(c) gives $y\in F$ with $\wp(y)=tz$. Then $\Pi(y)=tz(tz+t)=t^2\wp(z)=c$, so $y\in V\setminus\langle1,x\rangle$ by Lemma~\ref{lem:factor}, and $V=\langle1,x,y\rangle\subseteq F$. \end{proof} \begin{lemma}\label{lem:delta} $V=\{y\in F: L(\Phi(y))=0\}$, $Z(x)\subseteq V$ and $\Phi(V)\subseteq Z(x)$. The element $\delta(x)=\Phi(y_0)$ is the same for all $y_0\in V\setminus\langle1,x\rangle$ and lies in $\langle1,x\rangle$. If $\delta(x)=0$ then $Z(x)=V$; otherwise $Z(x)=\langle1,x\rangle$. \end{lemma} \begin{proof} Since $V\subseteq F$ (Lemma~\ref{lem:inF}) and $\rho$ is injective, Lemma~\ref{lem:key} shows that $V=\{y\in F: L(\Phi(y))=0\}$. If $L(y)=0$ then $\Phi(y)=0$ and $y\in V$, so $Z(x)\subseteq V$; and $\Phi(V)\subseteq Z(x)$ by the same description of $V$. Two choices of $y_0$ differ by an element of $\langle1,x\rangle\subseteq Z(x)$, on which $\Phi$ vanishes. As $\langle1,x\rangle\subseteq Z(x)\subseteq V$ and $\dim V=3$, either $Z(x)=V$ or $Z(x)=\langle1,x\rangle$. In the first case $\delta(x)=0$. In the second case $\delta(x)\in\Phi(V)\subseteq\langle1,x\rangle$, and $\delta(x)\ne0$ because $y_0\notin Z(x)$. \end{proof} \begin{lemma}\label{lem:first} $\delta(x)\in\{0,1\}$ if and only if $\Tr(1/d)=1$. \end{lemma} \begin{proof} Take $y_0\in V\setminus\langle1,x\rangle$ and let $z=\wp(y_0)/t$. By Lemma~\ref{lem:factor}, $t^2\wp(z)=\wp(y_0)(\wp(y_0)+t)=c$, so $\wp(z)=a$. Let $\ell_0=L(y_0)/t$, so that $\delta(x)=\rho(\ell_0)$, and $\delta(x)\in\{0,1\}$ if and only if $\ell_0\in\{0,1\}$, that is, $\wp(\ell_0)=0$. By Lemma~\ref{lem:Tm}, $\ell_0=T_m(tz)+dz$. By Lemma~\ref{lem:trace}(d), \eqref{eq:rho} and~\eqref{eq:t}, \[ \wp(\ell_0)=(tz)^{2^m}+tz+d^2z^2+dz=v\,z^{2^m}+(t+d^2+d)\,z+d^2a=v\bigl(z^{2^m}+z\bigr)+d^2a=v\,T_m(a)+d^2a . \] Put $\gamma=d^2a/v$, so that $\wp(\ell_0)=v\,(T_m(a)+\gamma)$. On the other hand $\delta(x)\in\langle1,x\rangle$ by Lemma~\ref{lem:delta}, so $\ell_0=\delta(x)^{2^m}\in\{0,1,u,u+1\}$ and $\wp(\ell_0)\in\{0,v\}$. Therefore $\varepsilon=T_m(a)+\gamma$ lies in $\F_2$, and $\varepsilon=0$ exactly when $\delta(x)\in\{0,1\}$. By Lemma~\ref{lem:trace}(b) and~(a), \[ \varepsilon=\Tr(\varepsilon)=\sum_{i=0}^{m-1}\Tr\bigl(a^{2^i}\bigr)+\Tr(\gamma)=m\Tr(a)+\Tr(\gamma)=\Tr(\gamma), \] using $\Tr(a)=0$ from Lemma~\ref{lem:inF}. By~\eqref{eq:t}, $\gamma=tv/(d^2(d+v))=v/d^2+v/(d+v)$. By~\eqref{eq:rho}, $\rho(t/d^2)=v^2/(d+v)^2$, so $\Tr(v/(d+v))=\Tr(t/d^2)$. Since $v+t=d^2+d$, \[ \Tr(\gamma)=\Tr\Bigl(\frac{v+t}{d^2}\Bigr)=\Tr(1)+\Tr\Bigl(\frac1d\Bigr)=1+\Tr\Bigl(\frac1d\Bigr). \] So $\delta(x)\in\{0,1\}$ if and only if $\Tr(1/d)=1$. \end{proof} \section{Proof of Theorem~\ref{thm:criterion}}\label{sec:criterion} The objects of Section~\ref{sec:fixed} depend on $x$; we write $L_x$, $t_x$, $\Phi_x$, $V_x$ and $d_x=x+x^{2^m}$ when the dependence matters. The second condition of Theorem~\ref{thm:criterion} comes from the substitution $x\mapsto 1/x$. \begin{lemma}\label{lem:inverse} For $x\in F\setminus\F_2$ and $y\in F$, $g(1/x,y/x)=x^{-(2^m+3)}g(x,y)$. Moreover $\Phi_{1/x}(y)=x^{-1}\Phi_x(xy)$, $V_{1/x}=x^{-1}V_x$ and $\delta(1/x)=x^{-1}\delta(x)$. \end{lemma} \begin{proof} With $u=x^{2^m}$, \[ u\,x^3\,g(1/x,y/x)=u\,x^3\Bigl(\frac{y^{2^m}}{u}\cdot\frac{1+x}{x^2}+\frac{y^2}{x^2}\cdot\frac{x+u}{xu}+\frac yx\cdot\frac{u+x^2}{x^2u}\Bigr)=g(x,y). \] So $L_{1/x}(y)=x^{-(2^m+3)}L_x(xy)$. Since $t_{1/x}=x^{-2}+x^{-1}=t_x/x^3$, \[ \Phi_{1/x}(y)=\rho\Bigl(\frac{L_x(xy)}{u\,t_x}\Bigr)=\rho(u^{-1})\,\Phi_x(xy)=x^{-1}\Phi_x(xy). \] By the first statement of Lemma~\ref{lem:delta}, applied to $x$ and to $1/x$, $V_{1/x}=\{y\in F:L_{1/x}(\Phi_{1/x}(y))=0\}=\{y\in F:L_x(\Phi_x(xy))=0\}=x^{-1}V_x$. If $y_0\in V_x\setminus\langle1,x\rangle$, then $y_0/x\in V_{1/x}\setminus\langle1,1/x\rangle$, and $\delta(1/x)=\Phi_{1/x}(y_0/x)=x^{-1}\Phi_x(y_0)=x^{-1}\delta(x)$. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:criterion}] By Lemma~\ref{lem:delta}, $\dim Z(x)\in\{2,3\}$, and $\dim Z(x)=3$ if and only if $\delta(x)=0$. Since $\delta(x)\in\langle1,x\rangle$ and $x\ne1$, $\delta(x)=0$ if and only if $\delta(x)\in\{0,1\}$ and $\delta(x)\in\{0,x\}$. By Lemma~\ref{lem:first}, the first condition is $\Tr(1/d_x)=1$. By Lemma~\ref{lem:inverse}, $\delta(x)\in\{0,x\}$ if and only if $\delta(1/x)\in\{0,1\}$, which by Lemma~\ref{lem:first} applied to $1/x$ means $\Tr(1/d_{1/x})=1$. Finally $d_{1/x}=1/x+1/u=(x+u)/(xu)$. \end{proof} \section{Counting}\label{sec:count} \begin{proof}[Proof of Theorem~\ref{thm:count}] For $x\in F\setminus\F_2$ let $d=x+x^{2^m}$, $u=x^{2^m}$, $\alpha(x)=(-1)^{\Tr(1/d)}$ and $\beta(x)=(-1)^{\Tr(xu/d)}$. By Theorem~\ref{thm:criterion}, \[ N_n=\frac14\sum_{x\in F\setminus\F_2}\bigl(1-\alpha(x)\bigr)\bigl(1-\beta(x)\bigr)=\frac14\Bigl(q-2-\sum\alpha-\sum\beta+\sum\alpha\beta\Bigr), \] all sums over $x\in F\setminus\F_2$. By Lemma~\ref{lem:trace}(c), $x\mapsto x+x^{2^m}$ maps $F\setminus\F_2$ two-to-one onto $T_0\setminus\{0\}$. Hence \[ \sum\alpha=2\sum_{w\in T_0\setminus\{0\}}(-1)^{\Tr(1/w)}=\sum_{w\in F^*}\bigl(1+(-1)^{\Tr(w)}\bigr)(-1)^{\Tr(1/w)}=-1+K_n , \] because $w\mapsto1/w$ permutes $F^*$ and $\sum_{w\in F^*}(-1)^{\Tr(w)}=-1$. The map $x\mapsto1/x$ permutes $F\setminus\F_2$, and $\beta(x)=\alpha(1/x)$ because $d_{1/x}=d/(xu)$; so $\sum\beta=K_n-1$. Finally, $x\mapsto x+1$ permutes $F\setminus\F_2$, leaves $d$ unchanged and sends $xu$ to $(x+1)(u+1)=xu+d+1$; since $\Tr(1)=1$, \[ \beta(x+1)=(-1)^{\Tr((xu+1)/d)+1}=-\alpha(x)\beta(x). \] So $\sum\alpha\beta=-\sum\beta=1-K_n$, and \[ N_n=\frac14\bigl(q-2-2(K_n-1)+1-K_n\bigr)=\frac{q+1-3K_n}{4}. \qedhere \] \end{proof} \begin{proof}[Proof of Corollary~\ref{cor:conj}] If $\dim Z(x)=3$, then $A=Z(x)$ is a three-dimensional subspace containing $1$ over which $W_n$ sums to zero, by Lemma~\ref{lem:reduction}(a). Conversely, let $A\ni1$ be such a subspace and $x'\in A\setminus\F_2$. Writing $A=\langle1,x',y'\rangle$, Lemma~\ref{lem:reduction}(a) gives $g(x',y')=0$, so $A\subseteq Z(x')$, and $Z(x')=A$ by Theorem~\ref{thm:criterion}. Hence the subspaces containing $1$ with zero sum correspond to the elements $x$ with $\dim Z(x)=3$, six elements to each subspace, and there are $N_n/6$ of them. Let $\mathcal S$ be the set of three-dimensional linear subspaces with zero sum. By Lemma~\ref{lem:reduction}(b), $A\in\mathcal S$ and $b\in F^*$ imply $bA\in\mathcal S$. The pairs $(A,b)$ with $A\in\mathcal S$ and $b\in A\setminus\{0\}$ number $7|\mathcal S|$, and $(A,b)\mapsto(b^{-1}A,b)$ maps them bijectively to the pairs $(B,b)$ with $B\in\mathcal S$, $1\in B$ and $b\in F^*$, which number $(q-1)N_n/6$. So $|\mathcal S|=N_n(q-1)/42$. By Lemma~\ref{lem:reduction}(b), an affine subspace has zero sum exactly when its linear part does, and each linear part has $2^{n-3}$ cosets. The numbers are positive for $n\ge7$ by the bound after Theorem~\ref{thm:count}, and $N_5=0$ because $K_5=11$. \end{proof} \begin{remark} For $n=5$, Corollary~\ref{cor:conj} gives a second proof that $W_5=X^7$ is third-order sum-free. For $n=3$ the polynomial $g(x,Y)$ vanishes identically, and indeed $N_3=6=2^3-2$ since $K_3=-5$. \end{remark} \section*{Declarations} \subsection*{Funding} No funding was received for this work. \subsection*{Conflict of interest} The author declares no conflict of interest. \subsection*{Data availability} The proofs in this paper are complete by hand. The source of the paper and programs that recompute Table~\ref{tab} and check Theorems~\ref{thm:criterion} and~\ref{thm:count} on small fields are in release \texttt{v2.0.0} of the repository named in the note under the abstract, archived as \href{https://doi.org/10.5281/zenodo.23663217}{10.5281/zenodo.23663217}; the paper does not depend on them. \begin{thebibliography}{9} \bibitem{Bha22a} D.~Bhattacharjee, Establishing equivariant class $[O]$ for hyperbolic groups, Asian Res. J. 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