text:
\begin{document}
	\begin{abstract}
		In studying rational points on elliptic K3 surfaces of the form
		$$f(t)y^2=g(x),$$
		where $f,g$ are cubic or quartic polynomials (without repeated roots), we introduce a condition on the quadratic twists of two elliptic curves having simultaneously positive Mordell-Weil rank, and we relate it to the Hilbert property. Applying to surfaces of Cassels-Schinzel type, we prove unconditionally that rational points are dense both in Zariski topology and in real topology.
	\end{abstract}
	\maketitle
	\tableofcontents

	\section{Introduction}
	\subsection{Topology of rational points on K3 surfaces -- an overview}
	Throughout this article, by a K3 surface, we mean a projective, smooth, geometrically integral surface $\CE$ defined over a field $k$ having trivial canonical class and $H^1(\CE,\mathcal{O}_\CE)=0$. Moreover it is called \emph{elliptic} if it admits over $k$ a fibration $\pi:\CE\to B$ into a curve $B$ whose generic fibre is a smooth genus one curve.

	Results on producing rational points over the ground field on K3 surfaces defined over number fields are rare in the current literature, although the result of Bogomolov-Tschinkel \cite{B-T} states that rational points become Zariski dense on elliptic K3 surfaces after a finite extension of the base field. Various conjectures have been made, which we shall discuss briefly.

	A conjecture of Mazur states the following:
	\begin{conjecture}[\cite{Mazur}, Conjecture 1]\label{conj:Mazur}
		Let $X$ be a smooth variety over $\BQ$. Suppose that $X(\BQ)$ is Zariski dense. Then the topological closure of $X(\BQ)$ in $X(\BR)$ is open (that is, a finite union of connected components of $X(\BR)$).
	\end{conjecture}
		In the context of elliptic K3 surfaces $\CE\to\BP^1_{\BQ}$, Conjecture \ref{conj:Mazur} implies (c.f. \cite[p. 40]{Mazur}):
	\begin{conjecture}[\cite{Mazur}, Conjecture 4]
		For $t\in \BP^1_\BQ$, let $\CE_t$ be the fibre $\pi^{-1}(t)$. Then the family $\{\CE_t\}_{t\in \BP^1_\BQ}$ verifies one of the following exclusive conditions.
		\begin{enumerate}
			\item The elliptic curves $\CE_t$ has Mordell-Weil rank equal to $0$ for all but a finite number of elements $t\in\BP^1(\BQ)$;
			\item The set $\{t\in\BP^1(\BQ):\rank(\CE_t(\BQ))>0\}$ is dense in $\BP^1(\BR)$.
		\end{enumerate}
	\end{conjecture}

	 A finer notion than the Zariski density is the so-called \emph{Hilbert property} (abbreviated as (HP) in the sequel) (c.f. \cite[Definition 3.1.2]{Serre}).
	 Recall that for $X$ a variety over a number field $k$, the set $X(k)$ is \emph{thin} if there exist a finite number of dominant rational maps $\pi_i:Y_i\dashrightarrow X,1\leqslant i\leqslant r$ over $k$ such that $X(k)\setminus \cup_{i=1}^r \pi_i(Y_i(k))$ is contained in a proper Zariski closed subset of $X$ (c.f. \cite[\S3.1]{Serre}, \cite[\S 1.1]{Corvaja-Zannier}). We say that $X$ verifies (HP) over $k$ if $X(k)$ is not thin.
	The following conjecture was raised by Corvaja and Zannier:
	\begin{conjecture}[\cite{Corvaja-Zannier}, Question-Conjecture 1]\label{conj:CZ}
		If $X$ is simply connected and $X(k)$ is Zariski dense, then $X$ verifies (HP) over the number field $k$.
	\end{conjecture}

	All the aforementioned conjectures for K3 surfaces are consequences of a stronger one due to Skorobogatov, predicting that they satisfy weak approximation with Brauer-Manin obstruction (c.f. \cite[p. 817]{H-Sk}).

	\subsection{Summary of results and layout of the paper}

	The object of study in this paper is a family of K3 surfaces of Cassels-Schinzel type \cite{C-S} arising as the isotrivial elliptic pencil over $\BQ$ (parametrized by the coordinate $T$)
\begin{equation}\label{eq:CSfamily}
	f(T)Y^2=g(X),
\end{equation}
	where $f,g$ are certain separable polynomials of degree respectively $4$ and $3$, and $f=0$ has no root over $\BQ$.
	Such elliptic surfaces admit a section over $\BQ$ but the Mordell-Weil rank of the group formed by all sections over $\BQ$ (sometimes called ``generic rank'') is zero. As they can have constant or varying root numbers, the Zariski density of rational points on such surfaces was previously conditional on the parity conjecture. One main result (Theorem \ref{th:mainthm1}) of this paper is to prove, in Section \ref{se:CSsurfaces}, unconditionally on standard conjectures on elliptic curves, Mazur's conjecture \ref{conj:Mazur} for them, thereby also confirming the Zariski density of rational points.
	Our result also fits into the empirical fact that if there exist extra rational points outside the evident smooth rational curves on a K3 surface, they tend to be Zariski dense (c.f. \cite{vanLuijk}). See also a recent result of Gvirtz \cite[\S4.1]{Gvirtz} for another family of the form \eqref{eq:CSfamily}.

	In general over a number field $k$, when $\deg f,\deg g\in \{3,4\}$, the surfaces \eqref{eq:CSfamily} are (twisted) Kummer surfaces associated to products of two (twisted) elliptic curves (c.f. Section \ref{se:Kummer}). In Section \ref{se:ICproperty}, in the direction of Conjecture \ref{conj:CZ}, we introduce a condition, denoted by ``\eqref{eq:bothrankpos}'' (with parameter $[C]\in\kmodksq$) in what follows, which arises in a natural way from our approach of parametrizing rational points on \eqref{eq:CSfamily} (c.f. Theorem \ref{thm:Z2torsors}) similar to \cite{CT-Sk-SD}, except that here we are taking unramified double covers over an open (non-proper) variety. Our second main result (Theorem \ref{thm:HP}) says that this condition with a single $[C]$ turns out to be equivalent to the Zariski density, and that with infinitely many different $[C]$'s is necessary for (HP). It also has consequence on jump of Mordell-Weil rank in elliptic fibrations (c.f. \S\ref{se:jumpmordellweil}), as inspired in recent work \cite{Loughran-Salgado,C-T}. Concerning the Cassels-Schinzel type surfaces, we point out that the validity of (HP), and hence that of ``\eqref{eq:bothrankpos}'' with infinitely many $[C]$, for the original surface constructed in \cite{C-S} has striking consequence on the \emph{congruent numbers} (c.f. \S\ref{se:a=1}). Based on current unconditional results towards the detection of congruent numbers, we establish ``\eqref{eq:bothrankpos}'' with infinitely many $[C]$ for a family of Cassels-Schinzel type surfaces (with slightly different coefficients in $f$, c.f. \S\ref{se:a=2}), therefore providing evidence in favour of (HP).

	It is worth mentioning that, when $(\deg f,\deg g)=(3,3)$ (so that the surface \eqref{eq:CSfamily} is untwisted) and under some extra assumption on the $j$-invariants of two elliptic curves defined by $f$ and $g$, the Zariski density is known unconditionally by the work of Kuwata-Wang \cite{K-W}, where the third elliptic fibration (in coordinate $Y$) comes into play in their method. More recently, Demeio \cite{Demeio} surprisingly shows that, the Zariski density is equivalent to (HP), whence one deduces the equivalence between \eqref{eq:bothrankpos} (with a single $[C]$ or infinitely many) and (HP) in this case. There again, the use of the elliptic fibration in $Y$ is essential. The extension of their results to cases where $\deg f$ or $\deg g$ is $4$ \emph{a priori} presents obstacle. It does not seem evident to find elliptic fibrations defined over the ground field other than the two existing ones, namely in $X$ and in $T$. We give a short discussion in Section \ref{se:remarks}.

\subsection*{Acknowledgement.} We thank Marc Hindry for drawing our attention to the Cassels-Schinzel surfaces as well as enlightening guidance. We are grateful to Mike Bennett, Julian Demeio, Julie Desjardins, Damián Gvirtz, Emmanuel Lecouturier, Marusia Rebolledo, Alexei Skorobogatov for fruitful discussions, and to Yang Cao for generous help.

We gratefully thank the hospitality of Max-Planck-Institut für Mathematik in Bonn and the organizers of the CMO-BIRS conference 18w5012 in Oaxaca, where part of this work was done and reported.

While working on this project the author was partially supported by the project ANR Gardio and a Riemann Fellowship.

\section{Quadratic twist type K3 surfaces}\label{se:Kummer}
Twisted Kummer surfaces \eqref{eq:CSfamily} are examples of Kummer varieties (in the sense of \cite{H-Sk}). We start by recalling some basic facts on Kummer varieties in \S\ref{se:Kummervar}. For more details, see \cite[\S6]{H-Sk}. In \S\ref{se:affinemodels}, we turn to giving explicit equations defining $\Kum(Y)$ when $Y$ is the product of two (twisted) elliptic curves. In \S\ref{se:parametrization} we give explicit parametrization of rational points on $\Kum(Y)$ by quadratic twists of $Y$. Throughout this section, we fix $k$ a field of characteristic $0$ and $\bar{k}$ an algebraic closure.
\subsection{Preliminaries on Kummer varieties}\label{se:Kummervar}

Let $A$ be an abelian variety over $k$. Let $\mathcal{T}$ be a $k$-torsor under $A[2]$, and $Y$ be a $k$-torsor under $A$ whose class is the image of $\mathcal{T} \in H^1(k,A[2])$ in $H^1(k,A)$. Then $Y$ is equipped with a \emph{$2$-covering map} $\Phi:Y\to A$  (c.f. \cite[\S3.4]{Skorobogatov}). Indeed, we have $Y\simeq (A\times \mathcal{T})/A[2]$, where $A[2]$ acts diagonally on $A\times \mathcal{T}$ and $\Phi$ is induced by the first projection to $A$. The antipodal involution $\iota_A:A\to A$ (i.e. the multiplication by $[-1]$ morphism) induces an involution $\iota_Y:Y\to Y$.
Let $$\pi_Y:Y\to \wKum(Y):=Y/\langle\iota_Y\rangle$$ be the quotient map. Note that the singular locus of the variety $\wKum(Y)$ comprises of precisely the fixed points of $\iota_Y$. The \emph{Kummer variety} $\Kum(Y)$ associated to $Y$ is the minimal desingulization of $\wKum(Y)$.

Let $C\in k^*$. We can view it as the $k$-torsor
\begin{equation}\label{eq:Ctor}
\mathcal{R}_C=\Spec (k[T]/(CT^2-1))
\end{equation}
under $\BZ/2$. By quadratic twist of $A$ (resp. $Y$) by $C$, denoted by $A^C$ (resp. $Y^C$), we mean the diagonal quotient of $A\times_k \mathcal{R}_C$ (resp. $Y\times_k \mathcal{R}_C$) by $\BZ/2$, where $\BZ/2$ acts on $A$ (resp. $Y$) as $\iota_A$ (resp. $\iota_Y$), equipped with the induced antipodal involution $\iota_{A^C}$ (resp. $\iota_{Y^C}$), so that we can define the quotient $\pi_{Y^C}:Y^C\to \widetilde{\operatorname{Kum}}(Y^C)=Y^C/\langle\iota_{Y^C}\rangle$ and the Kummer variety $\Kum(Y^C)$ associated to $Y^C$.  Since $A^C[2]\simeq_k A[2]$ canonically, we identify $Y^C$ as the torsor under $A^C$ defined by $[\mathcal{T}]\in H^1(k,A[2])\simeq H^1(k,A^C[2])$, each equipped with a $2$-covering map $\Phi_C:Y^C\to A^C$. This also allows to identify $\Kum(Y^C)$ with $\Kum(Y)$.
Observe that $\Phi_C$ commutes with the involutions on $A^C$ and $Y^C$: $$\Phi_C\circ\iota_{Y^C}=\iota_{A^C}\circ \Phi_C.$$
So $\Phi_C$ induces a covering map $\widetilde{\Phi}:\wKum(Y)\to \wKum(A)$ between (singular) Kummer varieties.

\subsection{Affine models of Kummer surfaces}\label{se:affinemodels}
The aim of this section is to provide affine models for Kummer surfaces and equations for the quotient maps.
\subsubsection{Hyperelliptic quartics}\label{se:Weierstrass}
Recall that a (smooth) hyperelliptic quartic has a smooth model as the intersection of two quadrics in $\BP^3$, whose affine plane model can be written as

\begin{equation}\label{eq:hyperelliptic}
\mathcal{H}:Y^2=G(X)=aX^4+cX^2+dX+e,\quad a,c,d,e\in k,a\neq 0.
\end{equation}
Its geometric genus is $1$.

It is classically known that the curve \eqref{eq:hyperelliptic} has the following $(\bar{k}/k)$-Weierstrass form (c.f. \cite[Proposition 3.3.6]{Skorobogatov})
\begin{equation}\label{eq:Weierstrassmod}
\mathbb{E}:u^2=v^3-27Iv-27J,
\end{equation}
where
$$I=12ae+c^2,\quad J=72ace-27ad^2-2c^3,$$
are the cubic invariants of \eqref{eq:hyperelliptic}, and $\CH$ is a $k$-torsor under $\mathbb{E}$, whose class is the image of that of the zero-dimensional scheme $(G(X)=0)$, as a torsor under the finite group scheme $\mathbb{E}[2]$. In particular, if $\mathcal{H}(k)\neq \varnothing$, we have $\mathcal{H}\simeq_k \mathbb{E}$ as elliptic curves by choosing some point as the origin.
If $G(X)=0$ has a root over $k$, by choosing this root as the origin, the isomorphism $\CH\simeq_k \mathbb{E}$ commutes with the involutions on both sides and identifies $(G(X)=0)$ with $\mathbb{E}[2]$.

For any $C\in k^*$, the quadratic twist $\mathcal{H}^C$ is defined by the affine equation $CY^2=G(X)$. As before, $[\CH^C]\in H^1(k,\mathbb{E}^C)$ is induced by the $k$-torsor $(G(X)=0)$ under $\mathbb{E}^C[2]\simeq \mathbb{E}[2]$. In this way $\mathcal{H}^C$ can be equipped with a $2$-covering map $\Phi_C:\mathcal{H}^C\to \mathbb{E}^C$ over $k$.

\subsubsection{Kummer surfaces associated to products of two hyperelliptic curves}
We fix two separable polynomials $f,g$. We assume that $\deg f,\deg g\in \{3,4\}$.
Define two hyper-elliptic curves

\begin{equation}\label{eq:twohyperellip}
\begin{split}
&E_1:y^2=g(x)\subset \mathbf{A}^2_{x,y};\\
&E_2:s^2=f(t)\subset \mathbf{A}^2_{s,t}.
\end{split}
\end{equation}
The quadratic twists of $E_1$ and $E_2$ by $C\in k^*$ are defined by
$$E_1^C:Cy^2=g(x),\quad E_2^C:Cs^2=f(t).$$
Their respective hyperelliptic involutions are
\begin{equation}\label{eq:ellipinvo}
\varrho_1:(x,y)\mapsto (x,-y),\quad \varrho_2:(t,s)\mapsto (t,-s).
\end{equation}

We now choose the affine model of $\Kum(E_1\times E_2)$ and the map $\pi_{E_1^C\times E_2^C}$ as follows. Let $\CS\subset \mathbf{A}^3_{X,Y,T}$ be defined by the equation \eqref{eq:CSfamily}.

Then outside the fixed point locus ($y=0$) of $\varrho_1$ and ($s=0$) of $\varrho_2$, the map
\begin{align}
\phi_C:E_1^C\times_k E_2^C&\dashrightarrow \mathcal{S},\label{eq:doublephi}\\
(x,y)\times(t,s)&\longmapsto (X,Y,T)=(x,\frac{y}{s},t)
\end{align}
is a (generically) double covering from the abelian surface $E_1\times E_2$ to $\mathcal{S}$.

When any one of the polynomials, say $f$, has degree $4$, we can suppose that $f=0$ has no root over $k$. Otherwise by the discussion in \S\ref{se:Weierstrass}, we have $\Kum(E_1\times E_2)\simeq_k \Kum(E_1\times E_2^\prime)$, where the elliptic curve $E_2^\prime$ is defined by the cubic resolvent \eqref{eq:Weierstrassmod} of $f$ (which is of degree $3$).

\subsection{Parametrization of rational points}\label{se:parametrization}
With the affine model above, we prove the following folklore result, saying that rational points on $\Kum(E_1\times E_2)$ are parametrized by those in a family of twisted abelian varieties.

The surface $\Kum(E_1\times E_2)$ has two genus one fibrations $\psi_X,\psi_T$ to $\BP^1$, that is, projections to $X$ and $T$ coordinates. They are respectively induced by $E_1\times E_2\to E_1 ,E_1\times E_2\to E_2$. Let $F_1$ (resp. $F_2$) be the image of $E_1[2]$ (resp. $E_2[2]$) in $\BP^1$. Then $\psi_X^{-1}(F_1)$ (resp. $\psi_T^{-1}(F_2)$) are four disjoint rational curves which are (geometrically) sections of $\psi_X$ (resp. $\psi_T$).
Let

 $$V=\Kum(E_1\times E_2)\setminus (\psi_X^{-1}(F_1) \cup\psi_T^{-1}(F_2)).$$ We identify the $V$ with the open set $\CS\setminus(Y=0,\infty)$ of the affine model $\CS$ \eqref{eq:CSfamily} in the previous section.
For each $[C]\in k^*/k^{*2}$, consider the open subset of $E_1^C\times E_2^C$:

\begin{equation}\label{eq:Uc}
U_C=(E_1^C\times E_2^C)\setminus \left((E_1^C[2]\times E_2^C)\cup (E_1^C\times E_2^C[2])\right).
\end{equation}
\begin{theorem}\label{thm:Z2torsors}
We have:
	\begin{enumerate}[label=(\roman*)]
		\item  $(U_C)_{[C]\in\kmodksq}$ is a family of $\BZ/2$-torsors over $V$;
		\item  	$$V(k)=\bigsqcup_{[C]\in\kmodksq} \phi_C(U_C(k)).$$
	\end{enumerate}

\end{theorem}
\begin{proof}
	The restriction of $\phi_C$ to $U_C\to V$ is clearly flat and étale.
	Moreover, the divisor $\psi_T^{-1}(F_2)$ is a \emph{double fibre} on $V$ (with respect to the fibration $\psi_T$) (c.f. \cite[\S2 p. 117]{CT-Sk-SD}). The function defining $\psi_T^{-1}(F_2)$ (an affine equation is given by $f$) is invertible on $V$. Applying \cite[Proposition 1.1 \& Theorem 2.1]{CT-Sk-SD} shows (i). The decomposition of rational points in (ii) now follows from the evaluation map
\begin{equation*}
	V(k)\longrightarrow \hetale(k,\BZ/2)=k^*/k^{*2}.
\end{equation*}
Indeed, for any $[C]\in\kmodksq$, the Kummer exact sequence twisted by $[C]$ gives
$$\xymatrix{1\ar[r]&\mathcal{R}_C\ar[r] &\mathbb{G}_{\operatorname{m}}\ar[r]^{\psi_C} &\mathbb{G}_{\operatorname{m}}\ar[r]& 1},$$
where $\mathcal{R}_C$ is defined by \eqref{eq:Ctor} and $\psi_C:\mathbb{G}_{\operatorname{m}}\to \mathbb{G}_{\operatorname{m}}$ is $t\mapsto Ct^2$.
The $\BZ/2$-torsor with class $[U_1]-[C]\in\hetale(X,\BZ/2)$ is isomorphic to the affine model $V\times_{\mathbb{G}_{\operatorname{m}},\psi_C} \mathbb{G}_{\operatorname{m}}$ where $V\to \mathbb{G}_{\operatorname{m}}$ given by $f$. It is then defined by (an open subset of) $$f(t)=Cs^2,\quad f(t)w^2=g(x).$$
And it is clearly isomorphic to $U_C$, under the change of variable $y=sw$.
\end{proof}
\begin{remark}\label{rmk:fibres}
	If we view the equation \eqref{eq:CSfamily} as a family of quadratic twists of $E_1$ by values of the polynomial $f(T)$, fibres of $\psi_T$ which are isomorphic to $E_1^C$ for certain $[C]\in\kmodksq$ are parametrized by rational points on the curve $f(t)=Cs^2$, namely $E_2^C$.
\end{remark}

\section{A necessary condition for the Hilbert property}\label{se:ICproperty}

In this section we fix $k$ a number field, and
$$E_1:y^2=g(x),\quad E_2:s^2=f(t).$$
where $3\leqslant \deg f,\deg g\leqslant 4$ without repeated roots.

 Recall that $E_1^C$ (resp. $E_2^C$) denotes the quadratic twist of $E_1$ (resp. $E_2$) by $[C]\in\kmodksq$. Consider the condition
\begin{equation}\label{eq:bothrankpos}
\rank(E_1^C(k))\rank(E_2^C(k))>0 \tag{SPR(C)}.
\end{equation}
	 Here by writing ``rank'' we implicitly mean that $E_i(k)\neq\varnothing$ for any $i=1,2$, so that they are equipped with some group law.

The aim of this section is to prove:
\begin{theorem}\label{thm:HP}
	\hfill
	\begin{enumerate}[label=(\roman*)]
		\item The set $\Kum(E_1\times E_2)(k)$ is Zariski dense if and only if there exists $[C]\in\kmodksq$ such that \eqref{eq:bothrankpos} holds.
		\item  If $\Kum(E_1\times E_2)$ verifies (HP) over $k$, then there exist infinitely many classes $[C]\in\kmodksq$ such that \eqref{eq:bothrankpos} holds.
	\end{enumerate}

\end{theorem}
\begin{remark}\label{rmk:oneCgivesinfinite}
	Suppose that Conjecture \ref{conj:CZ} holds for $\Kum(E_1\times E_2)$. We infer from Theorem \ref{thm:HP} (2) that the existence of one class $[C]\in\kmodksq$ with \eqref{eq:bothrankpos} implies the existence of infinitely many such $[C]$. See also the discussion in \S\ref{se:simulnonvan} below.
\end{remark}
\subsection{Proof of Theorem \ref{thm:HP}}
We begin with a basic observation towards the condition \eqref{eq:bothrankpos}.
\begin{lemma}\label{le:Zdense}
	Condition \eqref{eq:bothrankpos} holds for $[C]\in\kmodksq$ if and only if $(E_1^C\times E_2^C)(k)$ is Zariski dense.
\end{lemma}
\begin{proof}
	The ``if'' part is clear.
	Let us now assume \eqref{eq:bothrankpos} for certain $[C]\in\kmodksq$. Take $D\subset E_1^C\times E_2^C$ to be any integral curve. Then the genus $g(D)$ of $D$ is at least $1$. Consider the projections $p_1:D\to E_1^C,p_2:D\to E_2^C$, so that $D\simeq (p_1\times p_2)(D)$. If $g(D)=1$, then $D$ is an elliptic curve. So either one of the maps $p_1,p_2$ is constant and the other is étale (hence an isogeny), or all of them are étale. In the first case $D(k)\subset E_1^C(k)\times \{*\}$ or $\{*\}\times E_2^C(k)$. For the second case, the group $D(k)$ has the same (positive) rank as $E_1^C(k)$ and $E_2^C(k)$. Then clearly $(p_1\times p_2)(D(k))$ is not a subgroup of finite index of $E_1^C(k)\times E_2^C(k)$.

	 If $g(D)\geqslant 2$, then $\#D(k)<\infty$ by Faltings' theorem. Therefore $(E_1^C\times E_2^C)(k)$ is not contained in the union of a finite number of curves.

\end{proof}
The next finiteness result is a slight generalization of an observation due to Rohrlich \cite[\S8 Lemma, p. 147]{Rohrlich}, who proved the statement over $\BQ$.
\begin{lemma}\label{le:Rohrlich}
	Let $E$ be an elliptic curve over a number field $k$ with equation $v^2=h(u)$,
	where $h(u)\in k[u]$ is separable of degree $3$.
	We fix the equation for $E^C$ the quadratic twist by $C\in k^\times$ to be $Cv^2=h(u)$.
	Then
	$$\#\{u\in k:\exists [C]\in \kmodksq,\exists v\in k,(u,v)\in E^C(k)_{\tor}\}<\infty.$$
\end{lemma}
\begin{proof}
	We make use of Merel's uniformity theorem \cite{Merel} to extend the statement to arbitrary number fields. To simply the reading we give details.

	Since $E[2]\simeq E^C[2]$ for any $[C]\in\kmodksq$, the $u$-coordinates appearing in any $2$-torsion point (not the one at infinity) of $E^C$ for some $[C]$ satisfy $h(u)=0$.

	We next prove: for any fixed integer $N\geqslant 3$, if $E[N](k)\neq\varnothing$, then for any $[C]\in\kmodksq, [C]\neq [1]$, one has $E^C[N](k)=\varnothing$. Indeed, by choosing $P\in E[N](k)$ as part of a basis of $E[N]$, the image of the mod $N$ Galois representation $\varrho_{E,N}:\operatorname{Gal}(\bar{k}/k)\to \operatorname{GL}_2(\BZ/N)$ attached to $E$ is \begin{equation}\label{eq:galoisrep}
	\begin{pmatrix}
	1 & *  \\
	0 & \chi_N  \\
	\end{pmatrix},
	\end{equation}
	where $\chi_N$ is the mod $N$ cyclotomic character. And that of $\varrho_{E^C,N}$ attached to $E^C$ is conjugate inside $\operatorname{GL}_2(\BZ/N)$ to
	\begin{equation}\label{eq:galoisrep2}
	\psi_C\cdot\begin{pmatrix}
	1 & *  \\
	0 & \chi_N  \\
	\end{pmatrix},
	\end{equation}
	 where $\psi_C:\operatorname{Gal}(\bar{k}/k)\to\{\pm 1\}$ is the quadratic character associated to the extension $k(\sqrt{C})/k$. We see that elements of the form \eqref{eq:galoisrep2} cannot be conjugate to \eqref{eq:galoisrep}, unless $[C]=[1]$. This proves our claim.

	Finally, by \cite[Corollaire]{Merel}, there exists a constant $B=B(k)$ such that for any elliptic curve $\mathfrak{E}$ defined over $k$, any point in $\mathfrak{E}(k)_{\operatorname{tor}}$ has order  $<B$. We have therefore proven: there are only finitely classes $[C]\in k^*/k^{*2}$ such that $E^C$ has a torsion point over $k$ of order $>2$. This finishes the proof.

\end{proof}
We now deduce the following, which may be seen as a quantitative version of \cite[Lemma 3.2]{Demeio}.

Recall \eqref{eq:Uc} the open set $U_C\subset E_1^C\times E_2^C$.
\begin{lemma}\label{le:simultaneoustwist}
	The union of the image of $U_C(k),[C]\in\kmodksq$ under $\pi_{E_1^C\times E_2^C}$ such that
	\eqref{eq:bothrankpos} does not hold

	is contained in a proper Zariski closed subset of $\Kum(E_1\times E_2)$.
\end{lemma}
\begin{proof}
	We first suppose that $\deg f=\deg g=3$. Then for such $[C]\in\kmodksq$ verifying the hypothesis of lemma, at least one of the groups $E_1^C(k),E_2^C(k)$ contains only torsion points by Lemma \ref{le:Zdense}. According to the choice of affine model $\CS$ expression of the morphism $\phi_C$ \eqref{eq:doublephi}, on applying Lemma \ref{le:Rohrlich} to both curves $E_1,E_2$, we conclude that there exists a Zariski closed subset $Z$ of $\mathcal{S}$ consisting of points $(X,Y,T)$ with finitely many possible choices of $X$-coordinates or $T$-coordinates, so that
	$$\phi_C(U_C(k))\subset Z\cup\left( \left((Y=0)\cup (Y=\infty)\right)\cap \CS\right),$$
	the right-hand-side being a proper closed subset of $\CS$.

	Next, without loss of generality, let us assume $\deg f=4,\deg g=3$. Let $E_2^\prime$ be the Weierstrass form \eqref{eq:Weierstrassmod} of $E_2$. Fix $C\in k^*$. If $E_2^C(k)=\varnothing$ then $(E_1^C\times E_2^C)(k)=\varnothing$. So it suffices to consider the case where $E_2^C(k)\neq \varnothing$, then $E_2^C\simeq_k (E_2^\prime)^C$, with a choice of the zero element, and that \eqref{eq:bothrankpos} holds for $E_1^C\times E_2^C$ amounts to saying it holds for $E_1^C\times (E_2^\prime)^C$. Let us consider the diagram (the maps $\pi_{E_1^C\times E_2^C}$, $\pi_{E_1^C\times (E_2^\prime)^C}$ and $\widetilde{\Phi}$ are constructed in \S\ref{se:Kummervar})
	\begin{equation}\label{eq:coveringkum}
		\xymatrix{&E_1^C\times E_2^C \ar@{->}[r]^{\operatorname{Id}_{E_1}\times \Phi_C} \ar@{->}[d]^{\pi_{E_1^C\times E_2^C}} &E_1^C\times (E_2^\prime)^C\ar@{->}[d]^{\pi_{E_1^C\times (E_2^\prime)^C}}\\
		&\wKum(E_1\times E_2)\ar@{->}[r]^{\widetilde{\Phi}} &\wKum(E_1\times E_2^\prime),}
	\end{equation}
	where $\Phi_C:E_2^C\to (E_2^\prime)^C$ is the $2$-covering map (c.f. \S\ref{se:Weierstrass}).

	By the argument in the preceding paragraph, we know that the image of $(E_1^C\times (E_2^\prime)^C)(k)\setminus ((E_1^C[2]\times E_2^\prime) \cup (E_1\times (E_2^\prime)^C[2]))$ under $\pi_{E_1^C\times (E_2^\prime)^C}$ in $\wKum(E_1\times E_2^\prime)$ for all $[C]\in\kmodksq$, such that $(E_1^C\times (E_2^\prime)^C)(k)$ is not Zariski dense, is contained in a proper closed subset $W^\prime\subset\wKum(E_1\times E_2^\prime)$. Therefore the image of all $U_C(k)\subset (E_1^C\times E_2^C)(k)$ under $\pi_{E_1^C\times E_2^C}$ in $\wKum(E_1\times E_2)$ is contained in $W=\widetilde{\Phi}^{-1}(W^\prime)$, which is also proper Zariski closed.

	Finally, the case where $\deg f=\deg g=4$ is reduced to the above one, by considering the $2$-covering map between $E_1$ and its Weierstrass model.
\end{proof}
\begin{remark}
	Our proof above shows that, on the open subset $U=\Kum(E_1\times E_2)\setminus W$, for any $P\in U(k)$ (which might be empty), the fibres of $\psi_X(P)$ and $\psi_T(P)$ containing $P$ have positive Mordell-Weil rank.
\end{remark}
\begin{proof}[Proof of Theorem \ref{thm:HP}]
	It is clear that both Zariski density and (HP) do not depend on the choice of birational models. We may thus fix the model of $\Kum(E_1\times E_2)$ to be $\mathcal{S}$ \eqref{eq:CSfamily} as in \S\ref{se:Kummervar}. By Theorem \ref{thm:Z2torsors}, points in $\CS(k)$ outside a proper Zariski closed subset on are parametrized by the family $(U_C(k))_{[C]\in k^*/k^{*2}}$. Now (i) follows immediately from Lemmas \ref{le:Zdense} and \ref{le:simultaneoustwist}.
	 If $\Kum(E_1\times E_2)$ verifies (HP) over $k$, then $\mathcal{S}(k)$ is not dominated by rational points on any finite collection of  twisted abelian surfaces $E_1^C\times E_2^C$. On combining again Lemma \ref{le:simultaneoustwist}, such $C$'s verifying \eqref{eq:bothrankpos} are necessarily infinite.
\end{proof}

\subsection{Application to jumps of Mordell-Weil rank}\label{se:jumpmordellweil}
It has been observed by Colliot-Thélène \cite[Th\'eor\`eme 1.1]{C-T} that if an elliptic surface $\CE\to\BP^1$ verifies (HP), then the set of fibres whose Mordell-Weil rank jumps at least by one with respect to the generic rank is not thin. Regarding the condition \eqref{eq:bothrankpos}, we establish the following for the two fibrations $\psi_X$ and $\psi_T$.
\begin{corollary}\label{th:jump}
	Assuming that \eqref{eq:bothrankpos} holds for infinitely many $[C]\in k^*/k^{*2}$, then for each of the genus one fibrations $\psi_X$ and $\psi_T$, the set of fibres over $k$ whose rank is $>0$ is not thin.
\end{corollary}
\begin{proof}

 Consider for example the fibration $\psi_T$. If the set of $k$-fibres of positive rank were contained in the image of finitely many dominant maps $\phi_i:C_i\to\BP_k^1,i\in I$ of degree $\geqslant 2$, then take any $[C]\in\kmodksq$ such that \eqref{eq:bothrankpos} holds and that the composition map $\psi_T\circ \pi_{E_1^C\times E_2^C}|_{E_2^C\simeq \{P\}\times E_2^C\subset E_1^C\times E_2^C}:E_2^C\to \BP_k^1$ (where we take any $P\in E_1^C(k)$ and this composition is independent of $P$, c.f. Remark \ref{rmk:fibres}) is different from any $\phi_i,i\in I$, so that the maps $E_2^C\times_{\phi_i,\BP_k^1}C_i\to E_2^C$ are ramified (as each $\phi_i$ is). So the curves $E_2^C\times_{\phi_i,\BP_k^1}C_i$ have genus $\geqslant 2$. Thus the set $E_2^C(k)$ would also be contained in $\cup_{i\in I}(E_2^C\times_{\phi_i,\BP_k^1}C_i) (k)$, the former being infinite. This leads to a contradiction by Faltings' Theorem.
\end{proof}
\subsection{Remarks on simultaneous non-vanishing}\label{se:simulnonvan}
Let $j_{E_1}$ (resp. $j_{E_2}$) be the $j$-invariant of $E_1$ (resp. $E_2$). The following result had been discovered by Kuwata-Wang \cite[Theorem 5]{K-W} over $\BQ$ by making use of the existence of a rational curve on $\Kum(E_1\times E_2)$. We present a different proof based on Theorem \ref{thm:HP}.
\begin{corollary}\label{thm:degree3Kummer}
	Assume that $\deg f=\deg g=3$ and that $(j_{E_1},j_{E_2})\neq(0,0)$ or $(1728,1728)$. Then \eqref{eq:bothrankpos} holds for infinitely many classes $[C]\in\kmodksq$.

\end{corollary}
\begin{proof}
	The assumption fits into Kuwata-Wang's result \cite[Theorem 1]{K-W}. We briefly explain here how their method (stated over $\BQ$) can deduce that the rational points over $k$ an arbitrary number field are Zariski dense on $\Kum(E_1\times E_2)$. Indeed, after the base change $\zeta:\BP_k^1\to\BP_k^1,Y\mapsto Y^6$, we obtain a non-torsion section (the notation $\sigma_1$ in  \cite[p. 116]{K-W}) with respect to the new elliptic fibration $\CS\times_{\zeta,\BP_k^1}\BP_k^1\to\BP_k^1$ in coordinate $Y$. Therefore by Silverman's specialization theorem \cite{Silverman2}, almost all fibres have positive Mordell-Weil rank, whence the Zariski density (even the density in real topology if $k=\BQ$) holds. By Demeio's theorem \cite[Proposition 4.4]{Demeio}, the surface $\Kum(E_1\times E_2)$ verifies (HP) over $k$. The assertion now follows from Theorem \ref{thm:HP} (2).
\end{proof}

It would be interesting to compare with analytic approaches to this simultaneous non-vanishing condition. The recent result of Petrow \cite[Theorem 2.2]{Petrow} studies simultaneous non-vanishing at the central value of derivatives of $L$-functions attached to two modular forms. Conditionally on GRH and under some assumption on conductors and root numbers, there exist infinitely many square-free integers $C$ such that
$\rank(E_1^C(\BQ))\rank(E_2^C(\BQ))=1$.
See also a result Munshi \cite{Munshi} concerning simultaneous non-vanishing of $L$-functions at the central value, which has consequence in the whether $E_1^C$ and $E_2^C$ can have simultaneously rank zero.
\section{Cassels-Schinzel type K3 surfaces}\label{se:CSsurfaces}
From now on assume $k=\BQ$. We study in this section the isotrivial elliptic pencil over $\BQ$ of $j$-invariant $1728$ with affine model
\begin{equation}\label{eq:Cassels-Schinzel}
\mathfrak{S}^{d,a}:d(1+a^2T^4)Y^2=X^3-X,
\end{equation}
for which we refer to being of Cassels-Schinzel type.
Here $d,a\in\BZ_{\neq 0}$ are square-free parameters.
(The original one constructed in \cite{C-S} is given by $a=1,d=7$.)
Let $\psi_T:\mathfrak{S}^{d,a}\to\BP^1$ be the elliptic fibration in coordinate $T$, write $\mathfrak{S}^{d,a}_T$ for fibres and consider the set $$\mathcal{F}^{d,a}=\{T\in\BP^1(\BQ):\rank(\mathfrak{S}^{d,a}_T(\BQ))>0\}.$$ Cassels and Schinzel proved that the all sections with respect to $\psi_T$ are torsion (c.f. \cite[Theorem 1 \& Corollary]{C-S}. Their argument yields the same result for twists by the polynomial $(1+4t^4)$). So the N\'eron-Silverman specialization theorem \cite{Silverman2} does not provide useful information on the ranks of fibres. We shall focus on the validity of condition \eqref{eq:bothrankpos} and its consequence on density of rational points.
Our main theorems are concerned with two families of surfaces of Cassels-Schinzel type.

\begin{theorem}\label{th:mainthm1}
	\hfill
	\begin{enumerate}
		\item 	There exists an infinite set $\mathfrak{D}_1$ of square-free integers $d$ containing $7,41,\cdots$ such that \eqref{eq:bothrankpos} holds for at least one $[C]\in\BQ^*/\BQ^{*2}$ for the elliptic surface
		\begin{equation}\label{eq:CS1}
		\mathfrak{S}^{d,1}:d(1+T^4)Y^2=X^3-X.
		\end{equation}
		\item 	There exists an infinite set $\mathfrak{D}_2$ of square-free integers $d$ such that \eqref{eq:bothrankpos} holds for infinitely many $[C]\in\BQ^*/\BQ^{*2}$ for the elliptic surface
		\begin{equation}\label{eq:CS2}
		\mathfrak{S}^{d,2}:d(1+4T^4)Y^2=X^3-X.
		\end{equation}
		\item For any $d\in\mathfrak{D}_1$ (resp. $d\in\mathfrak{D}_2$) rational points on \eqref{eq:CS1} (resp. on \eqref{eq:CS2}) are dense in the real topology, and the set $\mathcal{F}^{d,1}$ (resp. $\mathcal{F}^{d,2}$) is dense in $\BP^1(\BR)$. The set $\mathcal{F}^{d,2}$ is moreover not thin in $\BP^1(\BQ)$.
	\end{enumerate}

\end{theorem}

\subsection{Topology of hyperelliptic quartics}\label{se:tophyp}

Let us define the family of hyperelliptic curves
\begin{equation}\label{eq:HaC}
\CH_a^C:Cs^2=1+a^2t^4
\end{equation}
with parameter $C>0,a\neq 0$.

The (affine) real locus $\CH_a^C(\BR)$ of \eqref{eq:HaC} has two symmetric branches (one on the upside of the $t$-axis and the other one below). We see that even though the curve \eqref{eq:HaC} is not connected, the involution $(s,t)\mapsto(-s,t)$ exchanges rational points between the two components of \eqref{eq:HaC}.

Consequently, if $\CH_a^C(\BQ)$ is infinite, then rational points are dense in real topology in all connected components.
\subsection{Root number of congruent number elliptic curves}
The \emph{root number} (denoted by $\omega(\cdot)$) of an elliptic curve appears in the functional equation of its $L$-function (c.f. \cite[\S C.16]{Silverman1}, where it is called ``sign of the functional equation'').
Although the connection between root number and rational points on elliptic curves remains mostly conjectural, we find helpful to give a short discussion here with regard to the variation of root numbers in families \eqref{eq:CS1} \eqref{eq:CS2}.

Let
\begin{equation}\label{eq:congruntnumbercurve}
\mathbf{E}:y^2=x^3-x
\end{equation}
be the congruent number elliptic curve.
As usual $\mathbf{E}^D$ denotes the quadratic twist of $\mathbf{E}$ by a square-free integer $D$. We say that $D>0$ is a \emph{congruent number} if $\rank(\mathbf{E}^D(\BQ))>0$.
A classical computation carried out by Birch and Stephens \cite{Birch-Stephens} shows that
\begin{equation}\label{eq:congruentnumbers}
\omega(\mathbf{E}^D)=\begin{cases}
1 & ~ D\equiv 1,2,3 \mod8;\\
-1 & ~ D\equiv 5,6,7 \mod8.
\end{cases}
\end{equation}
The parity conjecture (c.f. \cite[p. 119]{Rohrlich}) implies that all such integers $D$ satisfying $\omega(\mathbf{E}_D)=-1$ should be congruent numbers.

We proceed to compute the root numbers of the family \eqref{eq:CS1} (as in \cite[p.347]{C-S}). Write $T=\frac{l}{m}$, with $(l,m)\in\BZ,\gcd(l,m)=1$.  A change of variables yields the equation

\begin{equation*}\label{eq:CasselsSchinzel2}
d(l^4+m^4)Y^2=X^3-X
\end{equation*}
for \eqref{eq:CS1}. Since for any odd prime $p\mid l^4+m^4$, we have $p\equiv 1\mod 8$ (c.f. Proposition \ref{prop:a=1}).

We thus obtain a complete description of root numbers for fibres of \eqref{eq:CS1}, depending on the parity and class modulo $8$ of $d$. For instance, if $d\equiv 7\mod 8$ (resp. $d\equiv 1\mod 8$) then the surface \eqref{eq:CS1} has constant root number $-1$ (resp. $+1$). For the family \eqref{eq:CS1} however, fibres tend to have varying root numbers (c.f. Proposition \ref{prop:a=2}).
\subsection{Torsors under congruent number elliptic curves}\label{se:torsorundercong}

The purpose of this section is to discuss rational points on quadratic twists of hyperelliptic quartics \eqref{eq:HaC} with $C$ square-free and $a\in \BZ_{\neq 0}$.

By the discussion in \S\ref{se:Weierstrass}, we know that
$\mathcal{H}^C_a$ is a torsor under $\mathbf{E}^{2aC}$, the quadratic twist of $\mathbf{E}$ \eqref{eq:congruntnumbercurve}  by $2aC$.

Hence if $\mathcal{H}^C_a(\BQ)\neq\varnothing$, then $\mathcal{H}^C_a\simeq_\BQ \mathbf{E}^{2aC}$.
However it happens that even if $\mathcal{H}_a^C$ is everywhere locally soluble, it still has no $\BQ$-point. Indeed, consider the class $[\mathcal{H}^C_a]\in H^1(\BQ,\mathbf{E}^{2aC})$. According to the exact sequence of $\mathbb{F}_2$-vector spaces (c.f. \cite[X.4. Theorem 4.2]{Silverman1}) induced by multiplication by $[2]$ map,
\begin{equation}\label{eq:exactseq2}
0\to\mathbf{E}^{2aC}(\BQ)/2\mathbf{E}^{2aC}(\BQ)\to\Sel(\BQ,\mathbf{E}^{2aC})\to\Sha^1(\BQ,\mathbf{E}^{2aC})[2]\to 0,
\end{equation}
if $[\mathcal{H}^C_a]$ maps to a non-zero element in $\Sha^1(\BQ,\mathbf{E}^{2aC})[2]$, then $\mathcal{H}^C_a(\BQ)=\varnothing$.

It follows from a formula of Monsky (c.f. \cite[Appendix]{H-B}) that for $n\in\BN_{\geqslant 1}$ square-free,
\begin{equation}\label{eq:upperselmer}
	\dim_{\mathbb{F}_2}\Sel(\BQ,\BE^{n})\leqslant 2\Omega(n)+2,
\end{equation}
where $\Omega(n)$ is the number of odd prime divisors of $n$.

Rational points on $\mathcal{H}_a^C$ are closely related to integral solutions of Fermat-type equations $AX^4+BY^4+CZ^2=0$. Having Cassels-Schinzel's surface in mind, in what follows we shall be interested exclusively in cases where $a=1$ and $a=2$.
\subsubsection{Case $a=1$}\label{se:a=1}

\begin{proposition}[\cite{Cohen}, Proposition 6.5.2]\label{prop:a=1}
	The curve $\mathcal{H}^C_1$ is everywhere locally soluble if and only if
	\begin{center}
			(*) all odd prime factors of $C$ are $\equiv 1\mod 8 $.
	\end{center}
\end{proposition}
According to \eqref{eq:congruentnumbers}, under the hypothesis (*), the root number of $\BE^{2C}$ is zero, which conjecturally implies that $\rank(\BE^{2C}(\BQ))$ is even. Although most of such curves (are believed to) have rank zero, the following condition (**) guarantees the existence of infinitely many points on $\CH_1^C$.
\begin{corollary}\label{co:CohenC}
	Let $C\in\BN_{\geqslant 1}$ be square-free verifying (*), and
	$$(**)\quad 2\Omega(C)=\rank(\BE^{2C}).$$
	Then $\#\mathcal{H}^C_1(\BQ)=\infty$.
\end{corollary}
\begin{proof}
	Indeed, from the exact sequence \eqref{eq:exactseq2} and inequality \eqref{eq:upperselmer}, we have
	\begin{align*}
	\dim_{\mathbb{F}_2}(\mathbf{E}^{2aC}(\BQ)/2\mathbf{E}^{2aC}(\BQ))&=\rank(\BE^{2C})+\dim_{\mathbb{F}_2}(\BE^{2C}[2](\BQ))\\
	&=\rank(\BE^{2C})+2\leqslant \dim_{\mathbb{F}_2}\Sel(\BQ,\BE^{2C})\leqslant 2\Omega(C)+2.
	\end{align*}
	Hence under the assumption (**), the inequalities above are all equalities. So in particular $\Sha^1(\BQ,\mathbf{E}^{2C})[2]=0$.
	This implies that $\mathcal{H}_1^C\simeq_\BQ \BE^{2C}$.
\end{proof}
\begin{remark}
	We refer to Cohen's table \cite[p. 395]{Cohen} for all such $C$'s verifying (*) and (**) less than $10000$. However, such numbers are very sparse. To the best of the author's knowledge, it is not currently known whether there exist infinitely many congruent numbers verifying (*).
	Even though surfaces in the family \eqref{eq:CS1} can have constant root number $-1$, previous arguments indicate that it seems more difficult to produce rational points on them, as the hyperelliptic curves $\CH_1^C$ parametrizing fibres of \eqref{eq:CS1} have few rational points.
\end{remark}

\subsubsection{Case $a=2$}\label{se:a=2}
The curve $\CH^C_2$ is a $\BQ$-torsor under $\BE^{4C}\simeq_\BQ\BE^C$.
Let us begin by describing a sufficient criterion of local solubility.
\begin{proposition}\label{prop:a=2}
	Assume that $C\equiv 5\mod 8$, and that for any $p\mid C$, $-4$ is a fourth root in $\mathbb{F}_p$. Then $\CH_2^C$ is everywhere locally soluble.
\end{proposition}
\begin{proof}
	Non-trivial rational points on $\CH_2^C$ are in bijection with integral solutions of the Diophantine equation
	$$X^4+4Y^4-CZ^2=0.$$
	The sufficient condition above follows from \cite[Proposition 6.5.1 (2c)\&(3e)]{Cohen}.
\end{proof}

The condition $C\equiv 5\mod 8$ implies that $\omega(\BE^C)=1$. Standard conjectures predict that
$$\dim_{\mathbb{F}_2} \Sel(\BQ,\BE^C)\equiv \rank(\BE^C(\BQ))\equiv 1\mod 2.$$

The following unconditional result provides an infinite set of square-free integers $C$ such that $\CH_2^C$ has infinitely many rational points.
\begin{proposition}\label{prop:H2p}
	For any prime number $p\equiv 5\mod 8$, we have $\#\CH_2^p(\BQ)=\infty$.
\end{proposition}
\begin{proof}
	First we check local solubility. Write $\left(\frac{\cdot}{\cdot}\right)$ (resp. $\left(\frac{\cdot}{\cdot}\right)_4$) for the quadratic (resp. quartic) residue symbol. For any prime $p\equiv 5\mod 8$, we have
	$$\left(\frac{-1}{p}\right)=1,\quad \left(\frac{-1}{p}\right)_4=\left(\frac{2}{p}\right)=-1,$$
	and hence $$\left(\frac{4}{p}\right)_4=-1,\quad\left(\frac{-4}{p}\right)_4=1.$$
	So $\CH_2^p$ verifies the hypothesis of Proposition \ref{prop:a=2}.
	Now we make use of a classical result due to Heegner (c.f. \cite[Corollary 5.15 (1)]{Monsky}) affirming that
	$$\rank(\BE^p(\BQ))=1,\quad \dim_{\mathbb{F}_2} \Sel(\BQ,\BE^p)=3.$$
Combining the exact sequence \eqref{eq:exactseq2} we get $\Sha^1(\BQ,\BE^p)[2]=0$.
	So $\CH_2^p\simeq_\BQ\BE^p$ and $\rank(\CH_2^p(\BQ))=\rank(\BE^p(\BQ))=1$.
\end{proof}

\subsubsection{Remark on $2$-descent}

Following Cohen \cite[\S6.5]{Cohen}, one can give explicit criterion to determine whether the class $[\CH_a^C]\in \Sel(\BQ,\BE^{2aC})$ comes from some non-torsion point of $\BE^{2aC}(\BQ)$ using descent by $2$-isogeny.

Indeed, consider the elliptic curve
$\mathbb{E}^{aC}: Cy^2=x^3+x$.

There exists a $2$-isogeny $\psi:\BE^{2aC}\to \mathbb{E}^{aC}$, so that we get an exact sequence
$$\xymatrix{0\ar[r]&\BZ/2  \ar[r] &\mathbf{E}^{2aC}\ar[r]^{\psi} &\mathbb{E}^{aC}\ar[r]&0}.$$

This gives rise to
$$\xymatrix{
	&\mathbb{E}^{aC}(\BQ) \ar[r]^-\alpha &H^1(\BQ,\BZ/2)\ar[r] &H^1(\BQ,\mathbf{E}^{2aC}) \ar[r]^{\psi^\prime} &H^1(\BQ,\mathbb{E}^{aC})},$$
where $\alpha$ is a partial $2$-descent map and $\psi^\prime$ is induced by $\psi$.
There is an elementary way of transforming points on $\CH_a^C$ into $\mathbb{E}^{aC}$ (c.f. \cite[Proposition 6.5.5]{Cohen}). We conclude that $\psi^\prime[\CH_a^C]=0$ in $H^1(\BQ,\mathbb{E}^{aC})$. So whether $[\CH_a^C]\in H^1(\BQ,\BE^{2aC})$ is zero or not is characterised by the image of the map $\alpha$. See \cite[\S 6.5.3]{Cohen} for a simple description of $\alpha$.

\begin{comment}

To  we make use the following result of Tian.
\begin{theorem}[\cite{Tian}, Theorem 1.3]\label{thm:Tian}
	Let $C\in\BN_{\geqslant 1}$ be square-free and $C=p_0p_1\cdots p_k$ be a prime factorisation such that $p_0\equiv 5 \text{ or } 7\mod 8$ and $p_i\equiv 1\mod 8$ for all $1\leqslant i\leqslant k$. If the $4$-rank of the imaginary quadratic extension $\BQ(\sqrt{-C})$ is zero, then we have
	$$\rank(\BE^C(\BQ))=1 \quad \text{ and }\quad \Sha^1(\BE^C)[2^\infty]=0.$$
\end{theorem}

Assume in this section that $\deg f=\deg g=3$.
Demeio \cite[Theorem 1.1 \& Proposition 4.4]{Demeio} proved:
\begin{theorem}[Demeio]
	Assume $\CS(k)$ is Zariski dense. Then $CS$ verifies HP over $k$.
\end{theorem}
As an easy consequence, we obtain a stronger version of converse of Theorem \ref{thm:HP}.
\begin{corollary}
The surface $\CS$ verifies HP if and only if there exists one $[C]\in\kmodksq$ such that \eqref{eq:bothrankpos} holds.
\end{corollary}
\begin{proof}
	It remains to show that the hypothesis above is equivalent to $\CS(k)$ being Zariski dense. And this is a consequence of Lemma \ref{le:simultaneoustwist}.
\end{proof}
\end{comment}
\subsection{Proof of Theorem \ref{th:mainthm1}}
 By Theorem \ref{thm:Z2torsors}, outside a proper Zariski closed subset (comprising of $24$ rational curves), rational points on the surface $\mathfrak{S}^{d,a}$ \eqref{eq:Cassels-Schinzel} are parametrized by the family $((\CH_a^C\times \mathbf{E}^{Cd})(\BQ))_{C\text{ square-free}}$ (an open subset of each), where $\CH_a^C$ is defined by \eqref{eq:HaC} and $\mathbf{E}$ is given by \eqref{eq:congruntnumbercurve}.

\subsubsection{Proof of (1)}

We show that the infinite family of surfaces $\mathfrak{S}^{d,1}$ with $d=17p$ with $p$ prime $\equiv 5$ or $7\mod 8$ verifies \eqref{eq:bothrankpos} for at least one square-free $C>0$. We first extract the value $C=17$ from Cohen's table \cite[p. 395]{Cohen} satisfying the hypotheses of Corollary \ref{co:CohenC}, so that $\#\CH_1^{17}(\BQ)=\infty$. On the other hand, with this choice of $d$, we have $\mathbf{E}^{17\times 17p}\simeq_\BQ \mathbf{E}^p$. Thanks to \cite[Corollary 5.15 (2)]{Monsky}, any prime number $\equiv 5,7\mod 8$ is a congruent number. Therefore the condition (SPR(17)) holds for the infinite family $(\mathfrak{S}^{17p,1},p\equiv 5,7\mod 8)$.

To derive similar result for the surfaces $\mathfrak{S}^{7,1}$ and $\mathfrak{S}^{41,1}$ with constant root numbers, it suffices to take $C=113$ from Cohen's table \cite[p. 395]{Cohen} and a consultation of elliptic curve database shows that
$\rank(\mathbf{E}^{7\times 17}(\BQ))=1, \rank(\mathbf{E}^{41\times 113}(\BQ))=2$. So the condition (SPR(17)) (resp. (SPR(113))) holds for $\mathfrak{S}^{7,1}$ (resp. $\mathfrak{S}^{41,1}$).

We conclude that for any $d\in\mathfrak{D}_1$, where
$$\mathfrak{D}_1=\{7,41,17p ~(p\equiv 5,7\text{ mod }8)\},$$
the condition \eqref{eq:bothrankpos} holds for at least one square-free $C$.
\qed

\subsubsection{Proof of (2)}
Now we consider the family of surfaces $\mathfrak{S}^{d,2}$.
By a result of Monsky \cite[Corollary 5.15 (2)]{Monsky}, we have $\rank(\BE^C(\BQ))>0$ for any square-free integer $C$ of the form $pq$ or $2pq$ where $p\equiv 5\mod 8,q\equiv 3\text{ or }7\mod 8$ are prime numbers.
Therefore, for any $d=q$ or $2q$ with $q$ prime $\equiv 3\text{ or }7\mod 8$, we have
$\rank(\BE^{pd}(\BQ))\rank(\CH_2^p(\BQ))>0$ for any prime $p\equiv 5\mod 8$, using Proposition \ref{prop:H2p}.
We have shown that, for the infinite family $(\mathfrak{S}_{d,2})_{d\in\mathfrak{D}_2}$ of surfaces with $$\mathfrak{D}_2=\{q,2q,~(q\equiv 3\text{ or }7 \text{ mod } 8)\},$$
the condition (SPR($p$)) holds for any prime $p\equiv 5\mod 8$.

\qed

\subsubsection{Proof of (3)}

For any $C>0$, if $\rank(\BE^C(\BQ))>0$, then $\BE^C(\BQ)$ is dense in $\BE^C(\BR)$, since $\mathbf{E}^C[2]\simeq (\BZ/2)^2$ is defined over $\BQ$. The discussion in \S\ref{se:tophyp} shows that $\#\CH_a^C(\BQ)=\infty$ implies $\CH_a^C(\BQ)$ is dense in $\CH_a^C(\BR)$. Once the condition \eqref{eq:bothrankpos} holds for one square-free $C>0$, the set $(\BE^C\times\CH_a^C)(\BQ)$ is dense in real topology.
Since the map $\phi_C$ \eqref{eq:doublephi} is generically \'etale, this implies that $\mathfrak{S}^{d,a}(\BQ)$ is dense in $\mathfrak{S}^{d,a}(\BR)$.
Moreover, the fibres $\mathfrak{S}^{d,a}_T$ are parametrized by rational points on the family of hyperelliptic curves $(\CH_a^C)_{C\text{ square-free}}$.
Then $\#\CH_a^C(\BQ)=\infty$ shows that the image of rational points under the projection $\CH_a^C\to\BP^1$ (in coordinate $t$), as a subset of $\mathcal{F}^{d,a}$, is dense in $\BP^1(\BR)$, so is the set $\mathcal{F}^{d,a}$ itself. The fact that $\mathcal{F}^{d,2}$ is not a thin set follows from Corollary \ref{th:jump}. Thus the proof is achieved. \qed

\section{Remarks and questions}\label{se:remarks}
The property (HP) for a variety $X$ over $k$ amounts to saying that the complementary of the image of rational points on any finite collection of covers (that is, dominant generically finite rational maps of degree greater than two) in $X(k)$ is still Zariski dense. As K3 surfaces are simply connected, any cover has non-empty ramification locus. For elliptic K3 surfaces of quadratic twist type \eqref{eq:CSfamily}, there is one particular type of covers which merit consideration. That is, those whose ramification locus is within the sixteen exceptional lines. One can show that any such cover which does not factor through an abelian surface $E_1^C\times E_2^C$ is a cover between Kummer surfaces (c.f. the map $\widetilde{\Phi}$ constructed in \S\ref{se:Kummervar}). We have already seen a typical example in \eqref{eq:coveringkum}. The essential use of the third elliptic fibration in Demeio's argument, which does not exist for surfaces of Cassels-Schinzel type, successfully tackles with such covers, since it turns out that their generic fibres with respect to this fibration are curves of genus greater than two.
We conclude by asking whether the validity of \eqref{eq:bothrankpos} with infinitely many square-free integers $C$ suffices to guarantee (HP) for surfaces of Cassels-Schinzel type \eqref{eq:Cassels-Schinzel}, or in an equivalent manner, whether for any such $C$ the points in $(\BE^{Cd}\times\CH_a^C)(\BQ)$ are contained in the images of a finite number of families of isogenies (twisted by $C$).

\begin{comment}
		By our 2-descent result (Proposition ??), it suffices to construct an everywhere locally soluble class. For a fixed prime $p_0\equiv 5\mod 8$, let $p_1$ be another prime such that $p_1\equiv 1\mod 8$ and $\left(\frac{p_1}{p_0}\right)=1$ (here $\left(\frac{\cdot}{\cdot}\right)$ stands for the Legendre symbol).
	Consider the equation
	\begin{equation}\label{eq:dio1}
	X^4+4Y^4-p_0p_1Z^2=0.
	\end{equation}
	Its solutions are in bijection with that of $\mathcal{H}^{p_0p_1}$. By the discussion in \S\ref{se:Weierstrass} that $\mathcal{H}^{p_0p_1}$ is a $\mathbf{E}^{p_0p_1}$-torsor. We have $1+4-p_1p_0\equiv 0\mod 8$, and $-4$ is a forth root both modulo $p_0$ and $p_1$. Indeed, for $p_1$ we have $$\left(\frac{-1}{p_1}\right)_4=\left(\frac{2}{p_1}\right)=1$$
	whereas for $p_0$,
	$$\left(\frac{-1}{p_0}\right)_4=\left(\frac{2}{p_1}\right)=-1$$
	Thus the equation \eqref{eq:dio1} is everywhere locally soluble by \cite[Proposition 6.5.4]{Cohen}.

	Now the class $[\mathcal{H}^{p_0p_1}]\in\operatorname{Sel}^2(\mathbf{E}^{p_0p_1})$ and Theorem \ref{thm:Tian} guarantees that $\mathcal{H}^{p_0p_1}\simeq \mathbf{E}^{p_0p_1}$. Finally it is easy to see that there exist infinitely many such pairs $(p_0,p_1)$.
\end{comment}

config: arxiv


text:
\section{Function of Two Discrete Random Variables}
Tags: Probability Theory

\begin{theorem}
Let $\left({\Omega, \Sigma, \Pr}\right)$ be a [[Definition:Probability Space|probability space]].
Let $X$ and $Y$ be [[Definition:Discrete Random Variable|discrete random variables]] on $\left({\Omega, \Sigma, \Pr}\right)$.
Let $g: \R^2 \to \R$ be a [[Definition:Real-Valued Function|real-valued function]].
Then $Z = g \left({X, Y}\right)$, defined as:
:$\forall \omega \in \Omega: Z \left({\omega}\right) = g \left({X \left({\omega}\right), Y \left({\omega}\right)}\right)$
is also a [[Definition:Discrete Random Variable|discrete random variable]].
\end{theorem}

\begin{proof}
As $\operatorname{Im} \left({X}\right)$ and $\operatorname{Im} \left({Y}\right)$ are [[Definition:Countable|countable]], then so is $\operatorname{Im} \left({g \left({X, Y}\right)}\right)$.
Now consider $g^{-1} \left({Z}\right)$.
We have that:
* $\forall x \in \R: X^{-1} \left({x}\right) \in \Sigma$.
* $\forall y \in \R: Y^{-1} \left({x}\right) \in \Sigma$.
We also have that:
: $\displaystyle \forall z \in \R: g^{-1} \left({z}\right) = \bigcup_{\left({x, y}\right): g \left({x, y}\right) = z} \left\{{\left({x, y}\right)}\right\}$
But $\Sigma$ is a [[Definition:Sigma-Algebra|sigma-algebra]] and therefore closed for unions.
Thus $\forall z \in \R: g^{-1} \left({z}\right) \in \Sigma^2$.
Now consider $\left({v, w}\right) \in g^{-1} \left({Z}\right)$.
As $\Sigma$ is a [[Definition:Sigma-Algebra|sigma-algebra]] it follows directly that:
*$\forall x \in v: X^{-1} \left({x}\right) \in \Sigma$
*$\forall y \in w: Y^{-1} \left({y}\right) \in \Sigma$
Hence the result.
{{qed}}
=== Proof of Corollary ===
Follows straightforwardly by [[Principle of Mathematical Induction|induction]] from the main result.
{{qed}}
[[Category:Probability Theory]]
4n7jjhws4qt0p3jwk32xuo2q5klf416
\end{proof}

config: wiki


TITLE: Minimum values of multiple modulus function
QUESTION [0 upvotes]: Find the minimum value of the equation $|x-1|+|x-2|+|x+1|+|x+2|$.
I checked the values at x=1,x=2,x=-1 and x=-2 and getting the answer 6. Is there any other way to find the minimum values.

REPLY [2 votes]: For an even number of data their median minimizes the sum of the absolute deviations.  As in our case the median equals $0$, the minimum is $6$.


TITLE: differential equations defective eigenvalue problem
QUESTION [1 upvotes]: I am having trouble solving this Differential Equation proof/problem.
$y'= Ay$
For a 2x2 matrix A:
$y(t) = C_1 v e^{\lambda t} + C_2(u+tv)e^{\lambda t}   $
Where $\lambda$ is the defective eigenvalue, v is the corresponding eigenvector, and u is the generalized eigenvector.
Why is any choice of u alright? I'm uncertain as to how to proceed. How do I show that any generalized value of u will work? Any help will be greatly appreciated.

REPLY [1 votes]: The question is not complete as stated, you must require that $\def\v#1{{\bf#1}}\v v$ and $\v u$ are an eigenvector and a generalised eigenvector satisfying
$$(A-\lambda I)\v u=\v v\ .$$
(Note that this means any choice of $\v u$ is all right, but $\v v$ must then be given by the above formula.)  With this, just substitute back, remembering that for an eigenvector $\def\v#1{{\bf#1}}\v v$ we have
$$(A-\lambda I)\v v=\v0\ .$$
Therefore
$$\eqalign{RHS
  &=A\v y\cr
  &=(A-\lambda I)\v y+\lambda\v y\cr
  &=C_1e^{\lambda t}(A-\lambda I)\v v
    +C_2e^{\lambda t}(A-\lambda I)(\v u+t\v v)+\lambda\v y\cr
  &=C_1e^{\lambda t}\v 0+C_2e^{\lambda t}\v v+
    C_2e^{\lambda t}t\v 0+\lambda\v y\cr
  &=C_2e^{\lambda t}\v v+\lambda\v y\cr
  LHS
  &=\v y'\cr
  &=\lambda C_1e^{\lambda t}\v v+\lambda C_2e^{\lambda t}\v u
    +C_2(e^{\lambda t}+\lambda te^{\lambda t})\v v\cr
  &=C_2e^{\lambda t}\v v+\lambda(C_1e^{\lambda t}\v v+C_2e^{\lambda t}\v u
    +C_2te^{\lambda t}\v v)\cr
  &=RHS\ .\cr}$$


TITLE: Do photons lose energy after radiation pressure is applied to a perfect reflector?
QUESTION [1 upvotes]: I was readng Wikipedia article (English one) about radiation pressure because there is something I still cannot figure out. As I understand it, radiation pressure emerges from conservation of momentum. Photons or electromagnetic waves possess momentum and when they are absorbed, reflected or even emitted, the aborber/reflector/emitter experiences a pressure that is proportional to the irradiance (in watts per square meter).
OK. In the case of a perfect reflector that does nothing but reflecting incoming radiation, it would seem the reflector is pushed and the radiation is reflected in opposite directions. But the radiation has not lost anything, it has only changed direction. And I do not understand how something can be set in motion (the reflector), which amounts to do work, yet the source of this work does not lose energy. I mean, if the radiation, after a U turn, meets a second reflector, it would have pushed two reflectors a way, yet it would continue in its original direction as if nothing happened...? What am I missing here? Should not photons lose something?

REPLY [0 votes]: There will be a recoil momentum ($2p$ and energy ($2p^2/M$ of the mirror. The photon energy will therefore be $pc-2p^2/M$. For a mirror of one mole of glass (60g) and a photon of 500 nm (~2eV) that is a relative frequency change of $2p/Mc= 2/(6\cdot 10^{23}\cdot 10^9 \approx 3\cdot10^{-32}$. However this is not what will happen, as the mirror will not move as a rigid body. Instead a collective electronic excitation in a very thin layer of metal will absorb the momentum. As a consequence I guess that the momentum will be carried by some $10^9$ electrons and the effect will be a relative frequency change of somewhere around $10^{-10}$, give or take a few orders of magnitude.


text:
\begin{document}

\title{
Some Results on Bent-Negabent Boolean Functions over Finite Fields\thanks{This is a sufficiently revised and extended version of the paper \cite{Sseta12}. Section \ref{new} is a completely new contribution.}
}
\author{Sumanta Sarkar\\
Centre of Excellence in Cryptology\\
Indian Statistical Institute, Kolkata, INDIA\\
Sumanta.Sarkar@gmail.com
}

\date{}
\maketitle

\begin{abstract}
We consider negabent Boolean functions that have Trace representation.
We completely characterize quadratic negabent monomial functions.
We show the relation between negabent functions and bent functions
via a quadratic function.
Using this characterization, we give infinite classes of bent-negabent
Boolean functions over the finite field $\F_{2^n}$, with the maximum possible degree, $n \over 2$.
These are the first ever constructions of negabent functions with trace representation that have
optimal degree.

\end{abstract}

{\bf Keywords:} Negabent function,
bent function,
quadratic Boolean function,
Maiorana-McFarland function,
permutation,
complete mapping polynomial.

\section{Introduction}
Hadamard-Walsh transform is an important tool in characterizing Boolean functions.
For example, many cryptographic properties can be analyzed by the Hadamard-Walsh transform.
A function on even number of variables that has the maximum possible distance from the affine
functions is called a bent function. These functions have equal absolute spectral values under the
Hadamard-Walsh transform and was first introduced by Rothaus \cite{RO76}.
It is natural to investigate the spectral values of Boolean functions under some other
Fourier transform.
In 2007, Parker and Pott \cite{PP07}, considered the nega-Hadamard transform.
and introduced negabent functions.
These functions have equal absolute spectral values under the nega-Hadamard transform.
The periodic autocorrelation values of a bent function are all zero.
The negaperiodic autocorrelation value of a Boolean function under the nega-Hadamard transform
is the analogue of the periodic autocorrelation value.
The negaperiodic autocorrelation values are all zero for a negabent function.
These properties of a negabent function motivate us to study it further.
Negabent functions which are also bent are interesting as they
have extreme properties in terms of two different Fourier transforms.

Results on negabent functions can be found in \cite{PP07,SPP08,P00,RP06,Sugatanega,Sindocrypt09}.
As an example, the $6$-variable function
$
x_4(x_1 x_2 \oplus x_2 x_3 \oplus x_1 \oplus x_2) \oplus
x_5(x_1 x_2 \oplus x_2 x_3 + x_3) + x_6(x_1 \oplus x_3)$
is a cubic negabent function.

In \cite{PP07,SPP08} some classes of Boolean functions which are both
bent and negabent (bent-negabent) have been identified. In \cite{SPP08},
construction of negabent functions has been shown in the class of
Maiorana-McFarland bent functions.
It is interesting to note that all
the affine functions (both odd and even variables) are negabent
\cite[Proposition 1]{PP07}.
In \cite{Sindocrypt09}, symmetric negabent
functions have been characterized and shown to be all affine for both
odd and even number of variables.
The maximum degree of an $n$-variable bent-negabent function is
$n \over 2$.
Very recently construction of bent-negabent functions have been given in
\cite{Pottnega} with the optimal degree.

In this paper, we characterize
the negabent functions which are defined over finite fields,
{\em i.e.}, functions with Trace representation.

Let $\mathbb{F}_2^n$ be the vector space formed by the binary $n$-tuples
and $\mathbb{F}_{2^n}$ be the finite field with $2^n$ elements. For a set $E$,
the set of non zero elements of $E$ is denoted by $E^*$.

In \cite{PP07}, quadratic negabent Boolean functions defined over the vector space
$\mathbb F_2^n$  were characterized.
Any quadratic Boolean function can be written as
\begin{eqnarray*}
g(x_1, \ldots, x_n)& = & \sum_{1 \leq i < j \leq n} q_{i, j} x_i x_j
+ \sum_{1\leq i \leq n} l_i x_i + c\\
\ & = & x Q x^T + L x^T + c,
\end{eqnarray*}
where $Q=(q_{i,j})$ is an upper triangular binary matrix,
$L = (l_1, \ldots, l_n)$ is a binary vector and $c\in \{0, 1\}$.
Consider the binary symmetric matrix $B = Q + Q^T$ with the zero diagonal,
which is the symplectic matrix corresponding to the quadratic function $g$.
It is well known that a quadratic function $g$ is bent if and only if
the corresponding matrix $B$ has full rank. In \cite{PP07}, it was proved
that a quadratic function $g$ is negabent if and only if the matrix $B+I$
has full rank, where $B$ is the corresponding symplectic matrix and
$I$ is the identity matrix.

In this paper, first we consider quadratic monomials
defined over the field $\mathbb F_{2^n}$ and is of the form
\begin{equation}
\label{form-quad}
f: x \mapsto Tr_1^n(\lambda x^{2^k+1}), ~~~~\mbox{where } \lambda \in \mathbb F_{2^n}^*.
\end{equation}

The $\lambda$'s for which $f$ is bent is well known.
We characterize those $\lambda$'s for which $f$ is negabent.
We also give characterization of $\lambda$ for even $n$ such that $f$ is bent-negabent.
The existence of quadratic bent-negabent functions is known \cite{PP07}. We reprove
the existence by simple counting argument and using the characterization of
quadratic bent-negabent monomials.

We also study the negabent property of Maiorana-McFarland bent functions
$f: \mathbb F_{2^t}\times \mathbb F_{2^t} \mapsto \mathbb F_2$
defined by $f(x,y) = Tr_1^t(x \pi(y) + h(y))$, where $\pi(y)$ is a permutation
polynomial over $\mathbb{F}_{2^t}$ and $h(y)$ is any polynomial over $\F_{2^t}$.
We present a necessary and sufficient condition such that these functions are
negabent. As a consequence, we show that when the permutation $\pi$ is
$x \mapsto x^{2^i}$, the bent function $f$ is negabent if and only if
$h(y)$ is bent.
From this we show how a Maiorana-McFarland bent-negabent function of degree $n \over 4$
over $\mathbb{F}_{2^n}$ can be obtained.

Then we show that given a bent function $f$ over $\F_{2^n}$,
it is possible to obtain a negabent function by adding a quadratic function, and
vice versa. Using this result we are able to present infinite classes of bent-negabent
functions having the optimal degree $n \over 2$.

\section{Preliminary}
An $n$-variable Boolean function is a mapping
$f : \mathbb{F}_2^n \mapsto \mathbb{F}_2$.
The Hamming weight of a binary string $S$ is the number of $1$'s
in $S$ and it is denoted as $wt(S)$.
An $n$-variable Boolean function $f$ can be written as a function of
$x_1, \ldots, x_n$ variables as follows,
$$f(x_1,x_2,\ldots , x_n) =
\bigoplus_{a = (a_1, \ldots, a_n) \in \mathbb{F}_2^n} \mu_a
(\prod_{i=1}^n x_i^{a_i}), \mbox{ where } \mu_a \in \mathbb{F}_2.$$
This is called the algebraic normal form (ANF) of $f$.
The degree, $\deg(f)$, of $f$ is defined as
$\displaystyle\max _{a \in \mathbb{F}_2^n}\{wt(a) | \mu _a \ne 0\}$.

Let $\lambda = (\lambda_1, \ldots, \lambda_n)$ and
$x = (x_1, \ldots, x_n)$ be two vectors in $\mathbb{F}_2^n$
and $\lambda \cdot x = \lambda_1 x_1 \oplus \ldots \oplus \lambda_n x_n$.
Then the Hadamard-Walsh transform value of $f$ at
$\lambda$ is given by
\begin{equation}
\label{eqnhw}
\mathcal{H}_f(\lambda) = \frac{1}{2^\frac{n}{2}} \sum_{x \in \mathbb{F}_2^n} (-1)^{f(x) \oplus \lambda \cdot x}.
\end{equation}
The function is called bent if $|H_f(\lambda)| = 1$ for all $\lambda \in \mathbb{F}_2^n$.
After the introduction of bent functions in \cite{RO76}, there have been many
constructions of bent functions, for instance, \cite{Dillon}, \cite{Carlet93}, \cite{LHTK13}, and references therein.

For $a \in \mathbb{F}_2^n$, the periodic autocorrelation value of $f$
is computed as
$$\tau_a = \sum_{x \in \mathbb{F}_2^n}(-1)^{f(x) \oplus f(x \oplus a)}.$$
A Boolean function $f$ is bent if and only if $\tau_a = 0$ for all
$a \in {\mathbb{F}_2^n}^*$.

The nega-Hadamard transform value of
$f$ at $\lambda \in \mathbb{F}_2^n$
is given by
\begin{equation}
\label{eqnnh}
\mathcal{N}_f(\lambda) =
\frac{1}{2^{\frac{n}{2}}} \sum_{x \in \mathbb{F}_2^n} (-1)^{f(x) \oplus \lambda \cdot x}I^{wt(x)},
\end{equation}
where $I = \sqrt{-1}$, is the imaginary unit of the complex number.
Note that $\mathcal{N}_f(\lambda)$ is complex valued.

The function $f : \mathbb F_{2}^n \rightarrow \mathbb F_2$ is called {\em negabent} \cite{PP07} if
the magnitude of $\mathcal{N}_f(\lambda)$ is $1$, {\em i.e.},
$|\mathcal{N}_f(\lambda)| = 1$ for all $\lambda \in \mathbb F_2^n$.
In the following theorem we state an alternate characterization
of negabent functions in terms of their negaperiodic autocorrelation values
which has been shown in \cite[Theorem 2]{PP07} and \cite[Lemma 3]{Sugatanega}.

\begin{theorem}
A Boolean function $f$ is negabent if and only if
\begin{equation}
\sum_{x \in \mathbb F_2^n}
(-1)^{f(x) \oplus f(x \oplus y)}(-1)^{x \cdot y} = 0
\end{equation}
for all $y \in {\mathbb F_2^n}^*$.
\end{theorem}

From this theorem, we see that the correlation values between the function
$f(x)$ and $f(x \oplus y) \oplus y\cdot x$ are all zero
for all $y \in {\mathbb F_2^{n}}^*$. Moreover,
for even number of variables, if a negabent function is also a bent function,
then the correlation values of the function $f(x)$ and $f(x \oplus y)$ are also
equal to zero for all $y \in {\mathbb F_2^{n}}^*$.
Therefore, the functions which are both bent and negabent
are interesting to study. We call these functions \emph{bent-negabent}.
Bent functions can exist only on even number of variables and it has degree more than
$1$. However, all the affine functions are negabent \cite{PP07} which tells
that negabent functions exist for both even and odd number of variables.

\section{Characterization of negabent functions over the finite field $\mathbb F_{2^n}$}
Now we consider Boolean functions defined over the field
$\mathbb F_{2^n}$ and we characterize the negabent property of those functions.
The vector space $\mathbb F_2^n$ can be easily identified with the field
$\mathbb F_{2^n}$ by choosing a basis of $\mathbb F_{2^n}$ over $\mathbb F_2$.
The function
$Tr_1^n : \mathbb F_{2^n} \mapsto \mathbb F_2$ is defined
as $$Tr_1^n(x) = x + x^2 + \ldots + x^{2^{n-1}}.$$
We denote $Tr_1^n$ simply by $Tr$ and $``+"$ is the finite field addition.
If we choose the basis
$\{\alpha_1, \ldots, \alpha_n\}$ to be self dual then it can be shown that
$Tr(xy) = \sum_{i = 1}^n x_i y_i$.

Henceforth, in this paper we choose the basis to be self dual.
It is also notable that a linear function over $\mathbb F_{2^n}$
is given by $\ell(x) = Tr(ax)$, $a \ne 0$.

Given a polynomial $F(x)$ over $\F_{2^n}$, we can get a Boolean function
$f : \F_{2^n} \rightarrow \F_2$ defined as $f(x) = Tr(F(x))$.
The highest binary weight of the exponents of $F(x)$ is denoted as the algebraic degree of $F(x)$,
then the degree of $f(x) = Tr(F(x))$ is equal to the algebraic degree of $F(x)$.

With the above discussions it is now clear that we can characterize negabent
functions defined over the finite field as follows.
\begin{proposition}
The function $f : \mathbb F_{2^n} \rightarrow \mathbb F_2$ is negabent if and only if
\begin{equation}
\label{eq-main-nega}
\sum_{x \in \mathbb F_{2^n}} (-1)^{f(x) + f(x+a) + Tr(ax)} = 0
\end{equation}
for all $a \in \mathbb F_{2^n}^*$.
\end{proposition}

It is known that a bent function $f : \mathbb{F}_{2^n} \mapsto \mathbb{F}_{2}$
is bent if and only if

\begin{equation}
\label{eq-main-bent}
\sum_{x \in \mathbb F_{2^n}} (-1)^{f(x) + f(x+a)} = 0
\end{equation}
for all $a \in \mathbb F_{2^n}^*$.

A function for which both (\ref{eq-main-nega}) and (\ref{eq-main-bent})
hold is a bent-negabent function.

An $n$-variable Boolean function $\phi$ is called balanced if its weight is $2^{n-1}$.
Note that $\sum_{x \in \mathbb F_{2^n}} (-1)^{\phi(x)} = 0$ if and only if
$\phi$ is balanced.
Therefore, if $f$ is bent-negabent, then both
$f(x) + f(x+a)$ and $f(x) + f(x+a) + Tr(ax)$ are balanced
for all $a \in \mathbb F_{2^n}^*$.

\subsection{Linear structure of negabent functions}
As the sum (\ref{eq-main-nega}) involves derivatives of $f$, {\em i.e.}, $f(x) + f(x+a)$,
here we briefly discuss about the linear structures of negabent functions.
\begin{definition}
An $a \in \mathbb{F}_{2^n}^*$ is said to be a linear structure of a polynomial
$F(x)$ over $\mathbb{F}_{2^n}$ if the derivative $F(x) + F(x+a)$ is constant.
\end{definition}

From (\ref{eq-main-bent}), it is clear that a bent function can not have a linear structure.
However, a negabent function can have a linear structure. In fact, if $f$ is such that
any $a \in \mathbb{F}_{2^n}^*$ is a linear structure, then $f$ is negabent.
As in that case, the term $f(x)+ f(x+a)+ ax$ is $c+ax$, for some constant $c$,
which is affine, {\em i.e.}, balanced.
This happens when $f(x)$ is an affine polynomial,
which proves that if $f$ is affine it is negabent.

A polynomial $F(x)$ over $\F_{2^n}$ is called a complete mapping polynomial
if both $F(x)$ and $F(x) + x$ are permutation polynomials.
We use such permutation polynomials in our constructions.

\subsection{Quadratic negabent monomial functions}
We consider quadratic monomials and characterize when they are negabent.
\begin{proposition}
\label{quadprop}
Let $f: \mathbb F_{2^n} \rightarrow \mathbb F_2$ be a quadratic function of the form
$$f(x) = Tr(\lambda x^{2^k+1}).$$
Then $f$ is negabent if and only if
\begin{equation}
\label{ne-qu-co1}
\lambda^{2^{n-k}} a^{2^{n-k}} + \lambda a^{2^k} + a \ne 0
\end{equation}
for all $a \in \mathbb F_{2^n}^*$.
\end{proposition}
\begin{proof}
From (\ref{eq-main-nega}) we know that the function $f$ is negabent
if and only if for all $a \in \mathbb F_{2^n}^*$,
\begin{eqnarray}
\label{ne-qu-co}
\nonumber
\sum_{x \in \mathbb{F}_{2^n}}
(-1)^{Tr(\lambda x^{2^k+1}) + Tr(\lambda(x + a)^{2^k+1})
		+ Tr(ax)} &=& 0, \\
\mbox{ {\em i.e.}, } \sum_{x \in \mathbb{F}_{2^n}}
(-1)^{Tr\left((\lambda^{2^{n-k}} a^{2^{n-k}} + \lambda a^{2^k}
		+ a)x \right)} &=& 0.
\end{eqnarray}

Note that (\ref{ne-qu-co}) is true if and only if
$$\lambda^{2^{n-k}} a^{2^{n-k}} + \lambda a^{2^k} + a \ne 0,$$
for all $a \in \mathbb F_{2^n}^*$. Hence the result.

\end{proof}

\begin{proposition}
\label{prop-nega-perm}
The quadratic function $f: \mathbb F_{2^n} \rightarrow \mathbb F_2$
of the form $f(x) = Tr(\lambda x^{2^k+1})$
is negabent if and only if
$$P(x) = \lambda^{2^{n-k}} x^{2^{n-k}} + \lambda x^{2^k} + x$$
is a permutation polynomial over $\mathbb F_{2^n}$.
\end{proposition}
\begin{proof}
From Proposition \ref{quadprop}, we know that $f$ is negabent if and only if
$$\lambda^{2^{n-k}} a^{2^{n-k}} + \lambda a^{2^k} + a \ne 0$$
for all $a \in \mathbb F_{2^n}^*$. That means for such $\lambda \in \mathbb F_{2^n}^*$,
$P(x) = \lambda^{2^{n-k}} x^{2^{n-k}} + \lambda x^{2^k} + x$
has no non zero root in $\mathbb F_{2^n}$. Note that $P(x)$ is a linearized
polynomial and linearized polynomial is permutation if and only if
it has no non zero root. Hence the result.

\end{proof}

The polynomial $P(x)$ is a permutation over $\mathbb F_{2^n}$ if and only if
the polynomial $P(x)^{2^k}$ is a permutation over $\mathbb F_{2^n}$. Therefore,
$Tr(\lambda x^{2^k+1})$ is a negabent if and only if
$\lambda x + \lambda^{2^k} x^{2^{2k}} + x^{2^k}$
is a permutation. There are some results on the number of solutions
of the polynomial
$\lambda^{2^k} x^{2^{2k}} + x^{2^k} + \lambda x$
in \cite{HK10}.

Let us point out some results related to the equation
$$\lambda^{2^k} x^{2^{2k}} + x^{2^k} + \lambda x = 0$$
which has been extensively studied in \cite{HK10}.
Let $\gcd(k, n) = d \geq 1$ and $n = td$ for $t > 1$.
A particular sequence of polynomials over $\mathbb F_{2^n}$
is introduced as follows.
\begin{eqnarray}
\nonumber
C_1(x) &=& 1,\\
\nonumber
C_2(x) &=& 1,\\
C_{i+2}(x) &=& C_{i+1}(x) + x^{2^{ik}}C_i(x)
\mbox{~~for~~} 1 \leq i \leq t-1.
\end{eqnarray}

Another polynomial $Z_n(x)$ over $\mathbb F_{2^n}$ is defined as follows.
\begin{eqnarray}
\nonumber
Z_1(x) &=& 1,\\
Z_t(x) &=& C_{n+1}(x) + x C_{t-1}^{2^k}(x)
\mbox{~~for~~} t > 1.
\end{eqnarray}

Then we have the following result from \cite[Proposition 2]{HK10}.
\begin{proposition}
\label{z-root}
Let $\gcd(k, n) = d \geq 1$ and $n = td$ for $t > 1$.
The equation
$$\lambda^{2^k} x^{2^{2k}} + x^{2^k} + \lambda x = 0$$
defined over $\mathbb F_{2^n}$ has no non zero solution in
$\mathbb F_{2^n}$ if and only if
$Z_t(\lambda) \ne 0$.
\end{proposition}

The form of $\lambda$ for which
$Z_t(\lambda) = 0$ is known, which is as follows.
\begin{lemma}\cite[Corollary 1]{HK10}
\label{root-zt}
Let $\gcd(k, n) = d \geq 1$ and $n = td$ for $t > 1$.
Then $\alpha$ is a zero of $Z_t(x)$ in $\mathbb F_{2^n}$
if and only if it is of the form
\begin{equation}
\label{zero-form}
\frac{v_0^{2^{2k}+1}}{(v_0+v_1)^{2^k+1}},
\end{equation}
where $v_0 \in \mathbb F_{2^n} \setminus \mathbb F_{2^d}$ and $v_1 = v_0^{2^k}$.
The total number of distinct roots are
$$\begin{cases}
\frac{2^{n+d} - 2^d}{2^{2d} - 1} \mbox{~~~~ for even } t\\
\frac{2^{n+d} - 2^{2d}}{2^{2d} - 1} \mbox{~~~ for odd } t.
\end{cases}$$
\end{lemma}

Therefore, we have the following theorem which characterizes the quadratic
negabent monomials.
\begin{theorem}
The function $f : x \mapsto Tr(\lambda x^{2^k+1})$ is negabent if and
only if $\lambda$ can not be written as
$\frac{v_0^{2^{2k}+1}}{(v_0+v_1)^{2^k+1}}$ for
$v_0 \in \mathbb F_{2^n} \setminus \mathbb F_{2^d}$ and $v_1 = v_0^{2^k}$
where $\gcd(k, n) = d$ and $n = td$.
\end{theorem}
\begin{proof}
Proposition \ref{prop-nega-perm} and Proposition \ref{z-root}
imply that $f$ is negabent if and only if $\lambda$ is not a zero of
$Z_t(x)$ where $\gcd(k,n) = d$ and $n = td$.
From Lemma \ref{root-zt}, we know that $Z_t(\lambda) \ne 0$ if and only if
$\lambda$ is not of the form
$\frac{v_0^{2^{2k}+1}}{(v_0+v_1)^{2^k+1}}$
where $v_0 \in \mathbb F_{2^n} \setminus \mathbb F_{2^d}$ and $v_1 = v_0^{2^k}$.

\end{proof}

\subsection{Quadratic bent-negabent functions}
We recall the well known result on the quadratic bent monomials.
This is directly taken from \cite{DL04}.
\begin{lemma}\cite{DL04}
\label{gold-bent}
Let $\lambda \in \mathbb F_{2^n}$ and $n$ even.
The function
$f : \mathbb F_{2^n} \rightarrow \mathbb F_2$
with $$f(x) = Tr(\lambda x^{2^k+1})$$
is bent if and only if
$$\lambda \notin \{x^{2^k+1} \mid x \in \mathbb F_{2^n}\}.$$
\end{lemma}

Note that $\lambda^{2^{n-k}} x^{2^{n-k}} + \lambda x^{2^k}$
is a permutation if and only if
$\lambda \notin \{x^{2^k+1} \mid x \in \mathbb F_{2^n}\}$.
Therefore, if $f(x) = Tr(\lambda x^{2^k+1})$
is negabent, then Proposition \ref{prop-nega-perm} tells that
$\lambda^{2^{n-k}} x^{2^{n-k}} + \lambda x^{2^k} + x$
is also a permutation polynomial, {\em i.e.},
$\lambda^{2^{n-k}} x^{2^{n-k}} + \lambda x^{2^k}$
is a complete mapping polynomial.
We summarize these results as follows.

\begin{theorem}
\label{th-bent-negabent}
Let $\lambda \in \mathbb F_{2^n}$ where $n$ is even.
The function
$f : \mathbb F_{2^n} \rightarrow \mathbb F_2$
with $$f(x) = Tr(\lambda x^{2^k+1})$$
is bent negabent if and only if one of the following two
equivalent statements holds.
\begin{enumerate}
\item
$\lambda^{2^{n-k}} x^{2^{n-k}} + \lambda x^{2^k}$
is a complete mapping polynomial.
\item
$\lambda$ is neither of the form
$\frac{v_0^{2^{2k}+1}}{(v_0+v_1)^{2^k+1}}$ nor of the form
$v^{2^k+1}$ for
$v \in \mathbb F_{2^n}$,
$v_0 \in \mathbb F_{2^n} \setminus \mathbb F_{2^d}$ and $v_1 = v_0^{2^k}$,
where $\gcd(k, n) = d$ and $n = td$.
\end{enumerate}
\end{theorem}

The existence of quadratic bent-negabent functions is known
\cite[Theorem 5]{PP07}.
However, we reprove the same result by simple counting argument
and using the previous characterization of the bent-negabent functions.
\begin{theorem}
For all $n \geq 4$, quadratic bent-negabent functions always exist.
\end{theorem}
\begin{proof} We show that there always exists a $\lambda \in \mathbb F_{2^n}$
which satisfies the condition 2 of Theorem \ref{th-bent-negabent}.

If $\gcd(2^k+1, 2^n-1) = 1$, then $x \mapsto x^{2^k+1}$
is a bijection. Then for any $\lambda \in \mathbb F_{2^n}$ there exists
$x \in \mathbb F_{2^n}$ such that $\lambda = x^{2^k+1}$. Therefore, if
$Tr(\lambda x^{2^k+1})$ is bent then $\gcd(2^k+1, 2^n-1) > 1$.
Since $2$ does not divide both of $2^k+1$ and $2^n-1$. Therefore,
$\gcd(2^k+1, 2^n-1) \geq 3$.
Let $S_1 = \{x^{2^k+1} | x\in \mathbb F_{2^n}\},$ then $|S_1| \leq \frac{2^n-1}{3}$.
On the other hand, if $\gcd(k, n) = d$ and $n = td$, then
by Lemma \ref{root-zt}, we know that the number of possible
$\lambda \in \mathbb F_{2^n}$ such that $\lambda$ is of the form
$\frac{v_0^{2^{2k}+1}}{(v_0+v_1)^{2^k+1}}$ is
$\begin{cases}
\frac{2^{n+d} - 2^d}{2^{2d} - 1} \mbox{ for even } t\\
\frac{2^{n+d} - 2^{2d}}{2^{2d} - 1} \mbox{ for odd } t
\end{cases}$.
Let $S_2 = \{y \in \mathbb F_{2^n} | y = \frac{v_0^{2^{2k}+1}}{(v_0+v_1)^{2^k+1}}\}$.
Note that $|S_1 \cup S_2| = |S_1| + |S_2| - |S_1 \cap S_2|$. Then
$$|S_1 \cup S_2| \leq  \frac{2^n-1}{3}
+ \frac{2^{n+d}-2^{d}}{2^{2d}-1} - |S_1 \cap S_2|.$$
Therefore,
\begin{eqnarray*}
2^n - |S_1 \cup S_2| & \geq & 2^n - \frac{2^n-1}{3}
- \frac{2^{n+d} - 2^d}{2^{2d} - 1} + |S_1 \cap S_2|\\
\ & = & (2^n -1).\frac{2.2^{2d} - 3.2^d - 2}{3(2^{2d}-1)}
+ |S_1 \cap S_2| + 1\\
\ & \geq & |S_1 \cap S_2| + 1, ~~~~\mbox{ since } 2.2^{2d} - 3.2^d - 2 \geq 0.
\end{eqnarray*}

Therefore, we see that a $\lambda \in \mathbb F_{2^n}$ always exists
that satisfies Condition 2 of Theorem \ref{th-bent-negabent}.

This proves the theorem.

\end{proof}

Below we characterize bent-negabent functions when $n = 2k$.
\begin{proposition}
Let $n = 2k$ and $f: x \mapsto Tr(\lambda x^{2^k+1})$ be a quadratic function
defined over $\mathbb F_{2^n}$.
Then $f$ is negabent if and only if
$\lambda + \lambda^{2^k} \ne 1$. Moreover, $f$ is bent-negabent
if and only if
$\lambda + \lambda^{2^k} \notin \mathbb F_2$.
\end{proposition}
\begin{proof}
By Proposition \ref{prop-nega-perm} we have that
$f$ is negabent if and only if
$P(x) = \lambda^{2^{n-k}} x^{2^{n-k}} + \lambda x^{2^k} + x$ is a permutation,
{\em i.e.}, $P(x)^{2^k} = \lambda x + \lambda^{2^k} x^{2^{2k}} + x^{2^k}$
is a permutation. Since $n = 2k$, therefore,
$P(x)^{2^k} = (\lambda + \lambda^{2^k})x + x^{2^k}$. Now
$(\lambda + \lambda^{2^k})x + x^{2^k}$ is permutation if and only if
$(\lambda + \lambda^{2^k})x + x^{2^k} \ne 0$,
{\em i.e.}, $\lambda + \lambda^{2^k} \ne x^{2^k-1}$, for all $x \in \mathbb F_{2^n}^*$.
Note that $\lambda + \lambda^{2^k} \in \mathbb F_{2^k}$ for all
$\lambda \in \mathbb F^{2^n}$ and the mapping
$\lambda \mapsto \lambda + \lambda^{2^k}$ is onto.
Let us consider the group
$G = \{x^{2^k-1}| x \in \mathbb F_{2^n}^*\}$. The intersection
of $\mathbb F_{2^k}$ and $G$ is $\{1\}$.
Therefore, $Tr(\lambda x^{2^k+1})$ is negabent if and only if
$\lambda + \lambda^{2^k} \ne 1$.
We know that $f$ is bent if and only if $\lambda \ne x^{2^k+1}$
for some $x \in \mathbb F_{2^n}$. Note that if $\lambda = x^{2^k+1}$
then $\lambda \in \mathbb F_{2^k}$ and $\lambda + \lambda^{2^k} = 0$.
Therefore, $f$ is bent-negabent if and only if
$\lambda + \lambda^{2^k} \notin \mathbb F_{2}$.

\end{proof}

\section{Maiorana-McFarland bent-negabent functions}
Maiorana-McFarland is an important class of bent functions
which was extensively studied by Dillon \cite[pp.~90-95]{Dillon}. This
class is usually called the {\em class $\cal M$} of bent functions.
\begin{lemma}\label{mmf}
Let $n=2t$. Let us consider a Boolean function $f$ defined by
\begin{equation}\label{eq:mmf}
f~:~(x,y)\in \mathbb{F}_{2^t} \times \mathbb{F}_{2^t}
  ~~\mapsto ~~Tr^t_1\left(x\pi(y)+h(y)\right)
\end{equation}
where $\pi$ is a function over $\mathbb{F}_{2^t}$ and $h$ is any function
on $\mathbb{F}_{2^t}$. Then $f$ is a bent function if and only if
$\pi$ is a bijection.
\end{lemma}

\begin{theorem}
\label{th:mmf}
Let $f$ be a Maiorana-McFarland function as in Lemma \ref{mmf}.
Then $f$ is negabent if and only if for all $a, b \in \mathbb F_{2^t}^*$
\begin{equation}
\sum_{y \in Y_{a,b}} (-1)^{Tr_1^t(a \pi(y)) + h(y) + h(y+b) +by)} = 0,
\end{equation}
where $Y_{a,b} = \{ y \in \mathbb F_{2^t} | \pi(y) + \pi(y+b) = a\}$ such that
$Y_{a, b}$ is non empty.
\end{theorem}
\begin{proof}
From (\ref{eq-main-nega}) we have, $f(x, y)$ is negabent
if and only if for all $(a, b) \in \mathbb F_{2^t} \times \mathbb F_{2^t} \setminus \{(0,0)\}$
\begin{eqnarray}
\label{eq:mmf-nega}
\nonumber
\sum_{(x, y) \in \mathbb F_{2^t} \times \mathbb F_{2^t}}
(-1)^{f(x,y) + f(x+a, y+b) + Tr_1^t(ax) + Tr_1^t(by)} & = & 0\\
\nonumber
\sum_{(x, y) \in \mathbb F_{2^t} \times \mathbb F_{2^t}}
(-1)^{Tr_1^t(x(\pi(y) + \pi(y+b)+a)) + Tr_1^t(a \pi(y+b) + h(y) + h(y+b) + by)} & = & 0.
\end{eqnarray}
Let
$$S_{a,b} = \displaystyle \sum_{(x, y) \in \mathbb F_{2^t} \times \mathbb F_{2^t}}
(-1)^{Tr_1^t(x(\pi(y) + \pi(y+b)+a)) + Tr_1^t(a \pi(y+b) + h(y) + h(y+b) + by)}.$$
We treat the sum $S_{a,b}$ in the following cases.

\noindent
{\bf CASE 1:} For $a \ne 0$ and $b = 0$.
Then
\begin{eqnarray*}
S_{a,b} & = & \sum_{(x, y) \in \mathbb F_{2^t} \times \mathbb F_{2^t}}
(-1)^{Tr_1^t(ax) + Tr_1^t(a \pi(y))}\\
\ & = & \sum_{x \in \mathbb F_{2^t}} (-1)^{Tr_1^t(ax)}
\sum_{y \in \mathbb F_{2^t}} (-1)^{Tr_1^t(a\pi(y))}\\
\ & = & 0.
\end{eqnarray*}

\noindent
{\bf CASE 2:} For $a = 0$ and $b \ne 0$.
Then
\begin{eqnarray*}
S_{a,b} & = & \sum_{(x, y) \in \mathbb F_{2^t} \times \mathbb F_{2^t}}
(-1)^{Tr_1^t(x(\pi(y) + \pi(y+b))) + Tr_1^t(h(y) + h(y+b) + by)}\\
\ & = & \sum_{y \in \mathbb F_{2^t}} (-1)^{Tr_1^t(h(y) + h(y+b) + by)}
\sum_{x \in \mathbb F_{2^t}} (-1)^{Tr_1^t(x(\pi(y) + \pi(y+b)))}\\
\ & = & \sum_{y \in \mathbb F_{2^t}} (-1)^{Tr_1^t(h(y) + h(y+b) + by)} \times 0\\
\ & \ & ~~~~~~~~~~~~~\mbox{ since $\pi$ is a permutation, } \pi(y) \ne \pi(y+b)\\
\ & = & 0.
\end{eqnarray*}

\noindent
{\bf CASE 3:} For $a \ne 0$ and $b \ne 0$.
Then
\begin{eqnarray*}
S_{a,b} & = & \sum_{(x, y) \in \mathbb F_{2^t} \times \mathbb F_{2^t}}
(-1)^{Tr_1^t(a \pi(y+b) + h(y) + h(y+b) + by) + Tr_1^t(x(\pi(y) + \pi(y+b)+a))}\\
\ & = & \sum_{y \in \mathbb F_{2^t}} (-1)^{Tr_1^t(a \pi(y+b) + h(y) + h(y+b) + by)}
\sum_{x \in \mathbb F_{2^t}} (-1)^{Tr_1^t(x(\pi(y) + \pi(y+b)+a))}.
\end{eqnarray*}
If there exists some $y$ such that $y \notin Y_{a,b}$,
{\em i.e.}, $\pi(y) + \pi(y+b) \ne a$,
then $$\sum_{x \in \mathbb F_{2^t}} (-1)^{Tr_1^t(x(\pi(y) + \pi(y+b)+a))} = 0.$$
On the other hand if $y \in Y_{a,b}$, {\em i.e.}, $\pi(y) + \pi(y+b) = a$,
then $$\sum_{x \in \mathbb F_{2^t}} (-1)^{Tr_1^t(x(\pi(y) + \pi(y+b)+a))} = 2^t.$$
Therefore,
\begin{eqnarray*}
S_{a,b} & = & 2^t\sum_{y \in Y_{a,b}} (-1)^{Tr_1^t(a \pi(y+b) + h(y) + h(y+b) + by)}\\
\ & = & 2^t\sum_{y \in Y_{a,b}} (-1)^{Tr_1^t(a \pi(y) + a^2 + h(y) + h(y+b) + by)}\\
\ & \ & ~~~~~~~~~~~~ \mbox{since $\pi(y+b) = \pi(y) + a$ for $y \in Y_{a,b}$}\\
\ & = & 2^t (-1)^{Tr(a^2)} \sum_{y \in Y_{a,b}}(-1)^{Tr_1^t(a \pi(y) + h(y) + h(y+b) + by)}.
\end{eqnarray*}

Therefore, $S_{a,b} = 0$ if and only if
$$\displaystyle\sum_{y \in Y_{a,b}}(-1)^{Tr_1^t(a \pi(y) + h(y) + h(y+b) + by)} = 0.$$

Thus after discussing all the above cases it is clear that the Maiorana-McFarland
bent function $f$ is negabent if and only if
$$\sum_{y \in Y_{a,b}}(-1)^{Tr_1^t(a \pi(y) + h(y) + h(y+b) + by)} = 0.$$

\end{proof}

This Theorem gives us the clue to construct negabent functions over the finite
fields that belong to the class of Maiorana-McFarland bent functions.

\begin{definition}
A mapping $F : \mathbb{F}_{2^n} \rightarrow \mathbb{F}_{2^n}$ is
called homomorphic if $F(x + y) = F(x) + F(y)$ and $F(xy) = F(x) F(y)$
for all $x, y \in \mathbb{F}_{2^n}$.
\end{definition}
The only possible homomorphic permutation over $\mathbb{F}_{2^n}$
is of the form $x \mapsto x^{2^i}$. Note that $Tr_1^n(x) = Tr_1^n(x^{2^i})$,
therefore the mapping $x \mapsto Tr_1^n(x)$ is invariant under the
action of this permutation.
Using this observation we show an interesting consequence
of Theorem \ref{th:mmf}, when the permutation $\pi$ is chosen as
$\pi(x) = x^{2^i}$.

\begin{theorem}
\label{homo}
Let $f : (x,y) \in \mathbb F_{2^{t}} \times \mathbb F_{2^{t}} \mapsto
\mathbb{F}_{2}$ be a Maiorana-McFarland bent function
given by
\begin{equation}
f(x,y) = Tr_1^t(x y^{2^i} + h(y)),
\end{equation}
Then $f$ is negabent if and only if $Tr_1^t(h(y))$ is a
bent function over $\mathbb F_{2^t}$.
\end{theorem}
\begin{proof}
Let $\pi(y) = y^{2^i}$. Then $\pi(y)$ is a homomorphic
permutation polynomial over $\mathbb F_{2^{t}}$.
From the linearity of $\pi$ we have
$\pi(y)+\pi(y+b) = a$
if and only if $\pi(b) = a$.
Then \begin{equation*}
Y_{a,b} = \begin{cases}
\mathbb{F}_{2^t} & \text{when} ~\pi(b) = a\\
\text{empty} & \text{when} ~\pi(b) \ne a.
\end{cases}
\end{equation*}
Since $\pi$ is a permutation, for each $a$ there will be a $b$ such that
$\pi(b) = a$. For such $a$ and $b$
{\small
\begin{eqnarray*}
\sum_{y \in \mathbb F_{2^t}}(-1)^{Tr_1^t(a \pi(y) + h(y) + h(y+b) + by)}
& = & \sum_{y \in \mathbb F_{2^t}}(-1)^{Tr_1^t(\pi(b)\pi(y) + by
+ h(y) + h(y+b))}\\
& = & \sum_{y \in \mathbb F_{2^t}}(-1)^{Tr_1^t(\pi(by)) + Tr_1^t(by)
+ Tr_1^t(h(y) + h(y+b))}.
\end{eqnarray*}
}
Note that $Tr_1^t(y) = Tr_1^t(\pi(y))$, for all $y \in \mathbb{F}_2^t$.
So
$$
\sum_{y \in \mathbb F_{2^t}}(-1)^{Tr_1^t(a \pi(y) + h(y) + h(y+b) + by)}
 = \sum_{y \in \mathbb F_{2^t}}(-1)^{Tr_1^t(h(y) + h(y+b))}.
$$
Using Theorem \ref{th:mmf}, the function  $f$ is negabent if and only if
$$\sum_{y \in \mathbb F_{2^t}}(-1)^{Tr_1^t(h(y)) + Tr_1^t(h(y+b))} = 0,$$
for all $b \in \mathbb F_{2^t}^*$, {\em i.e.}, $Tr_1^t(h(y))$ is a bent function
over $\mathbb F_{2^t}$.

Thus the result follows.

\end{proof}

Similar kind of result was proved in \cite{Sugatanega}, where the
function was defined over the vector space $\mathbb{F}_2^n$ and the permutation
was  such that $wt(x + y) = wt(\pi(x) + \pi(y))$.
However, the result of Theorem \ref{homo} is quite distinct as it is
in the domain of finite fields.
Moreover, Theorem \ref{th:mmf} is a general characterization of bent-negabent
Maiorana-McFarland functions and several constructions of Maiorana-McFarland
bent-negabent functions can be obtained from this.
For instance, Theorem \ref{homo} allows us to construct bent-negabent
Maiorana-McFarland function of degree $n/4$ over $\mathbb{F}_{2^n}$
by choosing a bent function of degree $n/4$ as $h$, where $n = 2t$.

\section{Negabent functions from bent functions}
\label{new}
We show that given a negabent function over a finite field, one can construct a bent function,
and vice versa.
First we define
$Q : \mathbb F_{2^n} \rightarrow \mathbb F_2$ as
\begin{equation}
\label{def-Q}
Q(x) = \sum_{i = 1}^{{n \over 2} - 1}Tr_1^n(x^{2^i + 1}) + Tr_1^{n \over 2}(x^{2^{n \over 2} + 1})
\end{equation}

As mentioned earlier, for simplicity we write $Tr_1^n(x) = Tr(x)$.

We also mention a result
from \cite{decompose} and \cite{carlet-parseval} which will be useful in
proving our result. These results were proved for Boolean functions defined
over vector spaces, however, it is easy to see the equivalent results
when the Boolean function is defined by the trace representation.

\begin{lemma}{\cite[Corollary 1]{decompose}}
\label{pascale}
Suppose $H_{\beta} = \{x \in \F_{2^n} : Tr(\beta x) = 0\}$ is
a hyperplane. If $f: \F_{2^n} \rightarrow \F_2$ is bent, then for
any $a \notin H_{\beta}$,
\begin{eqnarray}
\sum_{x \in \mathbb{F}_{2^n}}
(-1)^{f(x) + f(x+a) + Tr(\beta x)} = 0.
\end{eqnarray}
\end{lemma}

\begin{lemma}{\cite[Theorem V.3]{carlet-parseval}}
\label{bent-hyp}
The Boolean function $f: \F_{2^n} \rightarrow \F_2$ is bent if and only
if there exists a hyperplane $\mathcal{H}$ such that $f(x) + f(x+a)$ is balanced
for every nonzero $a \in \mathcal{H}$.
\end{lemma}

\vspace{1cm}
\begin{theorem}
\label{bent-quad-th}
Suppose $f : \mathbb F_{2^n} \rightarrow \mathbb F_2$,
and $Q$ is as defined in (\ref{def-Q}).

\begin{enumerate}

\item
\label{b2n}
if $f$ is bent then $f+Q$ is negabent.

\item
\label{n2b}
If $f$ is negabent then $f+Q$ is bent,
\end{enumerate}
\end{theorem}

\begin{proof}
Suppose $a \in F_{2^n}^*$, then
\begin{eqnarray}
\label{der-Q}
\nonumber
Q(x) + Q(x+a) 	& = & \sum_{i = 1}^{{n \over 2}-1}Tr(a^{2^i} x + a x^{2^i}) + Tr_1^{n \over 2}(a x^{2^{n \over 2}} + a^{2^{n \over 2}} x) + \text{constant} \\
\nonumber
\		& = & \sum_{i = 1}^{{n \over 2}-1}Tr((a^{2^i} + a^{2^{n-i}})x) + Tr_1^n(a^{2^{n \over 2}} x) + \text{constant} \\
\nonumber
\		& = & \sum_{i = 1}^{{n \over 2}-1}Tr((a^{2^i} + a^{2^{n-i}})x)
			+ Tr(a^{2^{{n \over 2}}}x)+ Tr(ax)\\
\nonumber
		&   &~~~~~~~~~~	  + Tr(ax) +  \text{constant},\\
\nonumber
\		& = & Tr(Tr(a)x) + Tr(ax) + \text{constant} \\
\		& = & Tr(a)Tr(x) + Tr(ax) + \text{constant}.
\end{eqnarray}

\noindent
Without any loss of generality we ignore the constant term in $Q(x)+Q(x+a)$.

\vspace{1cm}
\noindent
{
\bf Case 1:}
We prove that if $f$ is bent, then $f+Q$ is negabent, for which we have to show that
\begin{center}
$ \begin{array}{l}
\displaystyle
\sum_{x \in \mathbb{F}_{2^n}}
(-1)^{f(x) + f(x+a) +Q(x) + Q(x+a) + Tr(ax)} = 0
\end{array}$
\end{center}

We have
\begin{center}
$ \begin{array}{l}
\displaystyle
\sum_{x \in \mathbb{F}_{2^n}}
(-1)^{f(x) + f(x+a) +Q(x) + Q(x+a) + Tr(ax)} \\
\displaystyle
= \sum_{x \in \mathbb{F}_{2^n}} (-1)^{f(x) + f(x+a)+ Tr(a)Tr(x)}.
\end{array}$
\end{center}

\vspace{1cm}
\noindent
{\bf Subcase 1.1:}
If $Tr(a) = 0$, then
\begin{eqnarray}
\nonumber
\sum_{x \in \mathbb{F}_{2^n}}
(-1)^{f(x) + f(x+a) + Tr(a)Tr(x)} = 0, \mbox{ since $f$ is bent}.
\end{eqnarray}

\vspace{1cm}
\noindent
{\bf Subcase 1.2:}
If $Tr(a) = 1$, then $a$ does not belong to the hyperplane
$$H_1 = \{x \in \F_{2^n} : Tr(1.x) = 0\}.$$ Therefore, by Lemma \ref{pascale},
\begin{eqnarray}
\nonumber
\sum_{x \in \mathbb{F}_{2^n}}
(-1)^{f(x) + f(x+a) + Tr(1.x)} = 0.
\end{eqnarray}
So for any $a \in F_{2^n}^*$,
\begin{eqnarray}
\nonumber
\sum_{x \in \mathbb{F}_{2^n}}
(-1)^{f(x) + f(x+a) +Q(x) + Q(x+a) + Tr(ax)} = 0.
\end{eqnarray}
This implies that is $f + Q$ is negabent.

\vspace{1cm}
\noindent
{\bf Case 2:}
Next we suppose that $f$ is negabent and prove that $g = f+Q$ is bent.
For any $a \in \F_{2^n}^*$ we have
\begin{center}
$\begin{array}{l}
\displaystyle \sum_{x \in \mathbb{F}_{2^n}}
(-1)^{g(x) + g(x+a)} \\
= \displaystyle \sum_{x \in \mathbb{F}_{2^n}}
(-1)^{f(x) + f(x+a) + Q(x) + Q(x+a)} \\
= \displaystyle \sum_{x \in \mathbb{F}_{2^n}} (-1)^{f(x) + f(x+a)+ Tr(a)Tr(x) + Tr(ax)}, \mbox{by } (\ref{der-Q})\\
= \displaystyle \sum_{x \in \mathbb{F}_{2^n}} (-1)^{f(x) + f(x+a) + Tr(ax)}, \mbox{if } a \in H_1, i.e. Tr(a) = 0\\
= 0, \mbox{since $f$ is negabent}.
\end{array}$
\end{center}

Therefore, we see that for any nonzero $a$ that belongs to the hyperplane $H_1$,
$g(x) + g(x+a)$ is balanced. Hence by Lemma \ref{bent-hyp}, we prove that $g$ is
bent.

\end{proof}

This theorem has interesting consequences.

\begin{corollary}
\label{btnbt-cor}
The Boolean function $f: \F_{2^n} \rightarrow \F_2$ is bent-negabent
if and only if both $f$ and $f+Q$ are bent.
\end{corollary}

\begin{corollary}
The Boolean function $f: \F_{2^n} \rightarrow \F_2$ is bent-negabent
if and only if $f+Q$ is bent-negabent.
\end{corollary}

\begin{corollary}
The function $Q$ is bent but not negabent.
\end{corollary}
\begin{proof}
It is easy to check that $Q$ is bent by looking at its derivative given
in (\ref{der-Q}).

Now on the contrary, assume that $Q$ is negabent. Then $g = Q+Q = 0$
is bent (by Theorem \ref{bent-quad-th}), which is a contradiction.

\end{proof}

We now use the result of Theorem \ref{bent-quad-th} to construct bent-negabent
functions.
Note that any two quadratic bent functions are affine equivalent.
It is clear that there is one-one correspondence between the
bent function defined over $\F_{2^{2t}}$ and $\F_{2^t} \times \F_{2^t}$.
With abuse of notation we use $Q(x,y)$ defined over $\F_{2^t} \times \F_{2^t}$
 as the corresponding bent function for $Q(x)$ which is defined in (\ref{def-Q}).
That means the bent function
$G : (x,y) \in \mathbb F_{2^{t}} \times \mathbb F_{2^{t}} \mapsto \mathbb{F}_{2}$
given by
\begin{equation}
\label{G}
G(x,y) = Tr_1^t(x y),
\end{equation}
is affine equivalent to the bent function $Q(x,y)$.
This also means by Theorem \ref{bent-quad-th} that if $f(x,y)$ is a bent function
then $f(x,y) + Q(x,y)$ is negabent and vice versa.

Suppose $G(x,y)$ and $Q(x,y)$ are related by the relation
\begin{eqnarray}
\label{q-g}
Q(x,y) &=& G(\alpha_1 x + \alpha_2,\alpha_3 y + \alpha_4)  +
	 Tr_1^t(\beta x) + Tr_1^t(\gamma y) + c,
\end{eqnarray}
for some $\alpha_1, \alpha_2, \alpha_3, \alpha_, \beta, \gamma$ in $\F_2^t$,
$c \in \F_2$.

\begin{theorem}
\label{comp-th}
Let $f : (x,y) \in \mathbb F_{2^{t}} \times \mathbb F_{2^{t}} \mapsto
\mathbb{F}_{2}$ be a Maiorana-McFarland bent function
given by
$$
f(x,y) = Tr_1^t(x \pi(y)) + Tr_1^t(h(y)),
$$
where $\pi(y)$ is a complete mapping polynomial over $\F_{2^t}$,
$h(y)$ is any polynomial over $\F_{2^t}$,
and $G : (x,y) \in \mathbb F_{2^{t}} \times \mathbb F_{2^{t}} \mapsto
\mathbb{F}_{2}$ defined by $G(x,y) = Tr_1^t(x y)$.
Then
\begin{eqnarray}
\label{comp-th-cons}
\nonumber
F(x,y) &=&
f(\alpha_1 x + \alpha_2, \alpha_3 y + \alpha_4) + G(\alpha_1 x + \alpha_2, \alpha_3 y + \alpha_4) \\
	& \ & + Tr_1^t(\beta x) + Tr_1^t(\gamma y) + c
\end{eqnarray}
is a bent-negabent function.
\end{theorem}

\begin{proof}
We have
\begin{eqnarray}
\nonumber
f(x,y) + G(x,y) 	& = &  Tr_1^t(x \pi(y)) + Tr_1^t(h(y)) +  Tr_1^t(x y) \\
\nonumber
\			& = &  Tr_1^t(x (\pi(y) + y)) + Tr_1^t(h(y)).
\end{eqnarray}
Since $\pi(y)$ is a complete mapping polynomial over $\F_{2^t}$, $(\pi(y) + y)$ is a permutation polynomial,
so $f+G$ is a bent function. This also implies that
$$
F(x,y) =
f(\alpha_1 x + \alpha_2, \alpha_3 y + \alpha_4)+G(\alpha_1 x + \alpha_2, \alpha_3 y + \alpha_4) + Tr_1^t(\beta x) +
Tr_1^t(\gamma y) + c
$$
is a bent function.
We have
\begin{center}

$\begin{array}{l}
F(x,y)\\
= f(\alpha_1 x + \alpha_2, \alpha_3 y + \alpha_4)+G(\alpha_1 x + \alpha_2, \alpha_3 y + \alpha_4) + Tr_1^t(\beta x)
+ Tr_1^t(\gamma y) + c \\
= f(\alpha_1 x + \alpha_2, \alpha_3 y + \alpha_4) + Q(x,y), \text{by} \ref{q-g}.
\end{array}$

\end{center}

Note that $F(x,y) + Q(x,y) = f(\alpha_1 x + \alpha_2, \alpha_3 y + \alpha_4)$ is also bent.
So both $F(x,y)$ and $F(x,y) + Q(x,y)$ are bent.
Therefore, by Corollary \ref{btnbt-cor}, $F(x,y)$ is bent-negabent.

\end{proof}

At this point, we would like to refer to \cite[Theorem 22]{Sugatanega}, which also states
a result that is similar to Theorem \ref{comp-th}. In that result, the Boolean
function is defined over the vector space $\F_2^n$. Note that
complete mapping polynomials are defined over finite fields, however, the proof of \cite[Theorem 22]{Sugatanega}
works in the vector space domain.
They claim that $\pi(x_1, \ldots, x_t)$ is a permutation of $\F_2^t$ that corresponds to
the permutation $\pi(x)$ over the field $\F_{2^t}$, as well as
$\pi(x_1, \ldots, x_t) \oplus (x_1, \ldots, x_t)$ is the permutation
of $\F_2^t$ that corresponds to the permutation $\pi(x)+x$ over the field $\F_{2^t}$.
But it is not clear how this correspondence is realized.
On the other hand, Theorem \ref{comp-th} can directly apply the complete mapping
polynomials as the underlying Boolean function is defined over a finite filed.

Now we construct infinite classes of $n$-variable bent-negabent function with the maximum degree $n \over 2$.
Our construction is similar to that of Theorem $5$ of \cite{Pottnega}.
Their proof works when there is a permutation polynomial $p(x)$ over the vector space $\F_2^n$ such that $p(x) + x$
is also a permutation polynomial over $\F_2^n$. However,
this kind of permutation over the vector space $\F_2^n$ is not characterized, on the other hand,
these kind of permutation polynomials (complete mapping polynomials) are well characterized over finite field.

\begin{corollary}
Suppose $n = 2t$. Then
the $n$-variable function $F(x,y)$ defined in Theorem \ref{comp-th}, where the polynomial $h(y)$
has algebraic degree $t = {n \over 2}$ is a bent-negabent function
of degree $n \over 2$.
\end{corollary}
\begin{proof}
The algebraic degree of $h(y)$ is $t = {n \over 2}$, which implies that
the degree of $f(x,y) + G(x,y)$ is also $n \over 2$.
That also implies that the degree
$
F(x,y) =
f(\alpha_1 x + \alpha_2, \alpha_3 y + \alpha_4)+G(\alpha_1 x + \alpha_2, \alpha_3 y + \alpha_4) + Tr_1^t(\beta x) +
Tr_1^t(\gamma y) + c
$ is $n \over 2$.

\end{proof}

Two infinite classes of complete mapping polynomial are given in \cite{yann}.

\begin{theorem}{\cite[Theorem 4.3]{yann}}
\label{yann-th}
 Let $p$ be a prime and $m$ and $\ell$ are two positive integers. Let $k$
 be the multiplicative order of $p$ in $\mathbb{Z}_m$.
 Assume $a \in \F_{p^{k \ell}}$ is such
 that $(-a)^m \ne 1$. Then the polynomials
 $$\pi_1(x) = x(x^{\frac{p^{k \ell m}-1}{m} } + a),$$
 and
 $$\pi_2(x) = ax^{\frac{p^{k \ell m}-1}{m} + 1},$$
 are complete mapping polynomials over $\F_{p^{k\ell m}}$.
\end{theorem}

\begin{theorem}
Suppose $n = 2 k \ell m$ and $\pi_1(y), \pi_2(y)$ are the complete mapping polynomials
as given in Theorem \ref{yann-th}.
Then the $n$-variable function $F(x,y)$ as given in (\ref{comp-th-cons}) is a bent-negabent
with degree $n \over 2$,
for $f(x,y) = Tr_1^{{n \over 2}}(x \pi_1(y)) + Tr_1^{{n \over 2}}(y^{2^{n \over 2}-1})$
and $f(x,y) = Tr_1^{{n \over 2}}(x \pi_2(y)) + Tr_1^{{n \over 2}}(y^{2^{n \over 2}-1})$.
\end{theorem}
\begin{proof}
 This follows easily as the algebraic degree of $Tr_1^{{n \over 2}}(y^{2^{n \over 2}-1})$ is $n \over 2$.

\end{proof}

\section{Conclusion}
We have presented some characterizations of negabent functions over the finite field.
The analysis done here is useful in order to obtain further results
on negabent functions over finite fields.
In this paper, we have characterized quadratic negabent monomials.
The characterization
of negabent monomials of higher degree will be interesting.
We also have characterized negabent functions which are Maiorana-McFarland bent.
Moreover. we have presented a construction of bent-negabent functions with optimal
degree. This is the second known construction of such functions. However,
it is interesting to see further classes of such functions.

\section{Acknowledgments}
The author would like to thank Pascale Charpin who helped in proving Case 2 of Theorem \ref{bent-quad-th}.
He is also thankful to Alexander Kholosha for helpful discussions.

\bibliographystyle{alpha}
\bibliography{negabent.bib}

\end{document}