text:
\section{Introduction}

Let ${\mathbb F}_{q^n}$ denote the finite field of $q^n$ elements where $q=p^h$ for some prime $p$. For $n>1$ and $s \mid n$ the trace and norm over ${\mathbb F}_{q^s}$ of elements of ${\mathbb F}_{q^n}$ are defined as $\Tr_{q^n/q^s}(x)=x+x^{q^s}+\ldots+x^{q^{n-s}}$ and $\N_{q^n/q^s}(x)=x^{1+q^s+\ldots+q^{n-s}}$, respectively. When $s=1$ then we will simply write $\Tr(x)$ and $\N(x)$. Every function $f \colon {\mathbb F}_{q^n} \rightarrow {\mathbb F}_{q^n}$ can be given uniquely as a polynomial with coefficients in ${\mathbb F}_{q^n}$ and of degree at most $q^n-1$.
The function $f$ is ${\mathbb F}_q$-linear if and only if it is represented by a \emph{$q$-polynomial}, that is,
\begin{equation}
\label{fff}
f(x)=\sum_{i=0}^{n-1}a_i x^{q^i}
\end{equation}
with coefficients in ${\mathbb F}_{q^n}$. Such polynomials are also called \emph{linearized}.
If $f$ is given as in \eqref{fff}, then its adjoint (w.r.t. the symmetric non-degenerate bilinear form defined by $\langle x, y\rangle=\Tr(xy)$) is
\[\hat f(x):=\sum_{i=0}^{n-1}a_i^{q^{n-i}} x^{q^{n-i}},\]
i.e. $\Tr(xf(y))=\Tr(y\hat f(x))$ for any $x,y\in{\mathbb F}_{q^n}$.

The aim of this paper is to study what can be said about two $q$-polynomials $f$ and $g$ over ${\mathbb F}_{q^n}$ if they satisfy
\begin{equation}
\label{start}
Im\,\left(\frac{f(x)}{x}\right)=Im\,\left(\frac{g(x)}{x}\right)\ \footnote{By $Im\,(f(x)/x)$ we mean the image of the rational function $f(x)/x$, i.e. $\{f(x)/x \colon x\in {\mathbb F}_{q^n}^*\}$.}.
\end{equation}

For a given $q$-polynomial $f$, the equality \eqref{start} clearly holds with $g(x)=f(\lambda x)/\lambda$ for each $\lambda\in {\mathbb F}_{q^n}^*$.
A less obvious choice when \eqref{start} holds is when $g(x)=\hat f(\lambda x)/\lambda$, see \cite[Lemma 2.6]{BGMP2015} and the first part of \cite[Section 3]{CSMP2016}.

\medskip

When one of the functions in \eqref{start} is a monomial then the answer to the question posed above follows from McConnel's generalization \cite[Theorem 1]{McConnel} of a result due to Carlitz \cite{Carlitz} (see also Bruen and Levinger \cite{BL}).

\begin{theorem}{\cite[Theorem 1]{McConnel}}
\label{Carlitz}
Let $p$ denote a prime, $q=p^h$, and $1<d$ a divisor of $q-1$. Also, let $F \colon {\mathbb F}_{q} \rightarrow {\mathbb F}_{q}$ be a function such that $F(0)=0$ and $F(1)=1$.
Then
\[\left(F(x)-F(y)\right)^{\frac{q-1}{d}}=\left(x-y\right)^{\frac{q-1}{d}}\]
for all $x,y\in {\mathbb F}_{q}$ if and only if $F(x)=x^{p^j}$ for some $0 \leq j < h$ and $d\mid p^j-1$.
\end{theorem}

Indeed, when the function $F$ of Theorem \ref{Carlitz} is ${\mathbb F}_q$-linear, we easily get the following corollary (see Section \ref{BPR} for the proof, or   \cite[Corollary 1.4]{Praha} for the case when $q$ is an odd prime).

\begin{corollary}
\label{norm}
Let $g(x)$ and $f(x)=\alpha x^{q^k}$, $q=p^h$, be $q$-polynomials over ${\mathbb F}_{q^n}$ such that
\begin{equation}\label{form:im}
Im\,\left(\frac{f(x)}{x}\right)=Im\,\left(\frac{g(x)}{x}\right).
\end{equation}
Denote $\gcd(k,n)$ by $t$. Then $g(x)=\beta x^{q^s}$ with $\gcd(s,n)=t$ for some $\beta$ with $\N_{q^n/q^t}(\alpha)=\N_{q^n/q^t}(\beta)$.
\end{corollary}

\noindent Another case for which we know a complete answer to our problem is when $f(x)=\Tr(x)$.

\begin{theorem}[{\cite[Theorem 3.7]{CSMP2016}}]
\label{trace}
Let $f(x)=\Tr(x)$ and let $g(x)$ be a $q$-polynomial over ${\mathbb F}_{q^n}$ such that
\[Im\,(f(x)/x)=Im \,(g(x)/x).\]
Then $g(x)=\Tr(\lambda x)/\lambda$ for some $\lambda \in {\mathbb F}_{q^n}^*$.
\end{theorem}

\noindent Note that in Theorem \ref{trace} we have $\hat f(x) = f(x)$ and the only solutions for $g$ are $g(x)=f(\lambda x)/\lambda$, while in Corollary \ref{norm} we have (up to scalars) $\varphi(n)$ different solutions for $g$, where $\varphi$ is the Euler's totient function.

The problem posed in \eqref{start} is also related to the study of the directions determined by an additive function. Indeed, when $f$ is additive, then
\[Im\,(f(x)/x)=\left\{\frac{f(x)-f(y)}{x-y} \colon x\neq y,\, x,y\in {\mathbb F}_{q^n}\right\},\]
is the \emph{set of directions} determined by the graph of $f$, i.e. by the point set ${\mathcal G}_f:=\{(x,f(x)) \colon x\in {\mathbb F}_{q^n}\}\subset \AG(2,q^n)$. Hence, in this setting, the problem posed in \eqref{start} corresponds to finding the ${\mathbb F}_q$-linear functions whose graph determines the same set of directions.
The maximum cardinality of the image set of $f(x)/x$, where $f$ is a $q$-polynomial
over ${\mathbb F}_{q^n}$, is $(q^n-1)/(q-1)$, whereas, if ${\mathbb F}_q$ is the maximum field of linearity of $f$, then by \cite{BBBSSz, B2003, Cs2017} $Im\,(f(x)/x)$ contains at least $q^{n-1}+1$ elements, hence
\begin{equation}
\label{directions}
q^{n-1}+1\leq |Im\,(f(x)/x)| \leq \frac{q^{n}-1}{q-1}.
\end{equation}

The classical examples which show the sharpness of these bounds are the monomial functions $x^{q^s}$, with $\gcd (s,n)=1$, and the $\Tr(x)$ function.
However, these bounds are also achieved by other polynomials which are not "equivalent" to these examples (see Section 2 for more details).
\bigskip

Two ${\mathbb F}_q$-linear polynomials $f(x)$ and $h(x)$ of ${\mathbb F}_{q^n}[x]$ are \emph{equivalent} if the two graphs ${\mathcal G}_f$ and ${\mathcal G}_h$ are equivalent under tha action of the group $\Gamma {\mathrm L}(2,q^n)$, i.e. if there exists an element $\varphi\in\Gamma {\mathrm L}(2,q^n)$ such that ${\mathcal G}_f^\varphi={\mathcal G}_h$. In such a case, we say that $f$ and $h$ are \emph{equivalent} (via $\varphi$) and we write $h=f_\varphi$. It is easy to see that in this way we defined an equivalence relation on the set of $q$-polynomials over ${\mathbb F}_{q^n}$. If $f$ and $g$ are two $q$-polynomials such that $Im\,(f(x)/x)=Im\,(g(x)/x)$, then $Im\,(f_\varphi(x)/x)=Im\,(g_\varphi(x)/x)$ for any admissible $\varphi\in\mathrm{\Gamma L}(2,q^n)$ (see Proposition \eqref{imtrans}). This means that the problem posed in \eqref{start} can be investigated up to equivalence.

\bigskip

For $n\leq 4$, the only solutions for $g$ in Problem \eqref{start} are the trivial ones, i.e. either $g(x)=f(\lambda x)/x$ or $g(x)=\hat f(\lambda x)/x$ (cf. Theorem \ref{n=4}).

For the case $n=5$, in Section \ref{section-proof}, we prove the following main result.

\begin{theorem}
\label{main}
Let $f(x)$ and $g(x)$ be two $q$-polynomials over ${\mathbb F}_{q^5}$, with maximum field of linearity ${\mathbb F}_q$, such that $Im\,(f(x)/x)=Im\,(g(x)/x)$.
Then either there exists $\varphi\in \mathrm{\Gamma L}(2,q^5)$ such that $f_\varphi(x)=\alpha x^{q^i}$ and $g_\varphi(x)=\beta x^{q^j}$ with $\N(\alpha)=\N(\beta)$ for some
$i,j\in \{1,2,3,4\}$, or there exists $\lambda\in {\mathbb F}_{q^5}^*$ such that $g(x)=f(\lambda x)/\lambda$ or $g(x)=\hat{f}(\lambda x)/\lambda$.
\end{theorem}

Finally, the relation between $Im\,(f(x)/x)$ and the linear sets of rank $n$ of the projective line $\PG(1,q^n)$ will be pointed out in Section \ref{geom}.
As an application of Theorem \ref{main} we get a criteria of $\mathrm{P}\Gamma\mathrm{L}(2,q^5)$-equivalence for linear sets in $\PG(1,q^5)$ and this allows us to prove that the family of (maximum scattered) linear sets of rank $n$ of size $(q^n-1)/(q-1)$ in $\PG(1,q^n)$ found by Sheekey in \cite{Sh} contains members which are not-equivalent to the previously known linear sets of this size.

\section{Background and preliminary results}
\label{BPR}

Let us start this section by the following immediate corollary of \eqref{directions}.

\begin{proposition}
	\label{fieldoflinearity}
	If $Im\,(f(x)/x)=Im\,(g(x)/x)$ for two $q$-polynomials $f$ and $g$ over ${\mathbb F}_{q^n}$, then their maximum field of linearity coincide.
\end{proposition}
\begin{proof}
	Let ${\mathbb F}_{q^m}$ and ${\mathbb F}_{q^k}$ be the maximum fields of linearity of $f$ and $g$, respectively. Suppose to the contrary $m<k$.
	Then $|Im\,(g(x)/x)|\leq (q^n-1)/(q^k-1)<q^{n-k+1}+1\leq q^{n-m}+1 \leq |Im\,(f(x)/x)|$, a contradiction.
\end{proof}

Now we are able to prove Corollary \ref{norm}.

\begin{proof}
The maximum field of linearity of $f(x)$ is ${\mathbb F}_{q^t}$, thus, by Proposition \ref{fieldoflinearity}, $g(x)$ has to be a $q^t$-polynomial as well.
Then for $t>1$ the result follows from the $t=1$ case (after substituting $q$ for $q^t$ and $n/t$ for $n$) and hence we can assume that $f(x)$ and $g(x)$ are strictly ${\mathbb F}_q$-linear. By \eqref{form:im}, we note that $g(1)=\alpha z_0^{q^k-1}$, for some $z_0\in{\mathbb F}_{q^n}^*$. Let $F(x):=g(x)/g(1)$, then $F$ is a $q$-polynomial over ${\mathbb F}_{q^n}$, with $F(0)=0$ and $F(1)=1$. Also, from \eqref{form:im}, for each $x\in{\mathbb F}_{q^n}^*$ there exists $z\in{\mathbb F}_{q^n}^*$ such that
\[\frac{F(x)}{x}=\left(\frac{z}{z_0}\right)^{q^k-1}.\] This means that for each $x \in{\mathbb F}_{q^n}^*$ we get $\N\left(\frac{F(x)}{x}\right)=1$. From Theorem \ref{Carlitz} (with $d=q-1$) it follows that $F(x)=x^{p^j}$ for some $0\leq j<nh$.
Then Proposition \ref{fieldoflinearity} yields $p^j=q^s$ with $\gcd(s,n)=1$. We get the first part of the statement by putting $\beta:=f(1)$. Again from \eqref{form:im} it follows that $\N(\alpha)=\N(\beta)$.
\end{proof}

Let  $f$ and $g$ be two equivalent $q$-polynomials over ${\mathbb F}_{q^n}$ via the element  $\varphi\in\Gamma{\mathrm L}(2,q^n)$ represented by the invertible matrix $\begin{pmatrix}
a&b\\
c&d\\
\end{pmatrix}$ and with companion automorphism $\sigma$ of ${\mathbb F}_{q^n}$. Then
\begin{equation}\label{form:equiv}
\left\{
\begin{pmatrix}
x\\
g(x)
\end{pmatrix}
\colon x\in {\mathbb F}_{q^n}
\right\}=
\left\{
\begin{pmatrix}
a&b\\
c&d\\
\end{pmatrix}
\begin{pmatrix}
x^\sigma\\
f(x)^\sigma
\end{pmatrix}
\colon x \in {\mathbb F}_{q^n}\right\}.
\end{equation}
By \eqref{form:equiv} we get that the function $k_f(x):=ax^\sigma+bf(x)^\sigma$ is invertible and, with $h_f(x):=cx^\sigma+df(x)^\sigma$, it follows that  $g(x)=h_f(k_f^{-1}(x))$. Hence, we have proved the following proposition.

\begin{proposition}\label{fvarphi}
Let $f$ and $g$ be $q$-polynomials over ${\mathbb F}_{q^n}$ such that $g=f_\varphi$ for some $\varphi\in\mathrm{\Gamma L}(2,q^n)$ with linear part represented by $\left(
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right)$ and with companion automorphism $\sigma$. Let $k_f(x)=ax^\sigma+bf(x)^\sigma$ and
$h_f(x)=cx^\sigma+df(x)^\sigma$, then $k_f$ is invertible and $g(x)=f_\varphi(x)=h_f(k_f^{-1}(x))$.
\end{proposition}

From \eqref{form:equiv} it is also clear that
\begin{equation}
\label{map}
Im\,\left(\frac{f_\varphi(x)}x\right)=\left\{\frac{c+dz^\sigma}{a+bz^\sigma}
\colon z \in Im\,\left(\frac{f(x)}x\right)\right\}
\end{equation}
and hence
\begin{equation}
\label{size}
|Im\,(f_\varphi(x)/x)|=|Im\,(f(x)/x)|.
\end{equation}

From Equations \eqref{size} and \eqref{directions} the next result easily follows.
\begin{proposition}
\label{prop:lin}
	If two $q$-polynomials over ${\mathbb F}_{q^n}$ are equivalent, then their maximum field of linearity coincide.\qed
\end{proposition}

Note that $|Im\,(g(x)/x)|=|Im\,(f(x)/x)|$ does not imply the equivalence of $f$ and $g$. In fact, in the last section we will list the known examples of $q$-polynomials $f$ which are not equivalent to monomials but the size of $Im\,(f(x)/x)$ is maximal. To find such functions was also proposed in \cite{Praha} and, as it was observed by Sheekey, they determine certain MRD-codes \cite{Sh}.

The following results will be useful later in the paper.

\begin{proposition}
\label{imtrans}
If $Im\, (f(x)/x)=Im\, (g(x)/x)$ for some $q$-polynomials over ${\mathbb F}_{q^n}$, then $Im\, (f_\varphi(x)/x)=Im\, (g_\varphi(x)/x)$ holds for each admissible \footnote{An element $\varphi\in \mathrm{\Gamma L}(2,q^n)$ represented by the invertible matrix $\begin{pmatrix}
a&b\\
c&d\\
\end{pmatrix}$ and with companion automorphism $\sigma$ of ${\mathbb F}_{q^n}$ is said to be \emph{admissible} w.r.t. a given $q$-polynomial $f$ if either $b=0$ or $-(a/b)^{\sigma^{-1}}\notin Im\, (f(x)/x)$.} $\varphi\in \mathrm{\Gamma L}(2,q^n)$.
\end{proposition}
\begin{proof}
From $Im\,(f(x)/x)=Im\, (g(x)/x)$ it follows that any $\varphi\in \mathrm{\Gamma L}(2,q^n)$ admissible w.r.t. $f$ is admissible w.r.t. $g$ as well. Hence $k_f$ and $k_g$ are both invertible and we may construct $f_\varphi$ and $g_\varphi$ as indicated in Proposition \ref{fvarphi}. The statement now follows from Equation \eqref{map}.
\end{proof}

\begin{proposition}
\label{lambda0}
Let $f$ and $g$ be $q$-polynomials over ${\mathbb F}_{q^n}$ and take some $\varphi\in \mathrm{\Gamma L}(2,q^n)$ with companion automorphism $\sigma$. Then
$g_\varphi(x)=f_\varphi(\lambda^\sigma x)/\lambda^\sigma$
for some $\lambda\in {\mathbb F}_{q^n}^*$ if and only if $g(x)=f(\lambda x)/\lambda$.
\end{proposition}

\begin{proof}
First we prove the "if" part. Since $g(x)=f(\lambda x)/\lambda=(\omega_{1/\lambda}\circ f \circ \omega_\lambda)(x)$, where $\omega_\alpha$ denotes the scalar map $x\in{\mathbb F}_{q^n}\mapsto\alpha x\in{\mathbb F}_{q^n}$, direct computations show that $h_g=\omega_{1/\lambda^\sigma}\circ h_f\circ\omega_\lambda$ and $k_g=\omega_{1/\lambda^\sigma}\circ k_f\circ\omega_\lambda$. Then $g_\varphi=\omega_{1/\lambda^\sigma}\circ f_\varphi\circ\omega_{\lambda^\sigma}$ and the first part of the statement follows. The "only if" part follows from the "if" part applied to $g_\varphi(x)=f_\varphi(\lambda^\sigma x)/\lambda^\sigma$ and $\varphi^{-1}$; and from $(f_\varphi)_{\varphi^{-1}}=f$ and $(g_\varphi)_{\varphi^{-1}}=g$.
\end{proof}

Next we summarize what is known about Problem \eqref{start} for $n\leq 4$.

\begin{theorem}
	\label{n=4}
	Suppose $Im\, (f(x)/x)=Im\, (g(x)/x)$ for some $q$-polynomials over ${\mathbb F}_{q^n}$, $n\leq 4$, with maximum field of linearity ${\mathbb F}_q$. Then there exist $\varphi\in \mathrm{GL}(2,q^n)$ and $\lambda \in {\mathbb F}_{q^n}^*$ such that the following holds.
	\begin{itemize}
		\item If $n=2$ then $f_\varphi(x) = x^q$ and $g(x) = f(\lambda x)/\lambda$.
		\item If $n=3$ then either
		\[f_\varphi(x)=\Tr(x) \mbox{ and } g(x)=f(\lambda x)/\lambda\]
		or
		\[f_\varphi(x) = x^q \mbox{ and } g(x)=f(\lambda x)/\lambda \mbox{ or } g(x)=\hat{f}(\lambda x)/\lambda.\]
		\item  If $n=4$ then $g(x) = f(\lambda x)/\lambda$ or $g(x) = \hat{f}(\lambda x)/\lambda$.
	\end{itemize}
\end{theorem}
\begin{proof}
	In the $n=2$ case $f(x)=ax+bx^q$, $b\neq 0$. Then $\varphi:=\begin{pmatrix}
	1 & 0 \\
	-a/b & 1/b \\
\end{pmatrix}$ maps $f(x)$ to $x^q$. Then Proposition \ref{imtrans} and Corollary \ref{norm} give $g_\varphi(x)=f_\varphi(\mu x)/\mu$ and hence Proposition \ref{lambda0} gives $g(x)=f(\lambda x)/\lambda$ for some $\lambda\in{\mathbb F}_{q^n}$.
	If $n=3$ then according to \cite[Theorem 5]{LaVa2010} and \cite[Theorem 1.3]{CSMP2016}  there exists $\varphi\in \mathrm{GL}(2,q^3)$ such that either $f_\varphi(x)=\Tr(x)$ or $f_\varphi(x)=x^q$.
	In the former case Proposition \ref{imtrans} and Theorem \ref{trace} give $g_\varphi(x)=f_\varphi(\mu x)/\mu$ and the assertion follows from Proposition \ref{lambda0}. In the latter case Proposition \ref{imtrans} and Corollary \ref{norm} give $g_\varphi(x)=\alpha x^{q^i}$ where $i\in \{1,2\}$ and $\N(\alpha)=1$. If $i=1$, then $g_\varphi(x)=f_\varphi(\mu x)/\mu$ where $\mu^{q-1}=\alpha$ and the assertion follows from Proposition \ref{lambda0}. Let now $i=2$ and denote by
	$\begin{pmatrix}
	A&B\\
	C&D\\
	\end{pmatrix}$ the matrix of $\varphi^{-1}$. Also, let $\Delta$ denote the determinant of this matrix. Since $\N(\alpha)=1$, direct computations show
	\[f(x)=(f_{\varphi})_{\varphi^{-1}}(x)=\frac{(A^{q+q^2}C+B^{q+q^2}D)x+A^{q^2}\Delta x^q-B^q\Delta x^{q^2}}{\N(A)+\N(B)},\]
	\[g(x)=(g_{\varphi})_{\varphi^{-1}}(x)=\frac{(A^{q+q^2}C+B^{q+q^2}D)x-B^{q^2}\Delta \alpha^{q^2+1}x^q+A^q\Delta\alpha x^{q^2}}{\N(A)+\N(B)},\]
	and hence $g(x)=\hat{f}(\lambda x)/\lambda$ for each $\lambda\in {\mathbb F}_{q^3}^*$ with  $\lambda^{q-1}=\Delta^{1-q}/\alpha^q$.

	The case $n=4$ is \cite[Proposition 4.2 ]{CSMP2016}.
\end{proof}

\begin{remark}
{\rm
Theorem \ref{n=4} yields that there is a unique equivalence class of $q$-polynomials, with maximum field of linearity ${\mathbb F}_q$, when $n=2$. For $n=3$ there are two non-equivalent classes and they correspond to the classical examples: $\Tr(x)$ and $x^q$. Whereas, for $n=4$, from \cite[Sec. 5.3]{CSMP2016} and \cite[Table p. 54]{BoPo2005}, there exist at least eight non-equivalent classes. The possible sizes for the sets of directions determined by these strictly ${\mathbb F}_q$-linear functions are $q^3+1$, $q^3+q^2-q+1$, $q^3+q^2+1$ and $q^3+q^2+q+1$ and each of them is determined by at least two non-equivalent $q$-polynomials. Also, by \cite[Theorem 3.4]{CSZ2017}, if $f$ is a $q$-polynomial over ${\mathbb F}_{q^4}$ for which the set of directions is of maximum size then $f$ is equivalent either to $x^q$ or to $\delta x^q+x^{q^3}$, for some $\delta\in{\mathbb F}_{q^4}^*$ with $N(\delta)\ne 1$ (see \cite{LP2001}).}
\end{remark}

\section{\texorpdfstring{Preliminary results about $\Tr(x)$ and the monomial $q$-polynomials over ${\mathbb F}_{q^5}$}{Preliminary results about Tr(x) and the monomial q-polynomials over Fq5}}

Let $q$ be a power of a prime $p$. We will need the following results.

\begin{proposition}
\label{trace5}
Let $f(x)=\sum_{i=0}^4a_ix^{q^i}$ and $g(x)=\Tr(x)$ be $q$-polynomials over ${\mathbb F}_{q^5}$. Then there is an element $\varphi\in\Gamma\mathrm{L}(2,q^5)$ such that $Im\,(f_\varphi(x)/x)=Im\,(g(x)/x)$ if and only if $a_1a_2a_3a_4\neq0$, $(a_1/a_2)^q=a_2/a_3$, $(a_2/a_3)^q=a_3/a_4$ and $\N(a_1)=\N(a_2)$.
\end{proposition}
\begin{proof}
Let $\varphi\in\Gamma\mathrm{L}(2,q^5)$ such that $Im\,(f_\varphi(x)/x)=Im\,(g(x)/x)$.  By Proposition \ref{prop:lin}, the maximum field of linearity of $f$ is ${\mathbb F}_q$ and by Theorem \ref{trace} there exists $\lambda\in{\mathbb F}_{q^5}^*$ such that $f_\varphi(x)=\Tr(\lambda x)/\lambda$
This is equivalent to say that there exist $a,b,c,d$, $ad-bc\neq 0$ and $\sigma=p^h$ such that
\[\left\{
\begin{pmatrix}
y\\
\Tr(y)
\end{pmatrix}
\colon y\in {\mathbb F}_{q^5}
\right\}=
\left\{
\begin{pmatrix}
a&b\\
c&d\\
\end{pmatrix}
\begin{pmatrix}
x^\sigma\\
f(x)^\sigma
\end{pmatrix}
\colon x \in {\mathbb F}_{q^5}\right\}.\]
Then $c x^\sigma+d f(x)^\sigma \in {\mathbb F}_q$ for each $x\in {\mathbb F}_{q^5}$. Let $z=x^{\sigma}$. Then
\[cz+d\sum_{i=0}^4a_i^{\sigma}z^{q^i}=c^q z^{q}+d^q\sum_{i=0}^4 a_i^{\sigma q}z^{q^{i+1}},\]
for each $z$. As polynomials of $z$ the left and right-hand sides of the above equation coincide modulo $z^{q^5}-z$ and hence comparing coefficients yield
\[c+da_0^\sigma = d^q a_4^{\sigma q},\]
\[d a_1^\sigma = c^q + d^q a_0^{\sigma q},\]
\[d a_{k+1}^\sigma=d^q a_k^{\sigma q},\]
for $k=1,2,3$.
If $d=0$, then $c=0$, a contradiction.
Since $d\neq 0$, if one of $a_1,a_2,a_3,a_4$ is zero, then all of them are zero and hence $f$ is ${\mathbb F}_{q^5}$-linear. This is not the case, so we have $a_1a_2a_3a_4\neq 0$.
Then the last three equations yield
\[\left(\frac{a_1}{a_2}\right)^q=\frac{a_2}{a_3},\]
\[\left(\frac{a_2}{a_3} \right)^q=\frac{a_3}{a_4},\]
and also $\N(a_1)=\N(a_2)$.

Now assume that the conditions of the assertion hold.
It follows that $a_3=a_2^{q+1}/a_1^q$ and $a_4=a_3^{q+1}/a_2^q=a_2^{q^2+q+1}/a_1^{q^2+q}$.
Let $\alpha_i=a_i/a_1$ for $i=0,1,2,3,4$. Then $\alpha_1=1$, $\N(\alpha_2)=1$, $\alpha_3=\alpha_2^{q+1}$ and $\alpha_4=\alpha_2^{1+q+q^2}$. We have $\alpha_2=\lambda^{q-1}$ for some $\lambda\in {\mathbb F}_{q^5}^*$.
If
\[
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}
=
\begin{pmatrix}
1&0\\
1-\lambda^{1-q^4}a_0/a_1& \lambda^{1-q^4}/a_1
\end{pmatrix},
\]
then
\[
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}
\begin{pmatrix}
x\\
f(x)
\end{pmatrix}
=\]
\[\begin{pmatrix}
x\\
x+\lambda^{1-q^4}x^q+\lambda^{q-q^4}x^{q^2}+\lambda^{q^2-q^4}x^{q^3}+\lambda^{q^3-q^4}x^{q^4}
\end{pmatrix}
=
\begin{pmatrix}
x\\
\Tr(x \lambda^{q^4})/\lambda^{q^4}
\end{pmatrix},\] i.e. $f_\varphi(x)=\Tr(\lambda^{q^4}x)/\lambda^{q^4}$, where $\varphi$ is defined by the matrix $\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}$.
\end{proof}

\begin{proposition}
\label{pseudoalg}
Let $f(x)=\sum_{i=0}^4a_ix^{q^i}$, with $a_1a_2a_3a_4\neq 0$.
Then there is an element $\varphi\in\Gamma\mathrm{L}(2,q^5)$ such that $Im\,(f_\varphi(x)/x)=Im\,(x^{q-1})$ if and only if one of the following holds:
\begin{enumerate}
	\item $(a_1/a_2)^q=a_2/a_3$, $(a_2/a_3)^q=a_3/a_4$ and $\N(a_1)\neq \N(a_2)$, or
	\item $(a_4/a_1)^{q^2}=a_1/a_3$, $(a_1/a_2)^{q^2}=a_3/a_4$ and $\N(a_1)\neq \N(a_3)$.
\end{enumerate}
In both cases, if the condition on the norms does not hold, then $Im\,(f_\varphi(x)/x)=Im\,(\Tr(x)/x)$.
\end{proposition}
\begin{proof}
We first note that the monomials $x^{q^i}$ and $x^{q^{5-i}}$ are equivalent via the map $\psi:=\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}$. Hence, by Corollary \ref{norm}, the statement holds if and only if there exist $a,b,c,d$, $ad-bc\neq 0$, $\sigma=p^h$ and $i\in \{1,2\}$ such that
\begin{equation}
\label{pseudoxx}
\left\{
\begin{pmatrix}
y\\
y^{q^i}
\end{pmatrix}
\colon y\in {\mathbb F}_{q^5}
\right\}
=
\left\{
\begin{pmatrix}
a&b\\
c&d\\
\end{pmatrix}
\begin{pmatrix}
x^\sigma\\
f(x)^\sigma
\end{pmatrix}
\colon x \in {\mathbb F}_{q^5}\right\}
.
\end{equation}

If Condition 1 holds then let
$\alpha_j=a_j/a_1$ for $j=0,1,2,3,4$. So $\alpha_1=1$, $\N(\alpha_2)\neq 1$, $\alpha_3=\alpha_2^{q+1}$, $\alpha_4=\alpha_2^{1+q+q^2}$ and \eqref{pseudoxx} is satisfied with
\[
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}
=
\begin{pmatrix}
1&\alpha_2^{q^4}\\
\alpha_2^{1+q+q^2+q^3}&1
\end{pmatrix}
\begin{pmatrix}
1&0\\
-\alpha_0 & 1/a_1
\end{pmatrix}.
\]

If Condition 2 holds then let $\alpha_j=a_j/a_3$ for $j=0,1,2,3,4$. So $\alpha_3=1$, $\N(\alpha_1)\neq 1$, $\alpha_2=\alpha_1^{1+q+q^3}$,  $\alpha_4=\alpha_1^{1+q^3}$ and \eqref{pseudoxx} is satisfied with
\[
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}
=
\begin{pmatrix}
\alpha_1^{1+q+q^3+q^4}&1\\
1&\alpha_1^{q^2}
\end{pmatrix}
\begin{pmatrix}
1&0\\
-\alpha_0 & 1/a_3
\end{pmatrix}.
\]

Suppose now that \eqref{pseudoxx} holds and put $z=x^\sigma$. Then
\[(za+b\sum_{j=0}^4 a_j^\sigma z^{q^j})^{q^i}=cz+d\sum_{j=0}^4a_j^\sigma z^{q^j}\]
for each $z\in {\mathbb F}_{q^5}$ and hence, as polynomials in $z$, the left-hand side and right-hand side of the above equation coincide modulo $z^{q^5}-z$.
The coefficients of $z$, $z^{q^i}$ and $z^{q^k}$ with $i\in \{1,2\}$ and $k\in \{1,2,3,4\}\setminus \{i\}$ give
\[b^{q^i} a_{-i}^{\sigma q^i} =c+da_0^\sigma,\]
\[a^{q^i} +b^{q^i} a_0^{\sigma q^i}=d a_i^\sigma,\]
\[b^{q^i} a_{k-i}^{\sigma q^i}=d a_{k}^\sigma,\]
respectively, where the indices are considered modulo 5.
Note that $db\neq 0$ since otherwise also $a=c=0$ and hence $ad-bc=0$.
With $\{r,s,t\}=\{1,2,3,4\} \setminus \{i\}$, the last three equations yield:
\[\left(\frac{a_{r-i}}{a_{s-i}}\right)^{q^i}=\frac{a_r}{a_s},\]
\[\left(\frac{a_{s-i}}{a_{t-i}} \right)^{q^i}=\frac{a_s}{a_t}.\]
First assume $i=1$.
Then we have
\[\left(\frac{a_1}{a_2}\right)^q=\frac{a_2}{a_3} \quad \text{and} \quad \left(\frac{a_2}{a_3} \right)^q=\frac{a_3}{a_4}.\]
If $\N(a_1)=\N(a_2)$, from Proposition \ref{trace5} and Equation \eqref{size} it follows that $|Im\,(x^{q-1})|=|Im\,(\Tr(x)/x)|$, a contradiction.

Now assume $i=2$. Then we have $(a_4/a_1)^{q^2}=a_1/a_3$ and
\begin{equation}
\label{form1}
\left(\frac{a_1}{a_2} \right)^{q^2}=\frac{a_3}{a_4}.
\end{equation}
Multiplying these two equations yields $a_4^{q^2+1}=a_1a_2^{q^2}$ and hence
\begin{equation}
\label{form2}
a_2=a_1^{1+q+q^3}/a_3^{q^3+q}.
\end{equation}
By \eqref{form1} this implies
\begin{equation}
\label{form3}
a_4=a_1^{q^3+1}/a_3^{q^3}.
\end{equation}
If $\N(a_1)=\N(a_3)$, then also $\N(a_1)=\N(a_2)=\N(a_3)=\N(a_4)$. We show that in this case $Im\,(f_\varphi(x)/x)=Im\,(\Tr(x)/x)$, so we must have $\N(a_1)\neq \N(a_3)$. According to Proposition \ref{trace5} it is enough to show $(a_1/a_2)^q=a_2/a_3$ and $(a_2/a_3)^q=a_3/a_4$.
By \eqref{form1} we have $(a_1/a_2)^q=(a_3/a_4)^{q^4}$, which equals $a_2/a_3$ if and only if $(a_2/a_3)^q=a_3/a_4$, i.e. $a_3^{1+q}=a_4a_2^q$.
Taking into account \eqref{form2} and \eqref{form3}, this equality follows from
$\N(a_1)=\N(a_3)$.
\end{proof}

\section{\texorpdfstring{Proof of the main theorem}{Proof of the main theorem}}
\label{section-proof}

In this section we prove Theorem \ref{main}. In order to do this, we use the following two results and the technique developed in \cite{CSMP2016}.

\begin{lemma}[{\cite[Lemma 3.4]{CSMP2016}}]
\label{lmain}
Let $f$ and $g$ be two linearized polynomials over ${\mathbb F}_{q^n}$. If $Im\,(f(x)/x)=Im \,(g(x)/x)$, then for each positive integer $d$ the following holds
\[\sum_{x\in {\mathbb F}_{q^n}^*} \left(\frac{f(x)}{x}\right)^d = \sum_{x\in {\mathbb F}_{q^n}^*} \left(\frac{g(x)}{x}\right)^d.\]
\end{lemma}

\begin{lemma}[see for example {\cite[Lemma 3.5]{CSMP2016}}]
\label{lfolk}
For any prime power $q$ and integer $d$ we have $\sum_{x\in {\mathbb F}_{q}^*} x^d=-1$ if $q-1 \mid d$ and
$\sum_{x\in {\mathbb F}_{q}^*} x^d=0$ otherwise.
\end{lemma}

\begin{proposition}
\label{prop}
Let $f(x)=\sum_{i=0}^4 a_i x^{q^i}$ and $g(x)=\sum_{i=0}^4 b_i x^{q^i}$ be two $q$-polynomials over ${\mathbb F}_{q^5}$ such that $Im\,(f(x)/x)=Im \,(g(x)/x)$.
Then the following relations hold between the coefficients of $f$ and $g$:
\begin{equation}
\label{e0}
a_0=b_0,
\end{equation}
\begin{equation}
\label{e1}
a_1a_4^q=b_1 b_4^q,
\end{equation}
\begin{equation}
\label{e2}
a_2a_3^{q^2}=b_2 b_3^{q^2},
\end{equation}
\begin{equation}
\label{e3}
a_1^{q+1}a_3^{q^2}+a_2a_4^{q+q^2}=b_1^{q+1}b_3^{q^2}+b_2b_4^{q+q^2},
\end{equation}
\begin{equation}
\label{e4}
a_1a_2^{q+q^3}+a_3^{1+q^3}a_4^q=b_1b_2^{q+q^3}+b_3^{1+q^3}b_4^q,
\end{equation}
\begin{equation}
\label{e5}
a_1^{1+q+q^2}a_2^{q^3}+a_2^{1+q}a_3^{q^2+q^3}+a_1^qa_3^{1+q^2+q^3}+a_1^{q^2}a_2a_3^{q^3}a_4^q+a_2^{1+q+q^3}a_4^{q^2}+
\end{equation}
\[a_1^qa_2^{q^3}a_3a_4^{q^2}+a_1a_2^qa_3^{q^2}a_4^{q^3}+a_1^{1+q^2}a_4^{q+q^3}+a_3a_4^{q+q^2+q^3}=\]
\[b_1^{1+q+q^2}b_2^{q^3}+b_2^{1+q}b_3^{q^2+q^3}+b_1^qb_3^{1+q^2+q^3}+b_1^{q^2}b_2b_3^{q^3}b_4^q+b_2^{1+q+q^3}b_4^{q^2}+\]
\[b_1^qb_2^{q^3}b_3b_4^{q^2}+b_1b_2^qb_3^{q^2}b_4^{q^3}+b_1^{1+q^2}b_4^{q+q^3}+b_3b_4^{q+q^2+q^3},\]
\begin{equation}
\label{e6}
\N(a_1)+\N(a_2)+\N(a_3)+\N(a_4)+\Tr(a_1^qa_2^{q^2+q^3+q^4}a_3+a_1^{q+q^3}a_2^{q^4}a_3^{1+q^2}+
\end{equation}
\[a_1^{q+q^2}a_2^{q^3+q^4}a_4+a_1^{q+q^2+q^4}a_3^{q^3}a_4+a_2^qa_3^{q^2+q^3+q^4}a_4+a_1^{q^2}a_3^{q^3+q^4}a_4^{1+q}+\]
\[a_2^{q+q^3}a_3^{q^4}a_4^{1+q^2}+a_1^{q^2}a_2^{q^4}a_4^{1+q+q^3})=\]
\[\N(b_1)+\N(b_2)+\N(b_3)+\N(b_4)+\Tr(b_1^qb_2^{q^2+q^3+q^4}b_3+b_1^{q+q^3}b_2^{q^4}b_3^{1+q^2}+\]
\[b_1^{q+q^2}b_2^{q^3+q^4}b_4+b_1^{q+q^2+q^4}b_3^{q^3}b_4+b_2^qb_3^{q^2+q^3+q^4}b_4+b_1^{q^2}b_3^{q^3+q^4}b_4^{1+q}+\]
\[b_2^{q+q^3}b_3^{q^4}b_4^{1+q^2}+b_1^{q^2}b_2^{q^4}b_4^{1+q+q^3}).\]
\end{proposition}
\begin{proof}

Equations \eqref{e0}--\eqref{e4} follow from \cite[Lemma 3.6]{CSMP2016}.
To prove \eqref{e5} we will use Lemma \ref{lfolk} together with Lemma \ref{lmain} with $d=q^3+q^2+q+1$. This gives us
\[\sum_{1\leq i,j,m,n \leq 4} a_ia_j^qa_m^{q^2}a_n^{q^3} \sum_{x\in {\mathbb F}_{q^5}^*} x^{q^i-1+q^{j+1}-q+q^{m+2}-q^2+q^{n+3}-q^3}=\]
\[\sum_{1\leq i,j,m \leq 4} b_ib_j^qb_m^{q^2}b_n^{q^3} \sum_{x\in {\mathbb F}_{q^5}^*} x^{q^i-1+q^{j+1}-q+q^{m+2}-q^2+q^{n+3}-q^3}.\]
We have $\sum_{x\in {\mathbb F}_{q^n}^*} x^{q^i-1+q^{j+1}-q+q^{m+2}-q^2+q^{n+3}-q^3}=-1$ if and only if
\begin{equation}
\label{eki}
q^i+q^{j+1}+q^{m+2}+q^{n+3} \equiv 1+q+q^2+q^3 \pmod {q^5-1},
\end{equation}
and zero otherwise. Suppose that the former case holds. The right-hand side of \eqref{eki} is smaller than the left-hand side, thus
\[q^i+q^{j+1}+q^{m+2}+q^{n+3} = 1+q+q^2+q^3 + k(q^5-1),\]
for some positive integer $k$.
We have $q^i+q^{j+1}+q^{m+2}+q^{n+3}\leq q^4+q^5+q^6+q^7< 1+q+q^2+q^3+(q^2+q+2)(q^5-1)$ and hence $k\leq q^2+q+1$.
If $i=1$, then $q^2 \mid 1-k$ and hence $k=1$, $j=m=1$ and $n=2$, or $k=q^2+1$, $n=4$ and either $j=2$ and $m=3$, or $j=4$ and $m=1$.
If $i>1$, then $q^2$ divides $q+1-k$ and hence $k=q+1$, or $k=q^2+q+1$.
In the former case $i=j=n=2$ and $m=4$, or $i=j=2$ and $n=m=3$, or $i=3$, $j=1$, $m=4$ and $n=2$, or $i=3$, $j=1$ and $m=n=3$,
or $m=1$, $i=2$, $j=4$ and $n=3$. In the latter case $i=3$ and $n=m=j=4$.
Then \eqref{e5} follows.

To prove \eqref{e6} we follow the previous approach with $d=q^4+q^3+q^2+q+1$.
We obtain
\[\sum a_ia_j^qa_m^{q^2}a_n^{q^3}a_r^{q^4}=\sum b_ib_j^qb_m^{q^2}b_n^{q^3}b_r^{q^4},\]
where the summation is on the quintuples $(i,j,m,n,r)$ with elements taken from $\{1,2,3,4\}$ such that  $L_{i,j,m,n,r}:=(q^i-1)+(q^{j+1}-q)+(q^{m+2}-q^2)+(q^{n+3}-q^3)+(q^{r+4}-q^4)$ is divisible by $q^5-1$.
Then
\[L_{i,j,m,n,r} \equiv K_{i,j',m',n',r'} \pmod {q^5-1},\]
where
\[K_{i,j',m',n',r'}=(q^i-1)+(q^{j'}-q)+(q^{m'}-q^2)+(q^{n'}-q^3)+(q^{r'}-q^4),\]
such that
\begin{equation}
\label{trex}
j'\in \{0,2,3,4\}, \quad m'\in \{0,1,3,4\}, \quad n'\in \{0,1,2,4\}, \quad r'\in \{0,1,2,3\}
\end{equation}
with $j'\equiv j+1,\, m'\equiv m+2,\, n'\equiv n+3,\, r'\equiv r+4 \pmod 5$.

For $q=2$ and $q=3$ we can determine by computer those quintuples $(i,j',m',n',r')$ for which $K_{i,j',m',n',r'}$ is divisible by $q^5-1$ and hence \eqref{e6} follows. So we may assume $q>3$. Then
\[3-q^2-q^3-q^4=(q-1)+(1-q)+(1-q^2)+(1-q^3)+(1-q^4) \leq \]
\[K_{i,j',m',n',r'}\leq \]
\[(q^4-1)+(q^4-q)+(q^4-q^2)+(q^4-q^3)+(q^3-q^4)=3q^4-1-q-q^2,\]
and hence $L_{i,j,m,n,r}$ is divisible by $q^5-1$ if and only if $K_{i,j',m',n',r'}=0$.
It follows that
\[q^i+q^{j'}+q^{m'}+q^{n'}+q^{r'}=1+q+q^2+q^3+q^4.\]
So $\sum_{h=0}^4 c_h q^h=1+q+q^2+q^3+q^4$ for some $0\leq c_h \leq 4$ with $\sum_{h=0}^4 c_h=5$. For $q>3$ this happens only if $c_h=1$ for $h=0,1,2,3,4$ thus
we have to find those quintuples $(i,j',m',n',r')$ for which $i\in \{1,2,3,4\}$, $\{i,j',m',n',r'\}=\{0,1,2,3,4\}$ and \eqref{trex} are satisfied. This can  be done by computer and the 44 solutions yield \eqref{e6}.
\end{proof}

\subsection*{Proof of Theorem \ref{main}}

Since $f$ has maximum field of linearity ${\mathbb F}_q$, we cannot have $a_1=a_2=a_3=a_4=0$. If three of $\{a_1,a_2,a_3,a_4\}$ are zeros, then $f(x)=a_0x+a_ix^{q^i}$, for some $i\in\{1,2,3,4\}$. Hence with $\varphi$ represented by $\begin{pmatrix}
	1 & 0 \\
	-a_0/a_i & 1/a_i \\
\end{pmatrix}$ we have $f_{\varphi}(x)=x^{q^i}$. Then Proposition \ref{imtrans} and Corollary \ref{norm} give $g_\varphi(x)=\beta x^{q^j}$ where $\N(\beta)=1$ and   $j\in\{1,2,3,4\}$.
Now, we distinguish three main cases according to the number of zeros among $\{a_1,a_2,a_3,a_4\}$.

\subsection*{Two zeros among $\{a_1, a_2, a_3, a_4\}$}

Applying Proposition \ref{prop} we obtain $a_0=b_0$.
The two non-zero coefficients can be chosen in six different ways, however the cases $a_1a_2\neq 0$ and $a_1a_3\neq 0$ correspond to $a_3a_4\neq 0$ and $a_2a_4\neq 0$, respectively,
since $Im\,(f(x)/x)=Im \,(\hat f(x)/x)$. Thus, after possibly interchanging $f$ with $\hat f$, we may consider only four cases.
\medskip

First let $f(x)=a_0x+a_1x^q+a_4x^{q^4}$, $a_1a_4\neq 0$.
Applying Proposition \ref{prop} we obtain
\begin{equation}
\label{f1}
a_1a_4^q=b_1 b_4^q,
\end{equation}
\[0=b_2 b_3^{q^2}.\]
Since $b_1b_4\neq 0$, from \eqref{e3} we get $b_2=b_3=0$ and hence \eqref{e6} gives
\[ \N(a_1)+\N(a_4)=\N(b_1)+\N(b_4).\]
Also, from \eqref{f1} we have $\N(a_1a_4)=\N(b_1b_4)$.
It follows that either $\N(a_1)=\N(b_1)$ and $\N(a_4)=\N(b_4)$, or $\N(a_1)=\N(b_4)$ and $\N(a_4)=\N(b_1)$.
In the first case $b_1=a_1\lambda^{q-1}$ for some $\lambda\in {\mathbb F}_{q^5}^*$ and by \eqref{f1}
we get $g(x)=f(\lambda x)/\lambda$. In the latter case  $b_1=a_4^q\lambda^{q-1}$ for some $\lambda\in {\mathbb F}_{q^5}^*$ and by \eqref{f1} we get  $g(x)=\hat{f}(\lambda x)/\lambda$.
\medskip

Now consider $f(x)=a_1x^q+a_3x^{q^3}$, $a_1a_3\neq 0$.
Applying Proposition \ref{prop} and arguing as above we have either $b_2=b_4=0$ or $b_1=b_3=0$.
In the first case from \eqref{e5}
we obtain
\[a_1^qa_3^{1+q^2+q^3}=b_1^qb_3^{1+q^2+q^3}\]
and together with \eqref{e3} this yields $\N(a_1)=\N(b_1)$ and $\N(a_3)=\N(b_3)$.
In this case $g(x)=f(\lambda x)/\lambda$ for some $\lambda\in {\mathbb F}_{q^5}^*$.
If $b_1=b_3=0$, then in $\hat{g}(x)$ the coefficients of $x^{q^2}$ and $x^{q^4}$ are zeros thus applying the result obtained in the former case we get $\lambda\hat{g}(x)=f(\lambda x)$ and hence after substituting $y=\lambda x$ and taking the adjoints of both sides we obtain $g(y)=\hat{f}(\mu y)/\mu$, where $\mu=\lambda^{-1}$.

\medskip

The cases $f(x)=a_1x^q+a_2x^{q^2}$ and $f(x)=a_2x^{q^2}+a_3x^{q^3}$ can be handled in a similar way, applying Equations \eqref{e1}--\eqref{e6} of Proposition \ref{prop}.

\subsection*{One zero among $\{a_1, a_2, a_3, a_4\}$}

Since $Im\,(f(x)/x)=Im \,(\hat f(x)/x)$, we may assume $a_1=0$ or $a_2=0$.
First suppose $a_1=0$. Then by \eqref{e1} either $b_1=0$ or $b_4=0$. In the former case
putting together Equations \eqref{e2}, \eqref{e3}, \eqref{e4} we get $\N(a_i)=\N(b_i)$ for $i\in \{2,3,4\}$ and hence there exists $\lambda\in {\mathbb F}_{q^5}^*$ such that $g(x)=f(\lambda x)/\lambda$.
If $a_1=b_4=0$, then in $\hat{g}(x)$ the coefficients of $x^q$ is zero thus applying the previous result we get $g(x)=\hat{f}(\mu x)/\mu$, where $\mu=\lambda^{-1}$.

Now suppose $a_2=0$. Then by \eqref{e2} either $b_2=0$ or $b_3=0$. Using the same approach but applying \eqref{e1}, \eqref{e3} and \eqref{e4} we obtain $g(x)=f(\lambda x)/\lambda$ or $g(x)=\hat{f}(\lambda x)/\lambda$.

\subsection*{Case $a_1a_2a_3a_4\neq 0$}

We will apply \eqref{e0}-\eqref{e5} of Proposition \ref{prop}.
Note that Equations \eqref{e1} and \eqref{e2} yield $a_1a_2a_3a_4\neq 0 \Leftrightarrow b_1b_2b_3b_4\neq 0$.
Multiplying \eqref{e3} by $a_2$ and applying \eqref{e2} yield
\[a_2^2a_4^{q+q^2}-a_2(b_1^{q+1}b_3^{q^2}+b_2b_4^{q+q^2})+a_1^{q+1}b_3^{q^2}b_2=0.\]
Taking \eqref{e1} into account, this is equivalent to \[(a_2a_4^{q+q^2}-b_1^{q+1}b_3^{q^2})(a_2a_4^{q+q^2}-b_2b_4^{q+q^2})=0.\]
Multiplying \eqref{e4} by $a_1$ and applying \eqref{e1} yield
\[a_1^2a_2^{q+q^3}-a_1(b_1b_2^{q+q^3}+b_3^{1+q^3}b_4^q)+a_3^{1+q^3}b_4^qb_1=0.\]
Taking \eqref{e2} into account, this is equivalent to
\[(a_1a_2^{q+q^3}-b_1b_2^{q+q^3})(a_1a_2^{q+q^3}-b_3^{1+q^3}b_4^q)=0.\]
We distinguish four cases:
\begin{enumerate}
	\item $a_2a_4^{q+q^2}=b_1^{q+1}b_3^{q^2}$ and $a_1a_2^{q+q^3}=b_1b_2^{q+q^3}$,
	\item $a_2a_4^{q+q^2}=b_1^{q+1}b_3^{q^2}$ and $a_1a_2^{q+q^3}=b_3^{1+q^3}b_4^q$,
	\item $a_2a_4^{q+q^2}=b_2b_4^{q+q^2}$ and $a_1a_2^{q+q^3}=b_1b_2^{q+q^3}$,
	\item $a_2a_4^{q+q^2}=b_2b_4^{q+q^2}$ and $a_1a_2^{q+q^3}=b_3^{1+q^3}b_4^q$.
\end{enumerate}
We show that these four cases produce the relations:
\begin{equation}
\label{1per4}
\N\left(\frac{b_1}{a_4}\right)=\frac{a_1a_2^{q+q^3}}{a_4^qa_3^{q^3+1}}=\frac{b_1b_2^{q+q^3}}{b_4^qb_3^{q^3+1}},
\end{equation}
\begin{equation}
\label{2per4}
\N\left(\frac{b_1}{a_4}\right)=1,
\end{equation}
\begin{equation}
\label{3per4}
\N\left(\frac{b_1}{a_1}\right)=1,
\end{equation}
\begin{equation}
\label{4per4}
\N\left(\frac{b_1}{a_1}\right)=\frac{a_3^{q^3+1}a_4^q}{a_1 a_2^{q+q^3}}=\frac{b_1b_2^{q+q^3}}{b_3^{q^3+1}b_4^q},
\end{equation}
respectively.
To see \eqref{1per4} observe that from $a_2a_4^{q+q^2}=b_1^{q+1}b_3^{q^2}$ and \eqref{e1} we get
\begin{equation}
\label{new}
\N\left(\frac{b_1}{a_4}\right)=\left(\frac{b_1^{q+1}}{a_4^{q+q^2}}\right)^{q^2+1}\frac{b_1^{q^4}}{a_4}=
\left(\frac{a_2^{q^2+1}}{b_3^{q^2+q^4}}\right)\frac{a_1^{q^4}}{b_4}=\frac{a_1a_2^{q+q^3}}{b_4^qb_3^{q^3+1}},
\end{equation}
and hence by $a_1a_2^{q+q^3}=b_1b_2^{q+q^3}$ and \eqref{e4} we get \eqref{1per4}.
Equation \eqref{2per4} immediately follows from \eqref{new} taking $a_1a_2^{q+q^3}=b_3^{1+q^3}b_4^q$ into account.
Similarly, using \eqref{e1} and \eqref{e4} we get \eqref{3per4} and \eqref{4per4}.

In Case 3 by \eqref{3per4} we get $b_1=a_1\lambda^{q-1}$ for some $\lambda\in {\mathbb F}_{q^5}^*$ and by \eqref{e1} and \eqref{e2} we have $g(x)=f(\lambda x)/\lambda$. Analogously, in Case 2 $g(x)=\hat{f}(\lambda x)/\lambda$.
Note that Case 4 is just Case 3 after replacing $g$ by $\hat{g}$ since $Im\,(g(x)/x)=Im \,(\hat g(x)/x)$.
This allows us to restrict ourself to Case 1. It will be useful to express $a_1$, $a_2$, $a_3$ as follows:
\begin{equation}
\label{ais}
a_1=\frac{b_1 b_4^q}{a_4^q}, \quad a_2=\frac{b_1^{q+1}b_3^{q^2}}{a_4^{q+q^2}}, \quad a_3=\frac{b_2^{q^3}b_4^{1+q^4}}{a_1^{q^3+q^4}}.
\end{equation}

We are going to simplify \eqref{e5}.
Using Equations \eqref{ais} and \eqref{e1} it is easy to see that $a_2^{1+q}a_3^{q^2+q^3}=b_2^{1+q}b_3^{q^2+q^3}$, $a_1^{1+q^2}a_4^{q+q^3}=b_1^{1+q^2}b_4^{q+q^3}$, $a_1^{q^2}a_2a_3^{q^3}a_4^q=b_1b_2^qb_3^{q^2}b_4^{q^3}$, $a_1^qa_2^{q^3}a_3a_4^{q^2}=b_1^qb_2^{q^3}b_3b_4^{q^2}$, $a_1a_2^qa_3^{q^2}a_4^{q^3}=b_1^{q^2}b_2b_3^{q^3}b_4^q$ and hence
\begin{equation}
\label{eq5b}
a_1^{1+q+q^2}a_2^{q^3}+a_1^qa_3^{1+q^2+q^3}+a_2^{1+q+q^3}a_4^{q^2}+a_3a_4^{q+q^2+q^3}=
\end{equation}
\[b_1^{1+q+q^2}b_2^{q^3}+b_1^qb_3^{1+q^2+q^3}+b_2^{1+q+q^3}b_4^{q^2}+b_3b_4^{q+q^2+q^3}.\]
The following equations can be proved applying \eqref{e1}, \eqref{e2} and \eqref{ais}:
\begin{equation}
\label{ea}
\N\left(\frac{b_1}{a_4}\right)b_3b_4^{q+q^2+q^3}=a_2^{q^3}a_1^{1+q+q^2},
\end{equation}
\begin{equation}
\label{eb}
\N\left(\frac{a_4}{b_1}\right)b_4^{q^2}b_2^{1+q+q^3}=a_1^qa_3^{1+q^2+q^3},
\end{equation}
\begin{equation}
\label{ec}
\N\left(\frac{b_1}{a_4}\right)b_1^qb_3^{1+q^2+q^3}=a_2^{1+q+q^3}a_4^{q^2},
\end{equation}
\begin{equation}
\label{ed}
\N\left(\frac{a_4}{b_1}\right)b_2^{q^3}b_1^{1+q+q^2}=a_3a_4^{q+q^2+q^3}.
\end{equation}

Then \eqref{eq5b} can be written as
\[(\N(b_1/a_4)-1)(b_3b_4^{q+q^2+q^3}+b_1^qb_3^{1+q^2+q^3})=\frac{\N(b_1/a_4)-1}{\N(b_1/a_4)}(b_4^{q^2}b_2^{1+q+q^3}+b_2^{q^3}b_1^{1+q+q^2}).\]
If $\N(b_1/a_4)=1$, then we are in Case 2 and hence the assertion follows. Otherwise dividing by $\N(b_1/a_4)-1$ and substituting $\N(b_1/a_4)=b_1b_2^{q+q^3}/b_4^qb_3^{q^3+1}$ we obtain
\[ b_1b_2^{q+q^3}(b_3b_4^{q+q^2+q^3}+b_1^qb_3^{1+q^2+q^3})=b_4^qb_3^{q^3+1}(b_4^{q^2}b_2^{1+q+q^3}+b_2^{q^3}b_1^{1+q+q^2}).\]
Substituting $\N(b_1/a_4) b_4^qb_3^{q^3+1}/b_2^{q+q^3}$ for $b_1$ and using the fact that $\N(b_1/a_4)\in {\mathbb F}_q$ we obtain
\[ \left(1-\N\left(\frac{b_1}{a_4}\right)^2\N\left(\frac{b_3}{b_2}\right)\right)\left(\N\left(\frac{b_1}{a_4}\right)b_4^{q+q^3}b_3-b_2^{1+q+q^3}\right)=0.\]
This gives us two possibilities:
\begin{equation}
\label{sec}
\N\left(\frac{b_1}{a_4}\right)b_4^{q+q^3}b_3=b_2^{1+q+q^3},
\end{equation}
or
\begin{equation}
\label{sec2}
\N\left(\frac{b_2}{b_3}\right)=\N\left(\frac{b_1}{a_4}\right)^2.
\end{equation}
First consider the case when \eqref{sec2} holds. We show $\N(a_1)=\N(b_1)$, that is, \eqref{3per4}.
We have $a_2a_4^{q+q^2}=b_1^{q+1}b_3^{q^2}$ from \eqref{ais} and hence $\N(a_2)\N(a_4)^2=\N(b_1)^2\N(b_3)$. It follows that
\[\N\left(\frac{b_1}{a_4}\right)^2=\N\left(\frac{a_2}{b_3}\right).\]
Combining this with \eqref{sec2} we obtain $\N(b_2)=\N(a_2)$. Then $\N(b_1)=\N(a_1)$ follows from $a_1a_2^{q+q^3}=b_1b_2^{q+q^3}$ since we are in Case 1.

From now on we can suppose that \eqref{sec} holds.
Then \eqref{1per4} yields
\begin{equation}
\label{elso}
\left(\frac{b_1}{b_2}\right)^{q^2}=\frac{b_3}{b_4}.
\end{equation}
Multiplying both sides of \eqref{sec} by $b_4^{q^2}$ and applying \eqref{ea} gives
\begin{equation}
\label{nuovo}
a_2^{q^3}a_1^{1+q+q^2}=b_2^{1+q+q^3}b_4^{q^2}.
\end{equation}
Then multiplying \eqref{ea} by \eqref{eb} and taking \eqref{nuovo} into account we obtain
\[a_1^qa_3^{1+q^2+q^3}=b_3b_4^{q+q^2+q^3}.\]
Multiplying \eqref{ec} and \eqref{ed} yield
\[(b_1^qb_3^{1+q^2+q^3})(b_2^{q^3}b_1^{1+q+q^2})=(a_2^{1+q+q^3}a_4^{q^2})(a_3a_4^{q+q^2+q^3}).\]
On the other hand, from \eqref{eq5b} it follows that
\[b_1^qb_3^{1+q^2+q^3}+b_2^{q^3}b_1^{1+q+q^2}=a_2^{1+q+q^3}a_4^{q^2}+a_3a_4^{q+q^2+q^3}.\]
Hence $b_1^qb_3^{1+q^2+q^3}=a_2^{1+q+q^3}a_4^{q^2}$ and $b_2^{q^3}b_1^{1+q+q^2}=a_3a_4^{q+q^2+q^3}$, or
$b_1^qb_3^{1+q^2+q^3}=a_3a_4^{q+q^2+q^3}$ and $b_2^{q^3}b_1^{1+q+q^2}=a_2^{1+q+q^3}a_4^{q^2}$.
In the former case \eqref{ec} yields $\N(b_1/a_4)=1$, which is \eqref{2per4}.
In the latter case \eqref{1per4} and \eqref{ed} gives
\[\frac{b_4^qb_3^{q^3+1}}{b_1b_2^{q+q^3}}b_2^{q^3}b_1^{1+q+q^2}=\N(a_4/b_1)b_2^{q^3}b_1^{1+q+q^2}=b_1^qb_3^{1+q^2+q^3},\]
and hence
\begin{equation}
\label{masodik}
\frac{b_4}{b_2}=\left(\frac{b_3}{b_1}\right)^q.
\end{equation}
Equation \eqref{elso} is equivalent to
\[b_4b_1^{q^2}=b_3b_2^{q^2},\]
while \eqref{masodik} is equivalent to
\[b_4b_1^q=b_3^qb_2.\]
Dividing these two equations by each other yield
\[b_2^{q^2-1}=b_3^{q-1}b_1^{q^2-q}.\]
It follows that there exists $\lambda\in {\mathbb F}_q^*$ such that
\begin{equation}
\label{lambda}
b_2^{q+1}=\lambda b_3 b_1^q,
\end{equation}
thus
\[b_3=b_2^{q+1}/(b_1^q\lambda)\]
and
\[b_4=b_2^{1+q+q^2}/(b_1^{q+q^2}\lambda).\]
Then \eqref{1per4} can be written as
\[\N\left(\frac{b_1}{a_4}\right)=\frac{b_1b_2^{q+q^3}}{b_4^qb_3^{q^3+1}}=\N\left(\frac{b_1}{b_2}\right)\lambda^3,\]
and hence
\[\N\left(\frac{b_2}{a_4}\right)=\lambda^3.\]
Then, using the previous expressions for $b_3$ and $b_4$ and taking  \eqref{e1} and  \eqref{e2} into account,  we can express $\N(a_i)$ for $i=1,2,3,4$ as follows
\begin{equation}
\label{bp1}
\N(a_1)=\N(b_2)^2/(\N(b_1)\lambda^2),
\end{equation}
\begin{equation}
\label{bp2}
\N(a_2)=\N(b_1)\lambda,
\end{equation}
\[\N(a_3)=\N(b_2)^3/(\N(b_1)^2\lambda^6),\]
\[\N(a_4)=\N(b_2)/\lambda^3.\]

Before we go further, we simplify \eqref{e6} and prove
\begin{equation}
\label{sumofnorms}
\N(a_1)+\N(a_2)+\N(a_3)+\N(a_4)=\N(b_1)+\N(b_2)+\N(b_3)+\N(b_4).
\end{equation}
It is enough to show
\[\Tr(\overbrace{\hbox{$a_1^qa_2^{q^2+q^3+q^4}a_3$}}^{A_1}+\overbrace{\hbox{$a_1^{q+q^3}a_2^{q^4}a_3^{1+q^2}$}}^{A_2}+
\overbrace{\hbox{$a_1^{q+q^2}a_2^{q^3+q^4}a_4$}}^{A_3}+\overbrace{\hbox{$a_1^{q+q^2+q^4}a_3^{q^3}a_4$}}^{A_4}+\]
\[\overbrace{\hbox{$a_2^qa_3^{q^2+q^3+q^4}a_4$}}^{A_5}+\overbrace{\hbox{$a_1^{q^2}a_3^{q^3+q^4}a_4^{1+q}$}}^{A_6}+
\overbrace{\hbox{$a_2^{q+q^3}a_3^{q^4}a_4^{1+q^2}$}}^{A_7}+\overbrace{\hbox{$a_1^{q^2}a_2^{q^4}a_4^{1+q+q^3}$}}^{A_8})=\]
\[\Tr(\overbrace{\hbox{$b_1^qb_2^{q^2+q^3+q^4}b_3$}}^{B_1}+\overbrace{\hbox{$b_1^{q+q^3}b_2^{q^4}b_3^{1+q^2}$}}^{B_7}+
\overbrace{\hbox{$b_1^{q+q^2}b_2^{q^3+q^4}b_4$}}^{B_3}+\overbrace{\hbox{$b_1^{q+q^2+q^4}b_3^{q^3}b_4$}}^{B_8}+\]
\[\overbrace{\hbox{$b_2^qb_3^{q^2+q^3+q^4}b_4$}}^{B_5}+\overbrace{\hbox{$b_1^{q^2}b_3^{q^3+q^4}b_4^{1+q}$}}^{B_6}+
\overbrace{\hbox{$b_2^{q+q^3}b_3^{q^4}b_4^{1+q^2}$}}^{B_2}+\overbrace{\hbox{$b_1^{q^2}b_2^{q^4}b_4^{1+q+q^3}$}}^{B_4}),\]
which can be done by proving $\Tr(A_i)=\Tr(B_i)$ for $i=1,2,\ldots,8$.
Expressing $a_3$ with $a_4$ in \eqref{ais} gives $a_3=b_2^{q^3}a_4^{q^4+1}/b_1^{q^3+q^4}$.
Then $a_1,a_2,a_3$ can be eliminated in all of the $A_i$, $i\in \{1,2\ldots,8\}$.
It turns out that this procedure eliminates also $a_4$ when $i\in\{2,4,7,8\}$ and we obtain
$A_2=B_2^{q^2}$, $A_4=B_4^{q^2}$, $A_7=B_7^{q^3}$ and $A_8=B_8^{q^2}$.
In each of the other cases what remains is $\N(a_4)$ times an expression in $b_1,b_2,b_3,b_4$.
Then by using \eqref{1per4} we can also eliminate $\N(a_4)$ and hence $A_i$ can be expressed in terms of $b_1,b_2,b_3,b_4$. This gives $A_1=B_1$ and $A_5=B_5$. Applying also $\eqref{elso}$ and $\eqref{masodik}$ we obtain $A_3=B_3^{q^2}$ and $A_6=B_6$.

Let $x=\N(b_2/b_1)$, then \eqref{sumofnorms} gives us the following equation
\[x^2\lambda^4+\lambda^7+x^3+x\lambda^3=\lambda^6+x\lambda^6+x^2\lambda+\lambda x^3.\]
After rearranging we get:
\[(1-\lambda)(x-\lambda)(x-\lambda^2)(x-\lambda^3)=0.\]
First suppose $\lambda\neq 1$, then we have three possibilities:
\[x=\lambda,\]
in which case $N(b_1)=N(a_1)$ follows from \eqref{bp1}, which is Case 3;
\[x=\lambda^3,\]
in which case $N(a_4)=N(b_1)$ follows from \eqref{bp2}, which is Case 2;
\[x=\lambda^2,\]
in which case we show
that there exists $\varphi\in\Gamma\mathrm{L}(2,q^5)$ such that either $Im\,(g_\varphi(x)/x)=Im\,(x^{q-1})$ or $Im\,(g_\varphi(x)/x)=Im\,(\Tr(x)/x)$. In the former case by Proposition \ref{imtrans} and Corollary \ref{norm} we get $f_\varphi(x)=\alpha x^{q^i}$ and $g_\varphi(x)=\beta x^{q^j}$ for some $i,j\in\{1,2,3,4\}$, with $\N(\alpha)=\N(\beta)=1$. In the latter case, by Theorem \ref{trace} and by Propositions \ref{imtrans} and \ref{lambda0}, there exists $\lambda\in{\mathbb F}_{q^5}^*$ such that $g(x)=f(\lambda x)/\lambda$.

According to Proposition \ref{pseudoalg} part 2, it is enough to show
\[(b_4/b_1)^{q^2}=b_1/b_3, \quad (b_1/b_2)^{q^2}=b_3/b_4.\]
The second equation is just \eqref{elso}, thus it is enough to prove the first one.
First we show
\begin{equation}
\label{meta}
b_2b_3^{q+q^3}=b_1^{1+q+q^3}.
\end{equation}
From \eqref{lambda} we have
\[\N\left(\frac{b_2}{b_1}\right)=\lambda^2=\left(\frac{b_2^{q+1}}{b_3b_1^q}\right)^2,\]
and hence after rearranging
\[\frac{b_2^{q^2+q^3+q^4}b_3}{b_1^{1+q^2+q^3+q^4}}=\frac{b_2^{q+1}}{b_3b_1^q}.\]
On the right-hand side we have $\lambda$, which is in ${\mathbb F}_q$, thus, after taking $q$-th powers on the left and $q^3$-th powers on the right, the following also holds
\[\frac{b_2^{q^3+q^4+1}b_3^q}{b_1^{q+q^3+q^4+1}}=\frac{b_2^{q^3+q^4}}{b_3^{q^3}b_1^{q^4}}.\]
After rearranging we obtain \eqref{meta}.
Now we show that $(b_4/b_1)^{q^2}=b_1/b_3$ is equivalent to \eqref{meta}.
Expressing $b_4$ from \eqref{elso} we get
\[(b_4/b_1)^{q^2}=b_1/b_3 \Leftrightarrow b_3^{1+q^2}b_2^{q^4}=b_1^{1+q^2+q^4},\]
where the equation on the right-hand side is just the $q^4$-th power of \eqref{meta}.

Now consider the case when $\lambda=1$. Then $b_3=b_2^{q+1}/b_1^q$, $b_4=b_2^{1+q+q^2}/b_1^{q+q^2}$ and it follows from Proposition \ref{pseudoalg}
that there exists $\varphi\in\Gamma\mathrm{L}(2,q^5)$ such that either $Im\,(g_\varphi(x)/x)=Im\,(x^{q-1})$ or $Im\,(g_\varphi(x)/x)=Im\,(\Tr(x)/x)$. As above, the assertion follows either from Proposition \ref{imtrans} and Corollary \ref{norm} or from Theorem \ref{trace} and by Propositions \ref{imtrans} and \ref{lambda0}.

This finishes the proof when $\prod_{i=1}^4 a_ib_i \neq 0$.

\qed

\section{\texorpdfstring{New maximum scattered linear sets of $\PG(1,q^5)$}{New maximum scattered linear sets of PG(1,q5)}}
\label{geom}

A point set $L$ of a line  $\Lambda=\PG(W,{\mathbb F}_{q^n})\allowbreak=\PG(1,q^n)$ is said to be an \emph{${\mathbb F}_q$-linear set} of $\Lambda$ of rank $n$ if it is
defined by the non-zero vectors of an $n$-dimensional ${\mathbb F}_q$-vector subspace $U$ of the two-dimensional ${\mathbb F}_{q^n}$-vector space $W$, i.e.
\[L=L_U:=\{\langle {\bf u} \rangle_{\mathbb{F}_{q^n}} \colon {\bf u}\in U\setminus \{{\bf 0} \}\}.\]
One of the most natural questions about linear sets is their equivalence. Two linear sets $L_U$ and $L_V$ of $\PG(1,q^n)$ are said to be \emph{$\mathrm{P\Gamma L}$-equivalent} (or simply \emph{equivalent}) if there is an element in $\mathrm{P\Gamma L}(2,q^n)$ mapping $L_U$ to $L_V$. In the applications it is crucial to have methods to decide whether two linear sets are equivalent or not. This can be a difficult problem and some results in this direction can be found in \cite{CSZ2015, CSMP2016}.
 If $L_U$ and $L_V$ are two equivalent ${\mathbb F}_q$-linear sets of rank $n$ in $\PG(1,q^n)$ and $\varphi$ is an element of $\mathrm{\Gamma L}(2,q^n)$ which induces a collineation mapping $L_U$ to $L_V$, then $L_{U^\varphi}=L_V$. Hence the first step to face with the equivalence problem for linear sets is to determine which ${\mathbb F}_q$-subspaces can define the same linear set.

For any $q$-polynomial $f(x)=\sum_{i=0}^{n-1} a_i x^{q^i}$ over ${\mathbb F}_{q^n}$, the graph ${\mathcal G}_f=\{(x,f(x)) \colon x\in {\mathbb F}_{q^n}\}$ is an ${\mathbb F}_q$-vector subspace of the 2-dimensional vector space $V={\mathbb F}_{q^n}\times{\mathbb F}_{q^n}$ and the point set
\[L_f:=L_{{\mathcal G}_f}=\{\langle (x,f(x))\rangle_{\mathbb{F}_{q^n}}  \colon x\in {\mathbb F}_{q^n}^*\}\] is an ${\mathbb F}_q$-linear set of rank $n$ of $\PG(1,q^n)$.
 In this context, the problem posed in \eqref{start} corresponds to find all ${\mathbb F}_q$-subspaces of $V$ of rank $n$ (cf. \cite[Proposition 2.3]{CSMP2016}) defining the linear set $L_f$. The maximum field of linearity of $f$ is the maximum field of linearity of $L_f$, and it is well-defined (cf. Proposition \ref{fieldoflinearity} and  \cite[Proposition 2.3]{CSMP2016}). Also, by the Introduction from any $q$-polynomial $f$ over ${\mathbb F}_{q^n}$, the linear sets $L_f$, $L_{f_\lambda}$ (with $f_\lambda(x):=f(\lambda x)/\lambda$ for each $\lambda\in{\mathbb F}_{q^n}^*$) and $L_{\hat f}$ coincide (cf. \cite[Lemma 2.6]{BGMP2015} and the first part of \cite[Section 3]{CSMP2016}).
 If $f$ and $g$ are two equivalent $q$-polynomials over ${\mathbb F}_{q^n}$, i.e. ${\mathcal G}_f$ an ${\mathcal G}_g$ are equivalent w.r.t. the action of the group $\Gamma{\mathrm L}(2,q^n)$, then the corresponding ${\mathbb F}_q$-linear sets $L_f$ and $L_g$ of $\PG(1,q^n)$ are ${\mathrm P}\Gamma{\mathrm L}(2,q^n)$-equivalent. The converse does not hold (see  \cite{CSZ2015} and \cite{CSMP2016} for further details). More precisely,
\begin{proposition}
\label{pgamma-equiv}
Let $L_f$ and $L_g$ be two ${\mathbb F}_q$-linear sets of rank $n$ of $\PG(1,q^n)$. Then $L_f$ and $L_g$ are $\mathrm{P\Gamma L}(2,q^n)$-equivalent if and only if there exists and element $\varphi\in\Gamma\mathrm{L}(2,q^n)$ such that $Im\,(f_\varphi(x)/x)=Im\,(g(x)/x)$.
\qed
 \end{proposition}

Linear sets of rank $n$ of $\PG(1,q^n)$ have size at most $(q^n-1)/(q-1)$.
A linear set $L_U$ of rank $n$ whose size achieves this bound is called \emph{maximum scattered}.
For applications of these objects we refer to \cite{OP2010} and \cite{Lavrauw}.

Following \cite{LuMaPoTr2014} and \cite{DoDu9} a maximum scattered ${\mathbb F}_q$-linear set $L_U$ of rank $n$ in $\PG(1,q^n)$ is of \emph{pseudoregulus type} if
it is $\mathrm{P}\Gamma\mathrm{L}(2,q^n)$-equivalent to $L_f$ with $f(x)=x^q$ or, equivalently, if there exists $\varphi\in\mathrm{GL}(2,q^n)$ such that
\[L_{U^\varphi}=\{\langle(x,x^q)\rangle_{{\mathbb F}_{q^n}}\colon x\in{\mathbb F}_{q^n}^*\}.\]
By Proposition \ref{pgamma-equiv} and Corollary \ref{norm}, it follows
\begin{proposition}
\label{peseud-equiv}
An ${\mathbb F}_q$-linear set $L_f$ of rank $n$ of $\PG(1,q^n)$ is
of pseudoregulus type if and only if $f(x)$ is equivalent to $x^{q^i}$ for some $i$ with $\gcd(i,n)=1$.
\qed
\end{proposition}
For the proof of the previous result see also \cite{LaShZa2013}.

The known pairwise non-equivalent families of $q$-polynomials over ${\mathbb F}_{q^n}$ which define maximum scattered linear sets of rank $n$ in $\PG(1,q^n)$ are

\begin{enumerate}
\item $f_{s}(x)= x^{q^s}$, $1\leq s\leq n-1$, $\gcd(s,n)=1$ (\cite{BL2000,CSZ2016}),
\item $g_{s,\delta}(x)= \delta x^{q^s} + x^{q^{n-s}}$, $n\geq 4$, $\N_{q^n/q}(\delta)\notin \{0,1\}$ \footnote{This condition  implies $q\neq 2$.}, $\gcd(s,n)=1$ (\cite{LP2001} for $s=1$, \cite{Sh,LTZ} for $s\neq 1$),
\item $h_{s,\delta}(x):= \delta x^{q^s}+x^{q^{s+n/2}}$, $n\in \{6,8\}$, $\gcd(s,n/2)=1$, $\N_{q^n/q^{n/2}}(\delta) \notin \{0,1\}$, for the precise conditions on $\delta$ and $q$ see \cite[Theorems 7.1 and 7.2]{CMPZ} \footnote{Also here $q>2$, otherwise the linear set defined by $h_{s,\delta}$ is never scattered.},
\item $k_b(x):=x^q+x^{q^3}+bx^{q^5}$, $n=6$, with $b^2+b=1$, $q\equiv 0,\pm 1 \pmod 5$ (\cite{CSMZ2017}).
\end{enumerate}

\begin{remark}
All the previous polynomials in cases 2.,3.,4. above are examples of functions which are not equivalent to monomials but the set of directions determined by their graph has size $(q^n-1)/(q-1)$, i.e. the corresponding linear sets are maximum scattered. The existence of such linearized polynomials is briefly discussed also in \cite[p.\ 132]{Praha}.
\end{remark}

For $n=2$ the maximum scattered ${\mathbb F}_q$-linear sets coincide with the Baer sublines. For $n=3$ the maximum scattered linear sets  are all of pseudoregulus type and the corresponding $q$-polynomials are all $\mathrm{GL}(2,q^3)$-equivalent to $x^q$  (cf. \cite{LaVa2010}). For $n=4$ there are two families of maximum scattered linear sets. More precisely, if $L_f$ is a maximum scattered linear set of rank $4$ of $\PG(1,q^4)$, with maximum field of linearity ${\mathbb F}_{q}$, then there exists $\varphi\in\mathrm{GL}(2,q^4)$ such that either $f_\varphi(x)=x^q$ or $f_\varphi(x)=\delta x^q+x^{q^3}$, for some $\delta\in{\mathbb F}_{q^4}^*$ with $\N_{q^4/q}(\delta)\notin\{0,1\}$ (cf. \cite{CSZ2017}). It is easy to see that $L_{f_{1}}=L_{f_s}$ for any $s$ with $\gcd(s,n)=1$, and $f_i$ is equivalent to $f_j$ if and only if $j\in\{i,n-i\}$. Also, the graph of $g_{s,\delta}$ is  $\mathrm{GL}(2,q^n)$-equivalent to the graph of $g_{n-s,\delta^{-1}}$.

In \cite[Theorem 3]{LP2001} Lunardon and Polverino proved that $L_{g_{1,\delta}}$ and $L_{f_1}$ are not $\mathrm{P}\Gamma \mathrm{L}(2,q^n)$-equivalent when $q>3$, $n\geq 4$. This was extended also for $q=3$ \cite[Theorem 3.4]{CSMZ2017}.
Also in \cite{CSMZ2017}, it has been proven that  for $n=6,8$ the linear sets $L_{f_1}$, $L_{g_{s,\delta}}$, $L_{h_{s',\delta'}}$ and $L_{k_b}$ are pairwise non-equivalent for any choice of $s,s',\delta, \delta',b$.

In this section we prove that one can find for each $q>2$ a suitable $\delta$ such that $L_{g_{2,\delta}}$ of $\PG(1,q^5)$ is not equivalent to the linear sets $L_{g_{1,\mu}}$ of $\PG(1,q^5)$ for each $\mu\in{\mathbb F}_{q^5}^*$, with $\N_{q^5/q}(\mu)\notin\{0,1\}$. In order to do this, we first reformulate Theorem \ref{main} as follows.

\begin{theorem}[Theorem \ref{main}]
\label{main1}
	Let $f(x)$ and $g(x)$ be two $q$-polynomials over ${\mathbb F}_{q^5}$ such that $L_f=L_g$. Then either  $L_f=L_g$ is of pseudoregulus type or there exists $\lambda\in {\mathbb F}_{q^5}^*$ such that $g(x)=f(\lambda x)/\lambda$ or $g(x)=\hat{f}(\lambda x)/\lambda$ holds.
\end{theorem}

\begin{theorem}
Let $g_{2,\delta}(x)=\delta x^{q^2}+x^{q^3}$ for some $\delta\in {\mathbb F}_{q^5}^*$ with $\N(\delta)^5 \neq 1$. Then $L_{g_{2,\delta}}$ is not ${\mathrm P}\Gamma{\mathrm L}(2,q^5)$-equivalent to any linear set $L_{g_{1,\mu}}$ and hence it is a new a maximum scattered linear set.
\end{theorem}
\begin{proof}
	Suppose, contrary to our claim, that  $L_{g_{2,\delta}}$ is ${\mathrm P}\Gamma{\mathrm L}(2,q^5)$-equivalent to a linear set $L_{g_{1,\mu}}$. From Proposition \ref{pgamma-equiv} and Theorem \ref{main1}, taking into account that $L_{g_{1,\mu}}$ is not of pseudoregulus type, it follows that there exist $\varphi\in\Gamma{\mathrm L}(2,q^5)$ and $\lambda\in{\mathbb F}_{q^5}^*$ such that either $(g_{2,\delta})_{\varphi}(x)=g_{1,\mu}(\lambda x)/\lambda$ or $(g_{2,\delta})_{\varphi}(x)=\hat g_{1,\mu}(\lambda x)/\lambda$. This is equivalent to say that
	there exist $A,B,C,D \in {\mathbb F}_{q^5}$ with $AD-BC \neq 0$ and a field automorphism $\tau$ of ${\mathbb F}_{q^5}$
such that
\[
\left\{
\begin{pmatrix}
A & B \\
C & D \\
\end{pmatrix}
\begin{pmatrix}
x^\tau \\
h(x)^\tau \\
\end{pmatrix}
\colon x\in {\mathbb F}_{q^5}
\right\}
=
\left\{
\begin{pmatrix}
z\\
\alpha z^q + \beta z^{q^4}\\
\end{pmatrix}
\colon z\in {\mathbb F}_{q^5}
\right\}
,\]
where  $N(\alpha) \neq N(\beta)$ and $\alpha \beta \neq 0$.
	We may substitute $x^{\tau}$ by $y$, then
	\[\alpha (Ay+B\delta^\tau y^{q^2}+By^{q^3})^q+\beta (Ay+B\delta^\tau y^{q^2}+By^{q^3})^{q^4} = \]
	\[ Cy+D\delta^\tau y^{q^2}+Dy^{q^3}\]
for each $y\in {\mathbb F}_{q^5}$.
	Comparing coefficients yields $C=0$ and
	\begin{equation}
	\label{1}
	\alpha A^q+\beta B^{q^4}\delta^{q^4 \tau}=0,
	\end{equation}
	\begin{equation}
	\label{2}
	\beta B^{q^4}=D\delta^\tau,
	\end{equation}
	\begin{equation}
	\label{3}
	\alpha B^q \delta^{q\tau} =D,
	\end{equation}
	\begin{equation}
	\label{4}
	\alpha B^q +\beta A^{q^4}=0.
	\end{equation}
	Conditions \eqref{2} and \eqref{3} give
	\begin{equation}
	\label{primo}
	B^{q^4-q}= \delta^{(q+1)\tau} \alpha/\beta.
	\end{equation}
	On the other hand from \eqref{4} we get $A^q=- B^{q^3}\alpha^{q^2}/\beta^{q^2}$ and substituting this into \eqref{1} we have
	\begin{equation}
	\label{secondo}
	B^{q^3-q^4}=\delta^{q^4 \tau} \beta^{q^2+1}/\alpha^{q^2+1}.
	\end{equation}
	Equations \eqref{primo} and \eqref{secondo} give $N(\beta/\alpha)=N(\delta)^{2\tau}$ and  $N(\alpha/\beta)^2=N(\delta)^{\tau}$, respectively. It follows that
	$N(\delta)^{5\tau}=1$ and hence $N(\delta)^5=1$, a contradiction.
\end{proof}

\section*{Open problems}

We conclude the paper by the following open problems.

\begin{enumerate}
	\item Is it true also for $n>5$ that for any pair of $q$-polynomials $f(x)$ and $g(x)$ of ${\mathbb F}_{q^n}[x]$, with maximum field of linearity ${\mathbb F}_q$, if $Im\,(f(x)/x)=Im\,(g(x)/x)$ then either there exists $\varphi\in \Gamma\mathrm{L}(2,q^n)$ such that $f_{\varphi}(x)=\alpha x^{q^i}$ and $g_{\varphi}(x)=\beta x^{q^j}$ with $\N(\alpha)=\N(\beta)$ and $\gcd(i,n)=\gcd(j,n)=1$, or there exists $\lambda\in {\mathbb F}_{q^n}^*$ such that $g(x)=f(\lambda x)/\lambda$ or $g(x)=\hat f(\lambda x)/\lambda$?
	\item Is it possible, at least for small values of $n>4$, to classify, up to equivalence, the $q$-polynomials $f(x)\in {\mathbb F}_{q^n}[x]$ such that $|Im\,(f(x)/x)|=(q^n-1)/(q-1)$? Find new examples!
	\item Is it possible, at least for small values of $n$, to classify, up to equivalence, the $q$-polynomials $f(x)\in {\mathbb F}_{q^n}[x]$ such that  $|Im\,(f(x)/x)|=q^{n-1}+1$? Find new examples!
	\item Is it possible, at least for small values of $n$, to classify, up to equivalence, the $q$-polynomials $f(x)\in {\mathbb F}_{q^n}[x]$ such that in the multiset  $\{f(x)/x \colon x\in {\mathbb F}_{q^n}^*\}$ there is a unique element which is represented more than $q-1$ times? In this case the linear set $L_f$ is an \emph{$i$-club} of rank $n$ and when $q=2$, then such linear sets correspond to translation KM-arcs cf. \cite{KMarcs2} (a KM-arc, or $(q+t)$-arc of type $(0,2,t)$, is a set of $q+t$ points of $\PG(2,2^n)$, such that each line meet the point set in 0,2 or in $t$ points, cf. \cite{KMarcs}). Find new examples!
	\item Determine the equivalence classes of the set of $q$-polynomials in ${\mathbb F}_{q^4}[x]$.
	\item Determine, at least for small values of $n$, all the possible sizes of $Im\,(f(x)/x)$ where $f(x)\in {\mathbb F}_{q^n}[x]$ is a $q$-polynomial.

\end{enumerate}

redpajama_set_name: RedPajamaArXiv


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Despite having seen the name Infestus on the Internet, I've never given them a chance, however upon discovering that they are signed to Debemur Morti Productions, I knew it was a release I had to check out; Debemur Morti is a label that consistently releases great material, so seeing Infestus being signed to them immediately gave me some inkling that this would be a strong album.
Infestus is a one-man project hailing from Germany, the brain-child of Andras and throughout their 15-year existence, his 4 albums to date all seem to have been well received from other reviews that I've seen on the Internet. So what about "Thrypsis"? It's been four years since his last album "The Reflecting Void" and one of the first impressions that I get from listening to "Thrypsis" is that it sounds like an album four years in the making, it is a well crafted and multi-layered piece of art. The attention that has been paid and effort involved is truly admirable, everything has its place within the album and feels vital to the body of work.
Musically, Infestus brings to mind a variety of bands like Secrets of the Moon, Shining, Schammasch and Blut Aus Nord, yet, a true testament to Andras is that it doesn't call to mind any direct comparison, yes, there are influences I can hear from elsewhere but they're so varied and sparse that this project can proudly stand on its feet as a unique and fully realized concept.
The songs for the most part are a mid-based affair, there's the occasional blast here and there but the controlled pace of the album conveys great emotion and manages intensity without resorting to all-out blast filled savagery, this is coupled with the amazing composition; throughout the record there are beautiful piano interludes, clean guitar sections, killer riffs and epic solos. The drums are played spectacularly, whilst they may not be the most technical chops, the variation and subtlety (at times) with which they are utilized really helps elevate the album into greatness.
The production on the album is similarly awesome, everything has space to be noticed, it has remarkably clean clarity when needed, but it can then easily shift into a more raw feel whilst still retaining all the qualities of each instrument involved. It's produced to a very high standard, not once does it sound sterile or lifeless, or needlessly lo-fi.
"Thrypsis" is great in every aspect, it constantly feels focused and planned, everything feels vital, each additional element only serves to add value to the songs, whether that be extra guitar overdubs or even the removal of instruments from the songs to bring space and atmosphere to the tracks, and all this from one guy; all too often one person projects get bogged down with narrow vision and repetition, it's impressive to see the scope at which he can visualise a project and bring it to fruition.
I had no idea what to expect from this release and I am more than pleasantly surprised, it is one of the best albums I have heard in recent months and potentially one of the best albums so far this year. I'll definitely be diving into Infestus' back catalogue to see what other gems I may have missed out on!
Infestus' "Thrypsis" is out now via Debemur Morti Productions. Order it here.


text:
Hip Hop News | Trending Hip Hop Stories
Barack Obama Gives Drake Greenlight to Portray Him in a Possible Biopic
Barack Obama and his entire family are some of the most famous people in the world. It won't be a surprise if the former President gets the biopic treatment.
Drake told Hip Hop DX in 2012 that he's interested in taking on the role of portraying our Forever President on the big screen.
"I hope somebody makes a movie about Obama's life soon because I could play him," he said at the time. "That's the goal. I watch all the addresses. Any time I see him on TV, I don't change the channel. I definitely pay attention and listen to the inflections of his voice. If you ask anyone who knows me, I'm pretty good at impressions. I've been reading scripts for a while. I want to do something great. I want to do something for my culture: the younger people who are still in tune with everything going on. I'm actually writing with my friends right now."
During an appearance on 360 with Speedy Morman, Barack Obama made it clear that he's perfectly fine with that. "I will say this: Drake seems to be able to do anything he wants," said Obama. "You know what, Drake has– more importantly, I think– my household's stamp of approval. I suspect Malia and Sasha would be just fine with it."
How do you guys feel about this match up?
Jiggy Jones
Mira B
Wendy Williams Suggest Reginae Carter Distance Herself From YFN Lucci Following Arrest
DaBaby Releases New Single "Masterpiece"
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Wendy Williams Alleges 80s R&B Singer Sherrick Raped Her
Today in Hip-Hip History: Gang Starr's Sophomore LP 'Step Into The Arena' Turns 30 Years Old!

redpajama_set_name: RedPajamaCommonCrawl


text:
DATAGROUP IT SERVICES HOLDING AG (FRA:D6H) DATAGROUP SE: DATAGROUP continues its growth trend in the second quarter of 2016/2017
DGAP-News: DATAGROUP SE / Key word(s): Half Year Results/Quarter Results
DATAGROUP SE: DATAGROUP continues its growth trend in the second quarter of 2016/2017
DATAGROUP continues its growth trend in the second quarter of 2016/2017
Figures for Q2 2016/2017 (01.01. - 31.03.2017)
Revenue up by 33%, EBITDA by 77%
Strong organic growth of 10%
Earnings per share skyrocketed from 13 auf 36 cents
Management raises guidance for the current fiscal year
Pliezhausen, May 8, 2017. DATAGROUP SE (WKN A0JC8S) has accelerated its strong and profitable growth in the second quarter of FY 2016/2017 once again. Revenues grew by one third, earnings per share almost tripled. Based on the very good business development and the recent acquisition of IT service provider HanseCom, the management board raises its guidance for FY 2016/2017.
In the second quarter of FY 2016/2017 (01.01. - 31.03.2017), DATAGROUP achieved a strong increase in revenues of 33% yoy. The IT service provider thus has been able to accelerate growth even further after an increase of 29% had been entered in the books in the first quarter. Q2 revenues amounted to EUR 54.6m (previous year EUR 41.1m). This was due to strong organic growth of 10% and the takeover of IT specialists from Hewlett-Packard Enterprise in September 2016. The significant expansion of the high-margin core business with Cloud and outsourcing services leads to a disproportionate increase in earnings. Quarterly earnings before interest, taxes, depreciation and amortisation (EBITDA) jumped 77% to EUR 6.3m (previous year EUR 3.5m). The EBITDA margin was up from 8.6% to 11.5%. Earnings before interest and taxes (EBIT) more than doubled to EUR 4.3m (previous year EUR 1.9m), corresponding to an increase of 123%. Earnings per share (EPS) almost tripled yoy from 13 cents to 36 cents.
In the first half of FY 2016/2017 (01.10.2016 - 31.03.2017) DATAGROUP boosted revenues by 31% to EUR 108m (previous year EUR 82.7m). The increase in EBITDA, which was up 92% yoy to EUR 11.5m (previous year EUR 6.0m), was even more pronounced. EPS improved from 17 cents to 59 cents. When accounting for the 10% capital increase on April 5, 2017, EPS amounted to 54 cents.
"We are very pleased with the course of the fiscal year up to now, especially with the strong organic growth of 10%", commented DATAGROUP CEO Max H.-H. Schaber. "The operating business with long-term service contracts and recurring income is developing extremely well. The acquisition of IT service provider HanseCom will provide an additional impetus and the successful capital increase creates financial scope for further strong growth. All of this makes us look to the future with great optimism and enthusiasm!"
The capital increase by 10% has substantially strengthened equity. Resulting from a significant extension of the balance sheet by 55% in the last fiscal year, the equity ratio had dropped to below 20% despite the very good operating business trend. The reasons were sharp increases in pension liabilities in the wake of the takeover of 306 IT specialists from Hewlett-Packard Enterprise as well as the issuance of a promissory note loan with a volume of EUR 30m to secure the low level of interest over the long term. At the reporting date on March 31, 2017, the equity ratio stood at 18% (including subordinated loan: 20%). When including the issue proceeds of the capital increase of over EUR 21m, the equity ratio would have been 28% at the reporting date (including subordinated loan: 29%). The sustainable improvement of the equity base enables DATAGROUP to continue to play an active and dynamic role in the consolidation of the market for IT services.
Since its IPO, DATAGROUP has acquired 19 companies or parts of companies, the most recent of which was Hamburg-based IT service provider HanseCom in early April. The company has some 70 employees and generates annual revenue of ca. EUR 16m, which will be consolidated in the annual financial statement of DATAGROUP SE from May 2017. In line with DATAGROUP's corporate strategy, HanseCom will act as an independent subsidiary in the market under the new name DATAGROUP Operate IT GmbH. At the same time, DATAGROUP's and HanseCom's service profiles will be interlinked so that customers benefit from the broader service offering and expertise within the group.
"An above-average customer satisfaction is one of our most important competitive advantages", said DATAGROUP board member and Chief Operating Officer Dirk Peters. "As our corporate structure is heavily influenced by the German Mittelstand, we are always close to our customers and at eye-level. At the same time, our structure guarantees a consistently high level of quality. This is achieved with our joint full-service portfolio CORBOX, which is partly produced locally and partly in central service factories, but always on the basis of our standardised production processes. This combination of proximity to our Mittelstand customers and certified process and service qualities is our core strength in the market."
Based on the very good development of the operating business and the successful acquisition of HanseCom, the management board raises its guidance for the current fiscal year 2016/2017 and now expects revenues of EUR 215 - 225m (previously EUR 210 - 220m ) and EBITDA of EUR 22.0 - 23.0m (previously EUR 21.5 - 22.5m).
The complete semi-annual report can be downloaded at https://www.datagroup.de/en/investor-relations/publikationen/
About DATAGROUP:
DATAGROUP is a leading German IT service company. Over 1,700 employees at locations across Germany design, implement, and operate IT infrastructure and business applications such as SAP. With CORBOX, DATAGROUP offers its customers all the IT services they need for a stable IT operation. DATAGROUP thus is a full-service provider, serving over 600,000 global IT workplaces for medium and large enterprises as well as for public-sector clients. In other words: "We manage IT." The company is growing organically and through acquisitions. The acquisition strategy is particularly noted for its optimal integration of the new companies. DATAGROUP is actively participating in the IT service market's consolidation process with its "buy and turn around" and its "buy and build" strategy.
DATAGROUP SE
Dr. Michael Klein
Wilhelm-Schickard-Str. 7
D-72124 Pliezhausen
T +49-7127-970-059
F +49-7127-970-033
Michael.Klein@datagroup.de
08.05.2017 Dissemination of a Corporate News, transmitted by DGAP - a service of EQS Group AG.
The DGAP Distribution Services include Regulatory Announcements, Financial/Corporate News and Press Releases.
Archive at www.dgap.de
Company: DATAGROUP SE
Internet: www.datagroup.de
ISIN: DE000A0JC8S7
WKN: A0JC8S
Listed: Regulated Unofficial Market in Berlin, Dusseldorf, Munich, Stuttgart, Tradegate Exchange; Open Market (Scale) in Frankfurt
End of News DGAP News Service
570757 08.05.2017
DATAGROUP SE: DATAGROUP Is One of the Best Employers for IT Professionals
DATAGROUP SE: DATAGROUP Asserts Itself Again at the Top for Customer...
DATAGROUP SE: DATAGROUP Posts Record Results - Guidance Significantly...
DATAGROUP SE: DATAGROUP Sets the Course for Future Growth and Stability
DATAGROUP SE: Chief Executive Officer Announces Resignation at the 2022...
DATAGROUP SE: DATAGROUP Presents Strong Quarter Results with Significant...
DATAGROUP SE: DATAGROUP enters into agreement to acquire majority stake in...
DATAGROUP SE: DATAGROUP entered into a contract to acquire assets and...
DATAGROUP SE: DATAGROUP wins major contract from IKB Deutsche...
DATAGROUP SE: DATAGROUP increases revenue and earnings outlook
Voir tous les communiqués de DATAGROUP IT SERVICES HOLDING AG


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text:
Q: ValueError: Error when checking input: expected cu_dnnlstm_22_input to have 3 dimensions, but got array with shape (2101, 17) I am new to machine learning. I am having trouble getting my data into my network.
This is the error that I am receiving:
ValueError: Error when checking input: expected cu_dnnlstm_22_input to have 3 dimensions, but got array with shape (2101, 17)
I have tried adding model.add(Flatten()) before the dense layer. I would really appreciate your help!

BATCH_SIZE = 64

test_size_length = int(len(main_df)*TESTING_SIZE)
training_df = main_df[:test_size_length]
validation_df = main_df[test_size_length:]

train_x, train_y = training_df.drop('target',1).to_numpy(), training_df['target'].tolist()
validation_x, validation_y = validation_df.drop('target',1).to_numpy(), validation_df['target'].tolist()

#train_x.shape is  (2101, 17)

model = Sequential()
# model.add(Flatten())
model.add(CuDNNLSTM(128, input_shape=(train_x.shape), return_sequences=True))
model.add(Dropout(0.2))
model.add(BatchNormalization())

model.add(CuDNNLSTM(128, return_sequences=True))
model.add(Dropout(0.1))
model.add(BatchNormalization())

model.add(CuDNNLSTM(128))
model.add(Dropout(0.2))
model.add(BatchNormalization())

model.add(Dense(32, activation='relu'))
model.add(Dropout(0.2))

model.add(Dense(2, activation='softmax'))

opt = tf.keras.optimizers.Adam(lr=0.001, decay=1e-6)

# Compile model
model.compile(
    loss='sparse_categorical_crossentropy',
    optimizer=opt,
    metrics=['accuracy']
)

tensorboard = TensorBoard(log_dir="logs/{}".format(NAME))

filepath = "RNN_Final-{epoch:02d}-{val_acc:.3f}"  # unique file name that will include the epoch and the validation acc for that epoch
checkpoint = ModelCheckpoint("models/{}.model".format(filepath, monitor='val_acc', verbose=1, save_best_only=True, mode='max')) # saves only the best ones

# Train model
history = model.fit(
    train_x, train_y,
    batch_size=BATCH_SIZE,
    epochs=EPOCHS,
    validation_data=(validation_x, validation_y),
    callbacks=[tensorboard, checkpoint],
)

# Score model
score = model.evaluate(validation_x, validation_y, verbose=0)
print('Test loss:', score[0])
print('Test accuracy:', score[1])
# Save model
model.save("models/{}".format(NAME))

A: The input to your LSTM layer (CuDNNLSTM) should have shape: (batch_size, timesteps, input_dim).
It looks like you are missing one of these dimensions.
Often we can oversee the last dimension in the case where the input dimension is 1. If this is the case with your model (if you are predicting from a sequence of single numbers), then you might consider expanding the dimensions before the CuDNNLSTM layer with something like this:
model.add(Lambda(lambda t: tf.expand_dims(t, axis=-1)))
model.add(CuDNNLSTM(128))

Without knowing the problem you are working on it's hard to know whether this is a valid way forward but certainly you should keep in mind the required shape of a LSTM layer and reshape/expand dims accordingly.

redpajama_set_name: RedPajamaStackExchange


text:
package kingpin

import (
	"bufio"
	"os"

	"strings"
)

type TokenType int

// Token types.
const (
	TokenShort TokenType = iota
	TokenLong
	TokenArg
	TokenEOL
)

var (
	TokenEOLMarker = Token{TokenEOL, ""}
)

type Token struct {
	Type  TokenType
	Value string
}

func (t *Token) IsFlag() bool {
	return t.Type == TokenShort || t.Type == TokenLong
}

func (t *Token) IsEOF() bool {
	return t.Type == TokenEOL
}

func (t *Token) String() string {
	switch t.Type {
	case TokenShort:
		return "-" + t.Value
	case TokenLong:
		return "--" + t.Value
	case TokenArg:
		return t.Value
	case TokenEOL:
		return "<EOL>"
	default:
		panic("unhandled type")
	}
}

type Tokens []*Token

func (t Tokens) String() string {
	out := []string{}
	for _, tok := range t {
		out = append(out, tok.String())
	}
	return strings.Join(out, " ")
}

func (t Tokens) Next() Tokens {
	if len(t) == 0 {
		return nil
	}
	return t[1:]
}

func (t Tokens) Return(token *Token) Tokens {
	if token.Type == TokenEOL {
		return t
	}
	return append(Tokens{token}, t...)
}

func (t Tokens) Peek() *Token {
	if len(t) == 0 {
		return &TokenEOLMarker
	}
	return t[0]
}

func Tokenize(args []string) *ParseContext {
	tokens := make(Tokens, 0, len(args))
	allowFlags := true
	for _, arg := range args {
		if allowFlags {
			if arg == "--" {
				allowFlags = false
				continue
			}
			if strings.HasPrefix(arg, "--") {
				parts := strings.SplitN(arg[2:], "=", 2)
				tokens = append(tokens, &Token{TokenLong, parts[0]})
				if len(parts) == 2 {
					tokens = append(tokens, &Token{TokenArg, parts[1]})
				}
				continue
			}
			if strings.HasPrefix(arg, "-") {
				for _, a := range arg[1:] {
					tokens = append(tokens, &Token{TokenShort, string(a)})
				}
				continue
			}
		}
		tokens = append(tokens, &Token{TokenArg, arg})
	}
	return &ParseContext{Tokens: tokens}
}

// ExpandArgsFromFiles expands arguments in the form @<file> into one-arg-per-
// line read from that file.
func ExpandArgsFromFiles(args []string) ([]string, error) {
	out := []string{}
	for _, arg := range args {
		if strings.HasPrefix(arg, "@") {
			r, err := os.Open(arg[1:])
			if err != nil {
				return nil, err
			}
			scanner := bufio.NewScanner(r)
			for scanner.Scan() {
				out = append(out, scanner.Text())
			}
			r.Close()
			if scanner.Err() != nil {
				return nil, scanner.Err()
			}
		} else {
			out = append(out, arg)
		}
	}
	return out, nil
}

redpajama_set_name: RedPajamaGithub


In the crime novella SECRETS?, fledgling private investigator Elin Bohlander takes on what looks like an easy assignment — at first: to determine if her client's boyfriend is having an affair with another woman.
"Phoenix, book one of The Peradon Fantasy Series, does not disappoint. It holds everything an avid reader of fantasy craves for. The scenes are luring, filled with intrigue and mystery. Your every emotion is drawn out of you as you move along the well plotted storyline.
"Oh my! This is another one of Mary J. Williams' wonderful books.


text:
Hobart and William Smith Colleges - Connection and Celebration at Homecoming and Family Weekend
Connection and Celebration at Homecoming and Family Weekend
By Hobart and William Smith Colleges on October 7th, 2019
It was a time for connection, celebration and community as students, families and alums gathered on campus for Homecoming and Family Weekend.
"Homecoming is my family's favorite weekend on campus," said Josh Hylkema '21, whose parents joined him for a barbecue dinner and live music from Black Tie Stereo and HWS a capella groups at Quad-a-Palooza. "It offers a chance for my parents to come on campus, engage with the HWS community and experience some of the amazing things I have available to me every day."
Hylkema took part in the annual house-decorating contest on Friday, helping to dress the Random Acts of Kindness theme house for autumn. The Guiding Eyes for the Blind house won the competition, displaying balloon figures walking balloon-dogs complete with drawn-on blue service vests. "The contest allowed us to show our pride for the Colleges and bond as a group — we're like family," said Courtney Page '22.
On Saturday morning, President Joyce P. Jacobsen delivered the State of the Colleges address, where she reflected on her first 90 days in office and touched on academic and athletic achievements as well as recent accolades the Colleges have received. "I want to encourage everyone — along with me and around me — to actively participate in the practice of learning and to see the Colleges from a new perspective" said Jacobsen, discussing the theme of her first year, Explore HWS.
[View more stories, videos, photo galleries and social media posts from Homecoming and Family Weekend.]
HWS officially dedicated the campus' newest athletics facility — a $3.5 million state-of-the-art, air-supported sports dome — in honor of Hobart soccer legend and one of the institution's longest serving trustees, Thomas B. Poole '61, P'91, L.H.D. '06.
"There's a phrase I'm sure everyone knows: it takes a village to raise a child. In this case, it is not a child. It is a needed facility and community resource for the Colleges," shared Poole during the ribbon-cutting ceremony for the Thomas B. Poole '61 and Family Sports Dome. "The structure has the Poole name on it, but in truth the village has made it possible."
On the field, the nationally-ranked William Smith soccer and Hobart football teams led the charge to victory. The No. 4 Herons had a 3-1 win over Skidmore on Cozzens Field. Phoebe Wade '20 delivered the game-winning goal in the 48th minute. On the gridiron, the No. 22 Statesmen rode a stout defense to a 16-3 win. Emmett Forde '21 recorded a game-high 12 tackles. On the tennis courts, Walker Anderson '21 and Alan Dubrovsky '20 returned to the finals of the ITA Northeast Regional doubles championship and will advance to the national ITA Cup in Rome, Ga. Dubrovsky also reached the semifinals of the singles bracket. Wiliam Smith field hockey fell to the University of Rochester on McCooey Field.
On Saturday afternoon, more than 50 students presented at the Student Research Symposium, the annual event that gives participants in the Colleges' Summer Research Program the opportunity to share their works with fellow students, parents, faculty and alums. Hugh "Nick" McKenny '20 worked with faculty from the mathematics, education and philosophy departments on his project, "Optimizing Fairness in British Parliamentary Debates." "It was a really unique experience to do this interdisciplinary research," he said. "These are questions that no one knows the answer to, so just trying to figure out the answers is exciting."
Julia Cilano '23 had a full weekend when six members of her family — including her brother's dog, Doc — made the trip to campus. Her mother Elizabeth summed up the meaning of Homecoming and Family Weekend: "It's about this here, right now, all of us together."
Celebrating the Legacy of Alger Adams '32
HWS Dedicates Muslim Life Center


Arrival to Devechi town. Devechi district is located to the north of Baku and its eastern border extends along the Caspian coast. The name of the district originates from the word "cameleer"; evidently, it is connected to the development of caravan roads. In ancient times residents of the town would hold camels and people from neighboring regions would come to local markets - not only to buy camels but also to purchase poultry, cattle, butter, milk products, honey, wool (of camels and sheep) and carpets and also to join a caravan or hire a personal cameleer.
Excursion to Chirag Gala Castle.
Located here in the Early Middles Ages was the second line of the Caspian defense zone. It has been included in the list of historical and cultural monuments of Azerbaijan under the name of Gilgilchai Defensive Structures and is protected by the state. As a part of "Caspian Defensive Structures" it has been nominated by UNESCO for the inclusion in the list of World Legacy Monuments. Archeologists revealed that Gilgilchai Defensive Construction was built in ancient times, before our era, and that it was constantly reinforced and fortified.
The wall originates in the very sea (level of the Caspian Sea varies from year to year) and extends many kilometers, rising to mountains where it terminates with the grand watch tower of Chirag Gala (5th century). The tower is worth ascending to: aside of historical importance, it features an astounding view from the mountain top over the sea and nearby mountains.
Here, between the sea and ridges of the Caucasus Mountains there is a very convenient passage from Azerbaijan to the north, to other countries. Since ancient times the passage has been named the Caspian Passage. It was used by invaders from neighboring countries and distant lands. They would invade Azerbaijan and ransack Baku ports, sometimes reaching as far as the inland areas of the country. Therefore, the walls served to defend and alarm Absheron (Baku) and further, all borderland regions of the ancient Azerbaijan. The magnitude of the construction and defensive significance of the wall and fortresses (they also secured the approaches to the Great Silk Road) make them a rival of the famous Great Wall in China. However, this object has not been restored yet.
Not far from the tower of Chirag Gala there is a resort complex Gala Alti ("Under the Tower") with a health center named "Chirag Gala", known for its mineral springs with medicinal sulfurated water, especially beneficial for patients with urolithiasis and cholelithiasis. On the road to the health resort there is a number of shashlik houses, cafes and small camps a majority of which operate only in summer.
Chirag Gala, which means lamp (or light) castle in Azerbaijani is a ruined ancient fortress overlooking the Caspian coastal plains north of Baku in Azerbaijan. Its located on the top of a mountain, in the Guba Forest. It was constructed by the Sassanid Persians in the 5th century, it was used as a defense for the khanate of Quba in the 18th century.
Today the Chirag Gala is a historic site and it frequently visited by tourists. Because of the large rocks and rough road the Chirag Gala is almost impossible to reach with any vehicle. Most tourists and people travel by foot to the top of the mountain.