The $\gamma b\overline{b}$ production via $\gamma^{*}\gamma^{*}$ collisions at
the ILC and LHC

Bui Thi Ha Gianga<EMAIL_ADDRESS>
a Hanoi National University of Education, 136 Xuan Thuy, Hanoi, Vietnam

###### Abstract

Taking into account of the mixing of Higgs-radion in the Randall-Sundrum model
and the vector anomalous couplings, we investigate the production of
$b\overline{b}$ associated with the photon through $\gamma^{*}\gamma^{*}$
collisions at the International Linear Collider (ILC) and Large Hadron
Collider (LHC). The total cross-section depends strongly on the vacuum
expectation value (VEV) of the radion field $\Lambda_{\phi}$, the radion mass
$m_{\phi}$, the parameters of anomalous couplings. The result shows that the
total cross-section in $\gamma b\overline{b}$ production at the LHC is much
larger than that at the ILC. The production cross-section gives the largest
value at the dominated state, $m_{\phi}=m_{h}=125$ GeV.

Keywords: Higgs production, radion production, cross-section, ILC, LHC.

##  I Introduction

The Standard model (SM) of particle physics is a theoretical framework that
describes the behavior of fundamental particles. SM is successful in
describing the picture of subatomic particles and the fundamental forces that
govern their interactions. However, it is not a complete theory, and there are
many open questions in physics that it does not addressed, such as the nature
of dark matter and dark energy. Therefore, there are some extended models
which provide new scenarios to solve the theoretical drawbacks in the SM. One
of the most attractive extended models is the Randall-Sundrum model (RS) [1].
Owing to the extra dimension, the additional scalar field called radion can be
mixed with Higgs boson and constitute two physical mixed states [2, 3, 4, 5].
The detection of the other radion-like mixed state in a collider experiment
will be an evidence for the new physics and Higgs-radion mixing.
The ILC experiment is one of the proposed next generation electron-positron
colliders [6, 7, 8]. A high energy $e^{-}e^{+}$ collider provides a clean
experimental environment to measure various observables related the SM Higgs
boson, gauge bosons and fermions, and further detect new physics effects [9,
10, 11, 12, 13]. Various ILC physics studies can have a great impact on
understanding a new physics around TeV scale. The high energy
$\gamma^{*}\gamma^{*}$ and $\gamma^{*}e^{-}$ collisions, with the photon beam
generated by the backward Compton scattering of incident electron and laser-
beams, might provide a good chance to test the SM and further to search new
particles or new physics.
Due to the new physics research at LHC, the subprocesses of gluon-gluon,
quark-quark, quark-gluon have not provided clean environment. Moreover, the
exclusive process and semi-elastic processes are much less examined [14]. One
important advantage of photon processes, not contain many QCD backgrounds and
uncertainties from proton dissociation, provide clean experimental channels.
Therefore, this makes it easy to determine any possible signal which may come
from new physics. The study of new physics via photon-photon and photon-proton
collisions at LHC embraces a wide range of models beyond the SM such as extra
dimensions, magnetic monopoles, supersymmetry,… [15] Both of the incoming
protons in the exclusive process $\gamma^{*}\gamma^{*}$, the cleanest channel,
remains intact and do not dissociate into partons. Based on the simpler final
states with respect to $pp$ processes, the $\gamma^{*}\gamma^{*}$ process
compensates for the advantages of $pp$ collisions such as having high center
of mass energy and high luminosity. The photon processes at the LHC pose a
considerable potential to probe new physics [16, 17, 18, 15].
In the present work, we analyze the $\gamma h/\gamma\phi$ associated
production, followed by Higgs boson and radion decaying $b\overline{b}$ at the
ILC and LHC. With the contribution of interactions in the RS model, the total
cross-sections are expected to experimentally detected. The organization of
this paper is as follows: in section II, we review the mixing of Higgs-radion
and the anomalous couplings. The total cross-sections for
$\gamma^{*}\gamma^{*}\to\gamma h/\gamma\phi\rightarrow\gamma b\overline{b}$
collisions is presented in section III. Finally, we summarize our results and
draw conclusions in section IV.

##  II The mixing of Higgs-radion and the anomalous couplings

The RS model consists of one extra dimension bounded by two 3-branes: UV-brane
and IR-brane [19]. The gravity-scalar mixing is described by the following
action[20]

$S_{\xi}=\xi\int d^{4}x\sqrt{g_{vis}}R(g_{vis})\hat{H}^{+}\hat{H},$ (1)

where $\xi$ is the mixing parameter, $R(g_{vis})$ is the Ricci scalar for the
metric $g_{vis}^{\mu\nu}=\Omega_{b}^{2}(x)(\eta^{\mu\nu}+\varepsilon
h^{\mu\nu})$ induced on the visible brane,
$\Omega_{b}(x)=e^{-kr_{c}\pi}(1+\frac{\phi_{0}}{\Lambda_{\phi}})$ is the warp
factor, $\hat{H}$ is the Higgs field in the 5D context before rescaling to
canonical normalization on the brane. With $\xi\neq 0$, there is neither a
pure Higgs boson nor pure radion mass eigenstate. This $\xi$ term mixes the
$h_{0}$ and $\phi_{0}$ into the mass eigenstates $h$ and $\phi$ as given by

$\left(\begin{array}[]{c}{h_{0}}\\\
{\phi_{0}}\end{array}\right)=\left(\begin{array}[]{cc}{1}&{6\xi\gamma/Z}\\\
{0}&{-1/Z}\end{array}\right)\left(\begin{array}[]{cc}{\cos\theta}&{\sin\theta}\\\
{-\sin\theta}&{\cos\theta}\end{array}\right)\left(\begin{array}[]{c}{h}\\\
{\phi}\end{array}\right)=\left(\begin{array}[]{cc}{d}&{c}\\\
{b}&{a}\end{array}\right)\left(\begin{array}[]{c}{h}\\\
{\phi}\end{array}\right),$ (2)

where
$Z^{2}=1+6\gamma^{2}\xi\left(1-\,\,6\xi\right)=\beta-36\xi^{2}\gamma^{2}$ is
the coefficient of the radion kinetic term after undoing the kinetic mixing,
$\gamma=\upsilon/\Lambda_{\phi},\upsilon=246$ GeV,
$a=-\dfrac{cos\theta}{Z},b=\dfrac{sin\theta}{Z},c=sin\theta+\dfrac{6\xi\gamma}{Z}cos\theta,d=cos\theta-\dfrac{6\xi\gamma}{Z}sin\theta$.
The mixing angle $\theta$ is

$\tan 2{\theta}=12{\gamma\xi
Z}\frac{m_{h_{0}}^{2}}{m_{\phi_{0}}^{2}-m_{h_{0}}^{2}\left(Z^{2}-36\xi^{2}\gamma^{2}\right)},$
(3)

where $m_{h_{0}}$ and $m_{\phi_{0}}$ are the Higgs and radion masses before
mixing.
The new physical fields h and $\phi$ in (2) are Higgs-dominated state and
radion, respectively

$m_{h,\phi}^{2}=\frac{1}{2Z^{2}}\left[m_{\phi_{0}}^{2}+\beta
m_{h_{0}}^{2}\pm\sqrt{(m_{\phi_{0}}^{2}+\beta
m_{h_{0}}^{2})^{2}-4Z^{2}m_{\phi_{0}}^{2}m_{h_{0}}^{2}}\right].$ (4)

There are four independent parameters
$\Lambda_{\phi},\,\,m_{h},\,\,m_{\phi},\,\,\xi$ that must be specified to fix
the state mixing parameters.
Feynman rules for the couplings in the RS model are shown as follows [19]

$\displaystyle C_{\gamma\gamma
h}=\dfrac{\alpha}{2\pi\upsilon_{0}}\left((d+\gamma
b)\sum_{i}e_{i}^{2}N_{c}^{i}F_{i}(\tau_{i})-(b_{2}+b_{Y})\gamma b\right)$ (5)
$\displaystyle
C_{\gamma\gamma\phi}=\dfrac{\alpha}{2\pi\upsilon_{0}}\left((c+\gamma
a)\sum_{i}e_{i}^{2}N_{c}^{i}F_{i}(\tau_{i})-(b_{2}+b_{Y})\gamma a\right)$ (6)
$\displaystyle C_{\gamma
Zh}=\dfrac{\alpha}{2\pi\nu_{0}}\left[2g_{h}^{r}\left(\dfrac{b_{2}}{tan\theta_{W}}-b_{Y}tan\theta_{W}\right)-g_{h}\left(A_{F}+A_{W}\right)\right],$
(7) $\displaystyle C_{\gamma
Z\phi}=\dfrac{\alpha}{2\pi\nu_{0}}\left[2g_{\phi}^{r}\left(\dfrac{b_{2}}{tan\theta_{W}}-b_{Y}tan\theta_{W}\right)-g_{\phi}\left(A_{F}+A_{W}\right)\right].$
(8)

where a, b, c, d are the state mixing parameters in RS model. $g_{h}=d+\gamma
b$, $g_{\phi}=c+\gamma a$, $g_{h}^{r}=\gamma b$, $g_{\phi}^{r}=\gamma a$, the
triangle loop functions $A_{F},A_{W}$ are given in Ref.[21]. The auxiliary
functions of the $h$ and $\phi$ are given by

$\displaystyle F_{1/2}(\tau_{i})=-2\tau_{i}[1+(1-\tau_{i})f(\tau_{i})],$ (9)
$\displaystyle F_{1}(\tau_{i})=2+3\tau_{i}+3\tau_{i}(2-\tau_{i})f(\tau_{i}),$
(10)

with

$\displaystyle
f(\tau_{i})=\left(\sin^{-1}\frac{1}{\sqrt{\tau_{i}}}\right)^{2}\,\,\,(for\,\,\,\tau_{i}>1),$
(11) $\displaystyle
f(\tau_{i})=-\frac{1}{4}\left(\ln\frac{\eta_{+}}{\eta_{-}}-i\pi\right)^{2}\,\,\,(for\,\,\,\tau_{i}<1),$
(12)
$\displaystyle\eta_{\pm}=1\pm\sqrt{1-\tau_{i}},\,\,\,\tau_{i}=\left(\frac{2m_{i}}{m_{s}}\right)^{2}.$
(13)

Here, $m_{i}$ is the mass of the internal loop particle (including quarks,
leptons and W boson), $m_{s}$ is the mass of the scalar state ($h$ or $\phi$),
$\tau_{f}=\left(\frac{2m_{f}}{m_{s}}\right)^{2},\tau_{W}=\left(\frac{2m_{W}}{m_{s}}\right)^{2}$
denote the squares of fermion and W gauge boson mass ratios, respectively.
Anomalous neutral gauge couplings have been actively searched for at LEP [22,
23], at the Tevatron [24, 25] and at the LHC [26, 27]. The anomalous trilinear
gauge boson couplings are given by [28, 29]

$\displaystyle\Gamma_{\gamma\gamma Z}^{\mu v\sigma}\left(qk_{1}k_{2}\right)=$
$\displaystyle\frac{g_{e}}{M_{Z}^{2}}\Biggl{[}h_{1}^{\gamma}\Biggl{(}q^{\mu}q^{v}k_{1}^{\sigma}+q^{\sigma}k_{1}^{\mu}k_{1}^{v}-\eta^{\mu
v}\left(q^{2}k_{1}^{\sigma}+k_{1}^{2}q^{\sigma}\right)$ (14)
$\displaystyle+\eta^{\mu\sigma}\left(k_{1}^{2}q^{v}-qk_{1}k_{1}^{v}\right)+\eta^{v\sigma}\left(q^{2}k_{1}^{\mu}-qk_{1}q^{\mu}\right)\Biggr{)}$
$\displaystyle-
h_{3}^{\gamma}\left(k_{1\beta}k_{1}^{v}q_{\alpha}\varepsilon^{\mu\sigma\alpha\beta}+q^{\mu}k_{1\alpha}q_{\beta}\varepsilon^{v\sigma\alpha\beta}+\left(q^{2}k_{1\alpha}-k_{1}^{2}q_{\alpha}\right)\varepsilon^{\mu
v\sigma\alpha}\right)\Biggr{]},$
$\Gamma^{\sigma\mu\nu}_{\gamma\gamma\gamma}(p_{1}p_{2}p_{3})=C_{\gamma\gamma\gamma}\Biggl{[}(p_{1}-p_{2})^{\nu}\eta^{\sigma\mu}+(p_{2}-p_{3})^{\sigma}\eta^{\mu\nu}+(p_{3}-p_{1})^{\mu}\eta^{\nu\sigma}\Biggr{]}.$
(15)

##  III The total cross-sections for the
$\gamma^{*}\gamma^{*}\to\gamma\phi/\gamma h\to\gamma b\overline{b}$ collisions

The transition amplitude representing the s – channel is given by

${M_{s}}={M_{s\gamma}}+{M_{sZ}},$ (16)

where

$\displaystyle{M_{s\gamma}}=\frac{{{C_{\gamma\gamma
X}}}}{{q_{s}^{2}}}{\varepsilon_{\mu}}({p_{1}})\Gamma_{\gamma\gamma\gamma}^{\mu\nu\sigma}({p_{1}}{p_{2}}{q_{s}})\,{\varepsilon_{\nu}}({p_{2}}){\eta_{\beta\sigma}}\varepsilon_{\rho}^{*}({k_{1}})\left({{\eta^{\beta\rho}}{k_{1}}{q_{s}}-k_{1}^{\beta}q_{s}^{\rho}}\right),$
(17a) $\displaystyle{M_{sZ}}=\frac{{{C_{\gamma
ZX}}}}{{q_{s}^{2}-m_{Z}^{2}}}{\varepsilon_{\mu}}({p_{1}})\Gamma_{\gamma\gamma
Z}^{\mu\nu\sigma}({p_{1}}{p_{2}}{q_{s}})\,{\varepsilon_{\nu}}({p_{2}})\left({{\eta_{\beta\sigma}}-\frac{{{q_{s\sigma}}{q_{s\beta}}}}{{m_{Z}^{2}}}}\right)\varepsilon_{\rho}^{*}({k_{1}})\left({{\eta^{\beta\rho}}{k_{1}}{q_{s}}-k_{1}^{\beta}q_{s}^{\rho}}\right).$
(17b)

Here, X stands for the Higgs boson or radion.
The transition amplitude representing the u-channel can be written as

${M_{u}}={M_{u\gamma}}+{M_{uZ}},$ (18)

where

$\displaystyle{M_{u\gamma}}=\frac{{{C_{\gamma\gamma
X}}}}{{q_{u}^{2}}}{\varepsilon_{\nu}}({p_{2}})\Gamma_{\gamma\gamma\gamma}^{\nu\rho\beta}({p_{2}}{k_{1}}{q_{u}})\varepsilon_{\rho}^{*}({k_{1}}){\eta_{\beta\sigma}}{\varepsilon_{\mu}}({p_{1}})\left({{\eta^{\sigma\mu}}{p_{1}}{q_{u}}-p_{1}^{\sigma}q_{u}^{\mu}}\right),$
(19a) $\displaystyle{M_{uZ}}=\frac{{{C_{\gamma
ZX}}}}{{q_{u}^{2}-m_{Z}^{2}}}{\varepsilon_{\nu}}({p_{2}})\Gamma_{\gamma\gamma
Z}^{\nu\rho\beta}({p_{2}}{k_{1}}{q_{u}})\varepsilon_{\rho}^{*}({k_{1}})\left({{\eta_{\beta\sigma}}-\frac{{{q_{u\sigma}}{q_{u\beta}}}}{{m_{Z}^{2}}}}\right){\varepsilon_{\mu}}({p_{1}})\left({{\eta^{\sigma\mu}}{p_{1}}{q_{u}}-p_{1}^{\sigma}q_{u}^{\mu}}\right).$
(19b)

The transition amplitude representing the t-channel is given by

${M_{t}}={M_{t\gamma}}+{M_{tZ}},$ (20)

where

$\displaystyle{M_{t\gamma}}=\frac{{{C_{\gamma\gamma
X}}}}{{q_{t}^{2}}}{\varepsilon_{\mu}}({p_{1}})\Gamma_{\gamma\gamma\gamma}^{\mu\rho\sigma}({p_{1}}{k_{1}}{q_{t}})\varepsilon_{\rho}^{*}({k_{1}}){\eta_{\beta\sigma}}{\varepsilon_{\nu}}({p_{2}})\left({{\eta^{\nu\beta}}{p_{2}}{q_{t}}-p_{2}^{\beta}q_{t}^{\nu}}\right),$
(21a) $\displaystyle{M_{tZ}}=\frac{{{C_{\gamma
ZX}}}}{{q_{t}^{2}-m_{Z}^{2}}}{\varepsilon_{\mu}}({p_{1}})\Gamma_{\gamma\gamma
Z}^{\mu\rho\sigma}({p_{1}}{k_{1}}{q_{t}})\varepsilon_{\rho}^{*}({k_{1}})\left({{\eta_{\beta\sigma}}-\frac{{{q_{t\sigma}}{q_{t\beta}}}}{{m_{Z}^{2}}}}\right){\varepsilon_{\nu}}({p_{2}})\left({{\eta^{\nu\beta}}{p_{2}}{q_{t}}-p_{2}^{\beta}q_{t}^{\nu}}\right).$
(21b)

The total cross-section for the whole processes can be calculated as follow

$\displaystyle\sigma=\sigma(\gamma^{*}\gamma^{*}\to\gamma X)\times
Br(X\rightarrow b\overline{b}),$ (22)

where X stands for Higgs boson or radion.

###  III.1 The total cross-sections for the $\gamma^{*}\gamma^{*}\to\gamma
h/\gamma\phi\to\gamma b\overline{b}$ collisions at the ILC

We consider a collision process in which the initial state contains two photon
generated by the backward Compton scattering of electron and laser beams, and
the final state contains photon and $b\overline{b}$. The effective cross-
section for the $\gamma^{*}\gamma^{*}\rightarrow\gamma h/\gamma\phi$
subprocess at the ILC can be calculated as follows [30]

$\sigma_{\gamma
h/\gamma\phi}=\int_{m_{X}^{2}/s}^{0.83}dxf_{\gamma/e}(x)\int_{(cos\psi)_{min}}^{(cos\psi)_{max}}dcos\psi\dfrac{d\widehat{\sigma}(\widehat{s})}{dcos\psi},$
(23)

where $x=\widehat{s}/s$ in which $\sqrt{\widehat{s}}$ is center of mass energy
of the $\gamma^{*}\gamma^{*}\rightarrow\gamma h/\gamma\phi$ subprocess,
$\sqrt{s}$ is center of mass energy of the ILC,
$x_{max}=\dfrac{\zeta}{1+\zeta}$. The photon distribution function
$f_{\gamma/e}$ is given by [31]

$f_{\gamma/e}=\dfrac{1}{D(\zeta)}\left[(1-x)+\dfrac{1}{1-x}-\dfrac{4x}{\zeta(1-x)}+\dfrac{4x^{2}}{\zeta^{2}(1-x)^{2}}\right],$
(24)

where

$D(\zeta)=\left(1-\dfrac{4}{\zeta}-\dfrac{8}{\zeta^{2}}\right)ln(1+\zeta)+\dfrac{1}{2}+\dfrac{8}{\zeta}-\dfrac{1}{2(1+\zeta)^{2}}.$
(25)

For $\zeta=4.8$, $x_{max}=0.83$. The parameters are chosen as
$h_{1}^{\gamma}=3.6\times 10^{-3}$, $h_{3}^{\gamma}=1.3\times 10^{-3}$ [28],
$m_{h}=125$ GeV (CMS), $\xi=1/6$ [19], $\sqrt{s}=500$ GeV (ILC),
$\Lambda_{\phi}=5000$ GeV [32], $m_{\phi}=125$ GeV. We estimate the production
cross-sections at the ILC as follows
i) In Fig.1, we evaluate the dependence of the total cross-section on the
vacuum expectation value (VEV) of the radion field $\Lambda_{\phi}$ at the
fixed collision energy $\sqrt{s}=500$ GeV (ILC). In the region of 1 TeV
$\leq\Lambda_{U}\leq$ 10 TeV, the total cross-sections increase gradually.
ii) In Fig.2, we evaluate the dependence of the total cross-section on the
energy $\sqrt{s}$. The energy is chosen in the range of 500
GeV$\leq\sqrt{s}\leq$ 1000 GeV. The figure shows that the total cross-sections
increase when the energy $\sqrt{s}$ increases. The total cross-section through
$\gamma h$ is much larger than that through $\gamma\phi$ under the same
conditions.
iii) In Table 2, some typical values for the total cross-section are measured.
The first line is the total cross-section through $\gamma h$ and the second
line is the total cross-section through $\gamma\phi$. The total cross-section
through $\gamma h$ is larger than that through $\gamma\phi$. We emphasize that
the $\gamma b\overline{b}$ production cross-sections achieve the maximum value
in the case of the radion-dominated state 125 GeV.
iv) In Fig.3, we evaluate the dependence of the total cross-section on the
parameter $h_{1}^{\gamma}$. The total cross-sections increase when
$h_{1}^{\gamma}$ increases and achieve the maximum value when
$h_{1}^{\gamma}=0.0036$. We also analysis the dependence of the total cross-
sections on the parameter $h_{3}^{\gamma}$, however, the total cross-sections
are unchanged when $h_{3}^{\gamma}$ increases.

|
---|---
(a) | (b)
Figure 1:  The total cross-section as a function of the $\Lambda_{\phi}$ in (a) $\gamma^{*}\gamma^{*}\to\gamma h\to\gamma b\overline{b}$, (b) $\gamma^{*}\gamma^{*}\to\gamma\phi\to\gamma b\overline{b}$ collisions at the ILC.  Table 1: Some typical values for the total cross-sections concerning the VEV of the radion field $\Lambda_{\phi}$ at the ILC. The parameters are chosen as $\xi=1/6$, $\sqrt{s}=500$ GeV, $h_{1}^{\gamma}=3.6\times 10^{-3}$, $h_{3}^{\gamma}=1.3\times 10^{-3}$, $m_{\phi}=125$ GeV, $m_{h}=125$ GeV. $\Lambda_{\phi}$ (TeV) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
---|---|---|---|---|---|---|---|---|---|---
$\sigma_{\gamma^{*}\gamma^{*}\to\gamma h\to\gamma b\overline{b}}$ | 0.877 | 4.253 | 7.196 | 9.401 | 11.222 | 12.803 | 14.216 | 15.506 | 16.698 | 17.812
($10^{-8}$ fb) |  |  |  |  |  |  |  |  |  |
$\sigma_{\gamma^{*}\gamma^{*}\to\gamma\phi\to\gamma b\overline{b}}$ | 0.105 | 0.424 | 0.725 | 0.950 | 1.137 | 1.298 | 1.443 | 1.575 | 1.697 | 1.811
($10^{-8}$ fb) |  |  |  |  |  |  |  |  |  |
|
---|---
(a) | (b)
Figure 2:  The cross-section as a function of the energy $\sqrt{s}$ in (a) $\gamma^{*}\gamma^{*}\to\gamma h\to\gamma b\overline{b}$, (b) $\gamma^{*}\gamma^{*}\to\gamma\phi\to\gamma b\overline{b}$ collisions at the ILC. Table 2: Some typical values for the total cross-sections concerning the radion mass $m_{\phi}$ at the ILC. The parameters are chosen as $\xi=1/6$, $\Lambda_{\phi}=5000$ GeV, $\sqrt{s}=500$ GeV, $h_{1}^{\gamma}=3.6\times 10^{-3}$, $h_{3}^{\gamma}=1.3\times 10^{-3}$, $m_{h}=125$ GeV. $m_{\phi}$ (GeV) | 10 | 30 | 50 | 70 | 90 | 110 | 125 | 130 | 150 | 170 | 190
---|---|---|---|---|---|---|---|---|---|---|---
$\sigma_{\gamma^{*}\gamma^{*}\to\gamma h\to\gamma b\overline{b}}$ | 0.1237 | 0.1236 | 0.1234 | 0.1231 | 0.122 | 0.116 | 11.222 | 3.279 | 0.119 | 0.122 | 0.123
($10^{-8}$ fb) |  |  |  |  |  |  |  |  |  |  |
$\sigma_{\gamma^{*}\gamma^{*}\to\gamma\phi\to\gamma b\overline{b}}$ | 0.016 | 0.0068 | 0.0051 | 0.0054 | 0.0079 | 0.02 | 11.367 | 3.465 | 0.017 | 0.101 | 0.086
($10^{-9}$ fb) |  |  |  |  |  |  |  |  |  |  |
Table 3: Some typical values for the total cross-sections concerning the parameter of the anomalous couplings $h_{1}^{\gamma}$ at the ILC. The parameters are chosen as $\xi=1/6$, $\Lambda_{\phi}=5000$ GeV, $\sqrt{s}=500$ GeV, $h_{3}^{\gamma}=1.3\times 10^{-3}$, $m_{\phi}=125$ GeV, $m_{h}=125$ GeV. $h_{1}^{\gamma}$ | 0 | 0.0005 | 0.001 | 0.0015 | 0.002 | 0.0025 | 0.003 | 0.0036
---|---|---|---|---|---|---|---|---
$\sigma_{\gamma^{*}\gamma^{*}\to\gamma h\to\gamma b\overline{b}}$($10^{-8}$ fb) | 0.0721 | 1.6210 | 3.1695 | 4.7181 | 6.2667 | 7.8153 | 9.3638 | 11.2222
$\sigma_{\gamma^{*}\gamma^{*}\to\gamma\phi\to\gamma b\overline{b}}$ ($10^{-8}$ fb) | 0.0256 | 0.1641 | 0.3202 | 0.4771 | 0.6340 | 0.7911 | 0.9481 | 1.1367

###  III.2 The total cross-sections for the
$\gamma^{*}\gamma^{*}\to\gamma\phi/\gamma h\to\gamma b\overline{b}$ collisions
at the LHC

We consider a collision process in which two quasi elastically incoming
protons fluctuate two photons. The emitted photons can collide and produce
photon and scalar particle (Higgs or radion), followed by scalar particle
decaying into $b\overline{b}$. The total production cross-section for the
subprocess at the LHC can be described [34]

$\sigma_{\gamma
h/\gamma\phi}=\int\limits_{{\omega_{\min}}}^{{\omega_{\max}}}{d\omega}\int\limits_{{y_{\min}}}^{{y_{\max}}}{\frac{\omega}{{2y}}dy}\int\limits_{Q_{1,\min}^{2}}^{Q_{\max}^{2}}{dQ_{1}^{2}}\int\limits_{Q_{2\min}^{2}}^{Q_{\max}^{2}}{dQ_{2}^{2}}f\left({\frac{{{\omega^{2}}}}{{4y}},Q_{1}^{2}}\right)f\left({y,Q_{2}^{2}}\right){\hat{\sigma}_{\gamma^{*}\gamma^{*}\to\gamma
h/\gamma\phi}}(Q_{1}^{2},Q_{2}^{2},y,\omega)$ (26)

Here, the photon spectrum is described as

$f\left({{E_{\gamma}},Q_{2}^{2}}\right)=\frac{{dN}}{{d{E_{\gamma}}d{Q^{2}}}}=\frac{{{\alpha_{e}}}}{\pi}\frac{1}{{{E_{\gamma}}{Q^{2}}}}\left[{\left({1-\frac{{{E_{\gamma}}}}{{{E_{p}}}}}\right)\left({1-\frac{{Q_{\min}^{2}}}{{{Q^{2}}}}}\right){F_{E}}+\frac{{E_{\gamma}^{2}}}{{2E_{p}^{2}}}{F_{M}}}\right],$
(27)

with

$\displaystyle
Q_{\min}^{2}=\frac{{E_{\gamma}^{2}m_{p}^{2}}}{{{E_{p}}({E_{p}}-{E_{\gamma}})}},$
(28) $\displaystyle{E_{\gamma}}={E_{p}}\xi^{*},{\alpha_{e}}=g_{e}^{2}/4\pi,$
(29)
$\displaystyle{F_{E}}=\frac{{4m_{p}^{2}G_{E}^{2}+{Q^{2}}G_{M}^{2}}}{{4m_{p}^{2}+{Q^{2}}}},$
(30) $\displaystyle{F_{M}}=G_{M}^{2},$ (31) $\displaystyle
G_{E}^{2}=\frac{{G_{M}^{2}}}{{\mu_{p}^{2}}}={\left({1+\frac{{{Q^{2}}}}{{Q_{0}^{2}}}}\right)^{-4}},Q_{0}^{2}=0.71GeV^{2}.$
(32)

${\alpha_{e}}$ is the fine-structure constant, ${m_{p}}$ is the proton mass.
$E_{p}$ is the energy of the incoming proton beam. $E_{\gamma}$ is the photon
energy, which is related to the loss energy of the emitted proton beam. Based
on the CMS standard running conditions, $\xi^{*}$ is chosen
$0.0015<\xi^{*}<0.5$ which provides the most sensitive interval to the
anomalous couplings. $F_{E},F_{M}$ are functions of the electric and magnetic
form factors $G_{E},G_{M}$ given in the dipole approximation, respectively.
The squared magnetic moment of the proton is taken to be a constant value
$\mu_{p}^{2}=7.78$. The invariant mass of centrally produced particles is
obtained from $\omega\simeq 2\sqrt{{E_{{\gamma_{1}}}}{E_{{\gamma_{2}}}}}$ .
The electric and magnetic proton form factors fall rapidly with the increase
of ${Q^{2}}$ , therefore, $Q_{\max}^{2}$ is chosen as 2 $GeV^{2}$ [30].
For numerical calculations, we choose the energy of the incoming proton beam
at the LHC $E_{p}=13$ TeV. The integration limits are chosen as

$\displaystyle{y_{\min}}=Max\left[{\frac{{{\omega^{2}}}}{{4{E_{p}}{\xi_{\max}}}},{E_{p}}{\xi_{\min}}}\right],$
(33)
$\displaystyle{y_{\max}}=Min\left[{\frac{{{\omega^{2}}}}{{4{E_{p}}{\xi_{\min}}}},{E_{p}}{\xi_{\max}}}\right],$
(34)
$\displaystyle{\omega_{\min}}=Max\left[{{m_{X}},2{E_{p}}{\xi_{\min}}}\right],{\omega_{\max}}=\sqrt{s}.$
(35)

The VEV of the radion field $\Lambda_{\phi}$ is chosen as 5000 GeV, which
makes the radion stabilization model most natural [32]. The radion mass has
been selected $m_{\phi}=125$ GeV, the Higgs mass $m_{h}=125$ GeV (CMS),
$h_{3}^{\gamma}=0.1$, $h_{1}^{\gamma}=0.001$ [35]. We next provide estimates
for the production cross-sections at the LHC as follows
i) In Fig.3, we evaluate the dependence of the total cross-sections on the VEV
of the radion field $\Lambda_{\phi}$. The VEV of the radion field is chosen in
the range of 1000 GeV$\leq\Lambda_{\phi}\leq$ 10000 GeV. The figures show that
the total cross-sections increase when $\Lambda_{\phi}$ increases.
ii) In Fig.4, we evaluate the dependence of the total cross-sections on the
parameter $h_{1}^{\gamma}$ in the case of $m_{\phi}=125$ GeV,
$\Lambda_{\phi}=5000$ GeV. The total cross-sections achieve the maximum value
when $h_{1}^{\gamma}=0.001$. We also analysis the dependence of the total
cross-sections on the parameter $h_{3}^{\gamma}$, however, the total cross-
sections are unchanged when $h_{3}^{\gamma}$ increases.
iii) In Table 4, some typical values for the total cross-sections are
measured. The first line is the total cross-section through $\gamma h$ and the
second line is the total cross-section through $\gamma\phi$. The total cross-
section in Higgs production is larger than that in radion production. We
emphasize that the $\gamma b\overline{b}$ production cross-sections in the
case of the radion-dominated state 125 GeV.

|
---|---
(a) | (b)
Figure 3:  The total cross-section as a function of the $\Lambda_{\phi}$ in (a) $\gamma^{*}\gamma^{*}\to\gamma h\to\gamma b\overline{b}$, (b) $\gamma^{*}\gamma^{*}\to\gamma\phi\to\gamma b\overline{b}$ collisions at the LHC.  |
---|---
(a) | (b)
Figure 4:  The total cross-section as a function of the $h_{1}^{\gamma}$ in (a) $\gamma^{*}\gamma^{*}\to\gamma h\to\gamma b\overline{b}$, (b) $\gamma^{*}\gamma^{*}\to\gamma\phi\to\gamma b\overline{b}$ collisions at the LHC. Table 4: Some typical values for the total cross-sections concerning the radion mass $m_{\phi}$ at the LHC. The parameters are chosen as $\xi=1/6$, $\Lambda_{\phi}=5000$ GeV, $h_{1}^{\gamma}=0.001$, $h_{3}^{\gamma}=0.1$, $m_{h}=125$ GeV. $m_{\phi}$ (GeV) | 10 | 30 | 50 | 70 | 90 | 110 | 125 | 130 | 150 | 170 | 190
---|---|---|---|---|---|---|---|---|---|---|---
$\sigma_{\gamma^{*}\gamma^{*}\to\gamma h\to\gamma b\overline{b}}$ | 0.103 | 0.1027 | 0.1025 | 0.1022 | 0.101 | 0.096 | 11.255 | 3.291 | 0.099 | 0.101 | 0.102
($10^{-1}$ fb) |  |  |  |  |  |  |  |  |  |  |
$\sigma_{\gamma^{*}\gamma^{*}\to\gamma\phi\to\gamma b\overline{b}}$ | 0.002 | 0.0007 | 0.0004 | 0.00032 | 0.00037 | 0.0009 | 1.140 | 0.344 | 0.0002 | 0.008 | 0.0007
($10^{-1}$ fb) |  |  |  |  |  |  |  |  |  |  |

##  IV Conclusion

In this work, we have evaluated the total cross-section in $\gamma
b\overline{b}$ production through $\gamma h/\gamma\phi$ at the ILC and LHC in
the RS model. In general, the total cross-section through $\gamma h$ is larger
than that through $\gamma\phi$ at the ILC and LHC. Moreover, with the higher
energy of incoming beams at the LHC, the total cross-sections at the LHC are
larger than that at the ILC. It is worth mentioning that the total cross-
sections achieve the maximum value in the case of the radion-dominated state
125 GeV. Finally, we emphasize that the mixing of Higgs-radion can be used to
open up the opportunity for a wider search for new physics.

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# New Results on Permutation Binomials of Finite Fields

Xiang-dong Hou  Department of Mathematics and Statistics, University of South
Florida, Tampa, FL 33620<EMAIL_ADDRESS>and  Vincenzo Pallozzi Lavorante
Università degli Studi di Modena e Reggio Emilia, Italy
<EMAIL_ADDRESS>
###### Abstract.

After a brief review of existing results on permutation binomials of finite
fields, we introduce the notion of equivalence among permutation binomials
(PBs) and describe how to bring a PB to its canonical form under equivalence.
We then focus on PBs of $\mathbb{F}_{q^{2}}$ of the form
$X^{n}(X^{d(q-1)}+a)$, where $n$ and $d$ are positive integers and
$a\in\mathbb{F}_{q^{2}}^{*}$. Our contributions include two nonexistence
results: (1) If $q$ is even and sufficiently large and $a^{q+1}\neq 1$, then
$X^{n}(X^{3(q-1)}+a)$ is not a PB of $\mathbb{F}_{q^{2}}$. (2) If $2\leq d\mid
q+1$, $q$ is sufficiently large and $a^{q+1}\neq 1$, then
$X^{n}(X^{d(q-1)}+a)$ is not a PB of $\mathbb{F}_{q^{2}}$ under certain
additional conditions. (1) partially confirms a recent conjecture by Tu et al.
(2) is an extension of a previous result with $n=1$.

###### Key words and phrases:

finite field, Hasse–Weil bound, permutation binomial

###### 2010 Mathematics Subject Classification:

11T06, 11T55, 14H05

##  1\. Introduction

Let $\mathbb{F}_{q}$ be the finite field with $q$ elements and
$\mathbb{F}_{q}^{*}$ be its multiplicative group. A polynomial
$f\in\mathbb{F}_{q}[X]$ is called a permutation polynomial (PP) of
$\mathbb{F}_{q}$ if it induces a permutation of $\mathbb{F}_{q}$. A
permutation binomial (PB) of $\mathbb{F}_{q}$ is a PP of the form
$aX^{m}+bX^{n}$, where $a,b\in\mathbb{F}_{q}^{*}$, $m\not\equiv 0$,
$n\not\equiv 0$ and $m\not\equiv n\pmod{q-1}$. Permutation binomials are an
active topic that has attracted much attention. We refer the reader to [1] for
a survey on PBs and to [2] for a survey on PPs. Permutation binomials are
complex objects; in general, one can not expect a simple criterion on the
parameters $q,m,n,a,b$ for $aX^{m}+bX^{n}$ to be a PB of $\mathbb{F}_{q}$. In
this paper, we focus on PBs of $\mathbb{F}_{q^{e}}$ of the form

(1.1) $f_{q,e,n,d,a}(X)=X^{n}(X^{d(q-1)}+a)\in\mathbb{F}_{q^{e}}[X],$

where $n,d\in\mathbb{Z}^{+}$, $n\not\equiv 0$, $d(q-1)\not\equiv 0$,
$n+d(q-1)\not\equiv 0\pmod{q^{e}-1}$, and $a\in\mathbb{F}_{q^{e}}^{*}$. Here
is an overview of current knowledge on such PBs.

######  Result 1.1 ([14, Corollary 5.3]).

Assume $e=2$ and $a^{q+1}=1$. Then $f_{q,2,n,d,a}=X^{n}(X^{d(q-1)}+a)$ is a PB
of $\mathbb{F}_{q^{2}}$ if and only if $\text{\rm gcd}(n,q-1)=1$, $\text{\rm
gcd}(n-d,q+1)=1$ and $(-a)^{(q+1)/\text{\rm gcd}(q+1,d)}\neq 1$.

######  Result 1.2 ([3, Theorem A]).

Assume $e=2$, $n=1$, $d=2$ and $a^{q+1}\neq 1$. Then
$f_{q,2,1,2,a}=X(X^{2(q-1)}+a)$ is a PB of $\mathbb{F}_{q^{2}}$ if and only if
$q$ is odd and $(-a)^{(q+1)/2}=3$.

######  Result 1.3 ([5, Theorem 1.1]).

Assume $e=2$, $n=1$, $d>2$, $a^{q+1}\neq 1$, and $q$ is large relative to $d$.
Then $f_{q,2,1,d,a}=X(X^{d(q-1)}+a)$ is not a PB of $\mathbb{F}_{q^{2}}$.

######  Result 1.4 ([6, 7]).

Assume $e=2$, $n=3$, $d=2$ and $a^{q+1}\neq 1$. Then
$f_{q,2,3,2,a}=X^{3}(X^{2(q-1)}+a)$ is a PB of $\mathbb{F}_{q^{2}}$ if and
only if $q$ is odd, $q\equiv-1\pmod{3}$ and $(-a)^{(q+1)/2}=1/3$.

######  Result 1.5 ([11, Theorem 1]).

Assume $e=2$, $q=2^{2m}$ and $d=3$. Then $f_{q,2,n,3,a}=X^{n}(X^{3(q-1)}+a)$
is a PB of $\mathbb{F}_{q^{2}}$ if and only if $\text{\rm gcd}(n,q-1)=1$,
$n\equiv 3\pmod{q+1}$ and $a^{q+1}\neq 1$.

(Note: In the original statement of Result 1.5 in [11], it is assumed that
$m\geq 2$. However, the result also holds for $m=1$; see Example 2.4.)

######  Result 1.6 ([4, Theorem 4.2]).

Assume $e=2$ and $d=1$. Then $f_{q,2,n,1,a}=X^{n}(X^{q-1}+a)$ is a PB of
$\mathbb{F}_{q^{2}}$ if and only if $\text{\rm gcd}(n,q-1)=1$, $n\equiv
1\pmod{q+1}$ and $a^{q+1}\neq 1$.

######  Result 1.7 ([8]).

Assume $e\geq 2$, $d=1$ and $n<q^{e}-q$. For the special cases
$(q,e)=(q,2),(q,3),(q,4),(p,5),(p,6)$, where $p$ is a prime, the following
statement is true: If $f_{q,e,n,1,a}=X^{n}(X^{q-1}+a)$ is a PB of
$\mathbb{F}_{q^{e}}$, then $f_{q,e,n,1,a}\equiv
X^{nq^{h}}+aX^{n}\pmod{X^{q^{e}}-X}$ for some integer $h>0$. It is conjectured
that the statement is true for all $q$.

(Note: In Result 1.7, when $q=2$, $f_{2,e,n,1,a}=X^{n}(X+a)$ is never a PB of
$\mathbb{F}_{2^{e}}$, so the statement is vacuously true.)

Through these results, we begin to understand the roles played by the
parameters in the PBs of the form (1.1). At the same time, as more results on
PBs gather, one feels a need for a properly defined notion of equivalence of
PBs that allows us to categorize existing results and channel future efforts
to PBs that are new under equivalence. Section 2 is included for this purpose.
We define the equivalence among all PBs (not just those of the form (1.1)). We
show that every PB can be brought to a canonical form which is uniquely
determined by a triple of invariants. In particular, we see that the PB in
Result 1.4 is equivalent to a PB in Result 1.2 and the PB in Result 1.5 is
equivalent to a PB in Result 1.6.

Regarding Result 1.5, if we assume $e=2$, $q=2^{2m+1}$, $d=3$ and $a^{q+1}\neq
1$, [11] conjectured that $f_{q,2,n,3,a}=X^{n}(X^{3(q-1)}+a)$ is not a PB of
$\mathbb{F}_{q^{2}}$ and provided strong evidence for this conjecture. Note
that in this case, $d\mid q+1$. As we will see in Section 2, when the PB in
(1.1) is brought to its canonical form, we always have $d\mid(q^{e}-1)/(q-1)$.

Let us further focus on the case $e=2$, and we assume $d\mid q+1$ by the above
comment. In this case, if $a^{q+1}=1$ or $d=1$, all PBs are given by Results
1.1 and 1.6. Therefore, we assume $e=2$, $2\leq d\mid q+1$ and $a^{q+1}\neq
1$. Under these assumptions and up to equivalence, the PBs in Result 1.2 form
the only known class that contains infinitely many $q$’s. This leads to the
following question.

######  Question 1.8.

Fix integers $n\geq 1$ and $d\geq 2$. If there are infinitely many pairs
$(q,a)$ such that $d\mid q+1$, $a\in\mathbb{F}_{q^{2}}^{*}$, $a^{q+1}\neq 1$,
and $f(X)=f_{q,2,n,d,a}(X)=X^{n}(X^{d(q-1)}+a)$ is a PB of
$\mathbb{F}_{q^{2}}$, can we conclude that when $q$ is sufficiently large, $f$
is equivalent the PB in Result 1.2?

In this paper, we prove two nonexistence results that support an affirmative
answer to the above question.

######  Theorem 1.9.

Let $q=2^{m}$, $n\geq 1$ and $a\in\mathbb{F}_{q^{2}}^{*}$ be such that
$q\geq(2\max\\{n,6-n\\})^{4}$ and $a^{q+1}\neq 1$. Then
$f(X)=f_{q,2,n,3,a}(X)=X^{n}(X^{3(q-1)}+a)$ is not a PB of
$\mathbb{F}_{q^{2}}$.

Theorem 1.9 proves the conjecture of [11] when $q$ is large relative to $n$.

######  Theorem 1.10.

Let $n\geq 1$, $d\geq 2$ and $a\in\mathbb{F}_{q^{2}}^{*}$ be such that $d\mid
q+1$, $q\geq(2\max\\{n,2d-n\\})^{4}$ and $a^{q+1}\neq 1$. Then
$f(X)=f_{q,2,n,d,a}(X)=X^{n}(X^{d(q-1)}+a)$ is not a PB of
$\mathbb{F}_{q^{2}}$ if one of the following conditions is satisfied.

  * (i)

$d-n>1$ and $\text{\rm gcd}(d,n+1)$ is a power of $2$.

  * (ii)

$d+2\leq n<2d$ and $\text{\rm gcd}(d,n-1)$ is a power of $2$.

  * (iii)

$n\geq 2d$, $\text{\rm gcd}(d,n-1)$ is a power of $2$, and $\text{\rm
gcd}(n-d,q-1)=1$.

######  Remark 1.11.

In Theorem 1.10, one can replace the assumption that $d\mid q+1$ with
$\text{gcd}(n,d)=1$. If the $f$ in Theorem 1.10 is a PB of
$\mathbb{F}_{q^{2}}$, then $d\mid q+1$ implies $\text{gcd}(n,d)=1$. However,
as we will see in Section 4, the proof of Theorem 1.10 only uses
$\text{gcd}(n,d)=1$. Moreover, the assumption that $\text{gcd}(n,d)=1$ causes
no loss of generality. If $f_{q,2,n,d,a}$ is a PB of $\mathbb{F}_{q^{2}}$ with
$\text{gcd}(n,d)=\delta$, then $\text{gcd}(\delta,q^{2}-1)=1$. Let
$\delta^{\prime}\in\mathbb{Z}^{+}$ be such that $\delta\delta^{\prime}\equiv
1\pmod{q^{2}-1}$. Then

$f_{q,2,n,d,a}(X^{\delta^{\prime}})\equiv
f_{q,2,n/\delta,d/\delta,a}(X)\pmod{X^{q^{2}}-X},$

where $\text{gcd}(n/\delta,d/\delta)=1$.

Result 1.3 is a special case of Theorem 1.10 (i) with $n=1$. Although the
conditions in Theorem 1.10 are rather restrictive, they do cover many
parameters that were not investigated previously. For example, (i) is
satisfied for all $d>n+1$ with $\text{gcd}(d,n+1)=1$.

Theorems 1.9 and 1.10 are proved in Sections 3 and 4, respectively. The method
is similar to that in [5]. Here we recall the basic strategy.

Let

(1.2) $f(X)=f_{q,2,n,d,a}(X)=X^{n}(X^{d(q-1)}+a)\in\mathbb{F}_{q^{2}}[X],$

where $n\geq 1$, $2\leq d\mid q+1$ and $a\in\mathbb{F}_{q^{2}}^{*}$. The
following theorem follows from a well-known folklore [9, 12, 13].

######  Theorem 1.12.

The binomial $f(X)$ in (1.2) is PB of $\mathbb{F}_{q^{2}}$ if and only if

  * (i)

$\text{\rm gcd}(n,d(q-1))=1$ and

  * (ii)

$X^{n}(X^{d}+a)^{q-1}$ permutes
$\mu_{q+1}:=\\{x\in\mathbb{F}_{q^{2}}^{*}:x^{q+1}=1\\}$.

Assume that $f(X)$ in (1.2) is a PB of $\mathbb{F}_{q^{2}}$. Then for
$x\in\mu_{q+1}$,

$x^{n}(x^{d}+a)^{q-1}=\frac{x^{n}(x^{dq}+a^{q})}{x^{d}+a}=\frac{x^{n}(a^{q}x^{d}+1)}{x^{d}(x^{d}+a)}=G(x),$

where

(1.3) $G(X)=\frac{a^{q}X^{n}+X^{n-d}}{X^{d}+a}.$

Write

$G(X)=\frac{P(X)}{Q(X)},$

where

$\begin{cases}P(X)=a^{q}X^{n}+X^{n-d},\cr
Q(X)=X^{d}+a,\end{cases}\qquad\text{if}\ n\geq d,$
$\begin{cases}P(X)=a^{q}X^{d}+1,\cr
Q(X)=X^{2d-n}+aX^{d-n},\end{cases}\qquad\text{if}\ n<d.$

We assume that $a^{q+1}\neq 1$, which implies that $\text{gcd}(P,Q)=1$. Thus

$\deg G=\begin{cases}n&\text{if}\ n\geq d,\cr 2d-n&\text{if}\ n<d.\end{cases}$

Let

(1.4) $N(G)=\frac{P(X)Q(Y)-P(Y)Q(X)}{X-Y}\in\mathbb{F}_{q^{2}}[X,Y],$

which is the numerator of $(G(X)-G(Y))/(X-Y)$. We have

$\deg N(G)\leq\begin{cases}n+d-1&\text{if}\ n\geq d,\cr 3d-n-1&\text{if}\
n<d.\end{cases}$

######  Theorem 1.13.

Assume that $f(X)$ in (1.2) is a PB of $\mathbb{F}_{q^{2}}$, where
$q\geq(2\max\\{n,2d-n\\})^{4}$. Then $N(G)$ in (1.4) is reducible in
$\overline{\mathbb{F}}_{q}[X,Y]$, where $\overline{\mathbb{F}}_{q}$ is the
algebraic closure of $\mathbb{F}_{q}$.

###### Proof.

We only give a sketch of the proof; the omitted details are given in [5, §3].

There exist $l_{1},l_{2}\in\mathbb{F}_{q^{2}}(X)$ of degree one such that
$H:=l_{1}\circ G\circ l_{2}$ permutes $\mathbb{F}_{q}$. Since $\deg H=\deg
G<q$, by [5, Lemma 3.2], $H\in\mathbb{F}_{q}(X)$. Let
$A(X,Y)=N(H)\in\mathbb{F}_{q}[X,Y]$, the numerator of $(H(X)-H(Y))/(X-Y)$.
Assume to the contrary that $N(G)$ is irreducible over
$\overline{\mathbb{F}}_{q}$. Then by [5, Lemma 3.1], $A(X,Y)$ is also
irreducible over $\overline{\mathbb{F}}_{q}$. We have

$\delta:=\deg A(X,Y)\leq 2\deg H-2=2\deg G-2.$

By the Hasse-Weil bound, the number of zeros of $A(X,Y)$ in the projective
plane $\mathbb{P}^{2}(\mathbb{F}_{q})$ is at least

$q-(\delta-1)(\delta-2)q^{1/2}.$

Excluding the zeros at infinity of $\mathbb{P}^{2}(\mathbb{F}_{q})$ and on the
diagonal $\\{(x,x):x\in\mathbb{F}_{q}\\}$ of the affine plane
$\mathbb{F}_{q}^{2}$, we have

$|\\{(x,y)\in\mathbb{F}_{q}^{2}:x\neq y,\ A(x,y)=0\\}|\geq
q-(\delta-1)(\delta-2)q^{1/2}-2\delta.$

The right side is positive since $q\geq\delta^{4}$. Hence there exist
$(x,y)\in\mathbb{F}_{q}^{2}$ with $x\neq y$ such that $A(x,y)=0$. Then
$H(x)=H(y)$, which is a contradiction. ∎

##  2\. Canonical Forms of Permutation Binomials

For our purpose, a binomial over $\mathbb{F}_{q}$ is a polynomial of the

$f(X)=aX^{m}+bX^{n}\in\mathbb{F}_{q}[X],$

where $a,b\in\mathbb{F}_{q}^{*}$, $m,n>0$, $m\not\equiv 0$, $n\not\equiv 0$
and $m\not\equiv n\pmod{q-1}$. We treat $f(X)$ as a function from
$\mathbb{F}_{q}$ to $\mathbb{F}_{q}$, that is, we identify $f(X)$ with its
image in the quotient ring $\mathbb{F}_{q}[X]/\langle X^{q}-X\rangle$. Let
$\mathcal{B}_{q}$ denote the set of all such binomials. Two members
$f,g\in\mathcal{B}_{q}$ are considered equivalent, denoted as $f\sim g$, if
one can be obtained from the other through a combination of the following
transformations of $\mathcal{B}_{q}$:

(2.1) $\displaystyle\alpha_{u}:\mathcal{B}_{q}\to\mathcal{B}_{q},\ f(X)\mapsto
uf(X),\quad u\in\mathbb{F}_{q}^{*},$ (2.2)
$\displaystyle\beta:\mathcal{B}_{q}\to\mathcal{B}_{q},\ f(X)\mapsto
f(X)^{p},\quad p=\text{char}\,\mathbb{F}_{q},$ (2.3)
$\displaystyle\gamma_{v,s}:\mathcal{B}_{q}\to\mathcal{B}_{q},\ f(X)\mapsto
f(vX^{s}),\quad v\in\mathbb{F}_{q}^{*},\ s\in\mathbb{Z}^{+},\
\text{gcd}(s,q-1)=1.$

If $f,g\in\mathcal{B}_{q}$ are equivalent, then $f$ permutes $\mathbb{F}_{q}$
if and only if $g$ does. It is clear that $\gamma_{v,s}$ commutes with
$\alpha_{u}$ and $\beta$, and $\beta\circ\alpha_{u}=\alpha_{u^{p}}\circ\beta$.
Therefore, for $f,g\in\mathcal{B}_{q}$, $f\sim g$ if and only if

(2.4) $g(X)=uf(vX^{s})^{p^{i}}$

for some $u,v\in\mathbb{F}_{q}^{*}$, $i\geq 0$ and $s>0$ with
$\text{gcd}(s,q-1)=1$.

For $d\mid q-1$, define

(2.5) $N_{d}=\\{1\leq n\leq q-1:n=n^{*}\\},$

where

$\displaystyle n^{*}=\min\bigl{\\{}1\leq n^{\prime}\leq q-1:\ $
$n^{\prime}\equiv tn\\!\\!\pmod{q-1}$ for some $t\in\mathbb{Z}_{q-1}^{\times}$
with $t\equiv 1\\!\\!\pmod{(q-1)/d}$ or $n^{\prime}\equiv
tn-d\\!\\!\pmod{q-1}$ for some $t\in\mathbb{Z}_{q-1}^{\times}$
$\displaystyle\text{with $t\equiv-1\\!\\!\pmod{(q-1)/d}$}\bigr{\\}}.$

(Here $\mathbb{Z}_{q-1}^{\times}$ denotes the multiplicative group of
$\mathbb{Z}_{q-1}$.) Let
$\theta:\mathbb{Z}_{q-1}^{\times}\to\mathbb{Z}_{(q-1)/d}^{\times}$ be the
natural homomorphism (which is onto). Then $G:=\theta^{-1}(\\{\pm 1\\})$ acts
on $\mathbb{Z}_{q-1}$ as follows: For $t\in G$ and $n\in\mathbb{Z}_{q-1}$,

$t(n)=\begin{cases}tn&\text{if}\ \theta(t)=1,\cr tn-d&\text{if}\
\theta(t)=-1.\end{cases}$

Write $\mathbb{Z}_{q-1}=\\{1,2,\dots,q-1\\}$. Then for $n\in\mathbb{Z}_{q-1}$,
$n^{*}$ is the least element in the $G$-orbit of $n$. Therefore $N_{d}$ is the
set of least elements of the $G$-orbits in $\mathbb{Z}_{q-1}$.

######  Example 2.1.

Let $q=2^{4}$ and $d=3$. We have
$\theta:\mathbb{Z}_{15}^{\times}\to\mathbb{Z}_{5}^{\times}$,
$\theta^{-1}(1)=\\{1,11\\}$, $\theta^{-1}(-1)=\\{-1,4\\}$ and
$G=\\{1,11,-1,4\\}$. The $G$-orbits of $\mathbb{Z}_{15}$ are $\\{1,11\\}$,
$\\{2,7,10,5\\}$, $\\{3,9\\}$, $\\{4,14,8,13\\}$, $\\{6\\}$, $\\{15\\}$. Hence
$N_{d}=\\{1,2,3,4,6,15\\}$.

For $d\mid q-1$ and $n\in N_{d}$, let

(2.6) $\displaystyle G_{d,n}=\,$ the subgroup of $\mathbb{Z}_{d}^{\times}$
generated by $\displaystyle\begin{cases}\\{p,-1\\}&\text{if
$d\equiv-2n\pmod{(q-1)/d}$ and
$\text{gcd}(n,q-1)=1$},\cr\\{p\\}&\text{otherwise},\end{cases}$

where $p=\text{char}\,\mathbb{F}_{q}$. Let $G_{d,n}$ act on
$\mathbb{F}_{q}^{*}/(\mathbb{F}_{q}^{*})^{d}$, where
$(\mathbb{F}_{q}^{*})^{d}=\\{x^{d}:x\in\mathbb{F}_{q}^{*}\\}$, as follows:

$\begin{array}[]{ccl}G_{d,n}\times\;\mathbb{F}_{q}^{*}/(\mathbb{F}_{q}^{*})^{d}&\longrightarrow&\mathbb{F}_{q}^{*}/(\mathbb{F}_{q}^{*})^{d}\vspace{0.5em}\\\
(s,a(\mathbb{F}_{q}^{*})^{d})&\longmapsto&a^{s}(\mathbb{F}_{q}^{*})^{d},\quad
a\in\mathbb{F}_{q}^{*}.\end{array}$

Let $A_{d,n}\subset\mathbb{F}_{q}^{*}$ be such that
$\\{a(\mathbb{F}_{q}^{*})^{d}:a\in A_{d,n}\\}$ is a system of representatives
of the $G_{d,n}$-orbits in $\mathbb{F}_{q}^{*}/(\mathbb{F}_{q}^{*})^{d}$.
Equivalently, let $G_{d,n}$ act on $\mathbb{Z}_{d}$ through multiplication and
let $\xi$ be a primitive element of $\mathbb{F}_{q}$. Then
$A_{d,n}=\\{\xi^{e}:e\in E_{d,n}\\}$, where $E_{d,n}$ is a system of
representatives of the $G_{d,n}$-orbits in $\mathbb{Z}_{d}$.

We now are ready to state and prove the main result of this section.

######  Theorem 2.2.

Assume that $f\in\mathcal{B}_{q}$ permutes $\mathbb{F}_{q}$. Then there is a
unique triple $(d,n,a)$, where $d\mid q-1$, $n\in N_{d}$ and $a\in A_{d,n}$,
such that

(2.7) $f(X)\sim X^{n}(X^{d}+a).$

We call the right side of (2.7) the canonical form of $f$.

###### Proof.

Existence of $(d,n,a)$

Write $f(X)=a_{0}X^{m_{0}}+b_{0}X^{n_{0}}$, where
$a_{0},b_{0}\in\mathbb{F}_{q}^{*}$ and $m_{0}>n_{0}$. Let
$d=\text{gcd}(m_{0}-n_{0},q-1)$. Let $r\in\mathbb{Z}^{+}$ be such that

$r\frac{m_{0}-n_{0}}{d}\equiv 1\pmod{\frac{q-1}{d}}.$

Since $\text{gcd}(r,(q-1)/d)=1$, there exists integer $k\geq 0$ such that
$s:=r+k(q-1)/d$ is relatively prime to $q-1$. (To see this, use Dirichlet’s
theorem on primes in arithmetic progression or the following simple argument:
Let $p_{1},\dots,p_{l}$ be the prime divisors of $q-1$ that do not divide $r$
and let $k=p_{1}\cdots p_{l}$.) Then

$f(X)\sim
f(X^{s})=X^{sn_{0}}(a_{0}X^{s(m_{0}-n_{0})}+b_{0})=X^{n_{1}}(a_{0}X^{d}+b_{0}),$

where $n_{1}=sn_{0}$. We now assume $f(X)=X^{n_{1}}(a_{0}X^{d}+b_{0})$.

Let $n=n_{1}^{*}\in N_{d}$. We claim that

(2.8) $f(X)\sim X^{n}(a_{1}X^{d}+b_{1})$

for some $a_{1},b_{1}\in\mathbb{F}_{q}^{*}$. To prove this claim, we consider
two cases.

Case 1. Assume that $n\equiv tn_{1}\pmod{q-1}$ for some
$t\in\mathbb{Z}_{q-1}^{\times}$ with $t\equiv 1\pmod{(q-1)/d}$. Then

$f(X)\sim f(X^{t})=X^{tn_{1}}(a_{0}X^{td}+b_{0})=X^{n}(a_{0}X^{d}+b_{0}).$

Case 2. Assume that $n\equiv tn_{1}-d\pmod{q-1}$ for some
$t\in\mathbb{Z}_{q-1}^{\times}$ with $t\equiv-1\pmod{(q-1)/d}$. Then

$\displaystyle f(X)\sim f(X^{t})\,$
$\displaystyle=X^{tn_{1}}(a_{0}X^{td}+b_{0})=a_{0}X^{tn_{1}+td}+b_{0}X^{tn_{1}}$
$\displaystyle=a_{0}X^{n}+b_{0}X^{n+d}=X^{n}(b_{0}X^{d}+a_{0}).$

Hence (2.8) is proved.

By (2.8), we may assume

$f(X)=X^{n}(X^{d}+c),$

where $c\in\mathbb{F}_{q}^{*}$. To prove that $f(X)\sim X^{n}(X^{d}+a)$ for
some $a\in A_{d,n}$, again, we consider two cases.

Case 1. Assume that $d\not\equiv-2n\pmod{(q-1)/d}$ or $\text{gcd}(n,q-1)\neq
1$. By (2.6), $G_{d,n}=\langle p\rangle<\mathbb{Z}_{d}^{\times}$. Then by the
definition of $A_{d,n}$, there exist $i\in\mathbb{N}$, $a\in A_{d,n}$ and
$b\in\mathbb{F}_{q}^{*}$ such that $c^{p^{i}}=ab^{d}$. Write
$b=b_{1}^{p^{i}}$, where $b_{1}\in\mathbb{F}_{q}^{*}$. Let
$s\in\mathbb{Z}^{+}$ be such that $sp^{i}\equiv 1\pmod{q-1}$. Then

$\displaystyle f(X)\,$ $\displaystyle\sim
f(b_{1}X^{s})^{p^{i}}=(b_{1}X^{s})^{np^{i}}((b_{1}X^{s})^{dp^{i}}+c^{p^{i}})$
$\displaystyle\sim
X^{n}(b_{1}^{dp^{i}}X^{d}+c^{p^{i}})=X^{n}(b^{d}X^{d}+c^{p^{i}})$
$\displaystyle\sim X^{n}(X^{d}+c^{p^{i}}b^{-d})=X^{n}(X^{d}+a).$

Case 2. Assume that $d\equiv-2n\pmod{(q-1)/d}$ and $\text{gcd}(n,q-1)=1$. Then
$G_{d,n}=\langle p,-1\rangle<\mathbb{Z}_{d}^{\times}$. So there exist
$i\in\mathbb{N}$, $a\in A_{d,n}$ and $b\in\mathbb{F}_{q}^{*}$ such that either
$c^{p^{i}}=ab^{d}$ or $c^{-p^{i}}=ab^{d}$. In the former case, the proof is
identical to Case 1. In the latter case, write $b=b_{1}^{p^{i}}$, where
$b_{1}\in\mathbb{F}_{q}^{*}$. Let $k\in\mathbb{Z}^{+}$ be such that $kn\equiv
1\pmod{q-1}$, and let $s=1+kd$. Then

$\displaystyle sn=n+nkd$ $\displaystyle\,\equiv n+d\pmod{q-1}$
$\displaystyle\equiv-n\pmod{(q-1)/d}.$

Hence $s\equiv-1\pmod{(q-1)/d}$. It follows that $\text{gcd}(s,(q-1)/d)=1$. We
also have $\text{gcd}(s,d)=\text{gcd}(1+kd,d)=1$. Therefore
$\text{gcd}(s,q-1)=1$. We have

$f(X)\sim f(X^{s})=X^{sn}(X^{sd}+c)=X^{sn+sd}+cX^{sn}.$

In the above,

$sn=n+nkd\equiv n+d\pmod{q-1}$

and

$sd=(1+kd)d\equiv(1+k(-2n))d\equiv-d\pmod{q-1}.$

Hence

$f(X)\sim X^{n}+cX^{n+d}\sim X^{n}(X^{d}+c^{-1}),$

where $(c^{-1})^{p^{i}}=ab^{d}$. It follows from Case 1 that

$X^{n}(X^{d}+c^{-1})\sim X^{n}(X^{d}+a).$

Uniqueness of $(d,n,a)$

Assume that

(2.9) $f(X)=X^{n}(X^{d}+a)\sim X^{n_{1}}(X^{d_{1}}+a_{1}),$

where $d\mid q-1$, $n\in N_{d}$, $a\in A_{d,n}$, $d_{1}\mid q-1$, $n_{1}\in
N_{d_{1}}$, $a_{1}\in A_{d_{1},n_{1}}$.

In general, for $bX^{m}+cX^{l}\in\mathcal{B}_{q}$, $\text{gcd}(m-l,q-1)$ is
invariant under equivalence. Therefore, in (2.9), we have $d=d_{1}$.

By (2.9),

(2.10) $X^{n_{1}}(X^{d}+a_{1})=uf(vX^{s})^{p^{i}}$

for some $u,v\in\mathbb{F}_{q}^{*}$, $i\geq 0$ and $s>0$ with
$\text{gcd}(s,q-1)=1$. Expanding (2.10) gives

$X^{n_{1}+d}+a_{1}X^{n_{1}}=\alpha X^{t(n+d)}+\beta X^{tn},$

where $t=sp^{i}$ and $\alpha,\beta\in\mathbb{F}_{q}^{*}$. It follows that

(2.11) $\begin{cases}n_{1}+d\equiv t(n+d)\pmod{q-1},\cr n_{1}\equiv
tn\pmod{q-1},\end{cases}$

or

(2.12) $\begin{cases}n_{1}+d\equiv tn\pmod{q-1},\cr n_{1}\equiv
t(n+d)\pmod{q-1}.\end{cases}$

Note that (2.11) is equivalent to

(2.13) $\begin{cases}t\equiv 1\pmod{(q-1)/d},\cr n_{1}\equiv
tn\pmod{q-1},\end{cases}$

and (2.12) is equivalent to

(2.14) $\begin{cases}t\equiv-1\pmod{(q-1)/d},\cr n_{1}\equiv
tn-d\pmod{q-1}.\end{cases}$

Since $n\in N_{d}$, it follows from (2.13), (2.14) and the definition of
$N_{d}$ ((2.5)) that $n\leq n_{1}$. By symmetry, $n_{1}\leq n$, whence
$n=n_{1}$.

Now (2.10) becomes

$\displaystyle X^{n+d}+a_{1}X^{n}\,$
$\displaystyle=u\bigl{[}(vX^{s})^{n+d}+a(vX^{s})^{n}\bigr{]}^{p^{i}}$
$\displaystyle=uv^{p^{i}(n+d)}X^{sp^{i}(n+d)}+ua^{p^{i}}v^{p^{i}n}X^{sp^{i}n}.$

Let $t=sp^{i}$. Then there are two possibilities.

Case 1. (2.13) holds with $n=n_{1}$ and

(2.15) $(uv^{p^{i}(n+d)},\,ua^{p^{i}}v^{p^{i}n})=(1,a_{1}).$

Case 2. (2.14) holds with $n=n_{1}$ and

(2.16) $(ua^{p^{i}}v^{p^{i}n},\,uv^{p^{i}(n+d)})=(1,a_{1}).$

It suffices to show that in both cases, $a$ and $a_{1}$ are in the same
$G_{d,n}$-orbit. (Then $a=a_{1}$.)

First, assume Case 1. We have

$a_{1}=\frac{ua^{p^{i}}v^{p^{i}n}}{uv^{p^{i}(n+d)}}=a^{p^{i}}v^{-p^{i}d},$

which is in the $G_{d,n}$-orbit of $a$.

Next, assume Case 2. (2.14) with $n=n_{1}$ gives

$\begin{cases}t\equiv-1\pmod{(q-1)/d},\cr n\equiv tn-d\pmod{q-1}.\end{cases}$

It follows that $n\equiv tn-d\equiv-n-d\pmod{(q-1)/d}$, i.e.,
$d\equiv-2n\pmod{(q-1)/d}$. Since $f(X)$ permutes $\mathbb{F}_{q}$, we have
$\text{gcd}(n,d)=1$. From $n\equiv tn-d\pmod{q-1}$, we have $(t-1)n-d\equiv
0\pmod{q-1}$, whence $d\mid t-1$ and

$\frac{t-1}{d}n-1\equiv 0\pmod{\frac{q-1}{d}}.$

In particular, $\text{gcd}(n,(q-1)/d)=1$. Combining this with
$\text{gcd}(n,d)=1$, we have $\text{gcd}(n,q-1)=1$. Therefore $G_{d,n}=\langle
p,-1\rangle$. Now by (2.16),

$a_{1}=\frac{uv^{p^{i}(n+d)}}{ua^{p^{i}}v^{p^{i}n}}=a^{-p^{i}}v^{p^{i}d},$

which is in the $G_{d,n}$-orbit of $a$. ∎

######  Example 2.3.

Assume that $n,d\in\mathbb{Z}^{+}$ are such that $d\mid q+1$, $n<2d$,
$\text{gcd}(n,q^{2}-1)=1$ and $\text{gcd}(2d-n,q-1)=1$, and let
$a\in\mathbb{F}_{q^{2}}^{*}$. Since
$\text{gcd}(dq-n+d,q-1)=\text{gcd}(2d-n,q-1)=1$ and
$\text{gcd}(dq-n+d,q+1)=\text{gcd}(n,q+1)=1$, we have
$\text{gcd}(dq-n+d,q^{2}-1)=1$. Then in $\mathcal{B}_{q}$,

$\displaystyle X^{n}(X^{d(q-1)}+a)\,$ $\displaystyle=X^{dq+n-d}+aX^{n}$
$\displaystyle\sim X^{(dq-n+d)(dq+n-d)}+aX^{(dq-n+d)n}\kern
20.00003pt(X\mapsto X^{dq-n+d})$
$\displaystyle=X^{d^{2}q^{2}-(n-d)^{2}}+aX^{(dq-n+d)n}$
$\displaystyle=X^{d^{2}-(n-d)^{2}}+aX^{(dq-n+d)n}$
$\displaystyle=X^{n(2d-n)}+aX^{(dq-n+d)n}$ $\displaystyle\sim
X^{2d-n}+aX^{dq-n+d}\kern 83.00015pt(X^{n}\mapsto X)$
$\displaystyle=X^{2d-n}(1+aX^{d(q-1)})$ $\displaystyle\sim
X^{2d-n}(X^{d(q-1)}+a^{-1}).$

In particular, when $n=1$, $d=2$, $q$ is odd and $q\not\equiv 1\pmod{3}$, we
have

$X(X^{2(q-1)}+a)\sim X^{3}(X^{2(q-1)}+a^{-1}).$

This shows that the PB in Result 1.4 is equivalent to a PB in Result 1.2.

######  Example 2.4.

We show that the PB in Result 1.5 is equivalent to a PB in Result 1.6. Let
$e=2$, $q=2^{2m}$, $n\in\mathbb{Z}^{+}$, $d=3$, $a\in\mathbb{F}_{q^{2}}^{*}$,
and consider $f=f_{q,2,n,3,a}=X^{n}(X^{3(q-1)}+a)$.

Let $s=(q+2)/3+k(q+1)$, where

$k=\begin{cases}0&\text{if}\ m\equiv 0,1\pmod{3},\cr 1&\text{if}\
m\equiv-1\pmod{3}.\end{cases}$

We claim that $\text{gcd}(s,q^{2}-1)=1$. Clearly, $\text{gcd}(s,q+1)=1$. We
have

$\displaystyle\text{gcd}(s,q-1)\,$
$\displaystyle=\text{gcd}\Bigl{(}\frac{q+2}{3}+2k,\,q-1\Bigr{)}$
$\displaystyle=\frac{1}{3}\text{gcd}(q+2+6k,\,3q-3)$
$\displaystyle=\frac{1}{3}\text{gcd}(q+2+6k,\,3(-2-6k)-3)$
$\displaystyle=\frac{1}{3}\text{gcd}(q+2+6k,\,9(2k+1)).$

In the above, $9(2k+1)=3^{2}$ or $3^{3}$, and

$\displaystyle q+2+6k\,$ $\displaystyle=(3-1)^{2m}+2+6k$ $\displaystyle\equiv
1-2m\cdot 3+2+6k\pmod{3^{2}}$ $\displaystyle=3+6(k-m)$
$\displaystyle\not\equiv 0\pmod{3^{2}}.$

So $\text{gcd}(s,q-1)=1$ and the claim is proved.

Now we have

$f(X)\sim f(X^{s})=X^{sn}(X^{s\cdot 3(q-1)}+a)=X^{sn}(X^{q-1}+a).$

By Result 1.6, $X^{sn}(X^{q-1}+a)$ permutes $\mathbb{F}_{q^{2}}$ if and only
if

$\text{gcd}(sn,q-1)=1,\quad sn\equiv 1\pmod{q+1},\quad\text{and}\ a^{q+1}\neq
1,$

i.e.,

$\text{gcd}(n,q-1)=1,\quad n\equiv 3\pmod{q+1},\quad\text{and}\ a^{q+1}\neq
1,$

which are precisely the conditions in Result 1.5.

##  3\. Proof of Theorem 1.9

Theorem 1.9. Let $q=2^{m}$, $n\geq 1$ and $a\in\mathbb{F}_{q^{2}}^{*}$ be such
that $q\geq(2\max\\{n,6-n\\})^{4}$ and $a^{q+1}\neq 1$. Then
$f(X)=f_{q,2,n,3,a}(X)=X^{n}(X^{3(q-1)}+a)$ is not a PB of
$\mathbb{F}_{q^{2}}$.

Assume to the contrary that $f$ is a PB of $\mathbb{F}_{q^{2}}$. If $m$ is
even, by Result 1.5, $n\geq q+4$, which is a contradiction. So $m$ is odd, and
$3\mid q+1$. By (1.3),

(3.1) $G(X)=\frac{a^{q}X^{n}+X^{n-3}}{X^{3}+a}.$

Let

(3.2) $N(X,Y)=\text{the numerator of}\ \frac{G(X)+G(Y)}{X+Y}.$

By Theorem 1.13, $N(X,Y)$ is reducible over $\overline{\mathbb{F}}_{q}$.
However, we will show that $N(X,Y)$ is irreducible over
$\overline{\mathbb{F}}_{q}$, hence creating a contradiction. We consider two
cases, $n\geq 3$ and $n\leq 2$, separately.

###  3.1. Case 1. $n\geq 3$

Since $\text{gcd}(n,3(q-1))=1$ (Theorem 1.12), we have $n>3$. We have

$\displaystyle N(X,Y)=\,$
$\displaystyle\frac{1}{X+Y}\Bigl{[}(a^{q}X^{n}+X^{n-3})(Y^{3}+a)+(a^{q}Y^{n}+Y^{n-3})(X^{3}+a)\Bigr{]}$
$\displaystyle=\,$ $\displaystyle
a\frac{X^{n-3}+Y^{n-3}}{X+Y}+\Bigl{[}a^{q+1}\frac{X^{n}+Y^{n}}{X+Y}+X^{3}Y^{3}\frac{X^{n-6}+Y^{n-6}}{X+Y}\Bigr{]}$
$\displaystyle+a^{q}X^{3}Y^{3}\frac{X^{n-3}+Y^{n-3}}{X+Y}.$

The homogenization of $N(X,Y)$ is

$\displaystyle N^{*}(X,Y,Z)=\,$ $\displaystyle
a\frac{X^{n-3}+Y^{n-3}}{X+Y}Z^{6}+\Bigl{[}a^{q+1}\frac{X^{n}+Y^{n}}{X+Y}+X^{3}Y^{3}\frac{X^{n-6}+Y^{n-6}}{X+Y}\Bigr{]}Z^{3}$
$\displaystyle+a^{q}X^{3}Y^{3}\frac{X^{n-3}+Y^{n-3}}{X+Y}$ $\displaystyle=\,$
$\displaystyle Q(Z^{3}),$

where

$\displaystyle Q(Z)=\,$ $\displaystyle
a\frac{X^{n-3}+Y^{n-3}}{X+Y}Z^{2}+\Bigl{[}a^{q+1}\frac{X^{n}+Y^{n}}{X+Y}+X^{3}Y^{3}\frac{X^{n-6}+Y^{n-6}}{X+Y}\Bigr{]}Z$
$\displaystyle+a^{q}X^{3}Y^{3}\frac{X^{n-3}+Y^{n-3}}{X+Y}.$

It suffices to show that $N^{*}(X,Y,Z)$ is irreducible over
$\overline{\mathbb{F}}_{q}$. We first show that $N^{*}(X,Y,Z)$, as a
polynomial in $Z$ over $\overline{\mathbb{F}}_{q}[X,Y]$, is primitive, i.e.,
the gcd of its coefficients is 1; that is,

(3.3)
$\text{gcd}\Bigl{(}\frac{X^{n-3}+Y^{n-3}}{X+Y},\,a^{q+1}\frac{X^{n}+Y^{n}}{X+Y}+X^{3}Y^{3}\frac{X^{n-6}+Y^{n-6}}{X+Y}\Bigr{)}=1.$

Since the polynomials in (3.3) are homogeneous, it suffices to prove (3.3)
with $Y=1$, i.e.,

(3.4)
$\text{gcd}\Bigl{(}\frac{X^{n-3}+1}{X+1},\,a^{q+1}\frac{X^{n}+1}{X+1}+X^{3}\frac{X^{n-6}+1}{X+1}\Bigr{)}=1.$

Let $\zeta\in\overline{\mathbb{F}}_{q}$ be a root of $(X^{n-3}+1)/(X+1)$. If
$\zeta\neq 1$, then $\zeta^{n-3}+1=0$. Thus

$\displaystyle\Bigl{(}a^{q+1}\frac{X^{n}+1}{X+1}+X^{3}\frac{X^{n-6}+1}{X+1}\Bigr{)}\Big{|}_{X=1}$
$\displaystyle=\,$
$\displaystyle\frac{1}{\zeta+1}\bigl{(}a^{q+1}(\zeta^{n}+1)+\zeta^{3}(\zeta^{n-6}+1)\bigr{)}$
$\displaystyle=\,$
$\displaystyle\frac{1}{\zeta+1}\bigl{(}a^{q+1}(\zeta^{3}+1)+1+\zeta^{3}\bigr{)}$
$\displaystyle=\,$ $\displaystyle\frac{1}{\zeta+1}(a^{q+1}+1)(\zeta^{3}+1)\neq
0.$

(Note: $\zeta^{3}\neq 1$ since $\zeta^{n-3}=1$ and $\text{gcd}(n,3(q-1))=1$.)
If $\zeta=1$, then $n$ must be odd, in which case,

$\Bigl{(}a^{q+1}\frac{X^{n}+1}{X+1}+X^{3}\frac{X^{n-6}+1}{X+1}\Bigr{)}\Big{|}_{X=1}=a^{q+1}n+n-6=n(a^{q+1}+1)\neq
0.$

This proves (3.4) and hence (3.3).

With (3.3), to prove that $N^{*}(X,Y,Z)$ is irreducible in
$\overline{\mathbb{F}}_{q}[X,Y,Z]$, it suffices to show that it is irreducible
in $\overline{\mathbb{F}}_{q}(X,Y)[Z]$. Let $w$ be a root of
$N^{*}(X,Y,Z)\in\overline{\mathbb{F}}_{q}(X,Y)[Z]$ and let $z=w^{3}$. Then $z$
is a root of $Q(Z)$. It suffices to show that
$[\overline{\mathbb{F}}_{q}(X,Y,z):\overline{\mathbb{F}}_{q}(X,Y)]=2$ and
$[\overline{\mathbb{F}}_{q}(X,Y,w):\overline{\mathbb{F}}_{q}(X,Y,z)]=3$.

..................................................................................................................$\overline{\mathbb{F}}_{q}(X,Y)$$\overline{\mathbb{F}}_{q}(X,Y,z)$$\overline{\mathbb{F}}_{q}(X,Y,w)$$\scriptstyle
2$$\scriptstyle 3$

####  3.1.1. Proof that
$[\overline{\mathbb{F}}_{q}(X,Y,z):\overline{\mathbb{F}}_{q}(X,Y)]=2$

Assume to the contrary that $Q(Z)$ is reducible over
$\overline{\mathbb{F}}_{q}(X,Y)$. Then there exists
$A/B\in\overline{\mathbb{F}}_{q}(X,Y)$
($A,B\in\overline{\mathbb{F}}_{q}[X,Y]$, $\text{gcd}(A,B)=1$) such that

(3.5) $\frac{\displaystyle
a^{q+1}X^{3}Y^{3}\Bigl{(}\frac{X^{n-3}+Y^{n-3}}{X+Y}\Bigr{)}^{2}}{\displaystyle\Bigl{(}a^{q+1}\frac{X^{n}+Y^{n}}{X+Y}+X^{3}Y^{3}\frac{X^{n-6}+Y^{n-6}}{X+Y}\Bigr{)}^{2}}=\Bigl{(}\frac{A}{B}\Bigr{)}^{2}+\frac{A}{B}=\frac{A(A+B)}{B^{2}}.$

In the above equation, the numerator and the denominator on the left side are
relatively prime (by (3.3)), so

(3.6) $B=a^{q+1}\frac{X^{n}+Y^{n}}{X+Y}+X^{3}Y^{3}\frac{X^{n-6}+Y^{n-6}}{X+Y}$

and

$A(A+B)=a^{q+1}X^{3}Y^{3}\Bigl{(}\frac{X^{n-3}+Y^{n-3}}{X+Y}\Bigr{)}^{2}.$

Since $\text{gcd}(A,A+B)=1$, we may assume that

(3.7) $\begin{cases}A=X^{3}U^{2},\cr A+B=Y^{3}V^{2},\end{cases}$

for some $U,V\in\overline{\mathbb{F}}_{q}[X,Y]$ with
$UV=(X^{n-3}+Y^{n-3})/(X+Y)$. Therefore,

(3.8) $B=X^{3}U^{2}+Y^{3}V^{2}.$

By (3.8), the coefficient of $XY^{n-2}$ in $B$ is $0$. However, by (3.6), the
coefficient of $XY^{n-2}$ in $B$ is either $a^{q+1}$ or $a^{q+1}+1$. We have a
contradiction.

####  3.1.2. Proof that
$[\overline{\mathbb{F}}_{q}(X,Y,w):\overline{\mathbb{F}}_{q}(X,Y,z)]=3$

Assume the contrary. Then $z$ is a third power in
$\overline{\mathbb{F}}_{q}(X,Y,z)$, that is, there exists
$A,B\in\overline{\mathbb{F}}_{q}(X,Y)$ such that

$z=(A+Bz)^{3},$

i.e.,

(3.9) $(A+BZ)^{3}-Z\equiv 0\pmod{Q(Z)}.$

Setting $Y=1$ in (3.9) gives

(3.10) $(A_{1}+B_{1}Z)^{3}-Z\equiv 0\pmod{Q_{1}(Z)},$

where $A_{1}(X)=A(X,1)$, $B_{1}(X)=B(X,1)$ and

(3.11) $\displaystyle Q_{1}(Z)=Q(Z)|_{Y=1}=$ $\displaystyle
a\frac{X^{n-3}+1}{X+1}Z^{2}+\Bigl{[}a^{q+1}\frac{X^{n}+1}{X+1}+X^{3}\frac{X^{n-6}+1}{X+1}\Bigr{]}Z+a^{q}X^{3}\frac{X^{n-3}+1}{X+1}.$

We find that

$(A_{1}+B_{1}Z)^{3}-Z\equiv\frac{f_{0}(X)}{a^{2}(X^{3}+X^{n})}+\frac{f_{1}(X)}{a^{2}(X^{3}+X^{n})^{2}}Z\pmod{Q_{1}(Z)},$

where

(3.12) $\displaystyle f_{0}(X)=\,$ $\displaystyle
a^{2}A_{1}^{3}X^{3}+a^{1+q}A_{1}B_{1}^{2}X^{6}+a^{1+2q}B_{1}^{3}X^{6}+a^{q}B_{1}^{3}X^{9}+a^{2}A_{1}^{3}X^{n}$
(3.13)
$\displaystyle+a^{1+q}A_{1}B_{1}^{2}X^{3+n}+a^{q}B_{1}^{3}X^{3+n}+a^{1+2q}B_{1}^{3}X^{6+n},$
(3.14) $\displaystyle f_{1}(X)=\,$ $\displaystyle
a^{2}X^{6}+a^{2}A_{1}^{2}B_{1}X^{6}+a^{2+q}A_{1}B_{1}^{2}X^{6}+a^{2+2q}B_{1}^{3}X^{6}+aA_{1}B_{1}^{2}X^{9}$
(3.15)
$\displaystyle+a^{1+q}B_{1}^{3}X^{9}+B_{1}^{3}X^{12}+a^{2}X^{2n}+a^{2}A_{1}^{2}B_{1}X^{2n}+aA_{1}B_{1}^{2}X^{2n}+B_{1}^{3}X^{2n}$
(3.16)
$\displaystyle+aA_{1}B_{1}^{2}X^{3+n}+a^{2+q}A_{1}B_{1}^{2}X^{3+n}+aA_{1}B_{1}^{2}X^{6+n}+a^{2+q}A_{1}B_{1}^{2}X^{6+n}$
$\displaystyle+a^{2+q}A_{1}B_{1}^{2}X^{3+2n}+a^{1+q}B_{1}^{3}X^{3+2n}+a^{2+2q}B_{1}^{3}X^{6+2n}.$

Therefore, $f_{0}(X)=f_{1}(X)=0$. (We will not need the fact that
$f_{1}(X)=0$.) From (3.10), $B_{1}\neq 0$. Then $f_{0}(X)=0$ implies
$A_{1}\neq 0$. Let $C=B_{1}/A_{1}$. Then $f_{0}(X)=0$ becomes

(3.17)
$(a^{2}+a^{1+q}X^{3}C^{2})(1+X^{n-3})=a^{q}X^{3}(a^{1+q}+X^{3}+X^{n-3}+a^{1+q}X^{n})C^{3}.$

In the above

$\displaystyle\text{gcd}(1+X^{n-3},\,a^{1+q}+X^{3}+X^{n-3}+a^{1+q}X^{n})$
$\displaystyle=\,$
$\displaystyle\text{gcd}(1+X^{n-3},\,a^{1+q}+X^{3}+1+a^{1+q}X^{3})$
$\displaystyle=\,$ $\displaystyle\text{gcd}(1+X^{n-3},\,(a^{1+q}+1)(1+X^{3}))$
$\displaystyle=\,$ $\displaystyle 1+X.$

Let $C=D/E$, where $D,E\in\overline{\mathbb{F}}_{q}[X]$, $E$ is monic and
$\text{gcd}(D,E)=1$. Then (3.17) becomes

(3.18)
$(a^{2}E^{3}+a^{1+q}X^{3}D^{2}E)\frac{1+X^{n-3}}{1+X}=a^{q}X^{3}D^{3}\,\frac{a^{1+q}+X^{3}+X^{n-3}+a^{1+q}X^{n}}{1+X}.$

It follows that

(3.19) $\frac{1+X^{n-3}}{1+X}\Bigm{|}D\quad\text{and}\quad
D\Bigm{|}\frac{1+X^{n-3}}{1+X}.$

(3.18) and (3.19) force $D\in\overline{\mathbb{F}}_{q}^{*}$ and $n=4$. So

$a^{2}E^{3}+a^{1+q}D^{2}X^{3}E=a^{q}D^{3}X^{3}(a^{1+q}(1+X)^{3}+X(1+X)).$

Then $X\mid E$, say $E=XE_{1}$. Thus

(3.20)
$a^{2}E_{1}^{3}+a^{1+q}D^{2}XE_{1}=a^{q}D^{3}(1+X)(a^{1+q}X^{2}+X+a^{1+q}).$

It follows that $\deg E_{1}=1$, say $E_{1}=X+\epsilon$,
$\epsilon\in\overline{\mathbb{F}}_{q}$. Comparing the coefficients of $X^{3}$
and $X^{0}$ in the above gives

(3.21) $\displaystyle a^{2}\,$ $\displaystyle=a^{1+2q}D^{3},$ $\displaystyle
a^{2}\epsilon^{3}\,$ $\displaystyle=a^{1+2q}D^{3}.$

Hence $\epsilon^{3}=1$. Then
$(a^{1+q}X^{2}+X+a^{1+q})|_{X=\epsilon}=a^{1+q}(1+\epsilon^{2})+\epsilon\neq
0$ since $a^{1+q}\neq 1$. It follows from (3.20) that $E_{1}\mid 1+X$, that
is, $E_{1}=X+1$. Now (3.20) becomes

$a^{2}(X+1)^{2}+a^{1+q}D^{2}X=a^{q}D^{3}(a^{1+q}X^{2}+X+a^{1+q}).$

Comparing the coefficients of $X$ in the above gives
$a^{1+q}D^{2}=a^{q}D^{3}$, i.e., $D=a$. But then (3.21) gives $a^{1+q}=1$,
which is a contradiction.

###  3.2. Case 2. $n\leq 2$

When $n=1$, the absolute irreducibility (irreducibility over
$\overline{\mathbb{F}}_{q}$) of $N(X,Y)$ follows from [5, §3]. So we assume
$n=2$. The arguments are similar to those in Case 1. We have

(3.22) $G(X)=\frac{a^{q}X^{3}+1}{X(X^{3}+a)},$ (3.23)
$N(X,Y)=a^{q}X^{3}Y^{3}+a^{q+1}XY(X+Y)+(X+Y)^{3}+a,$

and

(3.24) $Q(Z)=aZ^{2}+(a^{q+1}XY(X+Y)+(X+Y)^{3})Z+a^{q}X^{3}Y^{3}.$

When proving
$[\overline{\mathbb{F}}_{q}(X,Y,z):\overline{\mathbb{F}}_{q}(X,Y)]=2$,
equations (3.5), (3.6) and (3.7) are replaced by

$\frac{a^{q+1}X^{3}Y^{3}}{(a^{q+1}XY(X+Y)+(X+Y)^{3})^{2}}=\frac{A(A+B)}{B^{2}},$
(3.25) $B=a^{q+1}XY(X+Y)+(X+Y)^{3},$

and

$\begin{cases}A=uX^{3},\cr A+B=vY^{3},\end{cases}\quad
u,v\in\overline{\mathbb{F}}_{q}^{*}.$

Then $B=uX^{3}+vY^{3}$, which contradicts (3.25) since $a^{1+q}\neq 1$.

When proving
$[\overline{\mathbb{F}}_{q}(X,Y,w):\overline{\mathbb{F}}_{q}(X,Y,z)]=3$,
equation (3.12) is replaced by

(3.26) $\displaystyle f_{0}(X)=\,$ $\displaystyle
a^{2}A_{1}^{3}+a^{1+q}A_{1}B_{1}^{2}X^{3}+a^{q}B_{1}^{3}X^{3}+a^{q}B_{1}^{3}X^{4}+a^{1+2q}B_{1}^{3}X^{4}+a^{q}B_{1}^{3}X^{5}$
$\displaystyle+a^{1+2q}B_{1}^{3}X^{5}+a^{q}B_{1}^{3}X^{6}.$

Setting $E=A_{1}/B_{1}$, the equation $f_{0}(X)=0$ becomes

$a^{2}E^{3}+a^{1+q}X^{3}E+a^{q}X^{3}(1+X)(1+a^{1+q}X+X^{2})=0.$

It follows that $E\in\overline{\mathbb{F}}_{q}[X]$ and $X\mid E$. Write
$E=XE_{1}$. Then

(3.27) $a^{2}E_{1}^{3}+a^{1+q}XE_{1}+a^{q}(1+X)(1+a^{1+q}X+X^{2})=0.$

Thus $\deg E_{1}=1$, say $E_{1}=e(X+\epsilon)$,
$e\in\overline{\mathbb{F}}_{q}^{*}$, $\epsilon\in\overline{\mathbb{F}}_{q}$.
Comparing the coefficients of $X^{3}$ and $X^{0}$ in the above gives

(3.28) $\displaystyle a^{2}e^{3}+a^{q}=0,$ $\displaystyle
a^{2}e^{3}\epsilon^{3}+a^{q}=0.$

Hence $\epsilon^{3}=1$. Then $(1+a^{1+q}X+X^{2})|_{X=\epsilon}\neq 0$. It
follows form (3.27) that $E_{1}\mid 1+X$, whence $E_{1}=e(X+1)$. Then (3.27)
becomes

$a^{2}e^{3}(X+1)^{2}+a^{1+q}eX+a^{q}(1+a^{1+q}X+X^{2})=0.$

Comparing the coefficients of $X$ in the above gives $e=a^{q}$. But then
(3.28) gives $a^{1+q}=1$, which is a contradiction.

######  Remark 3.1.

Most likely, Theorem 1.9 also holds for odd $q$.

##  4\. Proof of Theorem 1.10

Theorem 1.10. Let $n\geq 1$, $d\geq 2$ and $a\in\mathbb{F}_{q^{2}}^{*}$ be
such that $d\mid q+1$, $q\geq(2\max\\{n,2d-n\\})^{4}$ and $a^{q+1}\neq 1$.
Then $f(X)=f_{q,2,n,d,a}(X)=X^{n}(X^{d(q-1)}+a)$ is not a PB of
$\mathbb{F}_{q^{2}}$ if one of the following conditions is satisfied.

  * (i)

$d-n>1$ and $\text{\rm gcd}(d,n+1)$ is a power of $2$.

  * (ii)

$d+2\leq n<2d$ and $\text{\rm gcd}(d,n-1)$ is a power of $2$.

  * (iii)

$n\geq 2d$, $\text{\rm gcd}(d,n-1)$ is a power of $2$, and $\text{\rm
gcd}(n-d,q-1)=1$.

Assume to the contrary that $f(X)$ is a PB of $\mathbb{F}_{q^{2}}$. Recall
that

$G(X)=\frac{a^{q}X^{n}+X^{n-d}}{X^{d}+a}.$

Let

$N(X,Y)=\text{the numerator of}\ \frac{G(X)-G(Y)}{X-Y}$

and

$N^{*}(X,Y,Z)=\text{the homogenization of}\ N(X,Y).$

Our objective is to show that $N^{*}(X,Y,Z)$ is irreducible over
$\overline{\mathbb{F}}_{q}$ under the conditions in Theorem 1.10. We consider
two cases: the case $d-n>1$, which corresponds to (i) in Theorem 1.10, and the
case $n-d>1$, which corresponds to (ii) and (iii) in Theorem 1.10.

###  4.1. The case $d-n>1$

We have

$G(X)=\frac{a^{q}X^{d}+1}{X^{d-n}(X^{d}+a)},$ $\displaystyle N(X,Y)=\,$
$\displaystyle-a\frac{X^{d-n}-Y^{d-n}}{X-Y}+\Bigl{[}a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}-Y^{n}}{X-Y}-\frac{X^{2d-n}-Y^{2d-n}}{X-Y}\Bigr{]}$
$\displaystyle-a^{q}X^{d}Y^{d}\frac{X^{d-n}-Y^{d-n}}{X-Y},$
$N^{*}(X,Y,Z)=Q(Z^{d}),$

where

(4.1) $\displaystyle Q(Z)=\,$
$\displaystyle-a\frac{X^{d-n}-Y^{d-n}}{X-Y}Z^{2}+\Bigl{[}a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}-Y^{n}}{X-Y}-\frac{X^{2d-n}-Y^{2d-n}}{X-Y}\Bigr{]}Z$
$\displaystyle-a^{q}X^{d}Y^{d}\frac{X^{d-n}-Y^{d-n}}{X-Y}.$

We claim that

(4.2)
$\text{gcd}\Bigl{(}\frac{X^{d-n}-Y^{d-n}}{X-Y},\,a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}-Y^{n}}{X-Y}-\frac{X^{2d-n}-Y^{2d-n}}{X-Y}\Bigr{)}=1.$

Since the polynomials in (4.2) are homogeneous, it suffices to prove (4.2)
with $Y=1$, i.e.,

(4.3)
$\text{gcd}\Bigl{(}\frac{X^{d-n}-1}{X-1},\,a^{q+1}X^{d-n}\frac{X^{n}-1}{X-1}-\frac{X^{2d-n}-1}{X-1}\Bigr{)}=1.$

Let $\zeta$ be a root of $(X^{d-n}-1)/(X-1)$. If $\zeta\neq 1$, then
$\zeta^{d-n}=1$. It follows that

$\displaystyle\Bigl{(}a^{q+1}X^{d-n}\frac{X^{n}-1}{X-1}-\frac{X^{2d-n}-1}{X-1}\Bigr{)}\Big{|}_{X=\zeta}$
$\displaystyle=\frac{1}{\zeta-1}\bigl{(}a^{q+1}(\zeta^{n}-1)-(\zeta^{n}-1)\bigr{)}=\frac{1}{\zeta-1}(a^{q+1}-1)(\zeta^{n}-1)\neq
0.$

(Note: $\zeta^{n}\neq 1$ since $\zeta^{d-n}=1$ and $\text{gcd}(n,d)=1$.) If
$\zeta=1$, then $d-n\equiv 0\pmod{p}$, where $p=\text{char}\,\mathbb{F}_{q}$,
whence

$\Bigl{(}a^{q+1}X^{d-n}\frac{X^{n}-1}{X-1}-\frac{X^{2d-n}-1}{X-1}\Bigr{)}\Big{|}_{X=1}=a^{q+1}n-(2d-n)=(a^{q+1}-1)n\neq
0.$

This proves (4.3) and hence (4.2). By (4.2), $N^{*}(X,Y,Z)$ is a primitive
polynomial in $Z$ over $\overline{\mathbb{F}}_{q}[X,Y]$, i.e., the gcd of its
coefficients in $\overline{\mathbb{F}}_{q}[X,Y]$ is 1. Thus, to prove that
$N^{*}(X,Y,Z)$ is irreducible in $\overline{\mathbb{F}}_{q}[X,Y,Z]$, it
suffices to show that it is irreducible in
$\overline{\mathbb{F}}_{q}(X,Y)[Z]$. Let $w$ be a root of $N^{*}(X,Y,Z)$ for
$Z$ and let $z=w^{d}$. Then $z$ is a root of $Q(Z)$, and it suffices to show
that $[\overline{\mathbb{F}}_{q}(X,Y,z):\overline{\mathbb{F}}_{q}(X,Y)]=2$ and
$[\overline{\mathbb{F}}_{q}(X,Y,w):\overline{\mathbb{F}}_{q}(X,Y,z)]=d$.

..................................................................................................................$\overline{\mathbb{F}}_{q}(X,Y)$$\overline{\mathbb{F}}_{q}(X,Y,z)$$\overline{\mathbb{F}}_{q}(X,Y,w)$$\scriptstyle
2$$\scriptstyle d$

####  4.1.1. Proof that
$[\overline{\mathbb{F}}_{q}(X,Y,z):\overline{\mathbb{F}}_{q}(X,Y)]=2$

Assume to the contrary that $Q(Z)$ is reducible over
$\overline{\mathbb{F}}_{q}(X,Y)$.

First assume that $q$ is odd. The discriminant of $Q$ is

$D=\Bigl{[}a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}-Y^{n}}{X-Y}-\frac{X^{2d-n}-Y^{2d-n}}{X-Y}\Bigr{]}^{2}-4a^{q+1}X^{d}Y^{d}\Bigl{(}\frac{X^{d-n}-Y^{d-n}}{X-Y}\Bigr{)}^{2}.$

By assumption, $D=\Delta^{2}$ for some
$\Delta\in\overline{\mathbb{F}}_{q}[X,Y]$. Then

(4.4) $\displaystyle
4a^{q+1}X^{d}Y^{d}\Bigl{(}\frac{X^{d-n}-Y^{d-n}}{X-Y}\Bigr{)}^{2}=$ (4.5)
$\displaystyle\Bigl{[}a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}-Y^{n}}{X-Y}-\frac{X^{2d-n}-Y^{2d-n}}{X-Y}+\Delta\Bigr{]}$
$\displaystyle\cdot\Bigl{[}a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}-Y^{n}}{X-Y}-\frac{X^{2d-n}-Y^{2d-n}}{X-Y}-\Delta\Bigr{]}.$

Let $\delta$ be the gcd of the two factors on the right side of (4.4). Then

$\delta\Bigm{|}a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}-Y^{n}}{X-Y}-\frac{X^{2d-n}-Y^{2d-n}}{X-Y}$

and

$\delta\bigm{|}\frac{X^{d-n}-Y^{d-n}}{X-Y}.$

It follows from (4.2) that $\delta=1$.

Now from (4.4), we have

$\begin{cases}\displaystyle
a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}-Y^{n}}{X-Y}-\frac{X^{2d-n}-Y^{2d-n}}{X-Y}+\Delta=X^{d}U,\vspace{0.4em}\cr\displaystyle
a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}-Y^{n}}{X-Y}-\frac{X^{2d-n}-Y^{2d-n}}{X-Y}-\Delta=Y^{d}V,\end{cases}$

for some $U,V\in\overline{\mathbb{F}}_{q}[X,Y]$. It follows that

(4.6)
$2a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}-Y^{n}}{X-Y}-2\frac{X^{2d-n}-Y^{2d-n}}{X-Y}=X^{d}U+Y^{d}V.$

The coefficient of $X^{d-1}Y^{d-n}$ on the left side of (4.6) is
$2(a^{q+1}-1)\neq 0$, while the coefficient of the same term on the right side
of (4.6) is 0. This is a contradiction.

Next, assume that $q$ is even. Since $Q(Z)$ is assumed to be reducible over
$\overline{\mathbb{F}}_{q}(X,Y)$, we have

$\frac{\displaystyle
a^{q+1}X^{d}Y^{d}\Bigl{(}\frac{X^{d-n}+Y^{d-n}}{X+Y}\Bigr{)}^{2}}{\displaystyle\Bigl{[}a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}+Y^{n}}{X+Y}+\frac{X^{2d-n}+Y^{2d-n}}{X+Y}\Bigr{]}^{2}}=\Bigl{(}\frac{A}{B}\Bigr{)}^{2}+\frac{A}{B}=\frac{A(A+B)}{B^{2}},$

where $A,B\in\overline{\mathbb{F}}_{q}[X,Y]$, $\text{gcd}(A,B)=1$. By (4.2),
the numerator and the denominator on the left side are relatively prime.
Therefore we may assume

(4.7)
$B=a^{q+1}X^{d-n}Y^{d-n}\frac{X^{n}+Y^{n}}{X+Y}+\frac{X^{2d-n}+Y^{2d-n}}{X+Y},$
$A(A+B)=a^{q+1}X^{d}Y^{d}\Bigl{(}\frac{X^{d-n}+Y^{d-n}}{X+Y}\Bigr{)}^{2}.$

Since $\text{gcd}(A,A+B)=1$, we have

$\begin{cases}A=X^{d}U^{2},\cr A+B=Y^{d}V^{2},\end{cases}$

where $U,V\in\overline{\mathbb{F}}_{q}[X,Y]$, $UV=(X^{d-n}+Y^{d-n})/(X+Y)$.
Then

(4.8) $B=X^{d}U^{2}+Y^{d}V^{2}.$

The coefficient of $X^{d-1}Y^{d-n}$ in (4.7) is $a^{q+1}+1\neq 0$. However,
the coefficient of $X^{d-1}Y^{d-n}$ in (4.8) is 0, which is a contradiction.

####  4.1.2. Proof that
$[\overline{\mathbb{F}}_{q}(X,Y,w):\overline{\mathbb{F}}_{q}(X,Y,z)]=d$

To prove this claim, it suffices to show that for each prime divisor $t$ of
$d$, $z$ is not a $t$th power in $\overline{\mathbb{F}}_{q}(X,Y,z)$. In (4.1),
divide $Q(Z)$ by its leading coefficient and set $Y=1$, the result is

(4.9) $Q_{1}(Z)=Z^{2}-\frac{\displaystyle
a^{q+1}X^{d-n}\frac{X^{n}-1}{X-1}-\frac{X^{2d-n}-1}{X-1}}{\displaystyle
a\frac{X^{d-n}-1}{X-1}}Z+a^{q-1}X^{d},$

which is irreducible in $\overline{\mathbb{F}}_{q}(X)[Z]$. Let $z_{1}$ be a
root of $Q_{1}(Z)$. By [5, §3.3, Claim II′], it suffices to show that for each
prime divisor $t$ of $d$, $z_{1}$ is not a $t$th power in
$\overline{\mathbb{F}}_{q}(X,z_{1})$.

Let $\overline{(\ )}$ denote the nonidentity automorphism in
$\text{Aut}(\overline{\mathbb{F}}_{q}(X,z_{1})/\overline{\mathbb{F}}_{q}(X))$.
We have

(4.10) $z_{1}\bar{z}_{1}=a^{q-1}X^{d},$ (4.11)
$z_{1}+\bar{z}_{1}=\frac{\displaystyle
a^{q+1}X^{d-n}\frac{X^{n}-1}{X-1}-\frac{X^{2d-n}-1}{X-1}}{\displaystyle
a\frac{X^{d-n}-1}{X-1}}.$

Let $d-n=p^{m}d^{\prime}$, where $p=\text{char}\,\mathbb{F}_{q}$, $p\nmid
d^{\prime}$. Let $\zeta\in\overline{\mathbb{F}}_{q}$ be a primitive
$d^{\prime}$th root of unity. Let $\mathfrak{p}$ be the place of the rational
function field $\overline{\mathbb{F}}_{q}(X)$ which is the zero of $X-\zeta$,
and let $\mathfrak{P}$ be a place of $\overline{\mathbb{F}}_{q}(X,z_{1})$ such
that $\mathfrak{P}\mid\mathfrak{p}$. Then $\mathfrak{P}$ is unramified over
$\mathfrak{p}$ ([10, III 7.3 (b) and 7.8 (b)]). From (4.10) and (4.11), we
have

(4.12) $\nu_{\mathfrak{p}}(z_{1}\bar{z}_{1})=0,$ (4.13)
$\nu_{\mathfrak{p}}(z_{1}+\bar{z}_{1})=\begin{cases}-p^{m}&\text{if}\
d^{\prime}>1,\cr-p^{m}+1&\text{if}\ d^{\prime}=1,\end{cases}$

where $\nu_{\mathfrak{p}}$ is the valuation of $\overline{\mathbb{F}}_{q}(X)$
at $\mathfrak{p}$. Equation (4.13) is derived as follows: First, note that in
(4.11),

(4.14)
$\nu_{\mathfrak{p}}\Bigl{(}\frac{X^{d-n}-1}{X-1}\Bigr{)}=\begin{cases}p^{m}&\text{if}\
d^{\prime}>1,\cr p^{m}-1&\text{if}\ d^{\prime}=1.\end{cases}$

Next, write

(4.15)
$a^{q+1}X^{d-n}\frac{X^{n}-1}{X-1}-\frac{X^{2d-n}-1}{X-1}=(a^{q+1}X^{d-n}-1)\frac{X^{n}-1}{X-1}-X^{n}\frac{X^{2(d-n)}-1}{X-1}.$

If $d^{\prime}>1$, the value of (4.15) at $X=\zeta$ is

$(a^{q+1}-1)\frac{\zeta^{n}-1}{\zeta-1}\neq 0.$

If $d^{\prime}=1$, we have $m>0$ (since $d-n>1$), whence $d-n\equiv
0\pmod{p}$. Then $n\not\equiv 0$ since $\text{gcd}(n,d)=1$. Therefore, the
value of (4.15) at $X=\zeta$ ($=1$) is

$(a^{q+1}-1)n-2(d-n)=(a^{q+1}-1)n\neq 0.$

Hence we always have

(4.16)
$\nu_{\mathfrak{p}}\Bigl{(}a^{q+1}X^{d-n}\frac{X^{n}-1}{X-1}-\frac{X^{2d-n}-1}{X-1}\Bigr{)}=0.$

Combining (4.11), (4.14) and (4.16) gives (4.13).

Write (4.12) and (4.13) as

$\nu_{\mathfrak{P}}(z_{1})+\nu_{\mathfrak{P}}(\bar{z}_{1})=0,$
$\nu_{\mathfrak{P}}(z_{1}+\bar{z}_{1})=\begin{cases}-p^{m}&\text{if}\
d^{\prime}>1,\cr-p^{m}+1&\text{if}\ d^{\prime}=1,\end{cases}$

where $\nu_{\mathfrak{P}}$ is the valuation of
$\overline{\mathbb{F}}_{q}(X,z_{1})$ at $\mathfrak{P}$. It follows that

(4.17)
$\\{\nu_{\mathfrak{P}}(z_{1}),\nu_{\mathfrak{P}}(\bar{z}_{1})\\}=\begin{cases}\\{\pm
p^{m}\\}&\text{if}\ d^{\prime}>1,\cr\\{\pm(p^{m}-1)\\}&\text{if}\
d^{\prime}=1.\end{cases}$

Assume to the contrary that $z_{1}$ is a $t$th power in
$\overline{\mathbb{F}}_{q}(X,z_{1})$. Then $t\mid\nu_{\mathfrak{P}}(z_{1})$.
If $d^{\prime}>1$, the by (4.17), $t\mid p^{m}$, whence $t\mid d-n$. This is
impossible since $t\mid d$ and $\text{gcd}(n,d)=1$. Therefore, $d^{\prime}=1$
and $d-n=p^{m}$. By (4.17), $t\mid p^{m}-1=d-n-1$. Since
$t\mid\text{gcd}(d,d-n-1)=\text{gcd}(d,n+1)$ and $\text{gcd}(d,n+1)$ is a
power of $2$, we have $t=2$. It follows that $p$ is odd.

Recall that $Q_{1}(z_{1})=0$, where $Q_{1}(Z)$ is given in (4.9). Using (4.15)
and $d-n=p^{m}$, the equation $Q_{1}(z_{1})=0$ can be written as

$u^{2}=\delta,$

where

$u=z_{1}-\gamma,$
$\gamma=\frac{1}{2}\frac{\displaystyle(a^{q+1}X^{p^{m}}-1)\frac{X^{n}-1}{X-1}-X^{n}(X+1)^{p^{m}}(X-1)^{p^{m}-1}}{a(X-1)^{p^{m}-1}},$

and

$\delta=\gamma^{2}-a^{q-1}X^{p^{m}+n}.$

By assumption, there exist $\alpha,\beta\in\overline{\mathbb{F}}_{q}(X)$ such
that

$(\alpha u+\beta)^{2}=u+\gamma,$

i.e.,

$\alpha^{2}\delta+\beta^{2}+2\alpha\beta u=u+\gamma.$

Since $u$ is of degree 2 over $\overline{\mathbb{F}}_{q}(X)$, we have

$\begin{cases}\alpha^{2}\delta+\beta^{2}=\gamma,\cr
2\alpha\beta=1.\end{cases}$

Letting $\tau=\alpha/\beta$, we have

(4.18) $1+\delta\tau^{2}-2\gamma\tau=0$

and

(4.19) $\tau=2\alpha^{2}.$

Fortunately, (4.18) has an explicit solution

$\tau=\frac{1}{\delta}(\gamma\pm
a^{(q-1)/2}X^{(p^{m}+n)/2})=\frac{1}{\gamma\mp a^{(q-1)/2}X^{(p^{m}+n)/2}}.$

In the above,

$\displaystyle\gamma\mp a^{(q-1)/2}X^{(p^{m}+n)/2}=$
$\displaystyle\frac{1}{2a(X-1)^{p^{m}-1}}\Bigl{[}(a^{q+1}X^{p^{m}}-1)\frac{X^{n}-1}{X-1}-X^{n}(X+1)^{p^{m}}(X-1)^{p^{m}-1}$
$\displaystyle\kern 74.00005pt\mp
2a^{(q+1)/2}X^{(p^{m}+n)/2}(X-1)^{p^{m}-1}\Bigr{]}.$

Since $\tau$ is square in $\overline{\mathbb{F}}_{q}(X)$ (by (4.19)),

$h:=(1-a^{q+1}X^{p^{m}})\frac{X^{n}-1}{X-1}+X^{n}(X+1)^{p^{m}}(X-1)^{p^{m}-1}+2\epsilon
X^{(p^{m}+n)/2}(X-1)^{p^{m}-1},$

where $\epsilon=\pm a^{(q+1)/2}$, is a square in
$\overline{\mathbb{F}}_{q}(X)$, say $h=g^{2}$, where
$g\in\overline{\mathbb{F}}_{q}[X]$ is monic of degree $p^{m}+(n-1)/2$. Note
that

$\displaystyle h=\,$
$\displaystyle\frac{X^{n}-1}{X-1}+(X^{n}+X^{p^{m}+n})\frac{X^{p^{m}}-1}{X-1}-a^{q+1}X^{p^{m}}\frac{X^{n}-1}{X-1}+2\epsilon
X^{(p^{m}+n)/2}\frac{X^{p^{m}}-1}{X-1}$ $\displaystyle=\,$
$\displaystyle(1+\cdots+X^{2p^{m}+n-1})$
$\displaystyle-a^{q+1}(X^{p^{m}}+\cdots+X^{p^{m}+n-1})$
$\displaystyle+2\epsilon(X^{(p^{m}+n)/2}+\cdots+X^{(3p^{m}+n)/2-1}),$

which is self-reciprocal. Hence $g^{*}=\pm g$, where $g^{*}$ is the reciprocal
polynomial of $g$. (In fact, If $g^{*}=g$, but we do not need to be precise.)
Let

$\displaystyle H\,$ $\displaystyle=(X-1)h$
$\displaystyle=(1-a^{q+1}X^{p^{m}})(X^{n}-1)+X^{n}(X+1)^{p^{m}}(X-1)^{p^{m}}+2\epsilon
X^{(p^{m}+n)/2}(X-1)^{p^{m}}.$

Then

$H^{\prime}=(1-a^{q+1}X^{p^{m}})nX^{n-1}+nX^{n-1}(X+1)^{p^{m}}(X-1)^{p^{m}}+\epsilon
nX^{(p^{m}+n)/2-1}(X-1)^{p^{m}}.$

(When computing $H^{\prime}$, we used the assumption that $m>0$.) Let

$\displaystyle K$
$\displaystyle=H-n^{-1}XH^{\prime}=-(1-a^{q+1}X^{p^{m}})+\epsilon
X^{(p^{m}+n)/2}(X-1)^{p^{m}}$ $\displaystyle=-1+a^{q+1}X^{p^{m}}-\epsilon
X^{(p^{m}+n)/2}+\epsilon X^{(p^{m}+n)/2+p^{m}}.$

The reciprocal of $K$ is

$K^{*}=\epsilon-\epsilon
X^{p^{m}}+a^{q+1}X^{(p^{m}+n)/2}-X^{(p^{m}+n)/2+p^{m}}.$

Since $g\mid K$ and $g$ is self-reciprocal, we also have $g=\pm g^{*}\mid
K^{*}$. Thus $g$ divides

$K+\epsilon K^{*}=-1+\epsilon^{2}+(-\epsilon+\epsilon
a^{q+1})X^{(p^{m}+n)/2}=(a^{q+1}-1)(1+\epsilon X^{(p^{m}+n)/2}).$

This is a contradiction since

$\frac{p^{m}+n}{2}<p^{m}+\frac{n-1}{2}=\deg g.$

###  4.2. The Case $n-d>1$

In this case,

$G(X)=\frac{a^{q}X^{n}+X^{n-d}}{X^{d}+a},$ $\displaystyle N(X,Y)=\,$
$\displaystyle
a\frac{X^{n-d}-Y^{n-d}}{X-Y}+\Bigl{[}a^{q+1}\frac{X^{n}-Y^{n}}{X-Y}+X^{d}Y^{d}\frac{X^{n-2d}-Y^{n-2d}}{X-Y}\Bigr{]}$
$\displaystyle+a^{q}X^{d}Y^{d}\frac{X^{n-d}-Y^{n-d}}{X-Y},$
$N^{*}(X,Y,Z)=Q(Z^{d}),$

where

$\displaystyle Q(Z)=\,$ $\displaystyle
a\frac{X^{n-d}-Y^{n-d}}{X-Y}Z^{2}+\Bigl{[}a^{q+1}\frac{X^{n}-Y^{n}}{X-Y}+X^{d}Y^{d}\frac{X^{n-2d}-Y^{n-2d}}{X-Y}\Bigr{]}Z$
$\displaystyle+a^{q}X^{d}Y^{d}\frac{X^{n-d}-Y^{n-d}}{X-Y}.$

We claim that

(4.20)
$\gcd\Big{(}\frac{X^{n-d}-Y^{n-d}}{X-Y},\,a^{q+1}\frac{X^{n}-Y^{n}}{X-Y}+X^{d}Y^{d}\frac{X^{n-2d}-Y^{n-2d}}{X-Y}\Big{)}=1,$

equivalently,

(4.21)
$\gcd\Bigl{(}\frac{X^{n-d}-1}{X-1},\,a^{q+1}\frac{X^{n}-1}{X-1}+X^{d}\frac{X^{n-2d}-1}{X-1}\Bigr{)}=1.$

Let $\zeta$ be a root of $(X^{n-d}-1)/(X-1)$. If $\zeta\neq 1$, then

$\displaystyle\Bigl{(}a^{q+1}\frac{X^{n}-1}{X-1}+X^{d}\frac{X^{n-2d}-1}{X-1}\Bigr{)}\big{|}_{X=\zeta}=\,$
$\displaystyle\frac{1}{\zeta-1}\bigl{(}a^{q+1}(\zeta^{n}-1)+\zeta^{d}(\zeta^{n-2d}-1)\bigr{)}$
$\displaystyle=\,$ $\displaystyle\frac{1}{\zeta-1}(a^{q+1}-1)(\zeta^{n}-1)\neq
0.$

If $\zeta=1$, then $n-d\equiv 0\pmod{p}$, and

$\Big{(}a^{q+1}\frac{X^{n}-1}{X-1}+X^{d}\frac{X^{n-2d}-1}{X-1}\Big{)}\Big{|}_{X=1}=a^{q+1}n+n-2d=n(a^{q+1}-1)\neq
0.$

So (4.21) and (4.20) hold. Therefore $Q(Z)$ is a primitive polynomial over
$\overline{\mathbb{F}}_{q}[X,Y]$.

Let

(4.22) $\displaystyle Q_{1}(Z)\,$
$\displaystyle=\Bigl{[}\Bigl{(}a\frac{X^{n-d}-Y^{n-d}}{X-Y}\Bigr{)}^{-1}Q(Z)\Bigr{]}\Big{|}_{Y=1}$
$\displaystyle=Z^{2}+\frac{\displaystyle
a^{q+1}\frac{X^{n}-1}{X-1}+X^{d}\frac{X^{n-2d}-1}{X-1}}{\displaystyle
a\frac{X^{n-d}-1}{X-1}}Z+a^{q-1}X^{d}\in\overline{\mathbb{F}}_{q}(X)[Z].$

Following the arguments in Section 4.1, we only have to prove the following
two claims:

Claim 1. $Q(Z)$ is irreducible in $\overline{\mathbb{F}}_{q}(X,Y)[Z]$.

Claim 2. Let $z$ be a root of $Q_{1}(Z)$ and $t$ be a prime divisor of $d$.
Then $z$ is not a $t$th power in $\overline{\mathbb{F}}_{q}(X,z)$.

####  4.2.1. Proof of Claim 1

Assume to the contrary that $Q(Z)$ is reducible in
$\overline{\mathbb{F}}_{q}(X,Y)[Z]$.

First, assume that $q$ is odd. The discriminant of $Q(Z)$ is

$D=\Bigl{[}\frac{a^{q+1}(X^{n}-Y^{n})}{X-Y}+\frac{X^{d}Y^{d}(X^{n-2d}-Y^{n-2d})}{X-Y}\Bigr{]}^{2}-\frac{4a^{q+1}X^{d}Y^{d}(X^{n-d}-Y^{n-d})^{2}}{(X-Y)^{2}}.$

By assumption, $D=\Delta^{2}$ for some
$\Delta\in\overline{\mathbb{F}}_{q}[X,Y]$. Then

$\displaystyle\frac{4a^{q+1}X^{d}Y^{d}\left(X^{n-d}-Y^{n-d}\right)^{2}}{(X-Y)^{2}}=\,$
$\displaystyle\Big{(}a^{q+1}\frac{X^{n}-Y^{n}}{X-Y}+X^{d}Y^{d}\frac{X^{n-2d}-Y^{n-2d}}{X-Y}+\Delta\Big{)}$
$\displaystyle\cdot\Big{(}a^{q+1}\frac{X^{n}-Y^{n}}{X-Y}+X^{d}Y^{d}\frac{X^{n-2d}-Y^{n-2d}}{X-Y}-\Delta\Big{)}.$

In the above, the two factors on the right side are relatively prime. (This
follows from (4.20).) Therefore, we may assume

(4.23) $\begin{cases}\displaystyle
a^{q+1}\frac{X^{n}-Y^{n}}{X-Y}+X^{d}Y^{d}\frac{X^{n-2d}-Y^{n-2d}}{X-Y}+\Delta=2a^{(q+1)/2}X^{d}U^{2},\vspace{0.5em}\cr\displaystyle
a^{q+1}\frac{X^{n}-Y^{n}}{X-Y}+X^{d}Y^{d}\frac{X^{n-2d}-Y^{n-2d}}{X-Y}-\Delta=2a^{(q+1)/2}Y^{d}V^{2},\end{cases}$

for some $U,V\in\overline{\mathbb{F}}_{q}[X,Y]$ with

(4.24) $UV=\frac{X^{n-d}-Y^{n-d}}{X-Y}.$

Then

(4.25)
$\frac{a^{q+1}\left(X^{n}-Y^{n}\right)}{X-Y}+\frac{X^{d}Y^{d}\left(X^{n-2d}-Y^{n-2d}\right)}{X-Y}=a^{(q+1)/2}(X^{d}U^{2}+Y^{d}V^{2}).$

Let $L$ denote the left side of (4.25). We have

$\displaystyle L=\,$ $\displaystyle a^{q+1}(Y^{n-1}+XY^{n-2}+\cdots+X^{n-1})$
$\displaystyle+\begin{cases}X^{d}Y^{n-d-1}+X^{d+1}Y^{n-d-2}+\cdots+X^{n-d-1}Y^{d}&\text{if}\
n\geq
2d,\vspace{0.2em}\cr-X^{n-d}Y^{d-1}-X^{n-d+1}Y^{d-2}-\cdots-X^{d-1}Y^{n-d}&\text{if}\
d+2\leq n<2d.\end{cases}$

If $d+2\leq n<2d$, the coefficient of $X^{d-1}Y^{n-d}$ in $L$ is
$a^{q+1}-1\neq 0$, while the the coefficient of $X^{d-1}Y^{n-d}$ on the right
side of (4.25) is 0, which is a contradiction. Hence Theorem 1.10 (iii) holds.
In particular, $\text{gcd}(n-d,q-1)=1$.

Since

$\Delta(Y,X)^{2}=D(Y,X)=D(X,Y)=\Delta(X,Y)^{2}.$

we have $\Delta(Y,X)=\pm\Delta(X,Y)$. If $\Delta(Y,X)=\Delta(X,Y)$, then by
(4.23), $X^{d}U(X,Y)^{2}=Y^{d}U(Y,X)^{2}$. Then $Y\mid U(X,Y)$, which is a
contradiction to (4.24). Hence $\Delta(Y,X)=-\Delta(X,Y)$, and by (4.23),

(4.26) $U(Y,X)^{2}=V(X,Y)^{2}.$

By (4.26) and (4.24), we have

(4.27) $\displaystyle U(X,Y)^{2}\,$
$\displaystyle=\alpha\prod_{i=1}^{(n-d-1)/2}(X-\epsilon_{i}Y)^{2},$ (4.28)
$\displaystyle V(X,Y)^{2}\,$
$\displaystyle=\alpha^{-1}\prod_{i=1}^{(n-d-1)/2}(X-\epsilon_{i}^{-1}Y)^{2},$

where $\alpha,\beta\in\overline{\mathbb{F}}_{q}$ and
$\epsilon_{i}\in\overline{\mathbb{F}}_{q}^{*}$ are such that

$\frac{X^{n-d}-Y^{n-d}}{X-Y}=\prod_{i=1}^{(n-d-1)/2}\bigl{[}(X-\epsilon_{i}Y)(X-\epsilon_{i}^{-1}Y)\bigr{]}.$

We have

$\displaystyle\alpha\,$ $\displaystyle=U(1,0)^{2}\kern 60.00009pt\text{(by
\eqref{U2})}$ $\displaystyle=V(0,1)^{2}\kern 60.00009pt\text{(by
\eqref{U2=V2})}$
$\displaystyle=\alpha^{-1}\prod_{i=1}^{(n-d-1)/2}\epsilon_{i}^{-2}\qquad\text{(by
\eqref{V2})}.$

It follows that

(4.29) $\alpha^{2}=\prod_{i=1}^{(n-d-1)/2}\epsilon_{i}^{-2}.$

On the other hand, comparing the coefficients of $X^{n-1}$ in (4.25) gives
$a^{q+1}=a^{(q+1)/2}\cdot\alpha$, i.e., $\alpha=a^{(q+1)/2}$. Since the
$\epsilon_{i}$’s are roots of $X^{n-d}-1$, we have

$a^{(q+1)(n-d)}=\alpha^{2(n-d)}=1\qquad\text{(by \eqref{alpha2})}.$

This, combined with $a^{(q+1)(q-1)}=1$ and $\text{gcd}(n-d,q-1)=1$, implies
that $a^{q+1}=1$, which is a contradiction.

Next, assume that $q$ is even. Since $Q(Z)$ is assumed to be reducible over
$\overline{\mathbb{F}}_{q}(X,Y)$, there are
$A,B\in\overline{\mathbb{F}}_{q}[X,Y]$, relatively prime, such that

$\frac{\displaystyle
a^{q+1}X^{d}Y^{d}\Bigl{(}\frac{X^{n-d}+Y^{n-d}}{X+Y}\Bigr{)}^{2}}{\displaystyle\Bigl{(}a^{q+1}\frac{X^{n}+Y^{n}}{X+Y}+X^{d}Y^{d}\frac{X^{n-2d}+Y^{n-2d}}{X+Y}\Bigr{)}^{2}}=\Big{(}\frac{A}{B}\Big{)}^{2}+\frac{A}{B}=\frac{A(A+B)}{B^{2}}.$

In the above, the numerator and the denominator on the left side are
relatively prime (by (4.20)). Thus

(4.30)
$B=a^{q+1}\frac{X^{n}+Y^{n}}{X+Y}+X^{d}Y^{d}\frac{X^{n-2d}+Y^{n-2d}}{X+Y}$

and

$A(A+B)=a^{q+1}X^{d}Y^{d}\Bigl{(}\frac{X^{n-d}+Y^{n-d}}{X+Y}\Bigr{)}^{2}.$

We may assume that

$\begin{cases}A=X^{d}U^{2},\\\ A+B=Y^{d}V^{2},\end{cases}$

for some $U,V\in\overline{\mathbb{F}}_{q}[X,Y]$ such that
$UV=(X^{n-d}+Y^{n-d})/(X+Y)$. Then

(4.31) $B=X^{d}U^{2}+Y^{d}V^{2}.$

By (4.30),

(4.32) $\displaystyle B=\,$ $\displaystyle
a^{q+1}(Y^{n-1}+XY^{n-2}+\cdots+X^{n-1})$
$\displaystyle+\begin{cases}X^{d}Y^{n-d-1}+X^{d+1}Y^{n-d-2}\cdots+X^{n-d-1}Y^{d}&\text{if}\
n\geq 2d,\cr X^{n-d}Y^{d-1}+X^{n-d+1}Y^{d-2}+\cdots+X^{d-1}Y^{n-d}&\text{if}\
d+2\leq n<2d.\end{cases}$

Since we assume $d>1$ and $n-d>1$, the coefficient of $XY^{n-2}$ in (4.32) is
$a^{q+1}\neq 0$. (Even if we allowed $d=1$ or $n-d=1$, the coefficient of
$XY^{n-2}$ in (4.32) would be $a^{q+1}+1$, which is still nonzero.) However,
the coefficient of $XY^{n-2}$ in (4.31) is 0, which is a contradiction.

####  4.2.2. Proof of Claim 2

Recall that $Q_{1}(Z)$ is given in (4.22). Let $z$ be a root of $Q_{1}(Z)$ and
$t$ be a prime divisor of $d$. Assume to the contrary that $z$ is a $t$th
power in $\overline{\mathbb{F}}_{q}(X,z)$. Let $\overline{(\ )}$ be the
nonidentity automorphism in
$\text{Aut}(\overline{\mathbb{F}}_{q}(X,z)/\overline{\mathbb{F}}_{q}(X))$.
Then

(4.33) $z\bar{z}=a^{q-1}X^{d},$ (4.34) $\displaystyle z+\bar{z}\,$
$\displaystyle=-\frac{\displaystyle
a^{q+1}\frac{X^{n}-1}{X-1}+X^{d}\frac{X^{n-2d}-1}{X-1}}{\displaystyle
a\frac{X^{n-d}-1}{X-1}}$ (4.35)
$\displaystyle=-\frac{\displaystyle(a^{q+1}-1)\frac{X^{d}-1}{X-1}+(a^{q+1}X^{d}+1)\frac{X^{n-d}-1}{X-1}}{\displaystyle
a\frac{X^{n-d}-1}{X-1}}.$

Write $n-d=p^{m}d^{\prime}$, where $p\nmid d^{\prime}$, and let $\zeta$ be a
primitive $d^{\prime}$th root of unity. Let $\mathfrak{p}$ be the place of the
rational function field $\overline{\mathbb{F}}_{q}(X)$ which is the zero of
$X-\zeta$, and let $\mathfrak{P}$ be a place of
$\overline{\mathbb{F}}_{q}(X,z)$ such that $\mathfrak{P}\mid\mathfrak{p}$.
Then $\mathfrak{P}$ is unramified over $\mathfrak{p}$ ([10, III 7.3 (b) and
7.8 (b)]). From (4.33) and (4.34), we have

(4.36) $\nu_{\mathfrak{p}}(z\bar{z})=0,$ (4.37)
$\nu_{\mathfrak{p}}(z+\bar{z})=\begin{cases}-p^{m}&\text{if}\
d^{\prime}>1,\cr-p^{m}+1&\text{if}\ d^{\prime}=1,\end{cases}$

(The proof of (4.37) is similar to that of (4.13) and uses the assumption
$n-d>1$ in the case $d^{\prime}=1$.) Therefore,

$\nu_{\mathfrak{P}}(z)+\nu_{\mathfrak{P}}(\bar{z})=0,$
$\nu_{\mathfrak{P}}(z+\bar{z})=\begin{cases}-p^{m}&\text{if}\
d^{\prime}>1,\cr-p^{m}+1&\text{if}\ d^{\prime}=1,\end{cases}$

and it follows that

$\\{\nu_{\mathfrak{P}}(z),\nu_{\mathfrak{P}}(\bar{z})\\}=\begin{cases}\\{\pm
p^{m}\\}&\text{if}\ d^{\prime}>1,\cr\\{\pm(p^{m}-1)\\}&\text{if}\
d^{\prime}=1.\end{cases}$

Since $z$ is $t$th power in $\overline{\mathbb{F}}_{q}(X,z)$, we have
$t\mid\nu_{\mathfrak{P}}(z)$. If $d^{\prime}>1$, then $t=p$. It follows from
$t\mid d$ and $t\mid n-d$ that $\text{gcd}(n,d)\neq 1$, which is a
contradiction. So we must have $d^{\prime}=1$ and $n-d=p^{m}$, $m>0$. Then
$t\mid p^{m}-1=n-d-1$. Since $t\mid\text{gcd}(n-d-1,d)=\text{gcd}(n-1,d)$,
where $\text{gcd}(n-1,d)$ is a power of 2 (by assumption), we have $t=2$.
Consequently, $p$ is odd.

The equation $Q_{1}(z)=0$ can be written as

$u^{2}=\delta,$

where

$u=z-\gamma,$
$\gamma=-\frac{1}{2}\frac{\displaystyle(a^{q+1}-1)\frac{X^{d}-1}{X-1}+(a^{q+1}X^{d}+1)(X-1)^{p^{m}-1}}{a(X-1)^{p^{m}-1}},$

and

$\delta=\gamma^{2}-a^{q-1}X^{d}.$

By assumption, there exist $\alpha,\beta\in\overline{\mathbb{F}}_{q}(X)$ such
that

$(\alpha u+\beta)^{2}=u+\gamma,$

i.e.,

$\alpha^{2}\delta+\beta^{2}+2\alpha\beta u=u+\gamma.$

So

$\begin{cases}\alpha^{2}\delta+\beta^{2}=\gamma,\cr
2\alpha\beta=1.\end{cases}$

Letting $\tau=\alpha/\beta$, we have

(4.38) $1+\delta\tau^{2}-2\gamma\tau=0$

and

(4.39) $\tau=2\alpha^{2}.$

Equation (4.38) has an explicit solution

$\tau=\frac{1}{\delta}(\gamma\pm a^{(q-1)/2}X^{d/2})=\frac{1}{\gamma\mp
a^{(q-1)/2}X^{d/2}}=\frac{-2a(X-1)^{p^{m}-1}}{h(X)},$

where

$h(X)=(a^{q+1}-1)\frac{X^{d}-1}{X-1}+(a^{q+1}X^{d}+1)(X-1)^{p^{m}-1}+2\epsilon
X^{d/2}(X-1)^{p^{m}-1}$

and $\epsilon=\pm a^{(q+1)/2}$. By (4.39), $h(X)$ is a square in
$\overline{\mathbb{F}}_{q}[X]$, say $h=g^{2}$ for some
$g\in\overline{\mathbb{F}}_{q}[X]$ with $\deg g=(d+p^{m}-1)/2$. Note that

$\displaystyle
h(X)=a^{q+1}\Bigl{(}\frac{X^{d}-1}{X-1}+X^{d}\frac{X^{p^{m}}-1}{X-1}\Bigr{)}+\Bigl{(}\frac{X^{p^{m}}-1}{X-1}-\frac{X^{d}-1}{X-1}\Bigr{)}+2\epsilon
X^{d/2}\frac{X^{p^{m}}-1}{X-1}$
$\displaystyle=a^{q+1}(1+\cdots+X^{p^{m}+d-1})+(X^{d}+\cdots+X^{p^{m}-1})+2\epsilon(X^{d/2}+\cdots+X^{p^{m}+d/2-1}),$

which is self-reciprocal. It follows that $g^{*}=\pm g$, where $g^{*}$ is the
reciprocal polynomial of $g$. Let

$H=(X-1)h=(a^{q+1}-1)(X^{d}-1)+(a^{q+1}X^{d}+1)(X-1)^{p^{m}}+2\epsilon
X^{d/2}(X-1)^{p^{m}}.$

Then

$H^{\prime}=(a^{q+1}-1)dX^{d-1}+a^{q+1}dX^{d-1}(X-1)^{p^{m}}+\epsilon
dX^{d/2-1}(X-1)^{p^{m}}.$

Let

$K=-H+d^{-1}XH^{\prime}=a^{q+1}-X^{p^{m}}+\epsilon X^{d/2}-\epsilon
X^{p^{m}+d/2}.$

The reciprocal of $K$ is

$K^{*}=-\epsilon+\epsilon X^{p^{m}}-X^{d/2}+a^{q+1}X^{p^{m}+d/2}.$

Since $g\mid K$, we have $g=\pm g^{*}\mid K^{*}$. Hence $g$ divides

$\epsilon K+K^{*}=(a^{q+1}-1)(\epsilon+X^{d/2}).$

This is a contradiction since

$\frac{d}{2}<\frac{d+p^{m}-1}{2}=\deg g.$

The proof of Theorem 1.10 is now complete.

###  4.3. Final Remarks

Theorem 1.10 leaves ample room for improvement, by which we mean nonexistence
results of PB under conditions that are weaker than or not covered by (i) –
(iii) in Theorem 1.10. While some improvements may be obtained by fine tuning
the techniques demonstrated in the present paper, breakthroughs may require
new methods or substantially new elements in the current approach.

The cases $d-n=\pm 1$ appear to be special. These are the two cases not
covered by Theorem 1.10 and there are indeed infinite classes of PBs in these
two cases with $e=2$ (Results 1.2 and 1.4). A natural question is this: When
$d-n=\pm 1$ and $e>2$, are there infinite classes of PBs of the form
$f_{q,e,n,d,a}(X)=X^{n}(X^{d(q-1)}+a)$ of $\mathbb{F}_{q^{e}}$?

## Acknowledgements

The research of Vincenzo Palozzi Lavorante was partially supported by the
Italian National Group for Algebraic and Geometric Structures and their
Applications (GNSAGA - INdAM).

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