# Reversed Dickson polynomials, Dembowski-Ostrom polynomials, and planar
functions

Neranga Fernando  Department of Mathematical Sciences, Carnegie Mellon
University, Pittsburgh, PA 15213, USA<EMAIL_ADDRESS>,  Sartaj Ul
Hasan  Department of Mathematics, Indian Institute of Technology Jammu, Jammu
181221, India<EMAIL_ADDRESS>and  Mohit Pal  Department of
Mathematics, Indian Institute of Technology Jammu, Jammu 181221, India
<EMAIL_ADDRESS>
###### Abstract.

We discuss the problem of classifying Dembowski-Ostrom polynomials from the
composition of reversed Dickson polynomials of arbitrary kind and monomials
over finite fields of odd characteristic. Moreover, by using a variant of Weil
bound for the number of points of affine algebraic curves over finite fields
or otherwise, we discuss the planarity of all such Dembowski-Ostrom
polynomials.

###### Key words and phrases:

Finite fields, reversed Dickson polynomials, Dembowski-Ostrom polynomials,
planar functions

###### 2010 Mathematics Subject Classification:

11T55, 05A10, 11T06

##  1\. Introduction

Denote, as usual, by $\mathbb{F}_{q}$ the finite field with $q=p^{e}$
elements, where $p$ is an odd prime number and $e$ is a positive integer, and
by $\mathbb{F}_{q}^{*}$ the multiplicative group of non-zero elements of
$\mathbb{F}_{q}$. For any nonnegative integer $k$, the $k$-th Dickson
polynomial of the first kind $D_{k}(X,a)$ over $\mathbb{F}_{q}$ was introduced
by Dickson [12] in 1897, and is defined as follows

$D_{k}(X,a):=\sum_{i=0}^{\lfloor\frac{k}{2}\rfloor}\frac{k}{k-i}\dbinom{k-i}{i}(-a)^{i}X^{k-2i},$

where $a\in\mathbb{F}_{q}$ is a parameter and $D_{0}(X,a)=2$. More than two
decades later, Schur [28] introduced a variant of the Dickson polynomial of
first kind in 1923, which is now known as Dickson polynomial of second kind.
For any nonnegative integer $k$, the $k$-th Dickson polynomial of second kind
$E_{k}(X,a)$ over $\mathbb{F}_{q}$ is defined as follows

$E_{k}(X,a):=\sum_{i=0}^{\lfloor\frac{k}{2}\rfloor}\dbinom{k-i}{i}(-a)^{i}X^{k-2i},$

where $a\in\mathbb{F}_{q}$ is a parameter and $E_{0}(X,a)=1$. The Dickson
polynomials of first and second kind over $\mathbb{F}_{q}$ were studied
extensively, especially in the context of their permutation behaviour. We
recall that a polynomial $f\in\mathbb{F}_{q}[X]$ is a permutation polynomial
over $\mathbb{F}_{q}$ if the associated mapping $X\mapsto f(X)$ is a bijection
from $\mathbb{F}_{q}$ to $\mathbb{F}_{q}$. For a non-zero element $a$ in
$\mathbb{F}_{q}$, Nöbauer [21] proved that the Dickson polynomial of first
kind $D_{k}(X,a)$ permutes the elements of $\mathbb{F}_{q}$ if and only if
$(k,q^{2}-1)=1.$ However, except for few cases, the permutation behaviour of
Dickson polynomial of second kind $E_{k}(X,a)$ remains unresolved. Further,
for interested readers, it may be worth mentioning that the results concerning
factorization of Dickson polynomials have been dealt in [1, 3, 8, 14, 25, 26],
and an existence result about permutation polynomials has recently been
discussed in [6]. One may also refer to the monograph [18] for more on Dickson
polynomials.

The notion of $k$-th reversed Dickson polynomial (RDP) of first kind was
introduced by Hou, Mullen, Sellers and Yucas [17] by simply reversing the
roles of the variable $X$ and the parameter $a$ in the $k$-th Dickson
polynomial of the first kind $D_{k}(X,a)$. Moreover, the authors also showed
that the reversed Dickson polynomials of first kind are closely related to
what are known as almost perfect nonlinear (APN) functions. The permutation
behaviour of RDP is also an interesting area of research. Hou and Ly [16] gave
the necessary conditions for RDP of first kind to be a permutation polynomial
over $\mathbb{F}_{q}$.

For any nonnegative integers $k$ and $m$, the notion of $k$-th Dickson
polynomial of $(m+1)$-th kind, denoted as $D_{k,m}(X,a)$, was introduced by
Wang and Yucas [29], and is defined as follows

(1.1)
$D_{k,m}(X,a):=\sum_{i=0}^{\lfloor\frac{k}{2}\rfloor}\frac{k-mi}{k-i}\dbinom{k-i}{i}(-a)^{i}X^{k-2i},$

where $0\leq m\leq p-1$, $a\in\mathbb{F}_{q}$ and $D_{0,m}(X,a)=2-m$. The
$k$-th RDP of $(m+1)$-th kind is also defined in a similar way by just
reversing the role of the variable $X$ and the parameter $a$ in (1.1). More
precisely, for any nonnegative integers $k$ and $m$, the $k$-th RDP of
$(m+1)$-th kind $D_{k,m}(a,X)$ is defined as follows

(1.2)
$D_{k,m}(a,X):=\sum_{i=0}^{\lfloor\frac{k}{2}\rfloor}\frac{k-mi}{k-i}\dbinom{k-i}{i}(-X)^{i}a^{k-2i},$

where $0\leq m\leq p-1$, $a\in\mathbb{F}_{q}$ and $D_{0,m}(a,X)=2-m$. The
$k$-th RDP of $(m+1)$-th kind also satisfies the following recurrence relation

(1.3) $D_{k,m}(a,X)=mE_{k}(a,X)-(m-1)D_{k}(a,X).$

It may be noted that the permutation behaviour of RDP of $(m+1)$-th kind has
been studied by Fernando [13].

A function $f:\mathbb{F}_{q}\rightarrow\mathbb{F}_{q}$ is called planar if the
mapping $X\mapsto f(X+\epsilon)-f(X)-f(\epsilon)$ induces a bijection from
$\mathbb{F}_{q}$ to $\mathbb{F}_{q}$ under evaluation for each
$\epsilon\in\mathbb{F}_{q}^{*}.$ Since any function from a finite field
$\mathbb{F}_{q}$ to itself can be represented by a polynomial of degree
$\leq(q-1)$, one may simply consider the planarity of polynomials over finite
fields. It is clear from the definition of the planar function itself that
there is no planar function in the even characteristic as $X$ and $X+\epsilon$
have the same image under the mapping $X\mapsto
f(X+\epsilon)-f(X)-f(\epsilon).$ Recently, Zhou [33] defined the planar
functions in even characteristic by tweaking the definition of planar
functions in odd characteristic. Moreover, Schmidt and Zhou [27] showed that
the planar functions in the even characteristic have the similar properties as
the planar functions in odd characteristic. The planar functions are very
important due to their wide range of applications. For example, the planar
functions are used to construct finite projective planes [11], relative
difference sets [15], error-correcting codes [5], and Paley type group schemes
[7] etc. It may also be interesting to note that a generalization of planar
functions has been recently proposed by Bartoli and Timpanella [2].

A Dembowski-Ostrom (DO) polynomial over finite field $\mathbb{F}_{q}$ is a
polynomial that admits the following shape

$\sum_{i,j}a_{ij}X^{p^{i}+p^{j}},$

where $a_{ij}\in\mathbb{F}_{q}$. The DO polynomials have been used in
designing a public key cryptosystem known as HFE [22]. Note that the DO
polynomials provide a very rich source of planar functions. It was conjectured
by Rónyai and SzÖnyi [24] (see also [20, Conjecture 9.5.19]) that all the
planar functions are of “DO type”. This conjecture is still open except in the
case of characteristic $3$ for which a counter example was given by Coulter
and Matthews [9]. In 2010, Coulter and Matthews [10] classified DO polynomials
from Dickson polynomials of first and second kind and they also discussed the
planarity of such DO polynomials.

In 2016, Zhang, Wu and Liu [32] classified DO polynomials from RDP of first
kind in even characteristic case and they also characterized APN functions
among all such DO polynomials. In this paper, we shall extend the results of
Zhang, Wu and Liu [32] to the odd characteristic case. In fact, we give a
complete classification of DO polynomials arising from the composition of RDP
of $(m+1)$-th kind and the monomial $X^{d}$, where $d$ is a positive integer,
in odd characteristic, and we further characterize the planar functions among
these DO polynomials. The motivation behind considering this composition
actually stems from the known fact that the exceptional planar polynomials
$X^{10}\pm X^{6}-X^{2}$ are essentially the composition of the Dickson
polynomials $D_{5}(X,\pm 1)$ and the monomial $X^{2}$. The DO polynomials do
not have any constant term. We shall, therefore, consider the polynomials
$\mathfrak{\widehat{D}}_{k,m}:=D_{k,m}(a,X^{d})-D_{k,m}(a,0)$ for the purpose
of classifying DO polynomials. Notice that $\mathfrak{\widehat{D}}_{k,m}$ is
given by

$\mathfrak{\widehat{D}}_{k,m}=\sum_{i=1}^{\lfloor\frac{k}{2}\rfloor}\frac{k-mi}{k-i}\dbinom{k-i}{i}(-X^{d})^{i}a^{k-2i}.$

For the sake of simplicity, we shall denote $\mathfrak{\widehat{D}}_{k,0}$,
$\mathfrak{\widehat{D}}_{k,1}$, $\mathfrak{\widehat{D}}_{k,2}$,
$\mathfrak{\widehat{D}}_{k,3}$, and $\mathfrak{\widehat{D}}_{k,4}$ by
$\mathfrak{\widehat{D}}_{k}$, $\mathfrak{\widehat{E}}_{k}$,
$\mathfrak{\widehat{F}}_{k}$, $\mathfrak{\widehat{G}}_{k}$ and
$\mathfrak{\widehat{H}}_{k}$, respectively. The paper has been organized as
follows. In Section 2, we state some lemmas that will be used in the
subsequent sections. In Sections 3, 4, 5 and 6, we classify DO polynomials
from $\mathfrak{\widehat{D}}_{k}$, $\mathfrak{\widehat{E}}_{k}$,
$\mathfrak{\widehat{G}}_{k}$ and $\mathfrak{\widehat{H}}_{k}$, respectively.
The case $m\geq 5$ has been considered in Section 7. In Section 8, we consider
the planarity of the DO polynomials obtained in the previous sections. The
complete list of DO polynomials derived from reversed Dickson polynomials is
given in Appendix A.

Throughout the paper, we always assume that $p$ is an odd prime, $d$ is a
positive integer, and $i,j,k,\ell,m,n,s,t,\alpha,\beta,\gamma,\delta$ are
nonnegative integers unless specified otherwise. The greatest common divisor
of positive integers $a$ and $b$ is denoted by $(a,b)$.

##  2\. Some useful lemmas

As alluded earlier, we shall first classify DO polynomials derived from the
composition of RDP of $(m+1)$-th kind and the monomial $X^{d}$, where $d$ is a
positive integer. Since DO polynomials are closed under the composition with
the monomial $X^{p}$, it would be sufficient to consider the cases when
$(d,p)=1$. One may also note that the monomial $X^{rd}$, where $r$ is positive
integer, is a DO monomial if and only if $rd=p^{\beta}(p^{\alpha}+1)$ for some
nonnegative integers $\alpha$ and $\beta$. Here, $\beta$ is the highest
exponent of $p$ such that $p^{\beta}\mid r$. It is obvious that whenever
$(r,p)=1$, we must have $\beta=0$. In what follows, we shall invoke these
assumptions and conventions as and when required.

We now present some lemmas which will be useful in the sequel.

######  Lemma 2.1.

Let $d$ be a positive integer and $p>3$ be a prime such that $(d,p)=1$. Assume
that the coefficients of $X^{d}$ and $X^{2d}$ in the polynomial
$\mathfrak{\widehat{D}}_{k,m}$ are non-zero. Then the polynomial
$\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial.

###### Proof.

Assume that $p>3$ and the coefficients of $X^{d}$ and $X^{2d}$ in the
polynomial $\mathfrak{\widehat{D}}_{k,m}$ are non-zero. Therefore, if
$\mathfrak{\widehat{D}}_{k,m}$ is a DO polynomial then $d=p^{\alpha}+1$ and
$2d=p^{\beta}+1$. Thus, we have $2p^{\alpha}+1=p^{\beta}$, which is true if
and only if $\alpha=0$, $\beta=1$ and $p=3$. This is a contradiction to our
assumption that $p>3$, hence $\mathfrak{\widehat{D}}_{k,m}$ is not a DO
polynomial. ∎

######  Lemma 2.2.

Let $d$ be a positive integer and $p>5$ be a prime such that $(d,p)=1$. Assume
that the coefficients of $X^{d}$ and $X^{3d}$ in the polynomial
$\mathfrak{\widehat{D}}_{k,m}$ are non-zero. Then the polynomial
$\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial.

###### Proof.

Assume that $p>5$ and the coefficients of $X^{d}$ and $X^{3d}$ in the
polynomial $\mathfrak{\widehat{D}}_{k,m}$ are non-zero. If
$\mathfrak{\widehat{D}}_{k,m}$ is a DO polynomial, then $d=p^{\alpha}+1$ and
$3d=p^{\beta}+1$. Thus, we have $3p^{\alpha}+2=p^{\beta}$, which forces
$\alpha=0$, $\beta=1$ and $p=5$. This contradicts the fact that $p>5$ and
hence $\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial. ∎

######  Lemma 2.3.

Let $d$ be a positive integer and $p>3$ be an odd prime such that $(d,p)=1$.
Assume that the coefficients of $X^{3d}$ and $X^{4d}$ in the polynomial
$\mathfrak{\widehat{D}}_{k,m}$ are non-zero. Then the polynomial
$\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial.

###### Proof.

Assume that $p>3$ and the coefficients of $X^{3d}$ and $X^{4d}$ in the
polynomial $\mathfrak{\widehat{D}}_{k,m}$ are non-zero. If
$\mathfrak{\widehat{D}}_{k,m}$ is a DO polynomial, then $3d=p^{\alpha}+1$ and
$4d=p^{\beta}+1$. Solving these equations, we have $4\cdot p^{\alpha}+1=3\cdot
p^{\beta}$, which forces $\alpha=0$ and $3p^{\beta}=5$, a contradiction. Thus,
$\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial. ∎

######  Lemma 2.4.

Let $p=3$ and $d$ be a positive integer such that $(d,3)=1$. Assume that the
coefficients of $X^{d}$ and $X^{4d}$ in the polynomial
$\mathfrak{\widehat{D}}_{k,m}$ are non-zero. Then the polynomial
$\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial.

###### Proof.

Assume that $p=3$ and the coefficients of $X^{d}$ and $X^{4d}$ in the
polynomial $\mathfrak{\widehat{D}}_{k,m}$ are non-zero. If
$\mathfrak{\widehat{D}}_{k,m}$ is a DO polynomial then $d=3^{\alpha}+1$ and
$4d=3^{\beta}+1$. Now solving these equations, we have $4\cdot
3^{\alpha}+3=3^{\beta}$, which forces $\alpha=1$ and $3^{\beta}=15$, a
contradiction. Thus, $\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial. ∎

######  Lemma 2.5.

[19, Proposition 6.39] For any nonnegative integer $\omega$ and any prime $p$,
$E_{p}(\omega!)$ the largest exponent of $p$ that divides $\omega!$ is given
by

$E_{p}(\omega!)=\sum_{i=1}^{\infty}\left\lfloor\frac{\omega}{p^{i}}\right\rfloor=\frac{\omega-\omega_{s}}{p-1},$

where $\omega_{s}$ is the sum of the digits in the representation of $\omega$
to the base $p$.

##  3\. DO polynomials from RDP of the first kind

Before we begin the classification of DO polynomials from RDP of the first
kind, we shall slightly deviate and prove the following proposition that
readily gives DO polynomials arising from RDP of $(m+1)$-th kind when the
parameter $a$ is zero.

######  Proposition 3.1.

The polynomial $D_{k,m}(0,X^{d})$ is DO if and only if $m\not\equiv 2\pmod{p}$
and $kd$ is of the form $2p^{j}(p^{i}+1)$, where $i,j\geq 0$.

###### Proof.

We know that

$\displaystyle D_{k,m}(0,X^{d})=\begin{cases}0&\mbox{if }k~{}\mbox{is odd};\\\
(2-m)(-X^{d})^{\frac{k}{2}}&\mbox{if }k~{}\mbox{is even}.\end{cases}$

If $k$ is even and $m\not\equiv 2\pmod{p}$, then $D_{k,m}(0,X^{d})$ is a DO
polynomial if and only if $\displaystyle\frac{kd}{2}=p^{j}(p^{i}+1)$, i.e.,
$kd=2p^{j}(p^{i}+1).$ ∎

In view of the Proposition 3.1, we shall assume that $a$ is non-zero for rest
of what follows across the sections.

We now consider RDP of first kind. Recall that $D_{k,0}(a,X^{d})-D_{k,0}(a,0)$
is denoted by $\mathfrak{\widehat{D}}_{k}$, where

$\mathfrak{\widehat{D}}_{k}=\sum_{i=1}^{\lfloor\frac{k}{2}\rfloor}\frac{k}{k-i}\dbinom{k-i}{i}(-X^{d})^{i}a^{k-2i}.$

Since $\mathfrak{\widehat{D}}_{kp}=\mathfrak{\widehat{D}}_{k}^{p}$, it would
be sufficient to consider the cases when $(k,p)=1$. The following theorem
gives a complete classification of DO polynomials from the polynomial
$\mathfrak{\widehat{D}}_{k}$.

######  Theorem 3.2.

Let $q$ be a power of odd prime $p$ and $a\in\mathbb{F}_{q}^{*}$. The
polynomial $\mathfrak{\widehat{D}}_{k}$ is a DO polynomial over
$\mathbb{F}_{q}$ if and only if one of the following holds.

  1. (i)

$p=3$, $d=p^{n}(p^{\alpha}+1)$, $k=2p^{\ell}$, where $\ell,n,\alpha\geq 0$.

  2. (ii)

$p=3$, $d=2p^{n}$, $k=4p^{\ell},5p^{\ell},7p^{\ell}$, where $\ell,n\geq 0$.

  3. (iii)

$p>3$, $d=p^{n}(p^{\alpha}+1)$, $k=2p^{\ell},3p^{\ell}$, where
$\ell,n,\alpha\geq 0$.

###### Proof.

The sufficiency of the theorem is straightforward. It only remains to show the
necessity.

Since $(k,p)=1$, therefore, the first term $-ka^{k-2}X^{d}$ in
$\mathfrak{\widehat{D}}_{k}$ will always exists. Thus, if
$\mathfrak{\widehat{D}}_{k}$ is a DO polynomial then $d=p^{j}(p^{i}+1)$. Since
$(d,p)=1$, we have $j=0$. Therefore, we shall always take $d=p^{i}+1$. We now
consider two cases, namely, $k$ odd and $k$ even.

Case 1 $k$ is odd.

In this case,

(3.1)
$\mathfrak{\widehat{D}}_{k}=-kX^{d}a^{k-2}+\frac{(k-3)k}{2!}X^{2d}a^{k-4}-\frac{(k-4)(k-5)k}{3!}X^{3d}a^{k-6}+\cdots\\\
+(-1)^{\frac{k-3}{2}}\,\frac{(k-1)k(k+1)}{24}\,X^{\frac{d(k-3)}{2}}a^{3}+(-1)^{\frac{k-1}{2}}\,k\,X^{\frac{d(k-1)}{2}}a.$

Now we consider two subcases, $k\not\equiv 3\pmod{p}$ and $k\equiv 3\pmod{p}$.

Subcase 1.1 $k\not\equiv 3\pmod{p}$.

In this subcase, the coefficient of the second term in (3.1) is non-zero.
Therefore, if $\mathfrak{\widehat{D}}_{k}$ is a DO polynomial then
$2d=p^{\beta}(p^{\alpha}+1)$ and $d=p^{i}+1$. Since $p$ is odd and $(d,p)=1$,
therefore, $\beta=0$. Hence, the first equation reduces to $2d=p^{\alpha}+1$.
On solving these equations, we obtain $2p^{i}+1=p^{\alpha}$, which is true if
and only if $p=3$, $\alpha=1$, $i=0$ and $d=2$. Therefore, in this subcase, we
shall always assume that $p=3$ and $d=2$. For $k=5$ and $k=7$, the polynomials
$\mathfrak{\widehat{D}}_{5}=a^{3}X^{2}+2aX^{4}$ and
$\mathfrak{\widehat{D}}_{7}=2a^{5}X^{2}+2a^{3}X^{4}+2aX^{6},$ are clearly DO
polynomials. Now we claim that when $p=3$ and $k>7$ is odd,
$\mathfrak{\widehat{D}}_{k}$ is never a DO polynomial. Since $(k,3)=1$, we
have only two cases to consider, namely, $k\equiv 2\pmod{3}$ and $k\equiv
1\pmod{3}$.

In the case of $k\equiv 2\pmod{3}$, consider the term before the last term in
(3.1) given by

(3.2) $(-1)^{\frac{k-3}{2}}\,\frac{(k-1)k(k+1)}{24}\,X^{k-3}a^{3}.$

Since $(k-1)k(k+1)=6\ell_{1}$ for some integer $\ell_{1}$, we have

$\frac{(k-1)k(k+1)}{24}=\frac{\ell_{1}}{4}\equiv\ell_{1}\pmod{3}.$

If $\ell_{1}\not\equiv 0\pmod{3}$, then we claim that $(k-3)$ can not be
written as $3^{i}+3^{j}$ for some nonnegative integers $i$ and $j$. On the
contrary assume that $k-3=3^{i}+3^{j}$, which implies $k-2=3^{i}+3^{j}+1$.
Since $k\equiv 2\pmod{3}$, $k-2=3^{i}+3^{j}+1$ if and only if $i=j=0$. But,
$i=j=0$ implies $k=5$, which is a contradiction to our assumption that $k>7$.
Therefore $\mathfrak{\widehat{D}}_{k}$ is not a DO polynomial in this case.

Now assume that $\ell_{1}\equiv 0\pmod{3}$. In this case, we shall show that
the fourth term always exists. Note that the fourth term contains the monomial
$X^{8}$ whose exponent can not be written as $3^{i}+3^{j}$ for some
nonnegative integers $i$ and $j$. The coefficient of the fourth term is given
by

(3.3)
$\frac{k}{k-4}\,\binom{k-4}{4}a^{k-8}=\frac{k(k-5)(k-6)(k-7)}{24}a^{k-8}.$

Since $(k,3)=1$, therefore $3\nmid k$ and $3\nmid(k-6)$. Now we show that
$(k-5)(k-7)$ is a multiple of 24. Recall that we are in the case $k\equiv
2\pmod{3}$, where $k$ is odd and greater than $7$. Since $k$ is odd therefore
$k=2\ell_{2}+1$, where $\ell_{2}$ is a positive integer. Thus
$(k-5)(k-7)=4(\ell_{2}-2)(\ell_{2}-3)$. Also since $k\equiv 2\pmod{3}$,
$\ell_{2}\equiv 2\pmod{3}$. Let $\ell_{2}=3n_{0}+2$ for some positive integer
$n_{0}$. Then this implies $(k-5)(k-7)=12n_{0}(3n_{0}-1)$. Notice that $n_{0}$
and $(3n_{0}-1)$ have different parity. Therefore $n_{0}(3n_{0}-1)$ is even
and hence $(k-5)(k-7)$ is a multiple of 24. Let $(k-5)(k-7)=24b$, where $b$ is
an integer. Then from equation (3.3), we have

$\frac{k(k-5)(k-6)(k-7)}{24}=k(k-6)b.$

Now we show that $3\nmid\,b$. On the contrary assume that $3\mid b$. Then we
have, $(k-5)(k-7)=72e$ for some integer $e$. Since $k\equiv 2\pmod{3}$, write
$k=3e_{1}-1$ for some integer $e_{1}$. Recall that we are in the case, where
the term before the last term as given in (3.2) is zero, i.e. $\ell_{1}\equiv
0\pmod{3}$. So

$\frac{(k-1)k(k+1)}{24}=\frac{(3e_{1}-2)(3e_{1}-1)(3e_{1})}{24}\equiv
e_{1}\equiv 0\pmod{3}.$

Let $e_{1}=3n_{1}$ for some integer $n_{1}$. Then
$k=3e_{1}-1=9n_{1}-1\equiv-1\pmod{9}$. From $(k-5)(k-7)=72e$ and
$k\equiv-1\pmod{9}$, we have $3\equiv 0\pmod{9}$, which is a contradiction.
Therefore our assumption that $3\mid b$ is wrong and hence the coefficient of
$X^{8}$ is nonzero. Therefore, when $\ell_{1}\equiv 0\pmod{3}$,
$\mathfrak{\widehat{D}}_{k}$ is not a DO polynomial.

In the case of $k\equiv 1\pmod{3}$, we first look at the fourth term. Recall
that the fourth term contains the monomial $X^{8}$ whose exponent can not be
written as $3^{i}+3^{j}$ for some nonnegative integers $i$ and $j$. In the
coefficient of the fourth term as given in (3.3), clearly
$(k-5)(k-6)(k-7)=6\ell_{3}$ for some integer $\ell_{3}$. Therefore

$\frac{k(k-5)(k-6)(k-7)}{24}=\frac{k\ell_{3}}{4}\equiv k\ell_{3}\pmod{3}.$

If $\ell_{3}\not\equiv 0\pmod{3}$, then clearly $\mathfrak{\widehat{D}}_{k}$
is not a DO polynomial. In the case of $\ell_{3}\equiv 0\pmod{3}$, we claim
that the coefficient of the 7th term is nonzero. The 7th term is given by

(3.4) $\frac{k(k-8)(k-9)(k-10)(k-11)(k-12)(k-13)}{7!}a^{k-14}(-X^{2})^{7}.$

It is clear that the exponent of the monomial $X^{14}$ can not be written as
$3^{i}+3^{j}$ for some nonnegative integers $i$ and $j$. We now claim that
$6\mid(k-13)$ and $3\mid(k-10)$. By division algorithm, we have
$k-13=6q_{1}+r_{1}$, where $0\leq r_{1}\leq 5$. Since $k\equiv 1\pmod{3}$,
$r_{1}\equiv 0\pmod{3}$ which implies $r_{1}=0\,\,\textnormal{or}\,\,3$. If
$r_{1}=3$, then $k-13=3(2q_{1}+1)$ which is a contradiction since the left
hand side is even and the right hand side is odd. Therefore $r_{1}=0$, which
implies that $6\mid(k-13)$. Similarly, by division algorithm, we have
$k-10=3q_{2}+r_{2}$, where $0\leq r_{2}\leq 2$. Since $k\equiv 1\pmod{3}$,
$r_{2}\equiv 0\pmod{3}$ which implies $r_{2}=0$, thus $3\mid(k-10)$. Now we
claim that $q_{1}$ and $q_{2}$ are not divisible by 3. On the contrary, assume
that $q_{1}=3n_{2}$ for some integer $n_{2}$, then $k-13=18n_{2}$, which
implies $k=13+18n_{2}$. Recall that we are in the case where coefficient of
the fourth term as given in (3.3) is zero, i.e., $\ell_{3}\equiv 0\pmod{3}$.
Straightforward computation yields that

$(k-5)(k-6)(k-7)=(18n_{2}+8)(18n_{2}+7)(18n_{2}+6)=6\cdot c,$

where $c$ is not divisible by 3. This implies that the coefficient of the
fourth term is nonzero, which is a contradiction. Similarly, on the contrary
assume that $q_{2}=3n_{3}$ for some integer $n_{3}$. Then $k-10=9n_{3}$ which
implies $k\equiv 1\pmod{9}$. Let $k=9n_{4}+1$ for some integer $n_{4}$, then
the coefficient of the fourth term as given in (3.3) is

$\begin{split}\frac{k(k-5)(k-6)(k-7)}{24}&=\frac{(9n_{4}+1)(9n_{4}-4)(9n_{4}-5)(9n_{4}-6)}{24}\cr&=\frac{(9n_{4}+1)(9n_{4}-4)(9n_{4}-5)(3n_{4}-2)}{8}\cr&\not\equiv
0\pmod{3},\end{split}$

which is a contradiction. Therefore, $q_{1}$ and $q_{2}$ are not divisible by
$3$. Now let us go back to the coefficient of the 7th term as given in (3.4)

$\begin{split}&\frac{k(k-8)(k-9)(k-10)(k-11)(k-12)(k-13)}{7!}\cr&=\frac{k(k-8)(k-9)(k-11)(k-12)\,q_{1}q_{2}}{7\cdot
5\cdot 4\cdot 2\cdot 1}\cr\end{split}$

Since we are in the case $k\equiv 1\pmod{3}$, therefore $3\nmid\,k$,
$3\nmid\,(k-9)$ and $3\nmid\,(k-12)$, otherwise $0\equiv 1\pmod{3}$. If
$3\mid(k-8)$, then $k-8=3e_{2}$ for some integer $e_{2}$. This implies
$2\equiv 0\pmod{3}$, which is a contradiction. If $3\mid(k-11)$, then
$k-11=3e_{3}$ for some integer $e_{3}$. This implies $2\equiv 0\pmod{3}$,
which is a contradiction. Therefore, the coefficient of the 7th term is
nonzero. Hence in the case of $\ell_{3}\equiv 0\pmod{3}$,
$\mathfrak{\widehat{D}}_{k}$ is not a DO polynomial.

Subcase 1.2 $k\equiv 3\pmod{p}$.

In this subcase, notice that if $p=3$ then $k\equiv 0\pmod{3}$, which is a
contradiction as $(k,p)=1$. Therefore we shall assume that $p>3$. For $k=3$,
the polynomial $\mathfrak{\widehat{D}}_{3}=-3aX^{d}$ is a DO polynomial if and
only if $d=p^{i}+1$. For $k>3$, consider the third term in (3.1), which
contains the monomial $X^{3d}$. Since $k\equiv 3\pmod{p}$ therefore
$k\not\equiv 4,5\pmod{p}$. Hence the coefficient of the third term is nonzero.
Thus, if $\mathfrak{\widehat{D}}_{k}$ is a DO polynomial then $d=p^{i}+1$ and
$3d=p^{j}+1$. Solving these equations, we have $3p^{i}+2=p^{j}$, which is true
if and only if $i=0$, $j=1$, $p=5$ and $d=2$. Notice that the coefficient of
last term in (3.1), which contains the monomial $X^{k-1}$ is nonzero. Thus, if
$\mathfrak{\widehat{D}}_{k}$ is a DO polynomial then $k-1=5^{j}(5^{i}+1)$.
Since $k\equiv 3\pmod{5}$ therefore $k\not\equiv 1\pmod{5}$ and hence $j=0$.
Also notice that if $i=0$ then $k=3$, which is a contradiction as $k>3$.
Therefore $k-1=5^{i}+1$ which implies that $k\equiv 2\pmod{5}$, a
contradiction as $k\equiv 3\pmod{5}$. Therefore for $k>3$,
$\mathfrak{\widehat{D}}_{k}$ is never a DO polynomial.

Case 2 $k$ is even.

In this case, we have

(3.5)
$\begin{split}\mathfrak{\widehat{D}}_{k}=-ka^{k-2}X^{d}+\frac{k(k-3)}{2}a^{k-4}X^{2d}-\frac{k(k-4)(k-5)}{6}a^{k-6}X^{3d}\\\
+\cdots+(-1)^{\frac{k}{2}-1}\,\frac{k^{2}}{4}a^{2}X^{d(\frac{k}{2}-1)}+(-1)^{\frac{k}{2}}\cdot
2\cdot X^{\frac{dk}{2}}.\end{split}$

In the case of $k=2$, the polynomials $\mathfrak{\widehat{D}}_{2}=-2X^{d}$ is
a DO polynomial if $d=p^{\alpha}+1$. For $k\geq 4$, we consider two subcases,
$k\not\equiv 3\pmod{p}$ and $k\equiv 3\pmod{p}$.

Subcase 2.1 $k\not\equiv 3\pmod{p}$.

In this subcase, the coefficient of the second term in (3.5) which contains
the monomial $X^{2d}$ is nonzero. Thus, if $\mathfrak{\widehat{D}}_{k}$ is a
DO polynomial then $2d=p^{\beta}(p^{\alpha}+1)$ and $d=p^{i}+1$. Since $p$ is
odd and $(d,p)=1$, we have $\beta=0$. Thus second equation reduces to
$2d=p^{\alpha}+1$. Solving these two equations, we have $2p^{i}+1=p^{\alpha}$.
Which is true if and only if $p=3,i=0,\alpha=1$ and $d=2$. Therefore in what
follows, we shall take $p=3$ and $d=2$. In the case of $k=4$, the polynomial
$\mathfrak{\widehat{D}}_{4}=2a^{2}X^{2}+2X^{4}$ is clearly DO polynomial.

Now for $k>4$, even and $k\not\equiv 3\pmod{3}$, we claim that
$\mathfrak{\widehat{D}}_{k}$ is not a DO polynomial. Consider the fourth term
which contains the monomial $X^{8}$. It is clear that $8$ can not be written
as $3^{i}+3^{j}$ for some nonnegative integers $i$ and $j$. In the coefficient
of the fourth term (3.3), $(k-5)(k-6)(k-7)=6\ell_{4}$ for some integer
$\ell_{4}$. Hence

$\frac{k(k-5)(k-6)(k-7)}{24}=\frac{k\ell_{4}}{4}\equiv k\ell_{4}\pmod{3}.$

If $\ell_{4}\not\equiv 0\pmod{3}$, then $\mathfrak{\widehat{D}}_{k}$ is not a
DO polynomial. Now consider the case when $\ell_{4}\equiv 0\pmod{3}$. Note
that the coefficient of the last term in (3.5), which contains the monomial
$X^{k}$ is always nonzero. Thus, if $\mathfrak{\widehat{D}}_{k}$ is a DO
polynomial then $k=3^{i}+1.$ Clearly $i\neq 0$, otherwise $k=2$, a
contradiction. If $i>0$, then $k\equiv 1\pmod{3}$. Now consider the term
before the the last term in (3.5), which contains the monomial $X^{k-2}$.
Clearly, the coefficient is nonzero as $(k,3)=1$. If
$\mathfrak{\widehat{D}}_{k}$ is a DO polynomial and $k\equiv 1\pmod{3}$, then
$k-2=3^{j}+1$. If $i=0$ then $k=4$, which is a contradiction since $k>4$. If
$i>0$, then $k=3^{i}+3$. This contradicts the assumption that $(k,3)=1$. Thus
$\mathfrak{\widehat{D}}_{k}$ is not a DO polynomial in this case.

Subcase 2.2 $k\equiv 3\pmod{p}$.

In this subcase, if $p=3$ then $k\equiv 0\pmod{3}$ which is a contradiction as
$(k,p)=1$. Therefore we shall always consider $p>3$. Notice that the
coefficient of the last term in (3.5), which contains the monomial
$\displaystyle X^{\frac{kd}{2}}$ is nonzero. Thus, if
$\mathfrak{\widehat{D}}_{k}$ is a DO polynomial then
$\displaystyle\frac{kd}{2}=p^{\beta}(p^{\alpha}+1)$ and $d=p^{i}+1$. Since
$(k,p)=1$ and $(d,p)=1$, therefore $\beta=0$. Hence the first equation reduces
to $kd=2p^{\alpha}+2$. Solving these equations, we get
$kp^{i}+k=2p^{\alpha}+2$. If $i=0$, then $k=p^{\alpha}+1$, which implies
$k\equiv 2\pmod{p}$ or $k\equiv 1\pmod{p}$ depending on whether $\alpha=0$ or
$\alpha>0$, respectively, a contradiction. If $i>0$, then $\alpha>0$ otherwise
$k(p^{i}+1)=4$ which is a contradiction as $p>3$. Therefore $k\equiv
2\pmod{p}$, a contradiction. This completes the proof. ∎

##  4\. DO polynomials from RDP of the second kind

Recall that $D_{k,1}(a,X^{d})-D_{k,1}(a,0)$ is denoted by
$\mathfrak{\widehat{E}}_{k}$, where

(4.1)
$\begin{split}\mathfrak{\widehat{E}}_{k}=(1-k)a^{k-2}X^{d}+\frac{(k-2)(k-3)}{2!}a^{k-4}X^{2d}\\\
-\frac{(k-3)(k-4)(k-5)}{3!}a^{k-6}X^{3d}+\cdots.\end{split}$

The following theorem gives the necessary and sufficient conditions for the
RDPs of second kind to be DO polynomials.

######  Theorem 4.1.

Let $q$ be a power of odd prime $p$ and $a\in\mathbb{F}_{q}^{*}$. The
polynomial $\mathfrak{\widehat{E}}_{k}$ is a DO polynomial over
$\mathbb{F}_{q}$ if and only if one of the following holds.

  1. (i)

$k=2,3$ and $d=p^{n}(p^{\alpha}+1)$, where $\alpha,~{}n\geq 0.$

  2. (ii)

$k=4$, $p=3$ and $d=p^{n}(p^{\alpha}+1)/2$, where $\alpha,~{}n\geq 0$.

  3. (iii)

$k=5,6$, $p=3$ and $d=p^{n}(p^{\alpha}+1)$, where $\alpha,~{}n\geq 0$.

  4. (iv)

$k=7$, $p=3,5$ and $d=2p^{n}$, where $n\geq 0$.

  5. (v)

$k=10,13,19$, $p=3$ and $d=2p^{n}$, where $n\geq 0$.

  6. (vi)

$k=15$, $p=3$ and $d=4p^{n}$, where $n\geq 0$.

###### Proof.

The sufficient part of the theorem is straightforward, therefore, we only
prove the necessary part.

If the polynomials $\mathfrak{\widehat{E}}_{2}=-X^{d}$ and
$\mathfrak{\widehat{E}}_{3}=-2aX^{d}$ are DO polynomial then $d$ is of the
form $p^{\alpha}+1$. For the polynomial
$\mathfrak{\widehat{E}}_{4}=-3a^{2}X^{d}+X^{2d}$, we consider two cases,
namely, $p=3$ and $p>3$. For the case $p=3$, if
$\mathfrak{\widehat{E}}_{4}=X^{2d}$ is a DO polynomial then $d$ is of the form
$(p^{\alpha}+1)/2$. If $p>3$, then coefficients of $X^{2d}$ and $X^{d}$ are
non-zero, therefore, $\mathfrak{\widehat{E}}_{4}$ is not a DO polynomial by
Lemma 2.1. For the polynomial
$\mathfrak{\widehat{E}}_{5}=-4a^{3}X^{d}+3aX^{2d}$, again we consider two
cases. For $p=3$, if $\mathfrak{\widehat{E}}_{5}=2a^{3}X^{d}$ is a DO
polynomial then $d$ is of the form $p^{\alpha}+1$. If $p>3$, then the
coefficients of $X^{d}$ and $X^{2d}$ are non-zero, therefore,
$\mathfrak{\widehat{E}}_{5}$ is not a DO polynomial by Lemma 2.1. For
$\mathfrak{\widehat{E}}_{6}=-5a^{4}X^{d}+6a^{2}X^{2d}-X^{3d}$, we consider two
cases. In the case $p=3$, $\mathfrak{\widehat{E}}_{6}=a^{4}X^{d}+2X^{3d}$ and
if it is a DO polynomial then $d=3^{\alpha}+1$ and $3d=3^{t}(3^{\beta}+1)$.
Since $3^{t}\mid 3$, $t=1$. Thus, the second equation reduces to
$d=3^{\beta}+1$. Therefore, if $\mathfrak{\widehat{E}}_{6}$ is a DO polynomial
then $d$ is of the form $p^{\alpha}+1$. For $p>3$, the coefficients of
$X^{2d}$ and $X^{3d}$ in $\mathfrak{\widehat{E}}_{6}$ are always non-zero.
Thus, if $\mathfrak{\widehat{E}}_{6}$ is a DO polynomial then
$2d=p^{\alpha}+1$ and $3d=p^{\beta}+1$. Solving this, we get
$3p^{\alpha}+1=2p^{\beta}$, which forces $\alpha=0$, $p^{\beta}=2$, a
contradiction. Therefore, for $p>3$, $\mathfrak{\widehat{E}}_{6}$ is not a DO
polynomial. For the polynomial
$\mathfrak{\widehat{E}}_{7}=-6a^{5}X^{d}+10a^{3}X^{2d}-4aX^{3d}$, we consider
three cases, namely $p=3$, $p=5$ and $p>5$. For $p=3$,
$\mathfrak{\widehat{E}}_{7}=a^{3}X^{2d}+2aX^{3d}$ and if it is a DO polynomial
then $2d=3^{\alpha}+1$ and $3d=3^{t}(3^{\beta}+1)$. Since $3^{t}\mid 3$,
$t=1$. Thus, the second equation reduces to $d=3^{\beta}+1$. Solving these
equations, we have $2\cdot 3^{\beta}+1=3^{\alpha}$, which forces $\beta=0$,
$\alpha=1$ and $d=2$. For the case $p=5$,
$\mathfrak{\widehat{E}}_{7}=4a^{5}X^{d}+X^{3d}$ and if it is a DO polynomial
then $d=5^{\alpha}+1$ and $3d=5^{\beta}+1$. Thus, we have $3\cdot
5^{\alpha}+2=5^{\beta}$, which forces $\alpha=0$, $\beta=1$ and $d=2$. If
$p>5$, then $\mathfrak{\widehat{E}}_{7}=-6a^{5}X^{d}+10a^{3}X^{2d}-4X^{3d}$.
Since the coefficients of $X^{d}$ and $X^{2d}$ are non-zero, therefore
$\mathfrak{\widehat{E}}_{7}$ is not a DO polynomial by Lemma 2.1.

For $k\geq 8$, we consider two cases, $p=3$ and $p\geq 5$.

Case 1 $p=3$.

We shall treat all possible cases depending upon the value of $k$ modulo $9$.
When $k\equiv 2,8\pmod{9}$, then $k\equiv 2\pmod{3}$, therefore, $k\not\equiv
1\pmod{3}$ and hence the coefficient of $X^{d}$ in (4.1) is non-zero. Now,
consider the fourth term

(4.2) $\frac{(k-4)(k-5)(k-6)(k-7)}{4!}a^{k-8}X^{4d}.$

It is clear that $3\nmid(k-4)$, $3\nmid(k-6)$ and $3\nmid(k-7)$. Also, since
$k\equiv 2,8\pmod{9}$ therefore $k\not\equiv 5\pmod{9}$ and hence the highest
exponent of $3$ which divides the numerator of the coefficient of fourth term
is $1$. By Lemma 2.5, the highest exponent of $3$ which divides $4!$ is $1$.
Therefore, coefficient of the fourth term is non-zero and
$\mathfrak{\widehat{E}}_{k}$ is not DO polynomial by Lemma 2.4. Similarly,
when $k\equiv 0,3\pmod{9}$ then $k\equiv 0\pmod{3}$ and hence, the coefficient
of $X^{d}$ in (4.1) is non-zero. Now consider the fourth term as given in
(4.2) again. It is clear that $3\nmid(k-4)$, $3\nmid(k-5)$ and $3\nmid(k-7)$.
Also, since $k\equiv 0,3\pmod{9}$ therefore $k\not\equiv 6\pmod{9}$ and hence
the highest exponent of $3$ which divides the numerator of the coefficient of
fourth term is $1$. By Lemma 2.5, the highest exponent of $3$ which divides
$4!$ is $1$. Therefore, fourth term will exist in this case too and
$\mathfrak{\widehat{E}}_{k}$ is not DO polynomial by Lemma 2.4.

In the case $k\equiv 1\pmod{9}$, if the polynomial
$\mathfrak{\widehat{E}}_{10}=a^{6}X^{2d}+a^{4}X^{3d}+2X^{5d}$ is a DO
polynomial then $2d=3^{\alpha}+1$, $3d=3^{t}(3^{\beta}+1)$ and
$5d=3^{\gamma}+1$. Since $3^{t}\mid 3$, $t=1$. Thus, second equation reduces
to $d=3^{\beta}+1$. Solving first two equations, we have $2\cdot
3^{\beta}+1=3^{\alpha}$, which forces $\beta=0$, $\alpha=1$ and $d=2$. Now,
putting these values in third equation, we have $3^{\gamma}=9$ and $\gamma=2$.
Similarly, if the polynomial
$\mathfrak{\widehat{E}}_{19}=a^{15}X^{2d}+a^{13}X^{3d}+2a^{9}X^{5d}+2aX^{9d}$
is a DO polynomial then $2d=3^{\alpha}+1$, $3d=3^{t}(3^{\beta}+1)$,
$5d=3^{\gamma}+1$ and $9d=3^{s}(3^{\delta}+1)$. Since $3^{t}\mid 3$ and
$3^{s}\mid 9$, we have $t=1$ and $s=2$. Therefore, the second and fourth
equation reduce to $d=3^{\beta}+1$ and $d=3^{\delta}+1$. Solving first and
second equations, we get $2\cdot 3^{\beta}+1=3^{\alpha}$, which forces
$\beta=0$, $\alpha=1$ and $d=2$. Now, putting these values in third equation,
we have $3^{\gamma}=9$ and $\gamma=2$. For $k\geq 28$, since $k\equiv
1\pmod{3}$, therefore, $k\not\equiv 0,2\pmod{3}$ and hence the coefficient of
$X^{2d}$ is non-zero. Now, consider the $11$th term

(4.3)
$\frac{(k-11)(k-12)(k-13)\cdots(k-19)(k-20)(k-21)}{11!}a^{k-22}(-X^{d})^{11}.$

By Lemma 2.5, the highest exponent of $3$ that divides $11!$ is $4$. In the
numerator of the coefficient of $11$th term, $(k-13),(k-16),(k-19)\equiv
0\pmod{3}$. Since $k\equiv 1\pmod{9}$, we have $(k-13),(k-16)\not\equiv
0\pmod{9}$. Now if $k\not\equiv 19\pmod{27}$, then the highest exponent of $3$
which divides the numerator is $4$. Hence the coefficient of $X^{11d}$ is non-
zero. Thus, if $\mathfrak{\widehat{E}}_{k}$ is a DO polynomial then
$2d=3^{\alpha}+1$ and $11d=3^{\beta}+1$. Solving these equations, we have
$11\cdot 3^{\alpha}+9=2\cdot 3^{\beta}$, which forces $\alpha=2$ and
$3^{\beta}=54$, a contradiction. Therefore, $\mathfrak{\widehat{E}}_{k}$ is
not a DO polynomial in this case. In the case of $k\equiv 19\pmod{27}$, we
have $k\geq 46$. In this case, consider the $20$th term

(4.4) $\frac{(k-20)(k-21)(k-22)\cdots(k-37)(k-38)(k-39)}{20!}a^{k-40}X^{20d}.$

In the numerator of the coefficient of the $20$th term,
$(k-22),(k-25),(k-28),(k-31),(k-34),(k-37)\equiv 0\pmod{3}$. Since $k\equiv
1\pmod{9}$, we have $(k-22),(k-25),(k-31),(k-34)\not\equiv 0\pmod{9}$. Also,
since $k\equiv 19\pmod{27}$, we have $(k-28),(k-37)\not\equiv 0\pmod{27}$.
Thus, the highest exponent of $3$ which divides the numerator is $8$. By Lemma
2.5, the highest exponent of $3$ which divides $20!$ is $8$. Hence the
coefficient of $X^{20d}$ is non-zero. Thus, if $\mathfrak{\widehat{E}}_{k}$ is
a DO polynomial then $2d=3^{\alpha}+1$ and $20d=3^{\beta}+1$. Solving these
equations, we have $10\cdot 3^{\alpha}+9=3^{\beta}$, which forces $\alpha=2$
and $3^{\beta}=99$, a contradiction. Therefore, $\mathfrak{\widehat{E}}_{k}$
is not a DO polynomial in this case.

For the case $k\equiv 4\pmod{9}$, if the polynomial
$\mathfrak{\widehat{E}}_{13}=a^{9}X^{2d}+a^{3}X^{5d}+aX^{6d}$ is a DO
polynomial then $2d=3^{\alpha}+1$, $5d=3^{\beta}+1$ and
$6d=3^{t}(3^{\gamma}+1)$. Since $3^{t}\mid 6$, $t=1$ and hence the third
equation reduces to $2d=3^{\gamma}+1$. Solving first two equations, we have
$5\cdot 3^{\alpha}+3=2\cdot 3^{\beta}$, which is true if and only if
$\alpha=1$, $\beta=2$ and $d=2$. Now, for $k\geq 22$, since $k\equiv
1\pmod{3}$, therefore, $k\not\equiv 0,2\pmod{3}$ and hence the coefficient of
$X^{2d}$ in (4.1) is non-zero. Now, consider the $11$th term as given in
(4.3). By Lemma 2.5, the highest exponent of $3$ that divides $11!$ is $4$. In
the numerator of the coefficient of $11$th term, $(k-13),(k-16),(k-19)\equiv
0\pmod{3}$. Since $k\equiv 4\pmod{9}$, we have $(k-16),(k-19)\not\equiv
0\pmod{9}$. Now if $k\not\equiv 13\pmod{27}$, then the highest exponent of $3$
which divides the numerator is $4$. Hence the coefficient of $X^{11d}$ is non-
zero. Thus if $\mathfrak{\widehat{E}}_{k}$ is a DO polynomial then
$2d=3^{\alpha}+1$ and $11d=3^{\beta}+1$. Solving these two equations, we have
$11\cdot 3^{\alpha}+9=2\cdot 3^{\beta}$, which forces $\alpha=2$ and
$3^{\beta}=54$, a contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$ is not
a DO polynomial in this case. In the case of $k\equiv 13\pmod{27}$, $k\geq 22$
is equivalent to $k\geq 40$. In this case, consider the $20$th term as given
in (4.4). By Lemma 2.5, the highest exponent of $3$ that divides $20!$ is $8$.
In the numerator of the coefficient of the $20$th term,
$(k-22),(k-25),(k-28),(k-31),(k-34),(k-37)\equiv 0\pmod{3}$. Since $k\equiv
1\pmod{9}$, we have $(k-25),(k-28),(k-34),(k-37)\not\equiv 0\pmod{9}$. Also,
since $k\equiv 13\pmod{27}$, we have $(k-22),(k-31)\not\equiv 0\pmod{27}$.
Thus the highest exponent of $3$ which divides the numerator is $8$. Hence the
coefficient of $X^{20d}$ is non-zero. Therefore, if
$\mathfrak{\widehat{E}}_{k}$ is a DO polynomial then $2d=3^{\alpha}+1$ and
$20d=3^{\beta}+1$. Solving these equations, we have $10\cdot
3^{\alpha}+9=3^{\beta}$, which forces $\alpha=2$ and $3^{\beta}=99$, a
contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$ is not a DO polynomial
in this case.

For the case $k\equiv 5\pmod{9}$, if the polynomial
$\mathfrak{\widehat{E}}_{14}=2a^{10}X^{2d}+a^{2}X^{6d}+X^{7d}$ is a DO
polynomial then $d=3^{\alpha}+1$, $6d=3^{t}(3^{\beta}+1)$ and
$7d=3^{\gamma}+1$. Since $3^{t}\mid 6$, $t=1$ and hence second equation
reduces to $2d=3^{\beta}+1$. Solving first two equations, we have $2\cdot
3^{\alpha}+1=3^{\beta}$, which forces $\alpha=0$, $\beta=1$ and $d=2$. Now
putting these values in third equation, we have $3^{\gamma}=13$, a
contradiction. Therefore $\mathfrak{\widehat{E}}_{14}$ is not a DO polynomial.
Now, for $k\geq 23$, consider the $10$th term

(4.5)
$\frac{(k-10)(k-11)(k-12)(k-13)\cdots(k-17)(k-18)(k-19)}{10!}a^{k-20}X^{10d}.$

By Lemma 2.5, the highest exponent of $3$ that divides $10!$ is $4$. In the
numerator of the coefficient of $10$th term, $(k-11),(k-14),(k-17)\equiv
0\pmod{3}$. Since $k\equiv 5\pmod{9}$, we have $(k-11),(k-17)\not\equiv
0\pmod{9}$. Now if $k\not\equiv 14\pmod{27}$, then the highest exponent of $3$
which divides the numerator is $4$. Hence the coefficient of $X^{10d}$ is non-
zero. Thus if $\mathfrak{\widehat{E}}_{k}$ is a DO polynomial then
$d=3^{\alpha}+1$ and $10d=3^{\beta}+1$. Solving these equations, we get
$10\cdot 3^{\alpha}+9=3^{\beta}$, which forces $\alpha=2$ and $3^{\beta}=99$,
a contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$ is not a DO polynomial
in this case. In the case of $k\equiv 14\pmod{27}$, $k\geq 23$ is equivalent
to $k\geq 41$. Now, consider the $16$th term

(4.6) $\frac{(k-16)(k-17)(k-18)\cdots(k-29)(k-30)(k-31)}{16!}a^{k-32}X^{16d}.$

By Lemma 2.5, the highest exponent of $3$ that divides $16!$ is $6$. In the
numerator of the coefficient of the $16$th term,
$(k-17),(k-20),(k-23),(k-26),(k-29)\equiv 0\pmod{3}$. Since $k\equiv
5\pmod{9}$, we have $(k-17),(k-20),(k-26),(k-29)\not\equiv 0\pmod{9}$. Also,
since $k\equiv 14\pmod{27}$, we have $k\not\equiv 23\pmod{27}$. Therefore, the
highest exponent of $3$ which divides the numerator is $6$. Hence the
coefficient of $X^{16d}$ is non-zero. Thus, if $\mathfrak{\widehat{E}}_{k}$ is
a DO polynomial then $d=3^{\alpha}+1$ and $16d=3^{\beta}+1$. Solving these
equations, we get $16\cdot 3^{\alpha}+15=3^{\beta}$, which forces $\alpha=1$
and $3^{\beta}=63$, a contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$ is
not a DO polynomial in this case.

In the case of $k\equiv 6\pmod{9}$, if the polynomial
$\mathfrak{\widehat{E}}_{15}=a^{13}X^{d}+2a^{9}X^{3d}+aX^{7d}$ is a DO
polynomial then $d=3^{\alpha}+1$, $3d=3^{t}(3^{\beta}+1)$ and
$7d=3^{\gamma}+1$. Since $3^{t}\mid 3$, $t=1$ and hence the second equation
reduces to $d=3^{\beta}+1$. Solving first and third equations, we have $7\cdot
3^{\alpha}+6=3^{\gamma}$, which forces $\alpha=1$, $\gamma=3$ and $d=4$. Now,
for $k\geq 24$, consider the $10$th term as given in (4.5). By Lemma 2.5, the
highest exponent of $3$ that divides $10!$ is $4$. In the numerator of the
coefficient of $10$th term, $(k-12),(k-15),(k-18)\equiv 0\pmod{3}$. Since
$k\equiv 6\pmod{9}$, we have $(k-12),(k-18)\not\equiv 0\pmod{9}$. Now if
$k\not\equiv 15\pmod{27}$, then the highest exponent of $3$ which divides the
numerator is $4$. Hence the coefficient of $X^{10d}$ is non-zero. Thus, if
$\mathfrak{\widehat{E}}_{k}$ is a DO polynomial then $d=3^{\alpha}+1$ and
$10d=3^{\beta}+1$. Solving these equations, we get $10\cdot
3^{\alpha}+9=3^{\beta}$, which forces $\alpha=2$ and $3^{\beta}=99$, a
contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$ is not a DO polynomial
in this case. In the case of $k\equiv 15\pmod{27}$, $k\geq 24$ is equivalent
to $k\geq 42$. In this case, consider the $16$th term as given in (4.6). By
Lemma 2.5, the highest exponent of $3$ that divides $16!$ is $6$. In the
numerator of the coefficient of the $16$th term,
$(k-18),(k-21),(k-24),(k-27),(k-30)\equiv 0\pmod{3}$. Since $k\equiv
6\pmod{9}$, we have $(k-18),(k-21),(k-27),(k-30)\not\equiv 0\pmod{9}$. Also,
since $k\equiv 15\pmod{27}$, we have $k\not\equiv 24\pmod{27}$. Therefore, the
highest exponent of $3$ which divides the numerator is $6$ and hence the
coefficient of $X^{16d}$ is non-zero. Thus, if $\mathfrak{\widehat{E}}_{k}$ is
a DO polynomial then $d=3^{\alpha}+1$ and $16d=3^{\beta}+1$. Solving these
equations, we get $16\cdot 3^{\alpha}+15=3^{\beta}$, which forces $\alpha=1$
and $3^{\beta}=63$, a contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$ is
not a DO polynomial in this case.

When $k\equiv 7\pmod{9}$, $k\geq 8$ is equivalent to $k\geq 16$. Also, since
$k\equiv 1\pmod{3}$, we have $k\not\equiv 0\,\,\textnormal{or}\,\,2\pmod{3}$
and hence the coefficient of $X^{2d}$ in (4.1) is non-zero. Now consider the
$8$th term, which is given by

(4.7)
$\frac{(k-8)(k-9)(k-10)(k-11)(k-12)(k-13)(k-14)(k-15)}{8!}a^{k-16}X^{8d}.$

By Lemma 2.5, the highest exponent of $3$ that divides $8!$ is $2$. In the
numerator of the coefficient of $8$th term, $(k-10),(k-13)\equiv 0\pmod{3}$.
Since $k\equiv 7\pmod{9}$, we have $(k-10),(k-13)\not\equiv 0\pmod{9}$. Thus,
the highest exponent of $3$ which divides the numerator is $2$. Hence the
coefficient of $X^{8d}$ is non-zero. Thus, if $\mathfrak{\widehat{E}}_{k}$ is
a DO polynomial then $2d=3^{\alpha}+1$ and $8d=3^{\beta}+1$. Solving these
equations, we have $4\cdot 3^{\alpha}+3=3^{\beta}$, which forces $\alpha=1$
and $3^{\beta}=15$, a contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$ is
not a DO polynomial in this case.

Case 2 $p\geq 5$.

Here we shall consider four cases, namely, $k\not\equiv 1,2,3\pmod{p}$,
$k\equiv 1\pmod{p}$, $k\equiv 2\pmod{p}$ and $k\equiv 3\pmod{p}$,
respectively.

Note that when $k\not\equiv 1,2,3\pmod{p}$, the coefficients of $X^{d}$ and
$X^{2d}$ in (4.1) are non-zero, therefore $\mathfrak{\widehat{E}}_{k}$ is not
DO polynomial by Lemma 2.1.

In the case of $k\equiv 1\pmod{p}$, we have $(k-2),(k-3),(k-4),(k-5)\not\equiv
0\pmod{p}$. Therefore, the coefficients of $X^{2d}$ and $X^{3d}$ in (4.1) are
non-zero. Thus, if $\mathfrak{\widehat{E}}_{k}$ is a DO polynomial then
$2d=p^{\alpha}+1$ and $3d=p^{\beta}+1$. Solving these equations, we have
$3p^{\alpha}+1=2p^{\beta}$, which forces $\alpha=0$ and $p^{\beta}=2$, a
contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$ is not a DO polynomial
in this case.

When $k\equiv 2\pmod{p}$, the coefficient of the first term in (4.1) which
contains the monomial $X^{d}$ is non-zero. Now we consider two cases, namely,
$p=5$ and $p>5$. In the case of $p=5$, $k\geq 8$ is equivalent to $k\geq 12$.
We now show that if $k\not\equiv 7\pmod{25}$ then the sixth term exists, which
is given by

$\frac{(k-6)(k-7)(k-8)(k-9)(k-10)(k-11)}{6!}a^{k-12}X^{6d}.$

Since $k\equiv 2\pmod{5}$, we have $(k-6),(k-8),(k-9),(k-10),(k-11)\not\equiv
0\pmod{5}$. Also if $k\not\equiv 7\pmod{25}$ then the highest exponent of $5$
which divides the numerator is 1. By Lemma 2.5, the highest exponent of $5$
that divides $6!$ is 1. Therefore the coefficient of $X^{6d}$ is non-zero.
Thus, if $\mathfrak{\widehat{E}}_{k}$ is a DO polynomial then $d=5^{\alpha}+1$
and $6d=5^{\beta}+1$. Solving these equations, we have $6\cdot
5^{\alpha}+5=5^{\beta}$, which forces $\alpha=1$ and $5^{\beta}=35$, a
contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$ is not a DO polynomial
in this case. Now if $k\equiv 7\pmod{25}$, then the condition $k\geq 12$ is
equivalent to $k\geq 32$. In this case, we claim that the 8th term as given in
(4.7) exists. Since we are in the case $k\equiv 2\pmod{5}$, therefore, we have
$(k-8),(k-9),(k-10),(k-11),(k-13),(k-14),(k-15)\not\equiv 0\pmod{5}$. Also,
since $k\equiv 7\pmod{25}$, we have $k\not\equiv 12\pmod{25}$ and hence the
highest exponent of $5$ which divides the numerator is 1. By Lemma 2.5, the
highest exponent of $5$ that divides $8!$ is 1. Therefore the coefficient of
$X^{8d}$ is non-zero. Thus, if $\mathfrak{\widehat{E}}_{k}$ is a DO polynomial
then $d=5^{\alpha}+1$ and $8d=5^{\beta}+1$. Solving these equations, we have
$8\cdot 5^{\alpha}+7=5^{\beta}$, which forces $\alpha=0$ and $5^{\beta}=15$, a
contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$ is not a DO polynomial
in this case. In the case of $p>5$, since $k\equiv 2\pmod{p}$, therefore, we
have $k\not\equiv 1,3,4,5\pmod{p}$. Hence the coefficients of $X^{d}$ and
$X^{3d}$ in (4.1) are non-zero, therefore $\mathfrak{\widehat{E}}_{k}$ is not
DO polynomial by Lemma 2.2

When $k\equiv 3\pmod{p}$, the first term $(1-k)X^{d}$ in (4.1) will exist. Now
we consider two cases, namely, $p=5$ and $p>5$. In the case of $p=5$, since
$k\equiv 3\pmod{5}$, therefore, we have $k\not\equiv 0,1,2,4\pmod{5}$ and
hence the fourth term as given in (4.2) will exist. Therefore, if
$\mathfrak{\widehat{E}}_{k}$ is a DO polynomial then $d=5^{i}+1$ and
$4d=5^{j}+1$. Solving these equations, we have $4\cdot 5^{i}+3=5^{j}$, which
forces $i=0$ and $5^{j}=7$, a contradiction. Therefore
$\mathfrak{\widehat{E}}_{k}$ is not a DO polynomial in this case. In the case
of $p>5$, since $k\equiv 3\pmod{p}$, we have
$(k-1),(k-4),(k-5),(k-6),(k-7),(k-8),(k-9)\not\equiv 0\pmod{p}$. Therefore,
the fourth term as given in (4.2) and the fifth term

$\frac{(k-5)(k-6)(k-7)(k-8)(k-9)}{5!}a^{k-10}(-X^{d})^{5},$

will exist. Thus, if $\mathfrak{\widehat{E}}_{k}$ is a DO polynomial then
$d=p^{\alpha}+1$, $4d=p^{\beta}+1$ and $5d=p^{\gamma}+1$. Solving first two
equations, we have $4p^{\alpha}+3=p^{\beta}$, which forces $\alpha=0$,
$\beta=1$, $p=7$ and $d=2$. Now putting these values in third equation, we
have $7^{\gamma}=9$, a contradiction. Therefore $\mathfrak{\widehat{E}}_{k}$
is not a DO polynomial in this case. ∎

One may recall from [29, Theorem 3.1] that the RDP of second kind and RDP of
third kind admit the following relationship

$D_{k,2}(a,X)=aD_{k-1,1}(a,X).$

Thus, it is obvious that $\mathfrak{\widehat{F}}_{k}$ is DO polynomial
whenever $\mathfrak{\widehat{E}}_{k-1}$ is DO polynomial. Consequently, the
classification of DO polynomials from RDP of third kind
$\mathfrak{\widehat{F}}_{k}$ follows immediately. In view of this, we shall
consider RDP of fourth kind in the next section.

##  5\. DO polynomials from RDP of the fourth kind

Recall that $D_{k,3}(a,X^{d})-D_{k,3}(a,0)$ is denoted by
$\mathfrak{\widehat{G}}_{k}$, where

(5.1)
$\begin{split}\mathfrak{\widehat{G}}_{k}=(3-k)a^{k-2}X^{d}+\frac{(k-3)(k-6)}{2}a^{k-4}X^{2d}\\\
-\frac{(k-4)(k-5)(k-9)}{3!}a^{k-6}X^{3d}+\cdots.\end{split}$

Also, from (1.3), it is easy to see that
$\mathfrak{\widehat{G}}_{k}=\mathfrak{\widehat{D}}_{k}\pmod{3}$. Therefore,
for $p=3$, $\mathfrak{\widehat{G}}_{k}$ is DO polynomial whenever
$\mathfrak{\widehat{D}}_{k}$ is DO polynomial and the classification of DO
polynomials from $\mathfrak{\widehat{D}}_{k}$ has already been discussed in
Section 3. Therefore, throughout this section, we take $p\geq 5$. The
following theorem gives a complete classification of DO polynomials derived
from $\mathfrak{\widehat{G}}_{k}$.

######  Theorem 5.1.

Let $q$ be a power of odd prime $p\geq 5$ and $a\in\mathbb{F}_{q}^{*}$. The
polynomial $\mathfrak{\widehat{G}}_{k}$ is a DO polynomial over
$\mathbb{F}_{q}$ if and only if one of the following holds.

  1. (i)

$k=2$ and $d=p^{n}(p^{\alpha}+1)$, where $\alpha,~{}n\geq 0$.

  2. (ii)

$k=6,11$, $p=5$ and $d=2p^{n}$, where $n\geq 0$.

###### Proof.

It is enough to prove only the necessary part. If the polynomial
$\mathfrak{\widehat{G}}_{2}=X^{d}$ is a DO polynomial, then $d=p^{\alpha}+1$.
The polynomial $\mathfrak{\widehat{G}}_{3}$ is a zero polynomial and hence it
is not a DO polynomial. The polynomials
$\mathfrak{\widehat{G}}_{4}=-a^{2}X^{d}-X^{2d}$,
$\mathfrak{\widehat{G}}_{5}=-2a^{3}X^{d}-aX^{2d}$ and
$\mathfrak{\widehat{G}}_{7}=-4a^{5}X^{d}+2a^{3}X^{2d}+2aX^{3d}$ are not DO
polynomials by Lemma 2.1. In the case of polynomial
$\mathfrak{\widehat{G}}_{6}=-3a^{4}X^{d}+X^{3d}$, we consider two cases,
namely, $p=5$ and $p>5$. In the case of $p=5$, if $\mathfrak{\widehat{G}}_{6}$
is a DO polynomial, then $d=5^{i}+1$ and $3d=5^{j}+1$. Solving these
equations, we have $3.5^{i}+2=5^{j}$, which is true if and only if $i=0$,
$j=1$ and $d=2$. When $p>5$, $\mathfrak{\widehat{G}}_{6}$ is not a DO
polynomial by Lemma 2.2.

For $k\geq 8$, we consider two cases, namely, $p=5$ and $p>5$.

Case 1 $p>5$.

Note that when $k\not\equiv 3,6\pmod{p}$, the coefficients of $X^{d}$ and
$X^{2d}$ in $\mathfrak{\widehat{G}}_{k}$ are non-zero. Therefore,
$\mathfrak{\widehat{G}}_{k}$ is not a DO polynomial by Lemma 2.1. In the case
of $k\equiv 3\pmod{p}$, the coefficient of $X^{3d}$ is non-zero and also, the
coefficient of $X^{4d}$ in $\mathfrak{\widehat{G}}_{k}$ given by

$\frac{(k-5)(k-6)(k-7)(k-12)}{4!}a^{k-8}X^{4d}$

is non-zero. Therefore, $\mathfrak{\widehat{G}}_{k}$ is not a DO polynomial by
Lemma 2.3. When $k\equiv 6\pmod{p}$, the coefficients of $X^{d}$ and $X^{3d}$
in $\mathfrak{\widehat{G}}_{k}$ are non-zero, therefore,
$\mathfrak{\widehat{G}}_{k}$ is not DO polynomial by Lemma 2.2.

Case 2 $p=5$.

Notice that when $k\not\equiv 1,3\pmod{5}$, the coefficients of $X^{d}$ and
$X^{2d}$ in $\mathfrak{\widehat{G}}_{k}$ are non-zero, therefore
$\mathfrak{\widehat{G}}_{k}$ is not a DO polynomial by Lemma 2.1. In the case
of $k\equiv 3\pmod{5}$, the coefficients of $X^{3d}$ and $X^{4d}$ in
$\mathfrak{\widehat{G}}_{k}$ are non-zero, therefore
$\mathfrak{\widehat{G}}_{k}$ is not a DO polynomial by Lemma 2.3. For $k\equiv
1\pmod{5}$, if the polynomial
$\mathfrak{\widehat{G}}_{11}=2a^{9}X^{d}+a^{5}X^{3d}+4aX^{5d}$ is a DO
polynomial, then $d=5^{\alpha}+1$, $3d=5^{\beta}+1$ and
$5d=5^{t}(5^{\gamma}+1)$. Since $5^{t}\mid 5$, $t=1$ and hence the third
equation reduces to $d=5^{\gamma}+1$. Solving first two equations, we get
$3\cdot 5^{\alpha}+2=5^{\beta}$, which is true if and only if $\alpha=0$,
$\beta=1$ and $d=2$. For $k\geq 16$, since $k\equiv 1\pmod{5}$, we have
$k\not\equiv 0,2,3,4\pmod{5}$ and hence the coefficient of $X^{3d}$ in
$\mathfrak{\widehat{G}}_{k}$ is non-zero. Now consider the $6$th term

$\frac{(k-7)(k-8)(k-9)(k-10)(k-11)(k-18)}{6!}a^{k-12}X^{6d}.$

By Lemma 2.5, the highest exponent of $5$ which divides $6!$ is $1$. Also if
$k\not\equiv 11\pmod{25}$, then highest exponent of $5$ that divides the
numerator of coefficient of $X^{6d}$ is $1$, hence the coefficient of $X^{6d}$
is non-zero. Thus, if $\mathfrak{\widehat{G}}_{k}$ is a DO polynomial, then
$3d=5^{\alpha}+1$ and $6d=5^{\beta}+1$. Solving these equations, we get
$2\cdot 5^{\alpha}+1=5^{\beta}$, which forces $\alpha=0$ and $5^{\beta}=3$, a
contradiction. Thus $\mathfrak{\widehat{G}}_{k}$ is not a DO polynomial in
this case. In the case of $k\equiv 11\pmod{25}$, consider the $11$th term

$\frac{(k-12)(k-13)\cdots(k-19)(k-20)(k-21)(k-33)}{11!}a^{k-22}(-X^{d})^{11}.$

By Lemma 2.5, the highest exponent of $5$ that divides $11!$ is $2$. In the
numerator of the coefficient of $11$th term, $(k-16),(k-21)\equiv 0\pmod{5}$.
Since $k\equiv 11\pmod{25}$, we have $k\not\equiv 16,21\pmod{25}$. Therefore
the highest power of $5$ which divides the numerator is $2$ and hence the
coefficient of $X^{11d}$ is non-zero. Thus, if $\mathfrak{\widehat{G}}_{k}$ is
a DO polynomial, then $3d=5^{\alpha}+1$ and $11d=5^{\beta}+1$. Solving these
equations, we have $11\cdot 5^{\alpha}+8=3.5^{\beta}$, which forces $\alpha=0$
and $3\cdot 5^{\beta}=19$, a contradiction. Thus $\mathfrak{\widehat{G}}_{k}$
is not a DO polynomial in this case. ∎

##  6\. DO polynomials from RDP of the fifth kind

Here we consider RDP of the fifth kind. Recall that
$D_{k,4}(a,X^{d})-D_{k,4}(a,0)$ is denoted by $\mathfrak{\widehat{H}}_{k}$,
where

(6.1)
$\begin{split}\mathfrak{\widehat{H}}_{k}=(4-k)a^{k-2}X^{d}+\frac{(k-3)(k-8)}{2}a^{k-4}X^{2d}\\\
-\frac{(k-4)(k-5)(k-12)}{3!}a^{k-6}X^{3d}+\cdots.\end{split}$

It is easy to see from (1.3) that
$\mathfrak{\widehat{H}}_{k}=\mathfrak{\widehat{E}}_{k}\pmod{3}$, thus for
$p=3$, $\mathfrak{\widehat{H}}_{k}$ is DO polynomial whenever
$\mathfrak{\widehat{E}}_{k}$ is DO polynomial. Thus, throughout this section,
we take $p\geq 5$.

######  Theorem 6.1.

Let $q$ be a power of odd prime $p\geq 5$ and $a\in\mathbb{F}_{q}^{*}$. The
polynomial $\mathfrak{\widehat{H}}_{k}$ is a DO polynomial over
$\mathbb{F}_{q}$ if and only if one of the following holds.

  1. (i)

$k=2,3$ and $d=p^{n}(p^{\alpha}+1)$, where $\alpha,~{}n\geq 0$.

  2. (ii)

$k=4$ and $d=p^{n}(p^{\alpha}+1)/2$, where $\alpha,~{}n\geq 0$.

###### Proof.

The sufficiency of the theorem is straightforward. It only remains to show the
necessity.

If the polynomials $\mathfrak{\widehat{H}}_{2}=2X^{d}$ and
$\mathfrak{\widehat{H}}_{3}=aX^{d}$ are DO polynomials, then $d=p^{\alpha}+1$.
Similarly, if the polynomial $\mathfrak{\widehat{H}}_{4}=-2X^{2d}$ is DO
polynomial, then $d=({p^{\alpha}+1})/2$. In the case of polynomials
$\mathfrak{\widehat{H}}_{5}=-a^{3}X^{d}-3aX^{2d}$,
$\mathfrak{\widehat{H}}_{6}=-2a^{4}X^{d}-3a^{2}X^{2d}+2X^{3d}$ and
$\mathfrak{\widehat{H}}_{7}=-3a^{5}X^{d}-2a^{3}X^{2d}+aX^{3d}$, the
coefficients of $X^{d}$ and $X^{2d}$ are non-zero. Therefore,
$\mathfrak{\widehat{H}}_{5}$, $\mathfrak{\widehat{H}}_{6}$ and
$\mathfrak{\widehat{H}}_{7}$ are not DO polynomials by Lemma 2.1. The
polynomial $\mathfrak{\widehat{H}}_{8}=-4a^{6}X^{d}+8a^{2}X^{3d}-2X^{4d}$ is
not a DO polynomial by Lemma 2.3. If the polynomial
$\mathfrak{\widehat{H}}_{9}=-5a^{7}X^{d}+3a^{5}X^{2d}+10a^{3}X^{3d}-7aX^{4d}$
is a DO polynomial, then $2d=5^{\alpha}+1$ and $4d=5^{\beta}+1$. Solving these
equations, we get $2\cdot 5^{\alpha}+1=5^{\beta}$, which forces $\alpha=0$,
$5^{\beta}=3$, a contradiction. Thus $\mathfrak{\widehat{H}}_{9}$ is not a DO
polynomial.

For $k\geq 10$, we consider two cases, $p=5$ and $p>5$.

Case 1 $p=5$.

Notice that when $k\not\equiv 3,4\pmod{5}$, the coefficients of $X^{d}$ and
$X^{2d}$ in $\mathfrak{\widehat{H}}_{k}$ are non-zero, therefore
$\mathfrak{\widehat{H}}_{k}$ is not a DO polynomial by Lemma 2.1. In the case
of $k\equiv 3\pmod{5}$, the coefficients of $X^{d}$ is clearly non-zero and
also, the coefficient of $X^{4d}$ in $\mathfrak{\widehat{H}}_{k}$ given by

$\frac{(k-4)(k-5)(k-6)(k-7)}{4!}a^{k-8}$

is non-zero. Thus, if $\mathfrak{\widehat{H}}_{k}$ is a DO polynomial, then
$d=5^{\alpha}+1$ and $4d=5^{\beta}+1$. Solving these equations, we have
$4\cdot 5^{\alpha}+3=5^{\beta}$, which forces $\alpha=0$ and $5^{\beta}=7$, a
contradiction. Thus $\mathfrak{\widehat{H}}_{k}$ is not a DO polynomial in
this case. When $k\equiv 4\pmod{5}$, the coefficient of $X^{2d}$ in
$\mathfrak{\widehat{H}}_{k}$ is always non-zero. Also, if $k\not\equiv
9\pmod{25}$, the coefficient of $X^{5d}$ in $\mathfrak{\widehat{H}}_{k}$ given
by

$\frac{(k-6)(k-7)(k-8)(k-9)(k-20)}{5!}a^{k-10}{(-1)}^{5d}$

is non-zero. Thus, if $\mathfrak{\widehat{H}}_{k}$ is a DO polynomial, then
$2d=5^{\alpha}+1$ and $5d=5^{t}(5^{\beta}+1)$. Since $5^{t}\mid 5$, therefore
$t=1$ and hence, the second equation reduces to $d=5^{\beta}+1$. Solving these
equations, we have $2\cdot 5^{\beta}+1=5^{\alpha}$, which forces $\beta=0$ and
$5^{\alpha}=3$, a contradiction. Thus $\mathfrak{\widehat{H}}_{k}$ is not a DO
polynomial. In the case of $k\equiv 9\pmod{25}$, the condition $k\geq 10$ is
equivalent to $k\geq 34$. Now consider the $9$th term

$\frac{(k-10)(k-11)(k-12)\cdots(k-16)(k-17)(k-36)}{9!}a^{k-18}(-X^{d})^{9}.$

Since $k\equiv 9\pmod{25}$, we have $k\not\equiv 14~{}(\mbox{mod}~{}25)$.
Hence the highest exponent of $5$, which divides the numerator is $1$. By
Lemma 2.5, highest exponent of $5$, which divides $9!$ is $1$. Therefore, the
coefficient of $X^{9d}$ is non-zero. Thus, if $\mathfrak{\widehat{H}}_{k}$ is
a DO polynomial then $2d=5^{\alpha}+1$ and $9d=5^{\beta}+1$. Solving these two
equations, we have $9\cdot 5^{\alpha}+7=2\cdot 5^{\beta}$, which forces
$\alpha=0$ and $5^{\beta}=8$, a contradiction. Thus
$\mathfrak{\widehat{H}}_{k}$ is not a DO polynomial.

Case 2 $p>5$.

Notice that when $k\not\equiv 3,4,8\pmod{p}$, the coefficients of $X^{d}$ and
$X^{2d}$ in $\mathfrak{\widehat{H}}_{k}$ are non-zero, therefore
$\mathfrak{\widehat{H}}_{k}$ is not a DO polynomial by Lemma 2.1. In the case
of $k\equiv 3,8\pmod{p}$, the coefficients of $X^{d}$ and $X^{3d}$ in
$\mathfrak{\widehat{H}}_{k}$ are non-zero, therefore
$\mathfrak{\widehat{H}}_{k}$ is not a DO polynomial by Lemma 2.2. When
$k\equiv 4\pmod{p}$, the coefficients of $X^{2d}$ and $X^{4d}$ in
$\mathfrak{\widehat{H}}_{k}$ are non-zero. Thus $\mathfrak{\widehat{H}}_{k}$
is DO polynomial if $2d=p^{\alpha}+1$ and $4d=p^{\beta}+1$. Solving these
equations, we have $2\cdot p^{\alpha}+1=p^{\beta}$, which forces $\alpha=0$
and $p^{\beta}=3$, a contradiction. Thus $\mathfrak{\widehat{H}}_{k}$ is not a
DO polynomial. This completes the proof. ∎

##  7\. The case $m\geq 5$

For $m\geq 5$, we shall classify the DO polynomials from the polynomial
$\mathfrak{\widehat{D}}_{k,m}$, where

(7.1)
$\begin{split}\mathfrak{\widehat{D}}_{k,m}=(m-k)a^{k-2}X^{d}+\frac{(k-3)(k-2m)}{2}a^{k-4}X^{2d}\\\
-\frac{(k-4)(k-5)(k-3m)}{3!}a^{k-6}X^{3d}+\cdots.\end{split}$

From (1.3), it is straightforward to see that for $p=3$,
$\mathfrak{\widehat{D}}_{k,m}=\mathfrak{\widehat{D}}_{k},\mathfrak{\widehat{E}}_{k},~{}\mbox{and}~{}\mathfrak{\widehat{F}}_{k}$,
whenever $m\equiv 0,1~{}\mbox{and}~{}2\pmod{3}$, respectively. Similarly, for
$p\geq 5$,
$\mathfrak{\widehat{D}}_{k,m}=\mathfrak{\widehat{D}}_{k},\mathfrak{\widehat{E}}_{k},\mathfrak{\widehat{F}}_{k},\mathfrak{\widehat{G}}_{k}~{}\mbox{and}~{}\mathfrak{\widehat{H}}_{k}$,
whenever $m\equiv 0,1,2,3~{}\mbox{and}~{}4\pmod{p}$, respectively. Thus the
only cases that remain to be considered are $p>5$ and $m\not\equiv
0,1,2,3,4\pmod{p}$ for which we have the following theorem.

######  Theorem 7.1.

Let $q$ be a power of odd prime $p>5$ and $a\in\mathbb{F}_{q}^{*}$. The
polynomial $\mathfrak{\widehat{D}}_{k,m}$ where $m\not\equiv
0,1,2,3,4\pmod{p}$ is a DO polynomial over $\mathbb{F}_{q}$ if and only if one
of the following holds.

  1. (i)

$k=2,3$ and $d=p^{n}(p^{\alpha}+1)$, where $\alpha,~{}n\geq 0$.

  2. (ii)

$k=5$, $m\equiv 5\pmod{p}$ and $d=p^{n}(p^{\alpha}+1)/2$, where
$\alpha,~{}n\geq 0$.

  3. (iii)

$k=5$, $2m\equiv 5\pmod{p}$ and $d=p^{n}(p^{\alpha}+1)$, where
$\alpha,~{}n\geq 0$.

###### Proof.

Only sufficiency of the theorem is required to be proved. If the polynomials
$\mathfrak{\widehat{D}}_{2,m}=(m-2)X^{d}$ and
$\mathfrak{\widehat{D}}_{3,m}=(m-3)aX^{d}$ are DO polynomial, then $d$ is of
the form $p^{\alpha}+1$. The polynomial
$\mathfrak{\widehat{D}}_{4,m}=(m-4)a^{2}X^{d}+(2-m)X^{2d}$ is not a DO
polynomial by Lemma 2.1. In the case of polynomial
$\mathfrak{\widehat{D}}_{5,m}=(m-5)a^{3}X^{d}+(5-2m)aX^{2d}$, we consider
three cases, namely, $m\equiv 5\pmod{p}$, $2m\equiv 5\pmod{p}$ and
$m,2m\not\equiv 5\pmod{p}$. In the case of $m\equiv 5\pmod{p}$, if
$\mathfrak{\widehat{D}}_{5,m}=-5aX^{2d}$ is a DO polynomial, then $d$ is of
the form $(p^{\alpha}+1)/2$. When $2m\equiv 5\pmod{p}$ and if
$\mathfrak{\widehat{D}}_{5,m}=(m-5)a^{3}X^{d}$ is a DO polynomial, then $d$ is
of the form $p^{\alpha}+1$. In the case of $m,2m\not\equiv 5\pmod{p}$,
$\mathfrak{\widehat{D}}_{5,m}=(m-5)a^{3}X^{d}+(5-2m)aX^{2d}$ is not a DO
polynomial by Lemma 2.1.

For $k\geq 6$, we consider four cases, namely, $k\not\equiv 3,m,2m\pmod{p}$,
$k\equiv 3\pmod{p}$, $k\equiv m\pmod{p}$ and $k\equiv 2m\pmod{p}$.

In the case of $k\not\equiv 3,m,2m\pmod{p}$, the coefficients of $X^{d}$ and
$X^{2d}$ in $\mathfrak{\widehat{D}}_{k,m}$ are non-zero and therefore
$\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial by Lemma 2.1.

In the case of $k\equiv 3\pmod{p}$, we have $k\not\equiv 4,5\pmod{p}$. Also,
note that $k\not\equiv m\pmod{p}$, otherwise $m\equiv 3\pmod{p}$. Similarly,
$k\not\equiv 3m\pmod{p}$, otherwise $m\equiv 1\pmod{p}$. Therefore, the
coefficients of $X^{d}$ and $X^{3d}$ in $\mathfrak{\widehat{D}}_{k,m}$ are
non-zero and hence $\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial by
Lemma 2.2.

In the case of $k\equiv m\pmod{p}$, $k\not\equiv 3,4\pmod{p}$. Also, note that
$k\not\equiv 2m,3m\pmod{p}$, otherwise $m\equiv 0\pmod{p}$. Therefore, the
coefficient of $X^{2d}$ in $\mathfrak{\widehat{D}}_{k,m}$ is non-zero. Also,
when $k\not\equiv 5\pmod{p}$, the coefficient of $X^{3d}$ in
$\mathfrak{\widehat{D}}_{k,m}$ is non-zero. Thus, if
$\mathfrak{\widehat{D}}_{k,m}$ is a DO polynomial, then $2d=p^{\alpha}+1$ and
$3d=p^{\beta}+1$. Solving these equations, we get $3p^{\alpha}+1=2p^{\beta}$,
which forces $\alpha=0$ and $p^{\beta}=2$, a contradiction. Therefore
$\mathfrak{\widehat{D}}_{k,m}$ not DO polynomials in this case. In the case of
$k\equiv 5\pmod{p}$, consider the fifth term

(7.2) $\frac{(k-6)(k-7)(k-8)(k-9)(k-5m)}{5!}a^{k-10}(-X^{d})^{5}.$

Since $k\equiv 5\pmod{p}$, we have $(k-6),(k-7),(k-8),(k-9)\not\equiv
0\pmod{p}$. Also note that $k\not\equiv 5m\pmod{p}$, otherwise $m\equiv
1\pmod{p}$. Therefore, the coefficients of $X^{2d}$ and $X^{5d}$ in
$\mathfrak{\widehat{D}}_{k,m}$ are non-zero. Thus, if
$\mathfrak{\widehat{D}}_{k,m}$ is a DO polynomial then $2d=p^{\alpha}+1$ and
$5d=p^{\beta}+1$. Solving these equations, we get $5p^{\alpha}+3=2p^{\beta}$,
which forces $\alpha=0$ and $p^{\beta}=4$, a contradiction. Therefore
$\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial.

In the case of $k\equiv 2m\pmod{p}$, we have $k\not\equiv m\pmod{p}$,
otherwise $m\equiv 0\pmod{p}$. Therefore the coefficient of $X^{d}$ in
$\mathfrak{\widehat{D}}_{k,m}$ is non-zero. Also note that $k\not\equiv
3m\pmod{p}$ and $k\not\equiv 4\pmod{p}$, otherwise $m\equiv 0\pmod{p}$ and
$m\equiv 2\pmod{p}$, respectively. When $k\not\equiv 5\pmod{p}$, the
coefficient of $X^{3d}$ in $\mathfrak{\widehat{D}}_{k,m}$ is non-zero.
Therefore $\mathfrak{\widehat{D}}_{k,m}$ is not a DO polynomial by Lemma 2.2.
In the case of $k\equiv 5\pmod{p}$, the condition $k\geq 6$ is equivalent to
$k\geq 13$. Now consider the fifth term as given in (7.2). Since $k\equiv
5\pmod{p}$, we have $(k-6),(k-7),(k-8),(k-9)\not\equiv 0\pmod{p}$. Also, note
that $k\not\equiv 5m\pmod{p}$, otherwise $m\equiv 1\pmod{p}$. Therefore, the
coefficient of $X^{5d}$ is non-zero. Thus, if $\mathfrak{\widehat{D}}_{k,m}$
is DO polynomial, then $d=p^{\alpha}+1$ and $5d=p^{\beta}+1$. Solving these
equations, we get $5p^{\alpha}+4=p^{\beta}$, which forces $\alpha=0$ and
$p^{\beta}=9$, a contradiction. Therefore $\mathfrak{\widehat{D}}_{k,m}$ is
not a DO polynomial. This completes the proof. ∎

##  8\. Discussion on planarity

We consider the planarity of the DO polynomials obtained from the RDPs of
$(m+1)$-th kind as listed in the Appendix A. First, we shall discuss the tools
and techniques that are needed to understand the planarity of the DO
polynomials. These tools and techniques are similar to the ones used in [10].
Recall that a polynomial function $f:\mathbb{F}_{q}\rightarrow\mathbb{F}_{q}$
is said to be planar if the difference function
$\Delta_{f}(X,\epsilon)=f(X+\epsilon)-f(X)-f(\epsilon)$ permutes the elements
of $\mathbb{F}_{q}$ for each $\epsilon\in\mathbb{F}_{q}^{*}$. If $f$ happens
to be a DO polynomial, the difference function $\Delta_{f}(X,\epsilon)$ for
each $\epsilon\in\mathbb{F}_{q}^{*}$, belongs to another well-known class of
polynomials called linearized polynomials. Therefore, a DO polynomial $f$ is
planar if and only if the linearized polynomial $\Delta_{f}(X,\epsilon)$ is a
permutation polynomial for each $\epsilon\in\mathbb{F}_{q}^{*}$. The
permutation behaviour of linearized polynomial is well-known. In fact, [19,
Theorem 7.9] tells us that a linearized polynomial is a permutation polynomial
over $\mathbb{F}_{q}$ if and only if its only root in $\mathbb{F}_{q}$ is $0$.
Therefore, in order to show that a DO polynomial $f$ is not planar, it is
sufficient to show that the difference function
$\Delta_{f}(X,Y)=f(X+Y)-f(X)-f(Y)$ has a root in
$\mathbb{F}_{q}^{*}\times\mathbb{F}_{q}^{*}$.

We recall that a DO polynomial function $f$ from $\mathbb{F}_{q}$ to itself is
called 2-to-1 function if the cardinality of the image set on
$\mathbb{F}_{q}^{*}$ is $(q-1)/2.$ Qiu et al. [23] showed that the size of the
image set on $\mathbb{F}_{q}^{*}$ of a planar polynomial $f$ over
$\mathbb{F}_{q}$ must be at least $(q-1)/2.$ For a DO polynomial $f$, Weng and
Zeng [31, Theorem 2.3] gave the following necessary and sufficient condition
for $f$ to be planar.

######  Lemma 8.1.

Let $f$ be a DO polynomial over $\mathbb{F}_{q}$. Then $f$ is planar if and
only if $f$ is 2-to-1.

The Lemma 8.1 has further consequences. First, if a DO polynomial $f$ has a
root $z\in\mathbb{F}_{q}^{*}$, then $-z$ is also a root of $f$. Therefore, the
cardinality of image set of $f$ on $\mathbb{F}_{q}^{*}$ is strictly less than
$(q-1)/2$ and hence, in such a case, $f$ is not planar.

For the second consequence, we begin with an easy observation that if $f(X)$
is a DO polynomial, then so is $f(X^{p^{n}})$. We know that $X^{p^{n}}$ is a
linearized permutation polynomial over $\mathbb{F}_{p^{e}}$. Therefore, the
cardinality of the image set of $f(X)$ and $f(X^{p^{n}})$ on
$\mathbb{F}_{q}^{*}$ is same. Hence if $f(X)$ is planar, then $f(X^{p^{n}})$
is also planar. Therefore in such situations, it would be sufficient to
consider the planarity of $f(X)$ alone.

Another important tool that we would require to study the planarity of the DO
polynomials is the following result of Weil [30].

######  Lemma 8.2.

Let $f(X,Y)$ be absolutely irreducible over $\mathbb{F}_{q}$. Then the number
$N_{f}$ of $(u,v)\in\mathbb{F}_{q}\times\mathbb{F}_{q}$ with $f(u,v)=0$
satisfies

$N_{f}\geq q-(d-1)(d-2)\sqrt{q}-d-1,$

where $d$ is the total degree of $f$.

We now describe the strategy for using Weil bound to understand the planarity
of certain DO polynomials. Let $f$ be a DO polynomial over $\mathbb{F}_{q}$
and consider the difference function $\Delta_{f}(X,Y)=f(X+Y)-f(X)-f(Y)$. If
this difference function has an absolutely irreducible factor, say $h(X,Y)$,
of total degree $d_{h}$, then the Lemma 8.2 gives a lower bound for the
cardinality $N_{h}$ of all the points
$(u,v)\in\mathbb{F}_{q}\times\mathbb{F}_{q}$ such that $h(u,v)=0$. If the
degree of the absolutely irreducible factor $h(X,Y)$ is not too large and $q$
is large enough, then we have many points on the affine algebraic curve
defined by $h(X,Y)=0$. Moreover, if $N_{h}$ is strictly larger than the number
of solutions to $h(X,Y)=0$ with either $X=0$ or $Y=0$. Then the Lemma 8.2
asserts that there exists a point $(u,v)$ in
$\mathbb{F}_{q}^{*}\times\mathbb{F}_{q}^{*}$ such that $h(u,v)=0$ and hence,
for such a point, we have $\Delta_{f}(u,v)=0$, i.e, $\Delta_{f}(X,Y)$ has a
root in $\mathbb{F}_{q}^{*}\times\mathbb{F}_{q}^{*}$. Thus, in order to show
that $f$ is not exceptional planar (i.e., planar over infinitely many
extensions of $\mathbb{F}_{q}$), it is sufficient to show that the difference
function of $f$ contains an absolutely irreducible component with a solution
in $\mathbb{F}_{q}^{*}\times\mathbb{F}_{q}^{*}$.

It is straightforward to see that for $b\in\mathbb{F}_{q}^{*}$, the RDP of
$(m+1)$-th kind admits the following relationship

(8.1) $\displaystyle b^{kd}D_{k,m}(a,X^{d})=D_{k,m}(ab^{d},(Xb^{2})^{d}).$

In view of (8.1), and due to the fact that the planarity property of a
function $f$ remains invariant under the linear transformations (i.e. if
$f(X)$ is planar so is $\alpha f(\lambda X+\mu)+\beta$ with
$\alpha,\lambda\neq 0$), we have the following lemma.

######  Lemma 8.3.

Let $D_{k,m}(a,X)$ be RDP of $(m+1)$-th kind. Then $D_{k,m}(a,X^{d})$ is
planar equivalent over $\mathbb{F}_{q}$ to $D_{k,m}(ab^{d},X^{d})$ for any
$b\in\mathbb{F}_{q}^{*}$.

Over the algebraic closure ${\overline{\mathbb{F}}_{q}}$ of $\mathbb{F}_{q}$,
we derive a useful consequence of Lemma 8.3. Note that one may always choose
$b\in\overline{\mathbb{F}}_{q}$ that satisfies the equation $aX^{d}=1$, and in
fact, such a choice of $b$ shall readily results in a crucial fact that the
factorizations of $\Delta_{D_{k,m}(a,X^{d})}$ and $\Delta_{D_{k,m}(1,X^{d})}$
over ${\overline{\mathbb{F}}_{q}}$ are linearly related. As a consequence, the
absolutely irreducible factors of $\Delta_{D_{k,m}(a,X^{d})}$ are of the same
form for all non-zero $a$. Thus, without loss of generality, one may always
take $a=1$, while checking the absolute irreducibility of certain polynomials.

Now we consider the planarity of the DO polynomials listed in the Appendix A
in three different cases.

Case 1 $p=3$.

The planarity of monomials $\mathfrak{\widehat{D}}_{2}=X^{3^{\alpha}+1}$,
$\mathfrak{\widehat{E}}_{2}=2X^{3^{\alpha}+1}$,
$\mathfrak{\widehat{E}}_{3}=aX^{3^{\alpha}+1}$,
$\mathfrak{\widehat{E}}_{4}=X^{3^{\alpha}+1}$, and
$\mathfrak{\widehat{E}}_{5}=2a^{3}X^{3^{\alpha}+1}$ is well-known by [9,
Theorem 3.3] and these monomials are planar over $\mathbb{F}_{3^{e}}$ if and
only if $e/(\alpha,e)$ is odd.

It is easy to see that $X=a$ is a root of the polynomials
$\mathfrak{\widehat{D}}_{5}=2aX^{4}+a^{3}X^{2}$,
$\mathfrak{\widehat{E}}_{7}=2aX^{6}+a^{3}X^{4}$,
$\mathfrak{\widehat{D}}_{7}=2aX^{6}+2a^{3}X^{4}+2a^{5}X^{2}$,
$\mathfrak{\widehat{E}}_{13}=aX^{12}+a^{3}X^{10}+a^{9}X^{4}$ and
$\mathfrak{\widehat{E}}_{19}=2aX^{18}+2a^{9}X^{10}+a^{13}X^{6}+a^{15}X^{4}$.
Therefore, these DO polynomials are not planar.

Now we consider the planarity of the rest of the DO polynomials one by one.

(i) In the case of binomial
$f(X)=\mathfrak{\widehat{D}}_{4}=2X^{4}+2a^{2}X^{2}$, consider the difference
function

$\Delta_{f}(X,Y)=f(X+Y)-f(X)-f(Y)=XYB(X,Y),$

where $B(X,Y)=X^{2}+Y^{2}-a^{2}$. Now since $(Y-a)\mid(Y^{2}-a^{2})$, and
$Y^{2}-a^{2}$ has no repeated factors, $B(X,Y)$ is absolutely irreducible by
Eisenstein’s criteria. Therefore, by Lemma 8.2, the number $N_{B}$ of
$(u,v)\in\mathbb{F}_{q}\times\mathbb{F}_{q}$ with $B(u,v)=0$ is greater than
or equal to $q-3$. Note that we can obtain at most $4$ solutions $(u,v)$ to
$B(X,Y)=0$ by putting either $X=0$ or $Y=0$. Therefore, when $q-3>4$, there
must exist a root $(u,v)\in\mathbb{F}_{q}^{*}\times\mathbb{F}_{q}^{*}$ of
$B(X,Y)$. Therefore, $\mathfrak{\widehat{D}}_{4}$ is not planar when $q>7$,
i.e., $e\geq 2$. For $e=1$, $\mathfrak{\widehat{D}}_{4}\equiv
X^{2}~{}(\mbox{mod}~{}X^{3}-X)$ which is clearly a planar function.

(ii) The DO binomial
$\mathfrak{\widehat{E}}_{6}=2X^{3(3^{\alpha}+1)}+a^{4}X^{3^{\alpha}+1}$ can be
written as composition of a linearized polynomial and a monomial as
$(2X^{3}+a^{4}X)\circ X^{3^{\alpha}+1}$. Now from [9, Theorem 2.3],
$\mathfrak{\widehat{E}}_{6}$ is planar if and only if $2X^{3}+a^{4}X$ is a
permutation polynomial and $X^{3^{\alpha}+1}$ is planar. Now, since $X=a^{2}$
is a root of the linearized polynomial $2X^{3}+a^{4}X$. Therefore,
$2X^{3}+a^{4}X$ is not a permutation polynomial. Hence,
$\mathfrak{\widehat{E}}_{6}$ is not planar.

(iii) In the case of DO polynomial
$f(X)=\mathfrak{\widehat{E}}_{10}=2X^{10}+a^{4}X^{6}+a^{6}X^{4}$, consider the
difference function $\Delta_{f}(X,Y)=XY~{}h(X,Y),$ where

$h(X,Y)=2(X^{8}+Y^{8})-a^{4}X^{2}Y^{2}+a^{6}(X^{2}+Y^{2}).$

The Magma algebra package [4] reveals that $h(X,Y)$ is absolutely irreducible.
Therefore, by Lemma 8.2, the number $N_{h}$ of solutions
$(u,v)\in\mathbb{F}_{q}\times\mathbb{F}_{q}$ of $h(X,Y)=0$ satisfies
$N_{h}\geq q-42\sqrt{q}-9.$ Now
$h(X,0)=2X^{8}+a^{6}X^{2}=X^{2}(a+X)^{3}(a-X)^{3}$ have total $8$ solutions in
$\mathbb{F}_{q}$. Similarly,
$h(0,Y)=2Y^{8}a+a^{6}Y^{2}=Y^{2}(a-Y)^{3}(a+Y)^{3}$ have total $8$ solutions
in $\mathbb{F}_{q}$. Therefore, total $64$ solutions can be obtained either by
putting $X=0$ or $Y=0$. Now if $q-42\sqrt{q}-9>64$, i.e., $q-42\sqrt{q}-73>0$
then $h(X,Y)$ will have a solution
$(u,v)\in\mathbb{F}_{q}^{*}\times\mathbb{F}_{q}^{*}$. This is true for $e\geq
7$, therefore, for $e\geq 7$, $\mathfrak{\widehat{E}}_{10}$ is not planar. For
$e=1$ $\mathfrak{\widehat{E}}_{10}=X^{2}\pmod{X^{3}-X}$, which is clearly a
planar polynomial. Computations show that for $2\leq e\leq 6$, the cardinality
of the image set of $\mathfrak{\widehat{E}}_{10}$ on $\mathbb{F}_{q}^{*}$ is
strictly less than $(3^{e}-1)/2$. Therefore, by Lemma 8.1,
$\mathfrak{\widehat{E}}_{10}$ is not planar in these cases.

(iv) Consider the DO polynomial,

$\begin{split}f(X)=\mathfrak{\widehat{E}}_{15}&=aX^{28}+2a^{9}X^{12}+a^{13}X^{4}=a(X^{7}+2a^{8}X^{3}+a^{12}X)\circ
X^{4}.\end{split}$

This polynomial is never planar over $\mathbb{F}_{3^{e}}$ when $e$ is even.
Because in this case $4\mid(q-1)$, so the cardinality of image set of $f(X)$
on $\mathbb{F}_{q}^{*}$ is at most $(q-1)/4$ and thus, by Lemma 8.1, $f(X)$ is
not planar.

When $e$ is odd, we consider the difference function of $f$,

$\Delta_{f}(X,Y)=aXY(X^{2}+Y^{2})~{}h(X,Y),$

where

$\begin{split}h(X,Y)=X^{24}-X^{22}Y^{2}+X^{20}Y^{4}-X^{18}Y^{6}+X^{16}Y^{8}-X^{14}Y^{10}+X^{12}Y^{12}\\\
-X^{10}Y^{14}+X^{8}Y^{16}-X^{6}Y^{18}+X^{4}Y^{20}-X^{2}Y^{22}+Y^{24}\\\
-a^{8}X^{6}Y^{2}+a^{8}X^{4}Y^{4}-a^{8}X^{2}Y^{6}+a^{12}.\end{split}$

Again, the Magma algebra package [4] shows that the polynomial
$h^{\prime}(X,Y)$ obtained from $h(X,Y)$ by putting $a=1$, is absolutely
irreducible. Therefore, by Lemma 8.2, the number $N_{h^{\prime}}$ of solutions
$(u,v)\in\mathbb{F}_{q}\times\mathbb{F}_{q}$ of $h^{\prime}(X,Y)=0$ satisfies
$N_{h^{\prime}}\geq q-506\sqrt{q}-25$. Also, $h^{\prime}(X,0)=X^{24}+1$ and
this has no root in odd degree extensions of $\mathbb{F}_{3}$. Similarly,
$h^{\prime}(0,Y)=Y^{24}+1$ has no root in odd degree extensions of
$\mathbb{F}_{3}$. Therefore, there is no solution to $h^{\prime}(X,Y)=0$
corresponding to $XY=0$. If $q-506\sqrt{q}-25>0$ then $h^{\prime}(X,Y)$ has a
root $(u,v)\in\mathbb{F}_{q}^{*}\times\mathbb{F}_{q}^{*}.$ This holds true for
all $e\geq 12$. Therefore, $\mathfrak{\widehat{E}}_{15}$ is not planar over
$\mathbb{F}_{3^{e}}$ for $e\geq 12$. In the case of $e=1$, the polynomial
$f(X)=\mathfrak{\widehat{E}}_{15}=aX^{2}\pmod{X^{3}-X}$ which is clearly a
planar function. Computations show that for $e=5,7,9,11$, the cardinality of
the image set of $\mathfrak{\widehat{E}}_{15}$ on $\mathbb{F}_{3^{e}}^{*}$ is
strictly less than $(3^{e}-1)/2$, therefore, $\mathfrak{\widehat{E}}_{15}$ is
not planar in these cases. In the case of $e=3$, $\mathfrak{\widehat{E}}_{15}$
is planar for every choice of $a\in\mathbb{F}_{27}^{*}$.

Case 2 $p=5$.

The planarity of DO monomials $\mathfrak{\widehat{D}}_{2}=3X^{5^{\alpha}+1}$,
$\mathfrak{\widehat{D}}_{3}=2aX^{5^{\alpha}+1}$,
$\mathfrak{\widehat{E}}_{2}=4X^{5^{\alpha}+1}$,
$\mathfrak{\widehat{E}}_{3}=3aX^{5^{\alpha}+1}$,
$\mathfrak{\widehat{G}}_{2}=X^{5^{\alpha}+1}$,
$\mathfrak{\widehat{H}}_{2}=2X^{5^{\alpha}+1}$,
$\mathfrak{\widehat{H}}_{3}=aX^{5^{\alpha}+1}$, and
$\mathfrak{\widehat{H}}_{4}=3X^{5^{\alpha}+1}$ is well-known by [9, Theorem
3.3] and these monomials are planar over $\mathbb{F}_{5^{e}}$ whenever
$e/(\alpha,e)$ is odd.

It is straightforward to see that $X=a$ is a root of the DO binomial
$\mathfrak{\widehat{E}}_{7}=4a^{5}X^{2}+aX^{6}$ and hence, it is not planar.

Now we consider the planarity of the rest of the DO polynomials one by one.

(i) For DO binomial $f(X)=\mathfrak{\widehat{G}}_{6}=2a^{4}X^{2}+X^{6}$,
consider the difference function of $f(X),$

$\Delta_{f}(X,Y)=XYB(X,Y),$

where $B(X,Y)=X^{4}+Y^{4}-a^{4}$. It is easy to see that $Y-a\mid Y^{4}-a^{4}$
and $Y^{4}-a^{4}$ has no repeated roots. Therefore, by Eisenstein’s criteria,
$B(X,Y)$ is absolutely irreducible. Thus, by Lemma 8.2, the number of
solutions $(u,v)\in\mathbb{F}_{q}\times\mathbb{F}_{q}$ of $B(X,Y)=0$ follows
the inequality $N_{B}\geq q-6\sqrt{q}-5.$ Now, at most $16$ roots of $B(X,Y)$
can be obtained by putting either $X=0$ or $Y=0$. Therefore if
$q-6\sqrt{q}-5>16$, $B(X,Y)$ will have a solution
$(u,v)\in\mathbb{F}_{q}^{*}\times\mathbb{F}_{q}^{*}$, which holds for all
$e\geq 3$. Therefore, $\mathfrak{\widehat{G}}_{6}$ is not planar over
$\mathbb{F}_{5^{e}}$ for $e\geq 3$. When $e=1$,
$f(X)=3X^{2}(\mbox{mod}~{}(X^{5}-X))$ which is clearly a planar function. For
$e=2$, the number of solutions of the equation $X^{4}+Y^{4}=a^{4}$ in
$\mathbb{F}_{5^{2}}\times\mathbb{F}_{5^{2}}$ is 40, which is greater than 16.
Therefore, $\mathfrak{\widehat{G}}_{6}$ is not planar in this case.

(ii) In the case of DO trinomial
$f(X)=\mathfrak{\widehat{G}}_{11}=-aX^{10}+a^{5}X^{6}+2a^{9}X^{2},$ consider
the difference function $\Delta_{f}(X,Y)=XY~{}h(X,Y),$ where

$h(X,Y)=3aX^{4}Y^{4}+a^{5}X^{4}+a^{5}Y^{4}+4a^{9}.$

The Magma algebra package [4] shows that $h(X,Y)$ is absolute irreducible.
Therefore, by Lemma 8.2, the number $N_{h}$ of solutions
$(u,v)\in\mathbb{F}_{q}\times\mathbb{F}_{q}$ of $h(X,Y)=0$ satisfies
$N_{h}\geq q-42\sqrt{q}-9$. Now, $h(X,0)=X^{4}-a^{4}=0$ can have at most $4$
solutions. Similarly, $h(0,Y)=Y^{4}-a^{4}=0$ can have at most $4$ solutions.
Therefore, at most $16$ solutions can be obtained by putting either $X=0$ or
$Y=0$. Now, if $q-42\sqrt{q}-9>16,$ i.e., $q-42\sqrt{q}-25>0$ then $h(X,Y)$
will have a solution $(u,v)\in\mathbb{F}_{q}^{*}\times\mathbb{F}_{q}^{*}$.
This is true for $e\geq 5$, therefore, for $e\geq 5$,
$\mathfrak{\widehat{G}}_{11}$ is not planar. For $e=1$,
$\mathfrak{\widehat{G}}_{11}=2aX^{2}$ is clearly a planar function. For
$e=2,4$, computations show that the cardinality of the image set of
$\mathfrak{\widehat{G}}_{11}$ on $\mathbb{F}_{5^{e}}^{*}$ is strictly less
than $(5^{e}-1)/2$. Therefore, $\mathfrak{\widehat{G}}_{11}$ is not planar in
these cases. For $e=3$, computations show that $\mathfrak{\widehat{G}}_{11}$
is planar for every choice of $a\in\mathbb{F}_{125}^{*}$.

Case 3 $p>5$.

In this case, the only DO polynomials we are getting are the monomials of the
form $bX^{p^{\alpha}+1}$ where $b\in\mathbb{F}_{q}^{*}$ and by [9, Theorem
3.3], these monomials are planar over $\mathbb{F}_{p^{e}}$ whenever
$e/(\alpha,e)$ is odd.

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##  Appendix A The complete list of DO polynomials

Here, we present the complete list of DO polynomials obtained from polynomial
$\mathfrak{\widehat{D}}_{k,m}$ over a finite field of odd characteristic.

  1. (1)

The case $p=3$.

    1. (a)

When $m\equiv 0\pmod{3}$

      1. (i)

$k=2\cdot 3^{\ell}$ then $X^{3^{n+\ell}(3^{\alpha}+1)}$ for nonnegative
integers $\alpha$, $n$ and $\ell$.

      2. (ii)

$k=4\cdot 3^{\ell}$ then $2a^{2}X^{2\cdot 3^{n+\ell}}+2X^{4\cdot 3^{n+\ell}}$
for nonnegative integers $n$ and $\ell$.

      3. (iii)

$k=5\cdot 3^{\ell}$ then $a^{3}X^{2\cdot 3^{n+\ell}}+2aX^{4\cdot 3^{n+\ell}}$
for nonnegative integers $n$ and $\ell$.

      4. (iv)

$k=7\cdot 3^{\ell}$ then $2a^{5}X^{2\cdot 3^{n+\ell}}+2a^{3}X^{4\cdot
3^{n+\ell}}+2aX^{2\cdot 3^{n+\ell+1}}$ for nonnegative integers $n$ and
$\ell$.

    2. (b)

When $m\equiv 1\pmod{3}$

      1. (i)

$k=2$ then $2X^{3^{n}(3^{\alpha}+1)}$ for nonnegative integers $\alpha$ and
$n$.

      2. (ii)

$k=3$ then $aX^{3^{n}(3^{\alpha}+1)}$ for nonnegative integers $\alpha$ and
$n$.

      3. (iii)

$k=4$ then $X^{3^{n}(3^{\alpha}+1)}$ for nonnegative integers $\alpha$ and
$n$.

      4. (iv)

$k=5$ then $2a^{3}X^{3^{n}(3^{\alpha}+1)}$ for nonnegative integers $\alpha$
and $n$.

      5. (v)

$k=6$ then $a^{4}X^{3^{n}(3^{\alpha}+1)}+2X^{3^{n+1}(3^{\alpha}+1)}$ for
nonnegative integers $\alpha$ and $n$.

      6. (vi)

$k=7$ then $a^{3}X^{4\cdot 3^{n}}+2aX^{2\cdot 3^{n+1}}$ for nonnegative
integer $n$.

      7. (vii)

$k=10$ then $a^{6}X^{4\cdot 3^{n}}+a^{4}X^{2\cdot 3^{n+1}}+2X^{10\cdot 3^{n}}$
for nonnegative integer $n$.

      8. (viii)

$k=13$ then $a^{9}X^{4\cdot 3^{n}}+a^{3}X^{10\cdot 3^{n}}+aX^{4\cdot 3^{n+1}}$
for nonnegative integer $n$.

      9. (ix)

$k=15$ then $a^{13}X^{4\cdot 3^{n}}+2a^{9}X^{4\cdot 3^{n+1}}+aX^{28\cdot
3^{n}}$ for nonnegative integer $n$.

      10. (x)

$k=19$ then $a^{15}X^{4\cdot 3^{n}}+a^{13}X^{2\cdot 3^{n+1}}+2a^{9}X^{10\cdot
3^{n}}+2aX^{2\cdot 3^{n+2}}$ for nonnegative integer $n$.

    3. (c)

When $m\equiv 2\pmod{3}$

      1. (i)

$k=3$ then $2aX^{3^{n}(3^{\alpha}+1)}$ for nonnegative integers $\alpha$ and
$n$.

      2. (ii)

$k=4$ then $a^{2}X^{3^{n}(3^{\alpha}+1)}$ for nonnegative integers $\alpha$
and $n$.

      3. (iii)

$k=5$ then $aX^{3^{n}(3^{\alpha}+1)}$ for nonnegative integers $\alpha$ and
$n$.

      4. (iv)

$k=6$ then $2a^{4}X^{3^{n}(3^{\alpha}+1)}$ for nonnegative integers $\alpha$
and $n$.

      5. (v)

$k=7$ then $a^{5}X^{3^{n}(3^{\alpha}+1)}+2aX^{3^{n+1}(3^{\alpha}+1)}$ for
nonnegative integers $\alpha$ and $n$.

      6. (vi)

$k=8$ then $a^{4}X^{4\cdot 3^{n}}+2a^{2}X^{2\cdot 3^{n+1}}$ for nonnegative
integer $n$.

      7. (vii)

$k=11$ then $a^{7}X^{4\cdot 3^{n}}+a^{5}X^{2\cdot 3^{n+1}}+2aX^{10\cdot
3^{n}}$ for nonnegative integer $n$.

      8. (viii)

$k=14$ then $a^{10}X^{4\cdot 3^{n}}+a^{4}X^{10\cdot 3^{n}}+a^{2}X^{4\cdot
3^{n+1}}$ for nonnegative integer $n$.

      9. (ix)

$k=16$ then $a^{14}X^{4\cdot 3^{n}}+2a^{10}X^{4\cdot 3^{n+1}}+a^{2}X^{28\cdot
3^{n}}$ for nonnegative integer $n$.

      10. (x)

$k=20$ then $a^{16}X^{4\cdot 3^{n}}+a^{14}X^{2\cdot 3^{n+1}}+2a^{10}X^{10\cdot
3^{n}}+2a^{2}X^{2\cdot 3^{n+2}}$ for nonnegative integer $n$.

  2. (2)

The case $p=5$.

    1. (a)

When $m\equiv 0\pmod{5}$

      1. (i)

$k=2\cdot 5^{\ell}$ then $3X^{5^{n+\ell}(5^{\alpha}+1)}$ for non negative
integers $\alpha$, $n$ and $\ell$.

      2. (ii)

$k=3\cdot 5^{\ell}$ then $2aX^{5^{n+\ell}(5^{\alpha}+1)}$ for non negative
integers $\alpha$, $n$ and $\ell$.

    2. (b)

When $m\equiv 1\pmod{5}$

      1. (i)

$k=2$ then $4X^{5^{n}(5^{\alpha}+1)}$ for nonnegative integers $\alpha$ and
$n$.

      2. (ii)

$k=3$ then $3aX^{5^{n}(5^{\alpha}+1)}$ for nonnegative integers $\alpha$ and
$n$.

      3. (iii)

$k=7$ then $4a^{5}X^{2\cdot 5^{n}}+aX^{6\cdot 5^{n}}$ for nonnegative integer
$n$.

    3. (c)

When $m\equiv 2\pmod{5}$

      1. (i)

$k=3$ then $4aX^{5^{n}(5^{\alpha}+1)}$ for nonnegative integers $\alpha$ and
$n$.

      2. (ii)

$k=4$ then $3a^{2}X^{5^{n}(5^{\alpha}+1)}$ for nonnegative integer $n$.

      3. (iii)

$k=8$ then $4a^{6}X^{2\cdot 5^{n}}+a^{2}X^{6\cdot 5^{n}}$ for nonnegative
integer $n$.

    4. (d)

When $m\equiv 3\pmod{5}$

      1. (i)

$k=2$ then $2X^{5^{n}(5^{\alpha}+1)}$ for nonnegative integers $\alpha$ and
$n$.

      2. (ii)

$k=6$ then $2a^{4}X^{2\cdot 5^{n}}+X^{6\cdot 5^{n}}$ for nonnegative integer
$n$.

      3. (iii)

$k=11$ then $2a^{9}X^{2\cdot 5^{n}}+a^{5}X^{6\cdot 5^{n}}+4aX^{2\cdot
5^{n+1}}$ for nonnegative integer $n$.

    5. (e)

When $m\equiv 4\pmod{5}$

      1. (i)

$k=2$ then $2X^{5^{n}(5^{\alpha}+1)}$ for nonnegative integers $\alpha$ and
$n$.

      2. (ii)

$k=3$ then $aX^{5^{n}(5^{\alpha}+1)}$ for nonnegative integer $n$.

      3. (iii)

$k=4$ then $3X^{5^{n}(5^{\alpha}+1)}$ for nonnegative integer $n$.

  3. (3)

The case $p>5$.
In this case, we are getting DO polynomials of the form $bX^{p^{\alpha}+1}$
where $b\in\mathbb{F}_{q}^{*}$.