# Low $c$-Differential Uniformity of the Swapped Inverse Function in Odd
Characteristic

Jaeseong Jeong1 Namhun Koo2 Soonhak Kwon1
Email<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
1Applied Algebra and Optimization Research Center, Sungkyunkwan University,
Suwon, Korea
2Institute of Mathematical Sciences, Ewha Womans University, Seoul, Korea

###### Abstract

The study of Boolean functions with low $c$-differential uniformity has become
recently an important topic of research. However, in odd characteristic case,
there are not many results on the ($c$-)differential uniformity of functions
that are not power functions. In this paper, we investigate the
$c$-differential uniformity of the swapped inverse functions in odd
characteristic, and show that their $c$-differential uniformities are at most
6 except for some special case.

Keywords. $c$-Differential Uniformity, Differential Uniformity, Permutation,

Mathematics Subject Classification(2020) 94A60, 06E30

##  1 Introduction

In this paper, we mainly focus on the finite fields in odd characteristic.
Throughout this paper, let

  * •

$\mathbb{F}_{p^{n}}$ be the finite field of $p^{n}$ elements and
$\mathbb{F}_{p^{n}}^{*}$ and
$\mathbb{F}_{p^{n}}^{*}=\mathbb{F}_{p^{n}}\setminus\\{0\\}$ be the
multiplicative group, where $p$ is an odd prime.

  * •

$Inv$ be the multiplicative inverse function on $\mathbb{F}_{p^{n}}$ such that
$Inv(x)=x^{-1}$ for all $x\in\mathbb{F}_{p^{n}}^{*}$ and $Inv(0)=0^{-1}=0$.

Recently, Ellinsen et. al.[6] introduced a new concept $c$-differential
uniformity, which is useful to estimate resistance against some variants of
differential attack[3].

######  Definition 1.1.

([6]) Let $F:\mathbb{F}_{p^{n}}\longrightarrow\mathbb{F}_{p^{n}}$ be a
function and $c\in\mathbb{F}_{p^{n}}$.
(i) We denote the $c$-differential of $F$ by ${}_{c}D_{a}F(x)=F(x+a)-cF(x)$.
(ii) Let $a,b\in\mathbb{F}_{p^{n}}$. We denote ${}_{c}\Delta_{F}(a,b)$ by the
number of solutions in $\mathbb{F}_{p^{n}}$ of ${}_{c}D_{a}F(x)=b$.
(iii) The $c$-differential uniformity of $F$ is defined by
${}_{c}\Delta_{F}=\max\\{_{c}\Delta_{F}(a,b):a,b\in\mathbb{F}_{p^{n}}\text{
and }a\neq 0\text{ if }c=1\\}$.
(iv) $F$ is called a perfect $c$-nonlinear(PcN) function if
${}_{c}\Delta_{F}=1$.
(v) $F$ is called a almost perfect $c$-nonlinear(APcN) function if
${}_{c}\Delta_{F}=2$.

Functions with low $c$-differential uniformity provide good resistance to
differential attacks of this type. Therefore, finding functions with low
$c$-differential uniformity has been a good research topic, and many new
classes of functions with low $c$-differential uniformity were proposed.
Please see [1, 2, 7, 11, 16, 19, 20, 21, 22, 23, 24, 26] for some new
progress.

In even characteristic, it is known that $Inv\circ(0,1)$ has good
cryptographic properties, for example, low differential uniformity[15], high
nonlinearity[15], low boomerang uniformity[14], and low differential-linear
uniformity[10]. Very recently, the $c$-differential uniformity and the
$c$-boomerang uniformity are investigated in [19].

However, the $c$-differential uniformity of $Inv\circ(0,1)$ in odd
characteristic has not been investigated, and the author of [19] suggested
that it would be a good research topic. In this paper, we investigate the
$c$-differential uniformity of the permutation obtained by swapping two points
from $Inv$, which can be expressed as $Inv\circ(\alpha,\beta)$ with
$\alpha\neq\beta\in\mathbb{F}_{p^{n}}$. In [11], we show that the
$c$-differential uniformity is not preserved under affine equivalence but is
preserved under some specific affine equivalence in even characteristic. The
next proposition shows that the $c$-differential uniformity is also preserved
under this specific affine equivalence in odd characteristic.

######  Proposition 1.2.

We call an affine permutation of the form $A(x)=a_{1}x+a_{0}$ where
$a_{0}\in\mathbb{F}_{p^{n}}$ and $a_{1}\in\mathbb{F}_{p^{n}}^{*}$ an affine
permutation of degree one. We call two functions $F$ and $F^{\prime}$ are
affine equivalent of degree one if $F=A_{1}\circ F^{\prime}\circ A_{2}$ where
$A_{1}$ and $A_{2}$ are affine permutations of degree one. If two functions
$F$ and $F^{\prime}$ defined on $\mathbb{F}_{p^{n}}$ are affine equivalent of
degree one, then ${}_{c}\Delta_{F}={}_{c}\Delta_{F^{\prime}}$.

###### Proof.

We write $F=A_{1}\circ F^{\prime}\circ A_{2}$ and
$F^{\prime\prime}=F^{\prime}\circ A_{2}$, where $A_{1}$ and $A_{2}$ are affine
permutations of degree one. $F^{\prime}$ and $F^{\prime\prime}$ have the same
$c$-differential uniformity in even characteristic[7], and the proof in [7]
can be naturally extended for odd characteristic case. It remains to show that
$F=A_{1}\circ F^{\prime\prime}$ and $F^{\prime\prime}$ has the same
$c$-differential uniformity. We denote $A_{1}(x)=a_{1}x+a_{0}$ where
$a_{1}\in\mathbb{F}_{p^{n}}^{*}$ and $a_{0}\in\mathbb{F}_{p^{n}}$. Let
$a,b\in\mathbb{F}_{p^{n}}$. Then ${}_{c}D_{a}F(x)=b$ implies that

$\displaystyle b$ $\displaystyle=F(x+a)-cF(x)=(A_{1}\circ
F^{\prime\prime})(x+a)-c(A_{1}\circ
F^{\prime\prime})(x)=A_{1}(F^{\prime\prime}(x+a))-cA_{1}(F^{\prime\prime}(x))$
$\displaystyle=a_{1}F^{\prime\prime}(x+a)+a_{0}-c(a_{1}F^{\prime\prime}(x)+a_{0})=a_{1}(F^{\prime\prime}(x+a)-cF^{\prime\prime}(x))+a_{0}(1-c).$

So we have
${}_{c}D_{a}F^{\prime\prime}(x)=a_{1}^{-1}\left(b+a_{0}(c-1)\right)$ and hence
${}_{c}\Delta_{F}(a,b)={}_{c}\Delta_{F^{\prime\prime}}\left(a,a_{1}^{-1}\left(b+a_{0}(c-1)\right)\right)$.
Since the map from $b$ to $a_{1}^{-1}\left(b+a_{0}(c-1)\right)$ is bijective
for all $c\in\mathbb{F}_{p^{n}}$, we obtain
${}_{c}\Delta_{F}={}_{c}\Delta_{F^{\prime\prime}}$. ∎

If $F^{\prime}=Inv\circ(\alpha,\beta)$, $A(x)=\alpha x$ where
$\alpha\in\mathbb{F}_{p^{n}}^{*}$ and $\beta\in\mathbb{F}_{p^{n}}$, then
$F=Inv\circ(1,\alpha^{-1}\beta)=A\circ F^{\prime}\circ A$ and $F^{\prime}$
have the same $c$-differential uniformity for all $c\in\mathbb{F}_{p^{n}}$ by
Proposition 1.2. Therefore, in this paper, we investigate the $c$-differential
uniformity of $Inv\circ(1,\gamma)$ where
$\gamma\in\mathbb{F}_{p^{n}}\setminus\\{1\\}$.

For even characteristic case, there are many non-power functions with low
differential uniformity (see [11] and references therein). The situation for
odd characteristic case is quite different. That is, though there are many
functions with low differential uniformity (see [4, 5, 8, 9, 12, 13, 18, 26,
27, 28] and the references therein), most of them are power functions. Also,
to the best of our knowledge, there is no published article on the
differential uniformity of $Inv\circ(1,\gamma)$ in odd characteristic.
Therefore, we will also study the special case $c=1$, i.e., the differential
uniformity of $Inv\circ(1,\gamma)$.

The rest of this paper is organized as follows. In section 2, we introduce
some preliminaries which are necessary in our subsequent sections. In section
3, we investigate the $c$-differential uniformity of $Inv\circ(0,1)$ in odd
characteristic, i.e., the case $\gamma=0$. In section 4, we investigate the
$c$-differential uniformity of $Inv\circ(1,\gamma)$ with $\gamma\neq 0,1$. In
section 5, we give a concluding remark.

##  2 Some Previous Results

In this section, we recall some previous results. First we briefly mention the
$c$-differential uniformity of $Inv$ and $Inv\circ(0,1)$ in even
characteristic. It is known [6] that $Inv$ is APcN if and only if
$\operatorname{Tr}(c)=\operatorname{Tr}\left(\frac{1}{c}\right)=1$, otherwise
${}_{c}\Delta_{Inv}=3$, where $p=2$ with $c\neq 0,1$ and $\operatorname{Tr}$
is the field trace. It is also known [19] that
${}_{c}\Delta_{Inv\circ(0,1)}=4$ when $n\geq 4$. Hence we can see that the
swapped inverse function has slightly higher $c$-differential uniformity than
the inverse function when $p=2$.

For odd characteristic case, if $c\neq 0,1$, then $Inv$ is APcN if and only if
$c\in\\{4,4^{-1}\\}$ or both $c^{2}-4c$ and $1-4c$ are not squares in
$\mathbb{F}_{p^{n}}$, otherwise, ${}_{c}\Delta_{Inv}=3$ (see [6]). The
differential uniformity of $Inv$ is as follows[5, 8] (as in usual notation, we
use the notation $D_{a}F(x)={}_{1}D_{a}F(x)$ and
$\Delta_{F}(a,b)={}_{1}\Delta_{F}(a,b)$, throughout this paper).

$\Delta_{Inv}=\begin{cases}2&\text{ if }p^{n}\equiv 2\pmod{3},\\\ 3&\text{ if
}p=3,\\\ 4&\text{ if }p^{n}\equiv 1\pmod{3}.\end{cases}$

To summarize, in odd characteristic, one has $2\leq\Delta_{Inv}\leq 4$ and
$2\leq{}_{c}\Delta_{Inv}\leq 3$ where $c\neq 0,1$. We will later show that the
swapped inverse function has slightly higher ($c$-)differential uniformities
than the inverse function in odd characteristic. For the details, see section
3 and section 4.

The quadratic character in $\mathbb{F}_{p^{n}}$ is defined as follows

$\chi(x)=\begin{cases}1&\text{ if }x\in\mathbb{F}_{p^{n}}^{*}\text{ is a
square,}\\\ -1&\text{ if }x\in\mathbb{F}_{p^{n}}^{*}\text{ is a nonsquare,}\\\
0&\text{ if }x=0.\end{cases}$

It is known that $\chi(x)=x^{\frac{p^{n}-1}{2}}$. The followings are also well
known and useful for our results.

$\begin{array}[]{ll}\chi(-1)=\begin{cases}1&\text{ if }p^{n}\equiv
1\pmod{4},\\\ -1&\text{ if }p^{n}\equiv 3\pmod{4},\end{cases}&\
\chi(5)=\begin{cases}1&\text{ if }p^{n}\equiv\pm 1\pmod{5},\\\ -1&\text{ if
}p^{n}\equiv\pm 2\pmod{5},\end{cases}\\\ &\\\ \chi(-3)=\begin{cases}1&\text{
if }p^{n}\equiv 1\pmod{3},\\\ -1&\text{ if }p^{n}\equiv
2\pmod{3},\end{cases}&\ \chi(2)=\begin{cases}1&\text{ if }p^{n}\equiv\pm
1\pmod{8},\\\ -1&\text{ if }p^{n}\equiv\pm 3\pmod{8}.\end{cases}\end{array}$

We also denote $\sqrt{x}$ by any square root of $x$ in $\mathbb{F}_{p^{n}}$ if
$\chi(x)=1$ throughout this paper. Next we introduce a well-known lemma about
the number of solutions of quadratic equations.

######  Lemma 2.1.

Let $a_{2}\in\mathbb{F}_{p^{n}}^{*}$ and $a_{1},a_{0}\in\mathbb{F}_{p^{n}}$
and $D=a_{1}^{2}-4a_{0}a_{2}$. Then,

$\\#\\{x\in\mathbb{F}_{p^{n}}:a_{2}x^{2}+a_{1}x+a_{0}=0\\}=\begin{cases}2&\text{
if }\chi(D)=1,\\\ 1&\text{ if }\chi(D)=0,\\\ 0&\text{ if }\chi(D)=-1.\\\
\end{cases}$

The following lemma is a generalization of Lemma 2.6 in [11] to the odd
characteristic case. This lemma is very simple and useful to our results.

######  Lemma 2.2.

Let $c\neq 0$. Then ${}_{c}\Delta_{F}(a,b)={}_{c^{-1}}\Delta_{F}(-a,-bc^{-1})$
for all $a\in\mathbb{F}_{p^{n}}$ and $b\in\mathbb{F}_{p^{n}}$, and hence
${}_{c}\Delta_{F}={}_{c^{-1}}\Delta_{F}$.

###### Proof.

Let $a\in\mathbb{F}_{p^{n}}$ and $b\in\mathbb{F}_{p^{n}}$. Then, it is clear
that $x$ is a solution of ${}_{c}D_{a}F(x)=b$ if and only if $x+a$ is a
solution of ${}_{c^{-1}}D_{-a}(x)=-bc^{-1}$. Hence we obtain
${}_{c}\Delta_{F}(a,b)={}_{c^{-1}}\Delta_{F}(-a,-bc^{-1})$. Therefore, we get
${}_{c}\Delta_{F}={}_{c^{-1}}\Delta_{F}$. ∎

##  3 The $c$-differential uniformity of $Inv\circ(0,1)$

Let $F=Inv\circ(0,1)$. Since $F$ is permutation, we have
${}_{c}\Delta_{F}(0,b)=1$ for all $b\in\mathbb{F}_{p^{n}}$ and
$c\in\mathbb{F}_{p^{n}}\setminus\\{1\\}$, and we also have
${}_{0}\Delta_{F}(a,b)=1$ for all $a,b\in\mathbb{F}_{p^{n}}$. Hence,
throughout this section we assume that

$F=Inv\circ(0,1),\ a\neq 0,\ c\neq 0.$ (1)

We also denote throughout this section that

$P=\\{x\in\mathbb{F}_{p^{n}}:F(x)\neq Inv(x)\\}=\\{0,1\\},\ \
P_{a}=P\cup\\{x-a:x\in P\\}=\\{0,1,-a,1-a\\}.$

It is easy to see that

${}_{c}\Delta_{F}(a,b)=\\#\\{x\in
P_{a}:{}_{c}D_{a}F(x)=b\\}+\\#\\{x\in\mathbb{F}_{p^{n}}\setminus
P_{a}:{}_{c}D_{a}F(x)=b\\}.$

The following lemma characterizes the case
$\\#\\{x\in\mathbb{F}_{p^{n}}\setminus P_{a}:D_{a}F(x)=b\\}=2$.

######  Lemma 3.1.

Under the same assumption as in (1) let $a\in\mathbb{F}_{p^{n}}^{*}$ and
$b\in\mathbb{F}_{p^{n}}$. Then ${}_{c}D_{a}F(x)=b$ has at most two solutions
in $\mathbb{F}_{p^{n}}\setminus P_{a}$. Furthermore, ${}_{c}D_{a}F(x)=b$ has
two solutions in $\mathbb{F}_{p^{n}}\setminus P_{a}$ if and only if
$(ab+c-1)^{2}-4abc$ is a nonzero square and $b\neq 0$, $(b+c)a+b+c-1\neq 0$,
$(1-b)a+b+c-1\neq 0$.

###### Proof.

If $x\in\mathbb{F}_{p^{n}}\setminus P_{a}$ then
$b={}_{c}D_{a}F(x)={}_{c}D_{a}Inv(x)$ implies that $bx^{2}+(ab+c-1)x+ac=0$
which has at most two solutions. Hence ${}_{c}D_{a}F(x)=b$ has at most two
solutions in $\mathbb{F}_{p^{n}}\setminus P_{a}$.
${}_{c}D_{a}F(x)=b$ has two solutions in $\mathbb{F}_{p^{n}}\setminus P_{a}$
implies that $bx^{2}+(ab+c-1)x+ac=0$ has two solutions which is equivalent to
$(ab+c-1)^{2}-4abc$ is a nonzero square and $b\neq 0$ by Lemma 2.1. If there
is a solution $x_{0}\in P_{a}$ of $bx^{2}+(ab+c-1)x+ac=0$ then $x_{0}$ cannot
be a solution of ${}_{c}D_{a}F(x)=b$. So we require that
$bx^{2}+(ab+c-1)x+ac=0$ has no solutions in $P_{a}$. We check this by
substituting every element in $P_{a}$ to $bx^{2}+(ab+c-1)x+ac=0$.

  * •

If $x=0$ or $x=-a$ then we have $a=0$ which is a contradiction to $a\neq 0$.

  * •

If $x=1$ then we have $(b+c)a+b+c-1=0$.

  * •

If $x=1-a$ then we have $(1-b)a+b+c-1=0$.

Thus $bx^{2}+(ab+c-1)x+ac=0$ has no solutions in $P_{a}$ if and only if
$(b+c)a+b+c-1\neq 0$ and $(1-b)a+b+c-1\neq 0$.
If all conditions in this theorem are satisfied, then $bx^{2}+(ab+c-1)x+ac=0$
has two solutions in $\mathbb{F}_{p^{n}}\setminus P_{a}$. Since all solutions
are not contained in $P_{a}$, they are also solutions of ${}_{c}D_{a}F(x)=b$.
∎

In the following lemma, we provide a lower bound on ${}_{c}\Delta_{F}$ where
$c\neq 0$. Unfortunately, the following lemma shows that $F$ is not APcN for
all $c\neq 0$, for every odd prime $p$ and $n>0$.

######  Lemma 3.2.

Under the same assumption as in (1), if $p^{n}>5$ then ${}_{c}\Delta_{F}\geq
3$.

###### Proof.

Let $b=a^{-1}-c={}_{c}D_{a}F(0)$. Then ${}_{c}D_{a}F(x)=b$ has at least one
solution $x=0$ in $P_{a}$. We investigate required conditions that
${}_{c}D_{a}F(x)=b$ has two solutions in $\mathbb{F}_{p^{n}}\setminus P_{a}$
using Lemma 3.1. We require that
$(ab+c-1)^{2}-4abc=(c-ca)^{2}-4c(1-ca)=c^{2}a^{2}+2c^{2}a+c^{2}-4c=c^{2}(a+1)^{2}-4c$
is a nonzero square. And we have $(b+c)a+b+c-1=a^{-1}\cdot a+a^{-1}-1=a^{-1}$
and $(1-b)a+b+c-1=(c+1-a^{-1})\cdot a+a^{-1}-1=a^{-1}((c+1)a^{2}-2a+1)$. So we
require $a\neq 0$ and $(c+1)a^{2}-2a+1\neq 0$. Since $(c+1)a^{2}-2a+1=0$ has
no solutions if $(-2)^{2}-4(c+1)=-4c$ is not a square by Lemma 2.1, it is
enough to check $-c$ is not a square.

(i) We assume that $-c$ is a square.

  * •

If $c\neq-4$, then we set $a=-1$. Then $c^{2}(a+1)^{2}-4c=-4c$ is a nonzero
square and $(c+1)a^{2}-2a+1=c+4\neq 0$.

  * •

If $c=-4$ and $p\neq 3,5,7$, then we choose $a=-\frac{1}{4}=c^{-1}$. Then
$c^{2}(a+1)^{2}-4c=25$ is a nonzero square and
$(c+1)a^{2}-2a+1=-\frac{3}{16}-\frac{1}{2}+1=\frac{21}{16}\neq 0$.

  * •

If $c=-4$ and $p=3,5,7$, then $c^{2}(a+1)^{2}-4c=16(a+1)^{2}+16=(a+1)^{2}+1$
when $p=3,5$ and $c^{2}(a+1)^{2}-4c=2\left((a+1)^{2}+1\right)$ when $p=7$.
Since $2$ is a square for every $n$ when $p=7$, it is enough to show that
there is $\alpha_{0}\in\mathbb{F}_{p^{n}}\setminus\mathbb{F}_{p}$ such that
both $\alpha_{0}$ and $\alpha_{0}+1$ are squares. We denote
$\mathbb{F}_{p^{n}}=\mathbb{F}_{p}[X]/\langle f(X)\rangle$ for some $f(X)$ and
$f(g)=0$ for some $g\in\mathbb{F}_{p^{n}}$. Then $\mathbb{F}_{p^{n}}$ is
partitioned by subsets of the form $\\{x+\alpha:\alpha\in\mathbb{F}_{p}\\}$
where $x\in\mathbb{F}_{p^{n}}\cdot g$. If all such subsets have at most
$\frac{p-1}{2}$ squares then there are at most $\frac{p-1}{2p}(p^{n}-p)$ which
is a contradiction to the fact that there are $\frac{1}{2}(p^{n}-p)$ squares
in $\mathbb{F}_{p^{n}}\setminus\mathbb{F}_{p}$. Hence there is
$\\{x+\alpha:\alpha\in\mathbb{F}_{p}\\}\neq\mathbb{F}_{p}$ containing
$\frac{p+1}{2}$ squares. If there are at least $\frac{p+1}{2}$ squares in
$\\{x_{0}+\alpha:\alpha\in\mathbb{F}_{p}\\}$ for some
$x_{0}\in\mathbb{F}_{p^{n}}\cdot g$, then there is
$\alpha_{0}\in\\{x_{0}+\alpha:\alpha\in\mathbb{F}_{p}\\}$ such that both
$\alpha_{0}$ and $\alpha_{0}+1$ are squares. Then there is
$a\in\mathbb{F}_{p^{n}}$ such that $\alpha_{0}=(a+1)^{2}$, and then
$(a+1)^{2}+1=\alpha_{0}+1$ is also a square. Since
$\alpha_{0}\not\in\mathbb{F}_{p}$ and hence $a\not\in\mathbb{F}_{p}$. When
$p=3$ we have $(c+1)a^{2}-2a+1=a+1\neq 0$. When $p=5$ we have
$(c+1)a^{2}-2a+1=2(a+1)(a-2)\neq 0$. When $p=7$ we have
$(c+1)a^{2}-2a+1=4(a+2)(a+1)\neq 0$.

(ii) We assume that $-c$ is not a square. As mentioned above, we only require
to check that there is $a\in\mathbb{F}_{p^{n}}$ such that $c^{2}(a+1)^{2}-4c$
is a nonzero square.

  * •

If $-1$ is a square, then one of $c-4$ and $-c^{2}+4c=-c(c-4)$ is a square.

    * –

If $c-4$ is a square then $4(c-4)=4c-16$ is also a square. If we choose
$a\in\mathbb{F}_{p^{n}}$ with $c^{2}(a+1)^{2}=4c-16$ then
$c^{2}(a+1)^{2}-4c=-16$ is a nonzero square.

    * –

If $-c^{2}+4c$ is a square then we choose $a\in\mathbb{F}_{p^{n}}$ with
$c^{2}(a+1)^{2}=-c^{2}+4c$. Then $c^{2}(a+1)^{2}-4c=-c^{2}$ is a nonzero
square.

  * •

If $-1$ is not a square, then $c$ is a square. Then one of $2c$ and $-2c$ is a
square.

    * –

If $2c$ is a square then $8c$ is a square. If we choose
$a\in\mathbb{F}_{p^{n}}$ with $c^{2}(a+1)^{2}=8c$ then $c^{2}(a+1)^{2}-4c=4c$
is a nonzero square.

    * –

If $-2c$ is a square then $-8c$ is a square. In this case, $p=3$ or one of $3$
and $-3$ is a nonzero square.

      * *

If $p=3$ and $n>1$ then $\mathbb{F}_{p^{n}}$ is partitioned by $p^{n-1}$
subsets of the form $\\{x+\alpha c:\alpha\in\mathbb{F}_{p}\\}$. By similar
argument with the case $c=-4$ and $p\in\\{3,5,7\\}$ above, there is a subset
$\\{x+\alpha c:\alpha\in\mathbb{F}_{p}\\}\neq\\{\alpha
c:\alpha\in\mathbb{F}_{p}\\}$ containing $2$ squares. Then there is
$\alpha_{0}\in\\{x_{0}+\alpha:\alpha\in\mathbb{F}_{p}\\}$ such that both
$\alpha_{0}$ and $\alpha_{0}+c$ are nonzero squares. We choose
$a\in\mathbb{F}_{p^{n}}$ such that $c^{2}(a+1)^{2}=\alpha_{0}+c$, then
$c^{2}(a+1)^{2}-4c=\alpha_{0}$ is also a nonzero square.

      * *

If $3$ is a nonzero square then we choose $a\in\mathbb{F}_{p^{n}}$ satisfying
$c^{2}(a+1)^{2}=-2c$. Then $c^{2}(a+1)^{2}-4c=-6c$ is also a nonzero square.

      * *

If $-3$ is a nonzero square then $-12$ is also a square. If we choose $a$
satisfying $c^{2}(a+1)^{2}=-8c$ then $c^{2}(a+1)^{2}-4c=-12c$ is also a
nonzero square.

Therefore, there is $a\in\mathbb{F}_{p^{n}}^{*}$ such that
${}_{c}D_{a}(x)=a^{-1}-c$ has two solutions in $\mathbb{F}_{2^{n}}\setminus
P_{a}$, and hence ${}_{c}\Delta_{F}(a,a^{-1}-c)\geq 3$ for all $p^{n}>7$. When
$p^{n}=7$ we already show that there is $a\in\mathbb{F}_{7}$ such that
${}_{c}\Delta(a,a^{-1}-c)\geq 3$ for all $c\neq-4$. Since $-4^{-1}=5\neq-4$,
we have ${}_{5}\Delta(a,a^{-1}-c)\geq 3$ and hence
${}_{-4}\Delta\left(-a,c^{-1}(c-a^{-1})\right)={}_{5}\Delta(a,a^{-1}-c)\geq
3$, by Lemma 2.2. Therefore, we have ${}_{c}\Delta_{F}\geq 3$ for all
$p^{n}>5$, which completes the proof. ∎

We obtain $\\#P_{a}<4$ only when $a=\pm 1$. In the following lemma we
investigate ${}_{c}\Delta_{F}(a,b)$ when $a=\pm 1$.

######  Lemma 3.3.

Under the same assumption as in (1) let $p^{n}>5$. Then
${}_{c}\Delta_{F}(1,b)\leq 3$ for all
$b\in\mathbb{F}_{p^{n}}\setminus\\{2^{-1}\\}$ and
$c\in\mathbb{F}_{p^{n}}^{*}\setminus\\{-2^{-1}\\}$, and
${}_{c}\Delta_{F}(-1,b)\leq 3$ for all
$b\in\mathbb{F}_{p^{n}}\setminus\\{1\\}$ and
$c\in\mathbb{F}_{p^{n}}^{*}\setminus\\{-2\\}$. Furthermore,

${}_{-2^{-1}}\Delta_{F}(1,2^{-1})={}_{-2}\Delta_{F}(-1,1)=\begin{cases}3&\text{
if }p^{n}\equiv\pm 3\pmod{8},\\\ 5&\text{ if }p^{n}\equiv\pm
1\pmod{8}.\end{cases}$

###### Proof.

If $a=1$ then $P_{a}=\\{0,1,-1\\}$ and then we have
${}_{c}D_{1}F(0)=F(1)-cF(0)=-c$, ${}_{c}D_{1}F(1)=F(2)-cF(1)=2^{-1}$ and
${}_{c}D_{1}(-1)=F(0)-cF(-1)=1+c$.

  * •

${}_{c}D_{1}F(0)={}_{c}D_{1}F(1)\Leftrightarrow-c=2^{-1}$ : Then we have
$c=-2^{-1}$.

  * •

${}_{c}D_{1}F(0)={}_{c}D_{1}F(-1)\Leftrightarrow-c=1+c$ : Then we have
$c=-2^{-1}$.

  * •

${}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)\Leftrightarrow 2^{-1}=1+c$ : Then $2c+2=1$
and hence we have $c=-2^{-1}$.

So if $c=-2^{-1}$ and $b=2^{-1}$ then ${}_{c}D_{1}F(x)=b$ has three solutions
in $P_{a}$. If $c\neq-2^{-1}$ or $b\neq 2^{-1}$ then ${}_{c}D_{1}F(x)=b$ has
at most one solution, and hence we have ${}_{c}\Delta_{F}(1,b)\leq 3$ for all
$b\in\mathbb{F}_{p^{n}}$, by Lemma 3.1. Now assume that $c=-2^{-1}$ and
$b=2^{-1}$. Then ${}_{-2^{-1}}D_{1}F(x)=2^{-1}$ has three solutions $x=0,\pm
1$ in $P_{a}$. If $a=1$, $b=2^{-1}$ and $c=-2^{-1}$ in Lemma 3.1 then

  * •

$(ab+c-1)^{2}-4abc=(-1)^{2}-4\cdot\left(-\frac{1}{4}\right)=2$ is a square if
and only if $p^{n}\equiv\pm 1\pmod{8}$.

  * •

$(b+c)a+b+c-1=-1\neq 0$ and $(1-b)a+b+c-1=-2^{-1}\neq 0$

Hence ${}_{-2^{-1}}D_{1}F(x)=2^{-1}$ has two solutions in
$\mathbb{F}_{p^{n}}\setminus P_{a}$ if and only if $p^{n}\equiv\pm 1\pmod{8}$,
by Lemma 3.1.

By Lemma 2.2, ${}_{c^{-1}}D_{-1}F(x)=-bc^{-1}$ and ${}_{c}D_{1}F(x)=b$ has the
same number of solutions. Hence
${}_{-2}\Delta_{F}(-1,1)={}_{-2^{-1}}\Delta_{F}(1,2^{-1})$, and
${}_{c}D_{-1}F(x)=b$ has at most three solutions for all
$b\in\mathbb{F}_{p^{n}}\setminus\\{1\\}$ and
$c\in\mathbb{F}_{p^{n}}^{*}\setminus\\{-2\\}$. ∎

######  Theorem 3.4.

Under the same assumption as in (1) if $p^{n}>5$ then
$3\leq{}_{c}\Delta_{F}\leq 5$. Moreover, ${}_{c}\Delta_{F}=5$ if and only if
at least one of the followings are satisfied :
(i) $p^{n}\equiv\pm 1\pmod{8}$ and $c\in\\{-2,-2^{-1}\\}$
(ii) $p^{n}\equiv 1\pmod{3}$ and one of the followings holds :

  * •

$c\in\left\\{\sqrt{-3},\frac{1}{\sqrt{-3}}\right\\}$, $\sqrt{-3}\neq 5$, and
$\frac{-3+5\sqrt{-3}}{6}$ is a nonzero square.

  * •

$c\in\left\\{-\sqrt{-3},-\frac{1}{\sqrt{-3}}\right\\}$, $\sqrt{-3}\neq-5$, and
$\frac{-3-5\sqrt{-3}}{6}$ is a nonzero square.

(iii) $p^{n}\equiv\pm 1\pmod{5}$ and one of the followings holds :

  * •

$c^{2}+4c-1=0$, and $7-30c$ is a nonzero square.

  * •

$c^{2}-4c-1=0$, and $7-30c^{-1}$ is a nonzero square.

###### Proof.

The case when $a=\pm 1$ is covered by Lemma 3.3 and we have (i). Hence it
remains to show (ii) and (iii) when $a\neq\pm 1$. We can see that

$\begin{array}[]{ll}{}_{c}D_{a}F(0)=F(a)-cF(0)=a^{-1}-c,&{}_{c}D_{a}F(1)=F(1+a)-cF(1)=(1+a)^{-1},\\\
{}_{c}D_{a}F(-a)=F(0)-cF(-a)=1+ca^{-1},&{}_{c}D_{a}F(1-a)=F(1)-cF(1-a)=c(a-1)^{-1}.\end{array}$
(2)

We investigate all possible cases that ${}_{c}D_{a}F(x)=b$ has at least two
solutions in $P_{a}$.

$\begin{array}[]{ll}{}_{c}D_{a}F(0)={}_{c}D_{a}F(1)\Leftrightarrow
ca^{2}+ca-1=0,&{}_{c}D_{a}F(0)={}_{c}D_{a}F(-a)\Leftrightarrow
a=\frac{1-c}{c+1}\text{ when }c\neq\pm 1,\\\
{}_{c}D_{a}F(0)={}_{c}D_{a}F(1-a)\Leftrightarrow
ca^{2}-a+1=0,&{}_{c}D_{a}F(1)={}_{c}D_{a}F(-a)\Leftrightarrow a^{2}+ca+c=0,\\\
{}_{c}D_{a}F(1)={}_{c}D_{a}F(1-a)\Leftrightarrow a=\frac{c+1}{1-c}\text{ when
}c\neq\pm 1,&{}_{c}D_{a}F(-a)={}_{c}D_{a}F(1-a)\Leftrightarrow
a^{2}-a-c=0.\end{array}$

If ${}_{c}D_{a}F(0)={}_{c}D_{a}F(-a)={}_{c}D_{a}F(1)={}_{c}D_{a}F(1-a)$ then
$a=\frac{1-c}{c+1}=\frac{c+1}{1-c}$ and hence we have $c=0$, a contradiction.
Hence ${}_{c}D_{a}F(x)=b$ has at most three solutions in $P_{a}$. We
investigate all possible cases that ${}_{c}D_{a}F(x)=b$ has three solutions in
$P_{a}$.

  * •

If ${}_{c}D_{a}F(0)={}_{c}D_{a}F(1)={}_{c}D_{a}F(-a)$ then $a=\frac{1-c}{c+1}$
is a solution of two equations $ca^{2}+ca-1=0$ and $a^{2}+ca+c=0$. Then we
have
$0=ca^{2}+ca-1=c\left(\frac{1-c}{c+1}\right)^{2}+c\cdot\frac{1-c}{c+1}-1=-\frac{3c^{2}+1}{(c+1)^{2}}$
and
$0=a^{2}+ca+c=\left(\frac{1-c}{c+1}\right)^{2}+c\cdot\frac{1-c}{c+1}+c=\frac{3c^{2}+1}{(c+1)^{2}}$.
Such $c$ exists when $-3$ is a nonzero square, if and only if $p^{n}\equiv
1\pmod{3}$, and then $c=\pm\frac{1}{\sqrt{-3}}$. Then $a=\frac{\sqrt{-3}\mp
1}{\sqrt{-3}\pm 1}=\frac{1\pm\sqrt{-3}}{2}$ and
$b={}_{c}D_{a}F(0)=a^{-1}-c=\frac{2}{1\pm\sqrt{-3}}\mp\frac{1}{\sqrt{-3}}=\frac{3\mp\sqrt{-3}}{6}$.
Since
$(ab+c-1)^{2}-4abc=\left(\frac{1\pm\sqrt{-3}}{2}\cdot\frac{3\mp\sqrt{-3}}{6}\pm\frac{1}{\sqrt{-3}}-1\right)^{2}\mp
4\cdot\frac{1\pm\sqrt{-3}}{2}\cdot\frac{3\mp\sqrt{-3}}{6}\cdot\frac{1}{\sqrt{-3}}=\left(\frac{-3\mp\sqrt{-3}}{6}\right)^{2}+\frac{-2\pm
2\sqrt{-3}}{3}=\frac{-3\pm 5\sqrt{-3}}{6}$. Using
$b+c=\frac{3\mp\sqrt{-3}}{6}\mp\frac{\sqrt{-3}}{3}=\frac{1\mp\sqrt{-3}}{2}$,
we have
$a(b+c)+b+c+1=\frac{1\pm\sqrt{-3}}{2}\cdot\frac{1\mp\sqrt{-3}}{2}+\frac{1\mp\sqrt{-3}}{2}+1=\frac{5\mp\sqrt{-3}}{2}$
and
$(1-b)a+b+c-1=\frac{3\pm\sqrt{-3}}{6}\cdot\frac{1\pm\sqrt{-3}}{2}+\frac{1\mp\sqrt{-3}}{2}-1=\frac{-3\mp\sqrt{-3}}{6}\neq
0$. So, we have ${}_{c}\Delta_{F}(a,b)=5$ if and only if $p^{n}\equiv
1\pmod{3}$, $\frac{5\mp\sqrt{-3}}{2}\neq 0$, and $\frac{-3\pm 5\sqrt{-3}}{6}$
is a square. Note that all double signs are in same order in above
computations this case.

  * •

If ${}_{c}D_{a}F(0)={}_{c}D_{a}F(1)={}_{c}D_{a}F(1-a)$ then
$a=\frac{c+1}{1-c}$ is a solution of two equations $ca^{2}+ca-1=0$ and
$ca^{2}-a+1=0$. Then we have
$0=c\cdot\left(\frac{c+1}{1-c}\right)^{2}+c\cdot\frac{c+1}{1-c}-1=\frac{c^{2}+4c-1}{(1-c)^{2}}$
and
$0=c\cdot\left(\frac{c+1}{1-c}\right)^{2}-\cdot\frac{c+1}{1-c}+1=\frac{c(c^{2}+4c-1)}{(1-c)^{2}}$,
and hence $c^{2}+4c-1=0$. Such $c$ exists if
$(-4)^{2}-4\cdot(-1)=20=2^{2}\cdot 5$ is a square, if and only if $p=5$ or
$p^{n}\equiv\pm 1\pmod{5}$. We set
$b={}_{c}D_{a}F(0)=a^{-1}-c=\frac{1-c}{c+1}-c=\frac{1-2c-c^{2}}{c+1}$. Using
the proof of Lemma 3.2, we have
$(ab+c-1)^{2}-4abc=c^{2}(a+1)^{2}-4c=c^{2}\cdot\left(\frac{2}{1-c}\right)^{2}-4c=-\frac{4(c^{3}-3c^{2}+c)}{(c-1)^{2}}=-\frac{4((c-7)(c^{2}+4c-1)+30c-7)}{(c-1)^{2}}=-\frac{4(30c-7)}{(c-1)^{2}}$
and $(b+c)a+b+c+1=a^{-1}\neq 0$ and
$(1-b)a+b+c-1=a^{-1}((c+1)a^{2}-2a+1)=a^{-1}\left((c+1)\cdot\left(\frac{c+1}{1-c}\right)^{2}-2\cdot\frac{c+1}{1-c}+1\right)=\frac{c(c^{2}+6c+1)}{a(c-1)^{2}}=\frac{2c(c+1)}{a(c-1)^{2}}\neq
0$ since $c\neq-1$. So, we have ${}_{c}\Delta_{F}(a,b)=5$ if and only if
$p^{n}\equiv 0,\pm 1\pmod{5}$ and $7-30c$ is a nonzero square. If $p=5$ then
we have $c=-2$, and $7-30c=2$ is a square if and only if $p^{n}\equiv\pm
1\pmod{8}$ and hence this case is already covered by (i).

  * •

If ${}_{c}D_{a}F(0)={}_{c}D_{a}F(-a)={}_{c}D_{a}F(1-a)$ then we have
${}_{c^{-1}}D_{-a}F(0)={}_{c^{-1}}D_{-a}F(1)={}_{c^{-1}}D_{-a}F(-(-a))$ by
Lemma 2.2. We set $c^{-1}=\pm\frac{1}{\sqrt{-3}}$ and
$-a=\frac{1-c^{-1}}{c^{-1}+1}=\frac{1\pm\sqrt{-3}}{2}$ or equivalently
$c=\pm\sqrt{-3}$ and $a=-\frac{1\pm\sqrt{-3}}{2}$, and
$b=(-a)^{-1}-c^{-1}=\frac{3\mp\sqrt{-3}}{6}$. By the above observation and
Lemma 2.2, $5={}_{c^{-1}}\Delta_{F}(-a,-bc^{-1})={}_{c}\Delta_{F}(a,b)$ if and
only if $p^{n}\equiv 1\pmod{3}$, $\frac{5\mp\sqrt{-3}}{2}\neq 0$, and
$\frac{-3\pm 5\sqrt{-3}}{6}$ is a square.

  * •

If ${}_{c}D_{a}F(1)={}_{c}D_{a}F(-a)={}_{c}D_{a}F(1-a)$ then we have
${}_{c^{-1}}D_{-a}F(0)={}_{c^{-1}}D_{-a}F(1)={}_{c^{-1}}D_{-a}F(1-(-a))$ by
Lemma 2.2. We assume that $(c^{-1})^{2}+4c^{-1}-1=0$ and
$-a=\frac{c^{-1}+1}{1-c^{-1}}$ or equivalently $c^{2}-4c-1=0$ and
$a=\frac{1+c}{1-c}$, and
$b=-a^{-1}-c^{-1}=\frac{1-2c^{-1}-(c^{-1})^{2}}{1+c^{-1}}$. By the above
observation and Lemma 2.2,
$5={}_{c^{-1}}\Delta_{F}(-a,b)={}_{c}\Delta_{F}(a,b)$ if and only if $p\equiv
0,\pm 1\pmod{5}$ and $7-30c^{-1}$ is a nonzero square. Similarly with the
above case, the case $p=5$ can be covered by (i).

Conversely, suppose each condition in (ii) and (iii) is satisfied. Using the
same $a$ and $b$ in previous analysis, we can see that ${}_{c}D_{a}F(x)=b$ has
three solutions in $P_{a}$. By Lemma 3.1, the given element is a nonzero
square, then ${}_{c}D_{a}F(x)=b$ has two solutions in
$\mathbb{F}_{p^{n}}\setminus P_{a}$. Hence we have ${}_{c}\Delta_{F}(a,b)=5$
which completes the proof. ∎

Next we further investigate the differential uniformity of $F$, the special
case $c=1$.

######  Theorem 3.5.

Let $F=Inv\circ(0,1)$ and $p^{n}>5$. Then $3\leq\Delta_{F}\leq 5$.
Furthermore,
(i) $\Delta_{F}=5$ if and only if $p=3$ and $n$ is even.
(ii) $\Delta_{F}=4$ if and only if at least one of the followings is satisfied
:

  * •

$p=5$ and $n\equiv 0\pmod{4}$.

  * •

$p^{n}\equiv\pm 1\pmod{5}$ with $p\neq 3$, and at least one of
$\frac{-5+\sqrt{5}}{2}$ and $\frac{-5-\sqrt{5}}{2}$ is a square.

  * •

$p^{n}\equiv 1\pmod{3}$ with $p\neq 5$, and at least one of
$\frac{-5+3\sqrt{-3}}{2}$ and $\frac{-5-3\sqrt{-3}}{2}$ is a nonzero square.

###### Proof.

By Theorem 3.4, $3\leq\Delta_{F}\leq 5$ is straightforward.

(i) We first show that $D_{a}F(x)=b$ has at most two solutions in
$\mathbb{F}_{p^{n}}\setminus P_{a}$ and hence $\Delta_{F}(a,b)\leq 4$, when
$a\in\mathbb{F}_{p^{n}}\setminus\\{0,\pm 1\\}$ and $b\in\mathbb{F}_{p^{n}}$.
We have the following to substitute $c=1$ in (2)

$\begin{array}[]{ll}D_{a}F(0)=F(a)-F(0)=a^{-1}-1,&D_{a}F(1)=F(1+a)-F(1)=(1+a)^{-1},\\\
D_{a}F(-a)=F(0)-F(-a)=1+a^{-1},&D_{a}F(1-a)=F(1)-F(1-a)=(a-1)^{-1}.\end{array}$

We can see that $D_{a}F(x-a)=F(x)-F(x-a)=-D_{-a}(x)$. If
$D_{a}F(0)=D_{a}F(-a)$ we have $1=-1$, a contradiction, and hence we have
$D_{a}F(0)\neq D_{a}F(-a)$. Similarly we have $D_{a}F(1)\neq D_{a}F(1-a)$.
Hence $D_{a}F(u)\neq D_{a}F(u-a)$ for $u\in P$. If there is
$b\in\mathbb{F}_{p^{n}}$ such that $D_{a}F(x)=b$ has at least three solutions
in $P_{a}$, then there is $u\in P_{a}$ such that both $u$ and $u-a$ are
solutions of $D_{a}F(x)=b$, which is a contradiction. Hence $D_{a}F(x)=b$ has
at most two solutions in $P_{a}$.

So we have $\Delta_{F}=5$ if and only if $\Delta_{F}(1,b)=5$ for some
$b\in\mathbb{F}_{p^{n}}$. We have $c=1=-2^{-1}$ which is equivalent to
$1=(-2)\cdot(-2)^{-1}=(-2)\cdot 1=-2$ if and only if $p=3$. Then
$3^{n}\equiv\pm 1\pmod{8}$ if and only if $n$ is even. Hence by Lemma 3.3,
$\Delta_{F}=5$ if and only if $p=3$ and $n$ is even.

(ii) We investigate all possible cases that $D_{a}F(x)=b$ has at least two
solutions in $P_{a}$ applying $D_{a}F(u)\neq D_{a}F(u-a)$ for $u\in P$. Since
$D_{a}F(x-a)=F(x)-F(x-a)=-D_{-a}(x)$, we have $D_{a}F(-a)=D_{a}F(u-a)$ if and
only if $D_{-a}F(0)=D_{-a}F(u)$ for $u\in\\{1,1-a\\}$. Hence it is enough to
investigate the following cases.

  * •

$D_{a}F(0)=D_{a}F(1)$ implies $a^{2}+a-1=0$. $D=(-1)^{2}-4\cdot(-1)=5$ is a
nonzero square if and only if $p\equiv\pm 1\pmod{5}$.

  * •

$D_{a}F(0)=D_{a}F(1-a)$ implies $a^{2}-a+1=0$. $D=(-1)^{2}-4\cdot 1=-3$ is a
nonzero square if and only if $p\equiv 1\pmod{3}$.

We set $b=D_{a}F(0)=a^{-1}-1$. Applying the proof of Lemma 3.2 we have
$(b+c)a+b+c-1=a^{-1}\neq 0$ and $(1-b)a+b+c-1=a^{-1}(2a^{2}-2a+1)$.

  * •

If $D_{a}F(0)=D_{a}F(1)$ then $a^{2}=-a+1$. Hence
$0=2a^{2}-2a+1=2(-a+1)-2a+1=-4a+3$ implies $a=\frac{3}{4}$. Then
$0=a^{2}+a-1=\left(\frac{3}{4}\right)^{2}+\frac{3}{4}-1=\frac{5}{16}$ requires
$p=5$.

  * •

If $D_{a}F(0)=D_{a}F(1-a)$ then $a^{2}-a=-1$. Hence we have
$2a^{2}-2a+1=2(a^{2}-a)+1=-1\neq 0$.

1\. If $p=3$ then it remains to show that $\Delta_{F}(a,b)\leq 3$ for all
$a\in\mathbb{F}_{p^{n}}\setminus\\{0,\pm 1\\}$ and $b\in\mathbb{F}_{p^{n}}$
when $n$ is odd, by Lemma 3.3. Then $a^{2}-a+1=0$ has one solution $a=-1$,
contradicts to the assumption. Furthermore we have $3^{n}\equiv\pm 2\pmod{5}$
if $n$ is odd. So $D_{a}F(x)=b$ has at most one solution in $P_{a}$ when
$a\in\mathbb{F}_{p^{n}}\setminus\\{0,1\\}$, and hence we have
$\Delta_{F}(a,b)\leq 3$ for all $a\in\mathbb{F}_{p^{n}}\setminus\\{0,1\\}$ and
$b\in\mathbb{F}_{p^{n}}$.

2\. If $p=5$ then $a^{2}+a-1=0$ has one solution. It is easy to see that $a=2$
is the only solution of $a^{2}+a-1=0$ and then $b=D_{2}F(0)=2^{-1}-1=2$. As
mentioned above, $D_{2}F(x)=2$ has two solutions $x=0,1$ in $P_{a}$. And then
we have $(ab+c-1)^{2}-4abc=0$ and hence $bx^{2}+(ab+c-1)x+ac=0$ has exactly
one solution, but by the above observation $x=1-a=-1\in P_{a}$ is the solution
of $bx^{2}+(ab+c-1)x+ac=0$ and hence it cannot be a solution of $D_{2}F(x)=2$.
Hence $D_{2}F(x)=2$ has no solutions in $\mathbb{F}_{p^{n}}\setminus P_{a}$.
So we have $\Delta_{F}(2,2)=2$.
If $n$ is odd then $5^{n}\equiv 2\pmod{3}$ and hence $D_{a}F(x)=b$ has at most
one solution in $P_{a}$. By Lemma 3.3 we have $\Delta_{F}=3$. If $n$ is even
then $5^{n}\equiv 1\pmod{3}$ so $a^{2}+a+1=0$ has two solutions. Then
$a^{3}=-1$ and hence $a^{-1}=-a^{2}$. So we have
$b=D_{a}F(0)=a^{-1}-1=-a^{2}-1=-a$ then
$(ab+c-1)^{2}-4abc=a^{4}+4a^{2}=-a+4(a-1)=3a+1$. Note that
$(3a+1)^{2}=9a^{2}+6a+1=-a^{2}+a+1=2$ and hence $3a+1$ is a square root of
$2$. Since the multiplicative order of $2$ in modulo $5$ is $4$, we have
$1=\chi(3a+1)=(3a+1)^{\frac{5^{n}-1}{2}}=2^{\frac{5^{n}-1}{4}}$ if and only if
$4\mid\frac{5^{n}-1}{4}$ if and only if $5^{n}\equiv 1\pmod{16}$. Since the
multiplicative order of $5$ in modulo $16$ is $4$, we have $5^{n}\equiv
1\pmod{16}$ if and only if $n\equiv 0\pmod{4}$. Hence, we have $\Delta_{F}=4$
if and only if $n\equiv 0\pmod{4}$.

3\. Now we investigate the case $p\neq 3,5$.

  * •

If $p^{n}\equiv\pm 1\pmod{5}$, then $5$ is a square and hence
$D_{a}F(0)=D_{a}F(1)$ where $a^{2}+a-1=0$. Then $a=\frac{-1\pm\sqrt{5}}{2}$
and $b=a^{-1}-1$ and
$(ab+c-1)^{2}-4abc=a^{2}+2a-3=a^{2}+a-1+a-2=a-2=\frac{-5\pm\sqrt{5}}{2}$.
Hence by Lemma 3.1 we have $\Delta_{F}(a,b)=4$ if and only if at least one of
$\frac{-5+\sqrt{5}}{2}$ and $\frac{-5-\sqrt{5}}{2}$ is a square. Note that
$\frac{-5\pm\sqrt{5}}{2}\neq 0$, since if $\frac{-5\pm\sqrt{5}}{2}=0$ then
$x=\pm\sqrt{5}$ is a solution of $x^{2}-x=0$ which implies $1=5$ or $0=5$,
which is a contradiction to $p\neq 5$.

  * •

If $p^{n}\equiv 1\pmod{3}$ then $-3$ is a square and hence
$D_{a}F(0)=D_{a}F(1-a)$ where $a^{2}-a+1=0$. Then $a=\frac{1\pm\sqrt{-3}}{2}$
and $b=a^{-1}-1$ and
$(ab+c-1)^{2}-4abc=a^{2}+2a-3=a^{2}-a+1+3a-4=3a-4=\frac{-5\pm 3\sqrt{-3}}{2}$.
Hence by Lemma 3.1 we have $\Delta_{F}(a,b)=4$ if and only if at least one of
$\frac{-5+3\sqrt{-3}}{2}$ and $\frac{-5-3\sqrt{-3}}{2}$ is a nonzero square.

Conversely, suppose each condition in (ii) is satisfied. Using the same $a$
and $b$ in previous analysis, we can see that ${}_{c}D_{a}F(x)=b$ has two
solutions in $P_{a}$. By Lemma 3.1, the given element in each case is a
nonzero square, then $D_{a}F(x)=b$ has two solutions in
$\mathbb{F}_{p^{n}}\setminus P_{a}$. Hence we have $\Delta_{F}(a,b)=4$ which
completes the proof. ∎

##  4 The $c$-differential uniformity of $Inv\circ(1,\gamma)$

Let $F=Inv\circ(1,\gamma)$ where
$\gamma\in\mathbb{F}_{p^{n}}\setminus\\{0,1\\}$. Similarly with section 3
since $F$ is permutation, we have ${}_{c}\Delta_{F}(0,b)=1$ for all
$b\in\mathbb{F}_{p^{n}}$ and $c\in\mathbb{F}_{p^{n}}\setminus\\{1\\}$, and we
also have ${}_{0}\Delta_{F}(a,b)=1$ for all $a,b\in\mathbb{F}_{p^{n}}$. Hence,
throughout this section we assume that

$F=Inv\circ(1,\gamma),\ \gamma\in\mathbb{F}_{p^{n}}\setminus\\{0,1\\},\ a\neq
0,\ c\neq 0$ (3)

We also denote throughout this section that

$P=\\{x\in\mathbb{F}_{p^{n}}:F(x)\neq Inv(x)\\}\cup\\{0\\}=\\{0,1,\gamma\\},\
\ P_{a}=P\cup\\{x-a:x\in P\\}=\\{0,1,\gamma,-a,1-a,\gamma-a\\}.$

It is easy to see that

${}_{c}\Delta_{F}(a,b)=\\#\\{x\in
P_{a}:{}_{c}D_{a}F(x)=b\\}+\\#\\{x\in\mathbb{F}_{p^{n}}\setminus
P_{a}:{}_{c}D_{a}F(x)=b\\}.$

The following lemma characterizes the case
$\\#\\{x\in\mathbb{F}_{p^{n}}\setminus P_{a}:D_{a}F(x)=b\\}=2$.

######  Lemma 4.1.

Under the same assumption as in (3) let $a\in\mathbb{F}_{p^{n}}^{*}$ and
$b\in\mathbb{F}_{p^{n}}$. Then ${}_{c}D_{a}F(x)=b$ has at most two solutions
in $\mathbb{F}_{p^{n}}\setminus P_{a}$. Furthermore, ${}_{c}D_{a}F(x)=b$ has
two solutions in $\mathbb{F}_{p^{n}}\setminus P_{a}$ if and only if
$(ab+c-1)^{2}-4abc$ is a nonzero square and $b\neq 0$, $(b+c)a+b+c-1\neq 0$,
$(1-b)a+b+c-1\neq 0$, $(b\gamma+c)a+\gamma(b\gamma+c-1)\neq 0$ and
$(1-b\gamma)a+\gamma(b\gamma+c-1)\neq 0$.

###### Proof.

Similarly with Lemma 3.1, since $F(x)=Inv(x)$ for all
$x\in\mathbb{F}_{2^{n}}\setminus P_{a}$, ${}_{c}D_{a}F(x)=b$ has at most two
solutions in $\mathbb{F}_{2^{n}}\setminus P_{a}$, and has two solutions if and
only if $(ab+c-1)^{2}-4abc$ is a nonzero square and $b\neq 0$. We require that
$bx^{2}+(ab+c-1)x+ac=0$ has no solutions in $P_{a}$. We check this by
substituting every element in $P_{a}$ to $bx^{2}+(ab+c-1)x+ac=0$.

  * •

If $x=0$ or $x=-a$ then we have $a=0$ which is a contradiction to $a\neq 0$.

  * •

If $x=1$ then we have $(b+c)a+b+c-1=0$.

  * •

If $x=1-a$ then we have $(1-b)a+b+c-1=0$.

  * •

If $x=\gamma$ then we have $(b\gamma+c)a+\gamma(b\gamma+c-1)=0$.

  * •

If $x=\gamma-a$ then we have $(1-b\gamma)a+\gamma(b\gamma+c-1)=0$.

Thus $bx^{2}+(ab+c-1)x+ac=0$ has no solutions in $P_{a}$ if and only if
$(b+c)a+b+c-1\neq 0$, $(1-b)a+b+c-1\neq 0$,
$(b\gamma+c)a+\gamma(b\gamma+c-1)\neq 0$ and
$(1-b\gamma)a+\gamma(b\gamma+c-1)\neq 0$.
Similarly with Lemma 3.1, if all conditions in this theorem are satisfied,
then ${}_{c}D_{a}F(x)=b$ has two solutions in $\mathbb{F}_{p^{n}}\setminus
P_{a}$. ∎

The following lemma shows symmetry of the $c$-differential uniformity with
respect to $\gamma$, and it is useful for our results.

######  Lemma 4.2.

Let $\gamma\neq 0,1$ and $F=Inv\circ(1,\gamma)$ and
$F^{\prime}=Inv\circ(1,\gamma^{-1})$. Then
${}_{c}\Delta_{F^{\prime}}(a,b)={}_{c}\Delta_{F}(\gamma a,\gamma^{-1}b)$ for
all $a,b\in\mathbb{F}_{p^{n}}$ and hence
${}_{c}\Delta_{F}={}_{c}\Delta_{F^{\prime}}$ for all $c\in\mathbb{F}_{p^{n}}$.

###### Proof.

Let $A(x)=\gamma x$. Then it is not difficult to see that $F^{\prime}=A\circ
F\circ A$. Then ${}_{c}D_{a}F^{\prime}(x)=b$ implies

$\displaystyle b$ $\displaystyle={}_{c}D_{a}F^{\prime}(x)=(A\circ F\circ
A)(x+a)-c(A\circ F\circ A)(x)=(A\circ F)(\gamma x+\gamma a)-c(A\circ F)(\gamma
x)$ $\displaystyle=\gamma\left(F(\gamma x+\gamma a)-F(\gamma
x)\right)=\gamma(F(X+\gamma a)-cF(X))=\gamma\cdot{}_{c}D_{\gamma a}F(X)$

if and only if $\gamma^{-1}b={}_{c}D_{\gamma a}F(X)$, where $X=\gamma x$ in
above. ∎

We can see that $\\#P_{a}<6$ if and only if $a\in\\{\pm
1,\pm\gamma,\pm(\gamma-1)\\}$. We investigate the number of solutions of
${}_{c}D_{a}F(x)=b$ in $P_{a}$ for this case.

######  Lemma 4.3.

Let $a\in\\{\pm 1,\pm\gamma,\pm(\gamma-1)\\}$. Then
(i) ${}_{c}D_{a}F(x)=b$ has five solutions in $P_{a}$ if and only if $p=5$,
$\gamma=-1$, $c=1$ and $(a,b)\in\\{(2,3),(3,2)\\}$.
(ii) ${}_{c}D_{a}F(x)=b$ has four solutions in $P_{a}$ if and only if $p=5$,
$\gamma=-1$, $c=1$ and $(a,b)\in\\{(1,-1),(-1,1)\\}$.
Otherwise, ${}_{c}D_{a}F(x)=b$ has at most three solutions in $P_{a}$.

###### Proof.

We investigate all possible cases, and details are in Appendix A. ∎

In Lemma 3.2 we show that the $c$-differential uniformity of $F$ is lower
bounded by $3$ for all $c\in\mathbb{F}_{p^{n}}^{*}$ when $\gamma=0$. But the
method used in the proof of Lemma 3.2 is not applicable when $\gamma\neq 0$,
and it is not easy to show a lower bound of the $c$-differential uniformity of
$F$. But we confirm that ${}_{c}\Delta_{F}\geq 3$ when $7<p^{n}<500$ by using
SageMath.

###  4.1 The differential uniformity

In this subsection we investigate the differential uniformity of $F$, that is,
the case $c=1$. We first investigate the number of solutions of $D_{a}F(x)=b$
in $P_{a}$ when $\\#P_{a}=6$.

######  Lemma 4.4.

Let $a\in\mathbb{F}_{p^{n}}\setminus\\{0,\pm 1,\pm\gamma,\pm(\gamma-1)\\}$ and
$b\in\mathbb{F}_{p^{n}}$. Then, $D_{a}F(x)=b$ has four solutions in $P_{a}$ if
and only if $\gamma=-1$ and at least one of the followings is satisfied:
(i) $p^{n}\equiv\pm 1\pmod{5}$, $a^{4}-3a^{2}+1=0$ and $b=a^{-1}$,
(ii) $p^{n}\equiv\pm 1\pmod{8}$, $a^{2}=2$ and $b=a$.
Otherwise, $D_{a}F(x)=b$ has at most three solutions.

###### Proof.

$D_{a}F(x)$ for each $x\in P_{a}$ is as follows:

$\begin{array}[]{ll}D_{a}F(0)=F(a)-F(0)=a^{-1},&D_{a}F(-a)=F(0)-F(-a)=a^{-1},\\\
D_{a}F(1)=F(1+a)-F(1)=(a+1)^{-1}-\gamma^{-1},&D_{a}F(1-a)=F(1)-F(1-a)=\gamma^{-1}-(1-a)^{-1},\\\
D_{a}F(\gamma)=F(\gamma+a)-F(\gamma)=(a+\gamma)^{-1}-1,&D_{a}F(\gamma-a)=F(\gamma)-F(\gamma-a)=1+(a-\gamma)^{-1}.\end{array}$

We first assume that $b=D_{a}F(0)=D_{a}F(-a)=a^{-1}$. We investigate
$D_{a}F(x)=a^{-1}$ has at least three solutions in $P_{a}$. First we
investigate equivalent condition for each case that $D_{a}F(x)=a^{-1}$ has at
least two solutions in $P_{a}$.

$\begin{array}[]{ll}a^{-1}=D_{a}F(1)\ \Leftrightarrow\
a^{2}+a+\gamma=0,&a^{-1}=D_{a}F(1-a)\ \Leftrightarrow\ a^{2}-a+\gamma=0,\\\
a^{-1}=D_{a}F(\gamma)\ \Leftrightarrow\ a^{2}+\gamma
a+\gamma=0,&a^{-1}=D_{a}F(\gamma-a)\ \Leftrightarrow\ a^{2}-\gamma
a+\gamma=0.\end{array}$

We can easily see that if $a^{-1}=D_{a}F(1)=D_{a}F(1-a)$ or
$a^{-1}=D_{a}F(\gamma)=D_{a}F(\gamma-a)$ then we have $a=0$, a contradiction.
Hence $D_{a}F(x)=a^{-1}$ has at most four solutions. If
$a^{-1}=D_{a}F(1)=D_{a}F(\gamma-a)$ then we have $\gamma=-1$ and hence
$a^{2}+a-1=0$, such $a$ exists if and only if $1^{2}-4\cdot(-1)=5$ is a square
which is equivalent to $p^{n}\equiv 0,\pm 1\pmod{5}$. But if $p=5$ then we
have $a=2=1-\gamma$ which contradicts to assumption, and hence we have $p\neq
5$ . Similarly, if $a^{-1}=D_{a}F(\gamma)=D_{a}F(1-a)$ we also have
$\gamma=-1$ and hence $a^{2}-a-1=0$, such $a$ exists if and only if
$(-1)^{2}-4\cdot(-1)=5$ is a square, and we have $p^{n}\equiv\pm 1\pmod{5}$.
Hence, $D_{a}F(x)=a^{-1}$ has four solutions if and only if $p^{n}\equiv\pm
1\pmod{5}$, $\gamma=-1$ and $0=(a^{2}+a-1)(a^{2}-a-1)=a^{4}-3a^{2}+1$.
Otherwise, $D_{a}F(x)=a^{-1}$ has at most three solutions. Conversely, if
$p\equiv\pm 1\pmod{5}$ then there exists $a\in\mathbb{F}_{p^{n}}$ with
$a^{2}+a-1=0$ or $a^{2}-a-1=0$ and then by the above analysis we can see that
$D_{a}F(x)=a^{-1}$ has four solutions in $P_{a}$ when $\gamma=-1$.

Next we investigate the case $b\neq a^{-1}$. In this case, $D_{a}F(x)=b$ has
four solutions in $P_{a}$ if and only if
$b=D_{a}F(1)=D_{a}F(\gamma)=D_{a}F(1-a)=D_{a}F(\gamma-a)$. First we
investigate equivalent condition for each case that $D_{a}F(x)=b$ has at least
two solutions in $\\{1,\gamma,1-a,\gamma-a\\}$.

$\begin{array}[]{ll}D_{a}F(1)=D_{a}F(\gamma)\Leftrightarrow
a^{2}+(\gamma+1)a+2\gamma=0,&D_{a}F(1)=D_{a}F(\gamma-a)\Leftrightarrow
a(\gamma+1)(a-\gamma+1)a=0,\\\ D_{a}F(1)=D_{a}F(1-a)\Leftrightarrow
a^{2}+\gamma-1=0,&D_{a}F(\gamma)=D_{a}F(1-a)\Leftrightarrow
a(\gamma+1)(a+\gamma-1)a=0,\\\ D_{a}F(\gamma)=D_{a}F(\gamma-a)\Leftrightarrow
a^{2}+\gamma(1-\gamma)=0,&D_{a}F(1-a)=D_{a}F(\gamma-a)\Leftrightarrow
a^{2}-(\gamma+1)a+2\gamma=0.\end{array}$

From $a^{2}+(\gamma+1)a+2\gamma=0$ and $a^{2}-(\gamma+1)a+2\gamma=0$ we have
$\gamma=-1$. Then we have $a^{2}=2$, such $a$ exists if and only if
$p^{n}\equiv\pm 1\pmod{8}$. Then we have
$b=D_{a}F(1)=\frac{1}{a+1}+1=\frac{a+2}{a+1}=\frac{a^{2}+a-2}{a^{2}-1}=a$.
Conversely, if $p^{n}\equiv 1\pmod{8}$ then there exists
$a\in\mathbb{F}_{p^{n}}$ with $a^{2}=2$ and then by the above analysis we can
see that $D_{a}F(x)=a$ has four solutions in $P_{a}$, when $\gamma=-1$. ∎

Next we show the main theorem of this subsection.

######  Theorem 4.5.

Let $F=Inv\circ(1,\gamma)$ with $\gamma\neq 0,1$. Then,
(i) $\Delta_{F}=7$ if and only if $\gamma=-1$, $p=5$ and $n$ is even.
(ii) $\Delta_{F}=6$ if and only if $\gamma=-1$, and $p^{n}\equiv 1\pmod{8}$ or
$p^{n}\equiv 1,4\pmod{15}$.
Otherwise, $\Delta_{F}\leq 5$.

###### Proof.

Using Lemma 4.3 and Lemma 4.4 we investigate all the cases that $D_{a}F(x)=b$
has at least four solutions, which requires $\gamma=-1$. By Lemma 4.1 with
$c=1$ and $\gamma=-1$, $D_{a}F(x)=b$ has two solutions in
$\mathbb{F}_{p^{n}}\setminus P_{a}$ if and only if $(ab)^{2}-4ab$ is a nonzero
square, $(b+1)a+b=ab+a+b\neq 0$ and $(1-b)a+b=a+b-ab\neq 0$.

  * •

If $p=5$ and $(a,b)\in\\{(2,3),(3,2)\\}$ then $D_{a}F(x)=b$ has five solutions
in $P_{a}$. Then $(ab)^{2}-4ab=1^{2}-4\cdot 1=-3$ is a square if and only if
$5^{n}\equiv 1\pmod{3}$ which is equivalent to $n$ is even. And, we have
$ab+a+b=1\neq 0$ and $a+b-ab=-1\neq 0$. Hence we have $\Delta_{F}(a,b)=7$ if
$n$ is even, $\Delta_{F}(a,b)=5$ if $n$ is odd.

  * •

If $p=5$ and $(a,b)\in\\{(1,-1),(-1,1)\\}$ then $D_{a}F(x)=b$ has four
solutions in $P_{a}$. Then we have $(ab)^{2}-4ab=1^{2}+4\cdot 1=5=0$ and hence
we have $\Delta_{F}(a,b)\leq 5$ in this case.

  * •

If $p^{n}\equiv\pm 1\pmod{5}$ and $a^{4}-3a^{2}+1=(a^{2}+a-1)(a^{2}-a-1)=0$
then $D_{a}(x)=a^{-1}$ has four solutions in $P_{a}$. Then we require
$(ab)^{2}-4ab=1^{2}-4\cdot 1=-3$ is a nonzero square if and only if
$p^{n}\equiv 1\pmod{3}$. If $a^{2}+a-1=0$ then
$ab+a+b=a^{-1}(a^{2}+a+1)=2a^{-1}\neq 0$ and
$a+b-ab=a^{-1}(a^{2}-a+1)=2a^{-1}(1-a)\neq 0$ since $a\neq 1$. Similarly, if
$a^{2}-a-1=0$ then $ab+a+b\neq 0$ and $a+b-ab\neq 0$. Hence we have
$\Delta_{F}(a,a^{-1})=6$ where $a^{4}-3a^{2}+1=0$, when $p^{n}\equiv\pm
1\pmod{5}$ and $p^{n}\equiv 1\pmod{3}$, or equivalently $p^{n}\equiv
1,4\pmod{15}$.

  * •

If $p^{n}\equiv\pm 1\pmod{8}$ and $a^{2}=2$ then $D_{a}(x)=a$ has four
solutions in $P_{a}$. Then we have $(ab)^{2}-4ab=a^{4}-4a^{2}=4-8=-4$ is a
square if and only if $p^{n}\equiv 1\pmod{4}$. Since $a\neq 1$ from $(\pm
1)^{2}=1\neq 2$, we have $ab+a+b=a^{2}+2a=2(a+1)\neq 0$ and
$a+b-ab=2a-2=2(a-1)\neq 0$. Hence we have $\Delta_{F}(a,a)=6$ where $a^{2}=2$,
when $p^{n}\equiv\pm 1\pmod{8}$ and $p^{n}\equiv 1\pmod{4}$, or equivalently
$p^{n}\equiv 1\pmod{8}$.

Otherwise, $D_{a}F(x)=b$ at most three solutions in $P_{a}$ by Lemma 4.3 and
Lemma 4.4, and hence we have $\Delta_{F}\leq 5$ by Lemma 4.1. ∎

###  4.2 The $c$-differential uniformity with $c\neq 0,1$

In the above subsection we investigate the differential uniformity of
$Inv\circ(1,\gamma)$. In this subsection we investigate the $c$-differential
uniformity of $Inv\circ(1,\gamma)$ where $c\neq 0,1$. First we investigate the
number of solutions of ${}_{c}D_{a}F(x)=b$ in $P_{a}$ when $\\#P_{a}=6$.

######  Lemma 4.6.

Let $F=Inv\circ(1,\gamma)$ with $\gamma\neq 0,1$ and $c\neq 0,1$. Let
$a\in\mathbb{F}_{p^{n}}\setminus\\{0,\pm 1,\pm\gamma,\pm(\gamma-1)\\}$ and
$b\in\mathbb{F}_{p^{n}}$. Then, ${}_{c}D_{a}F(x)=b$ has four solutions in
$P_{a}$ if and only if $\gamma c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma=0$,
$c\neq-1$, $a=\frac{\gamma^{2}-c}{\gamma+c}$ and $b=\frac{1-c}{1+\gamma}$.
Otherwise, ${}_{c}D_{a}F(x)=b$ has at most three solutions in $P_{a}$.

###### Proof.

${}_{c}D_{a}F(x)$ for each $x\in P_{a}$ is as follows :

$\begin{array}[]{ll}{}_{c}D_{a}F(0)=F(a)-cF(0)=a^{-1},&{}_{c}D_{a}F(-a)=F(0)-cF(-a)=ca^{-1},\\\
{}_{c}D_{a}F(1)=F(1+a)-cF(1)=(a+1)^{-1}-c\gamma^{-1},&{}_{c}D_{a}F(1-a)=F(1)-cF(1-a)=\gamma^{-1}-c(1-a)^{-1},\\\
{}_{c}D_{a}F(\gamma)=F(\gamma+a)-cF(\gamma)=(a+\gamma)^{-1}-c,&{}_{c}D_{a}F(\gamma-a)=F(\gamma)-cF(\gamma-a)=1+c(a-\gamma)^{-1}.\end{array}$

We investigate some possible cases to be used in this proof such that
${}_{c}D_{a}F(x)=b$ has at least two solutions in $P_{a}$ and we obtain an
equation on $a$ in each case.

$\begin{array}[]{ll}{}_{c}D_{a}F(0)={}_{c}D_{a}F(1)\Leftrightarrow
ca^{2}+ca+\gamma=0,&{}_{c}D_{a}F(0)={}_{c}D_{a}F(1-a)\Leftrightarrow
a^{2}+(c\gamma-\gamma-1)a+\gamma=0,\\\
{}_{c}D_{a}F(0)={}_{c}D_{a}F(\gamma)\Leftrightarrow ca^{2}+c\gamma
a+\gamma=0,&{}_{c}D_{a}F(0)={}_{c}D_{a}F(\gamma-a)\Leftrightarrow
a^{2}+(c-\gamma-1)a+\gamma=0,\\\
{}_{c}D_{a}F(-a)={}_{c}D_{a}F(1-a)\Leftrightarrow
a^{2}-a+c\gamma=0,&{}_{c}D_{a}F(1)={}_{c}D_{a}F(-a)\Leftrightarrow
ca^{2}+(c\gamma+c-\gamma)a+c\gamma=0,\\\
{}_{c}D_{a}F(-a)={}_{c}D_{a}F(\gamma-a)\Leftrightarrow a^{2}-\gamma
a+c\gamma=0,&{}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(-a)\Leftrightarrow
ca^{2}+(c\gamma+c-1)a+c\gamma=0.\end{array}$

We investigate possible cases such that ${}_{c}D_{a}F(x)=b$ has at least four
solutions in $P_{a}$. Since $c\neq 1$ we have
${}_{c}D_{a}F(0)\neq{}_{c}D_{a}F(-a)$, and hence ${}_{c}D_{a}F(x)=b$ has at
most one solution in $\\{0,-a\\}$. We first investigate the cases that
${}_{c}D_{a}F(x)=b$ has one solution in $\\{0,-a\\}$.

  * •

If ${}_{c}D_{a}F(0)={}_{c}D_{a}F(1)={}_{c}D_{a}F(\gamma)$ then
${}_{c}D_{a}F(0)={}_{c}D_{a}F(1)$ and ${}_{c}D_{a}F(0)={}_{c}D_{a}F(\gamma)$,
and hence we have $ca^{2}+ca+\gamma=0$ and $ca^{2}+c\gamma a+\gamma=0$. Then
we have $\gamma=1$, a contradiction, so this case cannot happen.

  * •

If ${}_{c}D_{a}F(0)={}_{c}D_{a}F(1-a)={}_{c}D_{a}F(\gamma-a)$ then
${}_{c}D_{a}F(0)={}_{c}D_{a}F(1-a)$ and
${}_{c}D_{a}F(0)={}_{c}D_{a}F(\gamma-a)$, and hence we have
$a^{2}+(c\gamma-\gamma-1)a+\gamma=0$ and $a^{2}+(c-\gamma-1)a+\gamma=0$. Then
we have $\gamma=1$, a contradiction, so this case cannot happen.

  * •

${}_{c}D_{a}F(1)={}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(-a)$ then
${}_{c}D_{a}F(1)={}_{c}D_{a}F(-a)$ and
${}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(-a)$, and hence
$ca^{2}+(c\gamma+c-\gamma)a+c\gamma=0$ and $ca^{2}+(c\gamma+c-1)a+c\gamma=0$.
Then we have $\gamma=1$, a contradiction, so this case cannot happen.

  * •

If ${}_{c}D_{a}F(-a)={}_{c}D_{a}F(1-a)={}_{c}D_{a}F(\gamma-a)$ then
${}_{c}D_{a}F(-a)={}_{c}D_{a}F(1-a)$ and
${}_{c}D_{a}F(-a)={}_{c}D_{a}F(\gamma-a)$, and hence we have
$a^{2}-a+c\gamma=0$ and $a^{2}-\gamma a+c\gamma=0$. Then we have $\gamma=1$, a
contradiction, so this case cannot happen.

In this case, ${}_{c}D_{a}F(x)=b$ has at least four solutions in $P_{a}$
requires it has at least three solutions in $\\{1,\gamma,1-a,\gamma-a\\}$.
Then at least one case that is considered above should be satisfied, which is
a contradiction. Hence the only possible case (that ${}_{c}D_{a}F(x)=b$ has
four solutions) is
$b={}_{c}D_{a}F(1)={}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(1-a)={}_{c}D_{a}F(\gamma-a)$.
We first investigate all possible cases that ${}_{c}D_{a}F(x)=b$ has at least
two solutions in $\\{1,\gamma,1-a,\gamma-a\\}$.

  * •

${}_{c}D_{a}F(1)={}_{c}D_{a}F(\gamma)\ \Leftrightarrow\
ca^{2}+c(\gamma+1)a+\gamma(c+1)=0$.

  * •

${}_{c}D_{a}F(1)={}_{c}D_{a}F(1-a)\ \Leftrightarrow\
(c+1)a^{2}+\gamma(c-1)a+(\gamma-1)(c+1)=0$.

  * •

${}_{c}D_{a}F(1)={}_{c}D_{a}F(\gamma-a)\ \Leftrightarrow\
(c+\gamma)a^{2}+(c-\gamma^{2})a=0$ and hence $a=\frac{\gamma^{2}-c}{\gamma+c}$
when $c\neq-\gamma$.

  * •

${}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(1-a)\ \Leftrightarrow\
(c\gamma+1)a^{2}+(c\gamma^{2}-1)a=0$ and hence
$a=\frac{1-c\gamma^{2}}{1+c\gamma}$ when $c\neq-\gamma^{-1}$.

  * •

${}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(\gamma-a)\ \Leftrightarrow\
(c+1)a^{2}+(c-1)a+\gamma(1-\gamma)(c+1)=0$

If $c=-1$, then ${}_{c}D_{a}F(1)={}_{c}D_{a}F(1-a)$ and
${}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(\gamma-a)$ imply $a=0$, a contradiction.
Hence we require $c\neq-1$. Next we investigate all possible cases that
${}_{c}D_{a}F(x)=b$ has at least three solutions in
$\\{1,\gamma,1-a,\gamma-a\\}$.

  * •

If ${}_{c}D_{a}F(1)={}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(1-a)$ then we have
$ca^{2}+c(\gamma+1)a+\gamma(c+1)=0$,
$(c+1)a^{2}+\gamma(c-1)a+(\gamma-1)(c+1)=0$ and
$(c\gamma+1)a^{2}+(c\gamma^{2}-1)a=0$. We substitute
$ca^{2}=-c(\gamma+1)a-\gamma(c+1)$ which is obtained from the first equation
to the second equation to have $a^{2}-(c+\gamma)a-(c+1)=0$. We substitute
$a=\frac{1-c\gamma^{2}}{c\gamma+1}$ to $a^{2}-(c+\gamma)a-(c+1)=0$ to have
$(\gamma^{3}-\gamma^{2})c^{3}+(2\gamma^{4}-3\gamma)c^{2}+(\gamma^{3}-3\gamma^{2}-2\gamma-2)c-\gamma=0$
when $c\neq-\gamma^{-1}$. But if $c=-\gamma^{-1}$, then we have
$0=(\gamma^{3}-\gamma^{2})c^{3}+(2\gamma^{4}-3\gamma)c^{2}+(\gamma^{3}-3\gamma^{2}-2\gamma-2)c-\gamma=(\gamma+1)^{2}$
and then $\gamma=-1$ and $c=1$, a contradiction.

  * •

If ${}_{c}D_{a}F(1)={}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(\gamma-a)$ then we have
$ca^{2}+c(\gamma+1)a+\gamma(c+1)=0$, $(c+\gamma)a^{2}+(c-\gamma^{2})a=0$ and
$(c+1)a^{2}+(c-1)a+\gamma(1-\gamma)(c+1)=0$. We substitute
$ca^{2}=-c(\gamma+1)a-\gamma(c+1)$ which is obtained from the first equation
to the third equation to have $a^{2}-(c\gamma+1)a-\gamma^{2}(c+1)=0$. We
substitute $a=\frac{\gamma^{2}-c}{\gamma+c}$ from the second equation to
$a^{2}-(c\gamma+1)a-\gamma^{2}(c+1)=0$ to have
$(\gamma^{2}-\gamma)c^{3}+(3\gamma^{3}-2)c^{2}+(2\gamma^{4}+2\gamma^{3}+3\gamma^{2}-\gamma)c+\gamma^{3}=0$
when $c\neq-\gamma$. But if $c=-\gamma$ then
$0=(\gamma^{2}-\gamma)c^{3}+(3\gamma^{3}-2)c^{2}+(2\gamma^{4}+2\gamma^{3}+3\gamma^{2}-\gamma)c+\gamma^{3}=-\gamma^{2}(\gamma+1)^{2}$
which implies $\gamma=-1$ and $c=1$, a contradiction.

  * •

If ${}_{c}D_{a}F(1)={}_{c}D_{a}F(1-a)={}_{c}D_{a}F(\gamma-a)$ then we have
${}_{c^{-1}}D_{-a}F(1-(-a))={}_{c^{-1}}D_{-a}F(1-a-(-a))={}_{c^{-1}}D_{-a}F(\gamma-a-(-a))$
by Lemma 2.2, or equivalently
${}_{c^{-1}}D_{a^{\prime}}F(1)={}_{c^{-1}}D_{a^{\prime}}F(\gamma)={}_{c^{-1}}D_{a^{\prime}}F(1-a^{\prime})$
where $a^{\prime}=-a$. By the above analysis, we have
$(\gamma^{3}-\gamma^{2})(c^{-1})^{3}+(2\gamma^{4}-3\gamma)(c^{-1})^{2}+(\gamma^{3}-3\gamma^{2}-2\gamma-2)c^{-1}-\gamma=0$
or equivalently $\gamma
c^{3}-(\gamma^{3}-3\gamma^{2}-2\gamma-2)c^{2}-(2\gamma^{4}-3\gamma)c-\gamma^{3}+\gamma^{2}=0$.

  * •

If ${}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(1-a)={}_{c}D_{a}F(\gamma-a)$ then we
have
${}_{c^{-1}}D_{a^{\prime}}F(1)={}_{c^{-1}}D_{a^{\prime}}F(\gamma)={}{c^{-1}}D_{a^{\prime}}F(\gamma-a^{\prime})$
where $a^{\prime}=-a$, similarly with the above case. Then we have
$(\gamma^{2}-\gamma)(c^{-1})^{3}+(3\gamma^{3}-2)(c^{-1})^{2}+(2\gamma^{4}+2\gamma^{3}+3\gamma^{2}-\gamma)c^{-1}+\gamma^{3}=0$
or equivalently
$\gamma^{3}c^{3}+(2\gamma^{4}+2\gamma^{3}+3\gamma^{2}-\gamma)c^{2}+(3\gamma^{3}-2)c+\gamma^{2}-\gamma=0$.

So we have the following four cubic equations on $c$.

$\displaystyle(\gamma^{3}-\gamma^{2})c^{3}+(2\gamma^{4}-3\gamma)c^{2}+(\gamma^{3}-3\gamma^{2}-2\gamma-2)c-\gamma$
$\displaystyle=0,$ (4)
$\displaystyle(\gamma^{2}-\gamma)c^{3}+(3\gamma^{3}-2)c^{2}+(2\gamma^{4}+2\gamma^{3}+3\gamma^{2}-\gamma)c+\gamma^{3}$
$\displaystyle=0,$ (5) $\displaystyle\gamma
c^{3}-(\gamma^{3}-3\gamma^{2}-2\gamma-2)c^{2}-(2\gamma^{4}-3\gamma)c-\gamma^{3}+\gamma^{2}$
$\displaystyle=0,$ (6)
$\displaystyle\gamma^{3}c^{3}+(2\gamma^{4}+2\gamma^{3}+3\gamma^{2}-\gamma)c^{2}+(3\gamma^{3}-2)c+\gamma^{2}-\gamma$
$\displaystyle=0.$ (7)

Then $\eqref{inv1r_eq2}-(\gamma-1)\times\eqref{inv1r_eq3}$ is the following
quadratic equation.

$\displaystyle 0$
$\displaystyle=(\gamma^{4}-\gamma^{3}+\gamma^{2})c^{2}+2(\gamma^{5}+\gamma^{3}+\gamma)c+\gamma^{4}-\gamma^{3}+\gamma^{2}$
$\displaystyle=\gamma(\gamma^{2}-\gamma+1)\left(\gamma
c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma\right)$ (8)

We also confirm that $-\eqref{inv1r_eq1}+\gamma\times\eqref{inv1r_eq2}$ and
$\left(-\eqref{inv1r_eq3}\times\gamma^{2}+\eqref{inv1r_eq4}\right)/(\gamma-1)$
are same with $\eqref{inv1r_eq23}\times(\gamma+1)/\gamma$. If
$\gamma^{2}-\gamma+1=0$ then equations (4), (5), (6) and (7) imply
$0=c^{3}+5c^{2}+5c+1=(c+1)(c^{2}+4c+1)$. Since $c\neq-1$, we have
$c^{2}+4c+1=0$ and then $\gamma
c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma=2c(\gamma^{2}-\gamma+1)=0$ when
$\gamma^{2}-\gamma+1=0$. Hence we only require that $\gamma
c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma=0$ whether $\gamma^{2}-\gamma+1=0$ or
not. In this case we set $a=\frac{\gamma^{2}-c}{\gamma+c}$ and
$b={}_{c}D_{a}F(1)=(a+1)^{-1}-c\gamma^{-1}=\frac{1-c}{1+\gamma}$.

Conversely, we assume that $\gamma c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma=0$
with $c=-1$. If $c=-\gamma$ then we have
$0=-\gamma^{3}-2\gamma^{2}-\gamma=-\gamma(\gamma+1)^{2}$ so $\gamma=-1$ and
hence $c=-\gamma=1$ which contradicts to the assumption, so we have
$c\neq-\gamma$. Similarly, we have $c\neq-\gamma^{-1}$. Since
$(\gamma^{2}-c)(1+c\gamma)-(\gamma+c)(1-c\gamma^{2})=(\gamma-1)\left(\gamma
c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma\right)=0$ we set
$a=\frac{\gamma^{2}-c}{\gamma+c}=\frac{1-c\gamma^{2}}{1+c\gamma}$. By the
above analysis, we have ${}_{c}D_{a}F(1)={}_{c}D_{a}F(\gamma-a)$ and
${}_{c}D_{a}F(\gamma)={}_{c}D_{a}F(1-a)$. Since
$a=\frac{\gamma^{2}-c}{\gamma+c}$, we have
$(c+1)a^{2}+\gamma(c-1)a+(\gamma-1)(c+1)=\frac{\gamma(\gamma-1)\left(\gamma
c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma\right)}{(c+\gamma)^{2}}=0$ and hence
${}_{c}D_{a}F(1)={}_{c}D_{a}F(1-a)$. Hence ${}_{c}D_{a}F(x)=b$ has four
solutions in $P_{a}$, which completes the proof. ∎

Next we show the main theorem of this subsection.

######  Theorem 4.7.

Let $F=Inv\circ(1,\gamma)$ with $\gamma\neq 0,1$ and $c\neq 0,1$. Then
${}_{c}\Delta_{F}=6$ if and only if $\gamma
c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma=0$, $c\neq-1$, and $-c$ is a square.
Otherwise, ${}_{c}\Delta_{F}\leq 5$.

###### Proof.

By Lemma 4.3 and Lemma 4.6, ${}_{c}D_{a}F(x)=b$ has four solutions in $P_{a}$
if and only if

$\gamma c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma=0,\ c\neq-1,\
a=\frac{\gamma^{2}-c}{\gamma+c},\ b=\frac{1-c}{1+\gamma},$ (9)

otherwise ${}_{c}D_{a}F(x)=b$ has at most three solutions in $P_{a}$. Hence it
remains to investigate required conditions that ${}_{c}D_{a}F(x)=b$ has two
solutions in $\mathbb{F}_{p^{n}}\setminus P_{a}$ using Lemma 4.1, under the
assumption as in (9). We require that

$\displaystyle(ab+c-1)^{2}-4abc$
$\displaystyle=\frac{\gamma^{2}c^{4}+4(\gamma^{4}-\gamma-1)c^{3}+2(2\gamma^{2}-\gamma^{2}+2)c^{2}-4(\gamma^{4}+\gamma^{3}-\gamma)c+\gamma^{2}}{(\gamma+1)^{2}(c+\gamma)^{2}}$
$\displaystyle=\frac{\left(\gamma
c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma\right)^{2}-4c(\gamma+1)^{2}(c+\gamma)^{2}}{(\gamma+1)^{2}(c+\gamma)^{2}}=-4c$

is a square. We can see that $c\neq-\gamma$ in the proof of Lemma 4.6. If
$\gamma=-1$ then we have $0=\gamma
c^{2}+2(\gamma^{2}+\gamma+1)c+\gamma=-(c-1)^{2}$, and then $c=1$ which is a
contradiction, so we have $\gamma\neq-1$. Hence we get
$(b+c)a+b+c-1=\frac{c(\gamma^{2}-1)}{c+\gamma}\neq 0$,
$(1-b)a+b+c-1=\frac{(\gamma-1)(c+\gamma)}{\gamma+1}\neq 0$,
$(b\gamma+c)a+\gamma(b\gamma+c-1)\frac{(\gamma-1)(c+\gamma)}{\gamma+1}\neq 0$
and $(1-b\gamma)a+\gamma(b\gamma+c-1)=\frac{c(\gamma^{2}-1)}{c+\gamma}\neq 0$.
Therefore, under the assumption (9) ${}_{c}D_{a}F(x)=b$ has two solutions in
$\mathbb{F}_{p^{n}}\setminus P_{a}$ if and only if $-c$ is a square, by Lemma
4.1. ∎

##  5 Concluding Remark

In this paper, we studied the $c$-differential uniformity of the swapped
inverse function $Inv\circ(1,\gamma)$. In section 3, we showed that the
$c$-differential uniformity of $Inv\circ(0,1)$ is at most $5$. In section 4,
we showed that the $c$-differential uniformity of $Inv\circ(1,\gamma)$ is at
most $6$ when $\gamma\neq 1$. For each case (i.e., $c=1$ or $c\neq 1$), we
gave a detailed classification of the ($c$-)differential uniformity of
$Inv\circ(1,\gamma)$, which has been unknown as far as we know. Unfortunately
we could not find (A)PcN permutations except when $p^{n}$ is very small. In
future studies, we will continue to find permutations $Inv\circ h$ with cyclic
permutation $h$ having low $c$-differential uniformities.

Acknowledgements : This work was supported by the National Research Foundation
of Korea (NRF) grant funded by the Korea government (MSIT) (No.
2021R1C1C2003888). Soonhak Kwon was supported by the National Research
Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (No.
2016R1A5A1008055, No. 2019R1F1A1058920 and 2021R1F1A1050721).

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##  Appendix A Proof of Lemma 4.3

By Lemma 2.2, we have
${}_{c}\Delta_{F}(-a,b)={}_{c^{-1}}\Delta_{F}(a,-bc^{-1})$. Hence it is enough
to investigate the case $a\in\\{1,\gamma,\gamma-1\\}$.

If $a=1$ with $\gamma=-1$ then $P_{a}=\\{0,\pm 1,-2\\}$ and we have
${}_{c}D_{1}F(0)=-1$, ${}_{c}D_{1}F(1)=2^{-1}+c$, ${}_{c}D_{1}F(-1)=-c$ and
${}_{c}D_{1}F(-2)=1+2^{-1}c$.

$\begin{array}[]{ll}{}_{c}D_{1}F(0)={}_{c}D_{1}F(1)\ \Leftrightarrow\
c=-\frac{3}{2},&{}_{c}D_{1}F(0)={}_{c}D_{1}F(-1)\ \Leftrightarrow\ c=1,\\\
{}_{c}D_{1}F(0)={}_{c}D_{1}F(-2)\ \Leftrightarrow\
c=-4,&{}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)\ \Leftrightarrow\ c=-\frac{1}{4},\\\
{}_{c}D_{1}F(1)={}_{c}D_{1}F(-2)\ \Leftrightarrow\
c=1,&{}_{c}D_{1}F(-1)={}_{c}D_{1}F(-2)\ \Leftrightarrow\
c=-\frac{2}{3}.\end{array}$

We investigate all possible cases such that ${}_{c}D_{1}F(x)=b$ has at least
three solutions in $P_{a}$.

$\begin{array}[]{l}{}_{c}D_{1}F(0)={}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)\
\Leftrightarrow\ c=-\frac{3}{2}=1=-\frac{1}{4}\ \Rightarrow 2=-3,\ 6=1\text{
and }1=-4\Rightarrow\ p=5.\\\
{}_{c}D_{1}F(0)={}_{c}D_{1}F(1)={}_{c}D_{1}F(-2)\ \Leftrightarrow\
c=-\frac{3}{2}=-4=1\ \Rightarrow-8=-3,\ 2=-3\text{ and }1=-4\Rightarrow\
p=5.\\\ {}_{c}D_{1}F(0)={}_{c}D_{1}F(-1)={}_{c}D_{1}F(-2)\ \Leftrightarrow\
c=1=-4=-\frac{2}{3}\ \Rightarrow 1=-4,\ 2=-3\text{ and }-12=-2\Rightarrow\
p=5.\\\ {}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)={}_{c}D_{1}F(-2)\ \Leftrightarrow\
c=-\frac{1}{4}=1=-4=-\frac{2}{3}\ \Rightarrow 1=-4=16\Rightarrow\
p=5.\end{array}$

So we have
${}_{c}D_{1}F(0)={}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)={}_{c}D_{1}F(-2)=-1$ when
$c=1$ and $p=5$.

If $a=1$ with $\gamma=2$ then $P_{a}=\\{0,\pm 1,2\\}$. If $p=3$ then we have
$P_{a}=\mathbb{F}_{3}$ and hence ${}_{c}D_{a}F(x)=b$ has at most three
solutions in $P_{a}$. If $p\neq 3$, then we have ${}_{c}D_{1}F(0)=2^{-1}$,
${}_{c}D_{1}F(1)=1-2^{-1}c$, ${}_{c}D_{1}F(-1)=c$ and
${}_{c}D_{1}F(2)=3^{-1}-c$. We investigate equivalent condition for each case
that $D_{a}F(x)=b$ has at least two solutions in $P_{a}$.

$\begin{array}[]{ll}{}_{c}D_{1}F(0)={}_{c}D_{1}F(1)\ \Leftrightarrow\
c=1,&{}_{c}D_{1}F(0)={}_{c}D_{1}F(-1)\ \Leftrightarrow\ c=2^{-1},\\\
{}_{c}D_{1}F(0)={}_{c}D_{1}F(2)\ \Leftrightarrow\
c=-\frac{1}{6},&{}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)\ \Leftrightarrow\
c=\frac{2}{3},\\\ {}_{c}D_{1}F(1)={}_{c}D_{1}F(2)\ \Leftrightarrow\
c=-\frac{4}{3},&{}_{c}D_{1}F(-1)={}_{c}D_{1}F(2)\ \Leftrightarrow\
c=\frac{1}{6}.\end{array}$

We investigate all possible cases such that ${}_{c}D_{1}F(x)=b$ has at least
three solutions in $P_{a}$.

$\begin{array}[]{l}{}_{c}D_{1}F(0)={}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)\
\Leftrightarrow\ c=1=2^{-1}\ \Rightarrow 2=1\text{, a contradiction.}\\\
{}_{c}D_{1}F(0)={}_{c}D_{1}F(1)={}_{c}D_{1}F(2)\ \Leftrightarrow\
c=1=-\frac{1}{6}=-\frac{4}{3}\ \Rightarrow 1=-6,\ 3=-4\text{ and
}3=24\Rightarrow\ p=7.\\\ {}_{c}D_{1}F(0)={}_{c}D_{1}F(-1)={}_{c}D_{1}F(2)\
\Leftrightarrow\ c=\frac{1}{6}=-\frac{1}{6}\text{, a contradiction.}\\\
{}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)={}_{c}D_{1}F(2)\ \Leftrightarrow\
c=\frac{2}{3}=-\frac{4}{3}\ \Rightarrow 2=0\text{, a
contradiction.}\end{array}$

So we can see that ${}_{c}D_{1}F(x)=b$ has at most three solutions.

If $a=1$ with $\gamma\neq-1,2$ then $P_{a}=\\{0,1,\gamma,-1,\gamma-1\\}$ and
we have ${}_{c}D_{1}F(0)=\gamma^{-1}$, ${}_{c}D_{1}F(1)=2^{-1}-c\gamma^{-1}$,
${}_{c}D_{1}F(\gamma)=(\gamma+1)^{-1}-c$, ${}_{c}D_{1}F(-1)=c$ and
${}_{c}D_{1}F(\gamma-1)=1-c(\gamma-1)^{-1}$. We investigate equivalent
condition for each case that $D_{a}F(x)=b$ has at least two solutions in
$P_{a}$.

$\begin{array}[]{ll}{}_{c}D_{1}F(0)={}_{c}D_{1}F(1)\ \Leftrightarrow\
c=\frac{\gamma-2}{2},&{}_{c}D_{1}F(0)={}_{c}D_{1}F(\gamma)\ \Leftrightarrow\
c=-\frac{1}{\gamma(\gamma+1)},\\\ {}_{c}D_{1}F(0)={}_{c}D_{1}F(-1)\
\Leftrightarrow\ c=\gamma^{-1},&{}_{c}D_{1}F(0)={}_{c}D_{1}F(\gamma-1)\
\Leftrightarrow\ c=\frac{(\gamma-1)^{2}}{\gamma},\\\
{}_{c}D_{1}F(1)={}_{c}D_{1}F(\gamma)\ \Leftrightarrow\
c=-\frac{\gamma}{2(\gamma+1)},&{}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)\
\Leftrightarrow\ c=\frac{2\gamma}{\gamma+1},\\\
{}_{c}D_{1}F(1)={}_{c}D_{1}F(\gamma-1)\ \Leftrightarrow\
c=\frac{\gamma(\gamma-1)}{2},&{}_{c}D_{1}F(\gamma)={}_{c}D_{1}F(-1)\
\Leftrightarrow\ c=\frac{1}{2(\gamma+1)},\\\
{}_{c}D_{1}F(\gamma)={}_{c}D_{1}F(\gamma-1)\ \Leftrightarrow\
c=-\frac{\gamma(\gamma-1)}{(\gamma+1)(\gamma-2)},&{}_{c}D_{1}F(-1)={}_{c}D_{1}F(\gamma-1)\
\Leftrightarrow\ c=\frac{\gamma-1}{\gamma}.\end{array}$

Next, we can see that ${}_{c}D_{1}F(x)=b$ cannot have four solutions in
$P_{a}$ by showing all possible cases cannot happen.

  * •

${}_{c}D_{1}F(0)={}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)\ \Leftrightarrow\
c=\frac{\gamma-2}{2}=\gamma^{-1}=\frac{2\gamma}{\gamma+1}$. Then we have
$\gamma^{2}-2\gamma-2=2\gamma^{2}-\gamma-1=\gamma^{2}-5\gamma-2=0$ and then
$\gamma=0$, a contradiction. Hence
${}_{c}D_{1}F(0)={}_{c}D_{1}F(1)={}_{c}D_{1}F(\gamma)={}_{c}D_{1}F(-1)$ and
${}_{c}D_{1}F(0)={}_{c}D_{1}F(1)={}_{c}D_{1}F(-1)={}_{c}D_{1}F(\gamma-1)$
cannot happen.

  * •

${}_{c}D_{1}F(1)={}_{c}D_{1}F(\gamma)={}_{c}D_{1}F(-1)\ \Leftrightarrow\
c=-\frac{\gamma}{2(\gamma+1)}=\frac{2\gamma}{\gamma+1}=\frac{1}{2(\gamma+1)}$.
This case cannot happen, since
$-\frac{\gamma}{2(\gamma+1)}=\frac{1}{2(\gamma+1)}$ implies $\frac{1}{2}=0$, a
contradiction. So
${}_{c}D_{1}F(0)={}_{c}D_{1}F(\gamma)={}_{c}D_{1}F(-1)={}_{c}D_{1}F(\gamma-1)$
is also impossible.

  * •

${}_{c}D_{1}F(1)={}_{c}D_{1}F(\gamma)={}_{c}D_{1}F(\gamma-1)\ \Leftrightarrow\
c=-\frac{\gamma}{2(\gamma+1)}=\frac{\gamma(\gamma-1)}{2}=-\frac{\gamma(\gamma-1)}{(\gamma+1)(\gamma-2)}$.
This case cannot happen, since
$-\frac{\gamma}{2(\gamma+1)}=\frac{\gamma(\gamma-1)}{2}$ implies $\gamma=0$, a
contradiction. Hence
${}_{c}D_{1}F(0)={}_{c}D_{1}F(1)={}_{c}D_{1}F(\gamma)={}_{c}D_{1}F(\gamma-1)$
and
${}_{c}D_{1}F(1)={}_{c}D_{1}F(\gamma)={}_{c}D_{1}F(-1)={}_{c}D_{1}F(\gamma-1)$
cannot happen, neither.

If $a=\gamma$ with $\gamma=-1$, then we have
${}_{c}D_{-1}F(1)={}_{c}D_{-1}F(2)={}_{c}D_{-1}F(0)={}_{c}D_{-1}F(-1)=1$ when
$c=1$ and $p=5$ by applying Lemma 2.2 for the already investigated case $a=1$
and $\gamma=-1$. If $a=\gamma$ with $\gamma=2^{-1}$ then
${}_{c}D_{2^{-1}}F_{1}(x)=b$ and ${}_{c}D_{1}F_{2}(x)=b$ have the same number
of solutions in $P_{a}$ where $F_{1}=Inv\circ(1,2^{-1})$ and
$F_{2}=Inv\circ(1,2)$, by applying $\gamma=2$ in Lemma 4.2. Hence
${}_{c}D_{2^{-1}}F_{1}(x)=b$ has at most three solutions in $P_{a}$. If
$a=\gamma$ with $\gamma\neq-1,2^{-1}$ then
$P_{a}=\\{0,1,\gamma,-\gamma,1-\gamma\\}$, and we have
${}_{c}D_{\gamma}F(0)=1$, ${}_{c}D_{\gamma}F(1)=(\gamma+1)^{-1}-c\gamma^{-1}$,
${}_{c}D_{\gamma}F(\gamma)=(2\gamma)^{-1}-c$,
${}_{c}D_{\gamma}F(-\gamma)=c\gamma^{-1}$ and
${}_{c}D_{\gamma}F(1-\gamma)=\gamma^{-1}+c(\gamma-1)^{-1}$. We investigate
equivalent condition for each case that $D_{a}F(x)=b$ has at least two
solutions in $P_{a}$.

$\begin{array}[]{ll}{}_{c}D_{\gamma}F(0)={}_{c}D_{\gamma}F(1)\
\Leftrightarrow\
c=-\frac{\gamma^{2}}{\gamma+1},&{}_{c}D_{\gamma}F(0)={}_{c}D_{\gamma}F(\gamma)\
\Leftrightarrow\ c=-\frac{2\gamma-1}{2\gamma},\\\
{}_{c}D_{\gamma}F(0)={}_{c}D_{\gamma}F(-\gamma)\ \Leftrightarrow\
c=\gamma,&{}_{c}D_{\gamma}F(0)={}_{c}D_{\gamma}F(1-\gamma)\ \Leftrightarrow\
c=\frac{(\gamma-1)^{2}}{\gamma},\\\
{}_{c}D_{\gamma}F(1)={}_{c}D_{\gamma}F(\gamma)\ \Leftrightarrow\
c=-\frac{1}{2(\gamma+1)},&{}_{c}D_{\gamma}F(1)={}_{c}D_{\gamma}F(-\gamma)\
\Leftrightarrow\ c=\frac{\gamma}{2(\gamma+1)},\\\
{}_{c}D_{\gamma}F(1)={}_{c}D_{\gamma}F(1-\gamma)\ \Leftrightarrow\
c=-\frac{\gamma-1}{(\gamma+1)(2\gamma-1)},&{}_{c}D_{\gamma}F(\gamma)={}_{c}D_{\gamma}F(-\gamma)\
\Leftrightarrow\ c=\frac{1}{2(\gamma+1)},\\\
{}_{c}D_{\gamma}F(\gamma)={}_{c}D_{\gamma}F(1-\gamma)\ \Leftrightarrow\
c=-\frac{\gamma-1}{2\gamma^{2}},&{}_{c}D_{\gamma}F(-\gamma)={}_{c}D_{\gamma}F(1-\gamma)\
\Leftrightarrow\ c=-\gamma+1.\end{array}$

Next, we can see that ${}_{c}D_{1}F(x)=b$ cannot have four solutions in
$P_{a}$ by showing all possible cases cannot happen.

  * •

${}_{c}D_{\gamma}F(0)={}_{c}D_{\gamma}F(-\gamma)={}_{c}D_{\gamma}F(1-\gamma)\
\Leftrightarrow\ c=\gamma=\frac{(\gamma-1)^{2}}{\gamma}=-\gamma+1$. Then we
have $c=\gamma=2^{-1}$, a contradiction. Hence
${}_{c}D_{\gamma}F(0)={}_{c}D_{\gamma}F(1)={}_{c}D_{\gamma}F(-\gamma)={}_{c}D_{\gamma}F(1-\gamma)$
and
${}_{c}D_{\gamma}F(0)={}_{c}D_{\gamma}F(\gamma)={}_{c}D_{\gamma}F(-\gamma)={}_{c}D_{\gamma}F(1-\gamma)$
are also impossible.

  * •

${}_{c}D_{\gamma}F(1)={}_{c}D_{\gamma}F(\gamma)={}_{c}D_{\gamma}F(-\gamma)\
\Leftrightarrow\
c=-\frac{1}{2(\gamma+1)}=\frac{\gamma}{2(\gamma+1)}=\frac{1}{2(\gamma+1)}$.
This case cannot happen, because we have $\frac{1}{2}=0$ from
$-\frac{1}{2(\gamma+1)}=\frac{\gamma}{2(\gamma+1)}$ a contradiction. Hence
${}_{c}D_{\gamma}F(0)={}_{c}D_{\gamma}F(1)={}_{c}D_{\gamma}F(\gamma)={}_{c}D_{\gamma}F(-\gamma)$
and
${}_{c}D_{\gamma}F(1)={}_{c}D_{\gamma}F(\gamma)={}_{c}D_{\gamma}F(-\gamma)={}_{c}D_{\gamma}F(1-\gamma)$
are also impossible.

  * •

${}_{c}D_{\gamma}F(1)={}_{c}D_{\gamma}F(\gamma)={}_{c}D_{\gamma}F(1-\gamma)\
\Leftrightarrow\
c=-\frac{1}{2(\gamma+1)}=-\frac{\gamma-1}{(\gamma+1)(2\gamma-1)}=-\frac{\gamma-1}{2\gamma^{2}}$.
This case cannot happen, because we have $-1=0$ from
$-\frac{1}{2(\gamma+1)}=-\frac{\gamma-1}{2\gamma^{2}}$, a contradiction. So,
${}_{c}D_{\gamma}F(0)={}_{c}D_{\gamma}F(1)={}_{c}D_{\gamma}F(\gamma)={}_{c}D_{\gamma}F(1-\gamma)$
is also impossible.

If $a=\gamma-1$ with $\gamma=2$ then we have $a=1$ which is already
investigated. If $a=\gamma-1$ with $\gamma=2^{-1}$ then we have $a=-2^{-1}$.
By Lemma 2.2, ${}_{c}D_{-2^{-1}}F(x)=b$ and ${}_{c^{-1}}D_{2^{-1}}F(x)=-b$
have the same number of solutions in $P_{a}$. By Lemma 4.2,
${}_{c^{-1}}D_{2^{-1}}F(x)=-b$ and ${}_{c^{-1}}D_{1}F_{2}(x)=-2^{-1}b$ have
the same number of solutions in $P_{a}$ where $F_{2}=Inv\circ(1,2)$. Hence
${}_{c}D_{-2^{-1}}F(x)=b$ has at most three solutions in $P_{a}$. If
$a=\gamma-1$ with $\gamma\neq 2,2^{-1}$ then
$P_{a}=\\{0,1,\gamma,1-\gamma,2-\gamma\\}$ and we have
${}_{c}D_{\gamma-1}F(0)=(\gamma-1)^{-1}$,
${}_{c}D_{\gamma-1}F(1)=1-c\gamma^{-1}$,
${}_{c}D_{\gamma-1}F(\gamma)=(2\gamma-1)^{-1}-c$,
${}_{c}D_{\gamma-1}F(1-\gamma)=c(\gamma-1)^{-1}$ and
${}_{c}D_{\gamma-1}F(2-\gamma)=\gamma^{-1}+c(\gamma-2)^{-1}$. We investigate
equivalent condition for each case that $D_{a}F(x)=b$ has at least two
solutions in $P_{a}$.

$\begin{array}[]{ll}{}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(1)\
\Leftrightarrow\
c=\frac{\gamma(\gamma-2)}{\gamma-1},&{}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(\gamma)\
\Leftrightarrow\ c=\frac{-\gamma}{(2\gamma-1)(\gamma-1)},\\\
{}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(1-\gamma)\ \Leftrightarrow\
c=1,&{}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(2-\gamma)\ \Leftrightarrow\
c=\frac{\gamma-2}{\gamma(\gamma-1)},\\\
{}_{c}D_{\gamma-1}F(1)={}_{c}D_{\gamma-1}F(\gamma)\ \Leftrightarrow\
c=-\frac{2\gamma}{2\gamma-1},&{}_{c}D_{\gamma-1}F(1)={}_{c}D_{\gamma-1}F(1-\gamma)\
\Leftrightarrow\ c=\frac{\gamma(\gamma-1)}{2\gamma-1},\\\
{}_{c}D_{\gamma-1}F(1)={}_{c}D_{\gamma-1}F(2-\gamma)\ \Leftrightarrow\
c=\frac{\gamma-2}{2},&{}_{c}D_{\gamma-1}F(\gamma)={}_{c}D_{\gamma-1}F(1-\gamma)\
\Leftrightarrow\ c=\frac{\gamma-1}{\gamma(2\gamma-1)},\\\
{}_{c}D_{\gamma-1}F(\gamma)={}_{c}D_{\gamma-1}F(2-\gamma)\Leftrightarrow
c=\frac{2-\gamma}{\gamma(2\gamma-1)},&{}_{c}D_{\gamma-1}F(1-\gamma)={}_{c}D_{\gamma-1}F(2-\gamma)\Leftrightarrow
c=-\frac{(\gamma-1)(\gamma-2)}{\gamma}.\end{array}$

  * •

${}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(1)={}_{c}D_{\gamma-1}F(\gamma)\
\Leftrightarrow\
c=\frac{\gamma(\gamma-2)}{\gamma-1}=\frac{-\gamma}{(2\gamma-1)(\gamma-1)}=-\frac{2\gamma}{2\gamma-1}$.
Then we have $\gamma=\frac{3}{2}$ and then $c=-\frac{3}{2}$.

  * •

${}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(1)={}_{c}D_{\gamma-1}F(1-\gamma)\
\Leftrightarrow\
c=\frac{\gamma(\gamma-2)}{\gamma-1}=\frac{\gamma(\gamma-1)}{2\gamma-1}=1$.
Then we have $\gamma^{2}-3\gamma+1=0$, such $\gamma$ exists if and only if
$(-3)^{2}-4=5$ is a square if and only if $p^{n}\equiv 0,\pm 1\pmod{5}$.

  * •

${}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(1)={}_{c}D_{\gamma-1}F(2-\gamma)\
\Leftrightarrow\
c=\frac{\gamma(\gamma-2)}{\gamma-1}=\frac{\gamma-2}{\gamma(\gamma-1)}=\frac{\gamma-2}{2}$.
Then we have $\gamma=-1$ and then $c=-\frac{3}{2}$.

  * •

${}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(\gamma)={}_{c}D_{\gamma-1}F(1-\gamma)\
\Leftrightarrow\
c=\frac{-\gamma}{(2\gamma-1)(\gamma-1)}=\frac{\gamma-1}{\gamma(2\gamma-1)}=1$.
Then we have $2\gamma^{2}-2\gamma+1=0$, such $\gamma$ exists if and only if
$(-2)^{2}-8=-4$ is a square if and only if $p^{n}\equiv 1\pmod{4}$.

  * •

${}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(\gamma)={}_{c}D_{\gamma-1}F(2-\gamma)\
\Leftrightarrow\
c=\frac{-\gamma}{(2\gamma-1)(\gamma-1)}=\frac{\gamma-2}{\gamma(\gamma-1)}=\frac{2-\gamma}{\gamma(2\gamma-1)}$.
Then we have $\gamma=\frac{2}{3}$ and then $c=6$.

  * •

${}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(1-\gamma)={}_{c}D_{\gamma-1}F(2-\gamma)\
\Leftrightarrow\
c=1=\frac{\gamma-2}{\gamma(\gamma-1)}=-\frac{(\gamma-1)(\gamma-2)}{\gamma}$.
Then we have $\gamma^{2}-2\gamma+2=0$, such $\gamma$ exists if and only if
$(-2)^{2}-8=-4$ is a square if and only if $p^{n}\equiv 1\pmod{4}$.

  * •

${}_{c}D_{\gamma-1}F(1)={}_{c}D_{\gamma-1}F(\gamma)={}_{c}D_{\gamma-1}F(1-\gamma)\
\Leftrightarrow\
c=-\frac{2\gamma}{2\gamma-1}=\frac{\gamma(\gamma-1)}{2\gamma-1}=\frac{\gamma-1}{\gamma(2\gamma-1)}$.
Then we have $\gamma=-1$ and then $c=-\frac{2}{3}$.

  * •

${}_{c}D_{\gamma-1}F(1)={}_{c}D_{\gamma-1}F(\gamma)={}_{c}D_{\gamma-1}F(2-\gamma)\
\Leftrightarrow\
c=-\frac{2\gamma}{2\gamma-1}=\frac{\gamma-2}{2}=\frac{2-\gamma}{\gamma(2\gamma-1)}$.
Then we have $2\gamma^{2}-\gamma+2=0$, such $\gamma$ exists if and only if
$(-1)^{2}-4\cdot 2\cdot 2=-15$ is a square if and only if $p^{n}\equiv
0,1,2,4,6,8,9,10\pmod{15}$.

  * •

${}_{c}D_{\gamma-1}F(1)={}_{c}D_{\gamma-1}F(1-\gamma)={}_{c}D_{\gamma-1}F(2-\gamma)\
\Leftrightarrow\
c=\frac{\gamma(\gamma-1)}{2\gamma-1}=\frac{\gamma-2}{2}=-\frac{(\gamma-1)(\gamma-2)}{\gamma}$.
Then we have $\gamma=\frac{2}{3}$ and then $c=-\frac{2}{3}$.

  * •

${}_{c}D_{\gamma-1}F(\gamma)={}_{c}D_{\gamma-1}F(1-\gamma)={}_{c}D_{\gamma-1}F(2-\gamma)\
\Leftrightarrow\
c=\frac{\gamma-1}{\gamma(2\gamma-1)}=\frac{2-\gamma}{\gamma(2\gamma-1)}=-\frac{(\gamma-1)(\gamma-2)}{\gamma}$.
Then we have $\gamma=\frac{3}{2}$ and then $c=\frac{1}{6}$.

From above we can see that ${}_{c}D_{\gamma-1}F(x)=b$ has at least four
solutions only when $p=5$, $\gamma=-1$ and $c=1$. For example, if
${}_{c}D_{\gamma-1}F(0)={}_{c}D_{\gamma-1}F(1)={}_{c}D_{\gamma-1}F(\gamma)={}_{c}D_{\gamma-1}F(1-\gamma)$
then $c=-\frac{3}{2}=1=-\frac{2}{3}$ and $\gamma=\frac{3}{2}=-1$ and
$\gamma^{2}-3\gamma+1=2\gamma^{2}-2\gamma+1=0$ can happen only when $p=5$.
Hence we can see that all element in $P_{a}$ is a solution of
${}_{c}D_{\gamma-1}F(x)=b$ if and only if $p=5$, $\gamma=-1$ and $c=1$, and
note that $a=3$ and $b=2$ in this case. By Lemma 2.2, ${}_{1}D_{2}F(x)=3$ also
have five solutions in $P_{a}$.