# The second-order zero differential uniformity of the swapped inverse
functions over finite fields

Jaeseong Jeong1 Namhun Koo2 Soonhak Kwon3
Email<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
1Department of Innovation Center for Industrial Mathematics, National
Institute for Mathematical Sciences, Seongnam, Republic of Korea
2Institute of Mathematical Sciences, Ewha Womans University, Seoul, Republic
of Korea
3Department of Mathematics, Sungkyunkwan University, Suwon, Republic of Korea

###### Abstract

The Feistel Boomerang Connectivity Table (FBCT) was proposed as the feistel
counterpart of the Boomerang Connectivity Table. The entries of the FBCT are
actually related to the second-order zero differential spectrum. Recently,
several results on the second-order zero differential uniformity of some
functions were introduced. However, almost all of them were focused on power
functions, and there are only few results on non-power functions. In this
paper, we investigate the second-order zero differential uniformity of the
swapped inverse functions, which are functions obtained from swapping two
points in the inverse function. We also present the second-order zero
differential spectrum of the swapped inverse functions for certain cases. In
particular, this paper is the first result to characterize classes of non-
power functions with the second-order zero differential uniformity equal to
$4$, in even characteristic.

Keywords. Second-order zero differential uniformity, Swapped inverse function,
Permutation

Mathematics Subject Classification(2020) 94A60, 06E30

##  1 Introduction

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 a prime.

  * •

$Inv(x)=x^{p^{n}-2}$ be the multiplicative inverse function on
$\mathbb{F}_{p^{n}}$.

The substitution box(S-box) plays an important role in the security of block
ciphers, since it is the only nonlinear part in block ciphers. Hence,
vectorial Boolean functions to be used as S-boxes in block ciphers should have
good cryptographic properties. For example, S-boxes need to have low
differential uniformity, where the differential uniformity[15] is a
cryptographic parameter to measure the resistance against differential attack,
which is one of the most well-known attacks on block ciphers. The definition
of the differential uniformity is as follows.

######  Definition 1.1.

Let $f$ be a function over $\mathbb{F}_{p^{n}}$ and
$a,b\in\mathbb{F}_{p^{n}}$. Then we define $\Delta_{f}(a,b)$ by the number of
solutions of $f(x+a)-f(x)=b$. The differential uniformity of $f$ is defined by

$\Delta_{f}=\displaystyle\max_{a\in\mathbb{F}_{p^{n}}^{*},b\in\mathbb{F}_{p^{n}}}\Delta_{f}(a,b).$

Furthermore, we say that $f$ is second-order zero differentially
$\Delta_{f}$-uniform. We say $f$ is Perfect Nonlinear (PN) if $\Delta_{f}=1$,
and $f$ is Almost Perfect Nonlinear (APN) if $\Delta_{f}=2$.

The boomerang attack, an extension of the differential attack, was proposed by
Wagner[17]. At Eurocrypt 2018, Cid et al. [3] introduced the Boomerang
Connectivity Table (BCT), as a tool to analyze the boomerang attack. The BCT
only addressed the case of Substitution Permutation Networks, and does not
cover the case of block ciphers with feistel structures. Boukerrou et al.[1]
proposed the Feistel Boomerang Connectivity Table (FBCT), the feistel
counterpart of the BCT. They also introduced the Feistel Boomerang Uniformity,
which is defined by the maximum value in the FBCT except the first row, the
first column, and diagonal part. In [11], Li et al. observed that the entries
of the FBCT are actually related to the second-order zero differential spectra
of functions. The definition of the second-order zero differential uniformity
is given as follows.

######  Definition 1.2.

Let $f$ be a function over $\mathbb{F}_{p^{n}}$ and
$a,b\in\mathbb{F}_{p^{n}}$. Then we define $\nabla_{f}(a,b)$ by the number of
solutions of

$f(x+a+b)-f(x+a)-f(x+b)+f(x)=0.$ (1)

We also define the second-order zero differential uniformity by

$\nabla_{f}=\begin{cases}\displaystyle\max\\{\nabla_{f}(a,b):a,b\in\mathbb{F}_{2^{n}}^{*},a\neq
b\\}&\text{ if }p=2,\\\
\displaystyle\max\\{\nabla_{f}(a,b):a,b\in\mathbb{F}_{p^{n}}^{*}\\}&\text{ if
}p>2.\end{cases}$

Furthermore, we say that $f$ is second-order zero differentially
$\nabla_{f}$-uniform. We denote

$\omega_{i}=\\#\\{(a,b)\in\mathbb{F}_{p^{n}}\times\mathbb{F}_{p^{n}}:\nabla_{f}(a,b)=i\\},$

for $i\leq\nabla_{f}$. Then, the second-order zero differential spectrum of
$f$ by

$DS_{f}=\\{\omega_{i}\ :\ i\leq\nabla_{f}\\}.$

Recently, several researchers studied functions with low second-order zero
differential uniformity[4, 5, 6, 11, 13, 14]. The majority of those functions
are power functions, and there are only few results on the second-order zero
differential uniformity of non-power functions. Table 1 describes all known
non-power functions for which the second-order zero differential uniformity is
known.

$F(x)$ | $p$ | Conditions | $\nabla_{f}$ | Ref.
---|---|---|---|---
$x^{2^{n}-2}+\operatorname{Tr}\left(\frac{x^{2}}{x+1}\right)$ | $2$ |  $n$ : even | $8$ | [5]
$1/\left(x+\gamma\operatorname{Tr}(x^{2^{k}+1})\right)$ | $2$ |  $k<n$, $\gamma\in\mathbb{F}_{2^{n}}\cap\mathbb{F}_{2^{2k}}^{*}$, $\operatorname{Tr}(\gamma^{2k+1})=0$ | $\leq 8$
$x^{p^{n}-1}+ux^{2}$ | $p=3$ | $u\neq 0$ | $2$ | [6]
$p>3$ | $4$
Table 1: Known non-power functions studied for their second-order zero
differential uniformity

For a transposition $(\alpha,\beta)$ on $\mathbb{F}_{p^{n}}$ where
$\alpha,\beta\in\mathbb{F}_{p^{n}}$ with $\alpha\neq\beta$, we call functions
of the form $Inv\circ(\alpha,\beta)$ by the _swapped inverse functions_. Many
researchers were interested in cryptographic properties of the swapped inverse
functions, and it is known that the swapped inverse functions have good
cryptographic properties, for example, low differential uniformity[10, 8], low
$c$-differential uniformity[16, 8], low boomerang uniformity[9, 2], low
$c$-boomerang uniformity[16], low differential-linear uniformity[7] and high
nonlinearity[10]. In this paper, we study the second-order zero differential
uniformity of the swapped inverse functions. As observed in [8], if
$\alpha\neq 0$, then the swapped inverse function $F=Inv\circ(\alpha,\beta)$
is linear equivalent to

$Inv\circ(1,\alpha^{-1}\beta)=L_{\alpha}\circ F\circ L_{\alpha},$

where $L_{\alpha}(x)=\alpha x$. Since the second-order zero differential
uniformity is invariant under extended-affine equivalence[1], $F$ and
$Inv\circ(1,\alpha^{-1}\beta)$ have the same second-order zero differential
uniformity. Hence, in this paper, we investigate the second order zero
differential uniformity of $Inv\circ(1,\gamma)$, where
$\gamma\in\mathbb{F}_{p^{n}}\setminus\\{1\\}$.

We divided our investigation into two sections: Section 3 for the case
$\gamma=0$, and Section 4 for the case $\gamma\neq 0$. Especially when $p=2$,
we study the second-order zero differential spectrum of $Inv\circ(1,\gamma)$.
For $p$ odd, we find an upper bound of the second-order zero differential
uniformity of $Inv\circ(0,1)$. We could not completely determine the second-
order zero differential uniformity of $Inv\circ(1,\gamma)$ for $\gamma\neq 0$,
but we give several experimental results via SageMath and state some
conjectures on second-order zero differential uniformity of
$Inv\circ(1,\gamma)$ when $p=3$. This case ($\gamma\neq 0$) is notably
extensive, posing many challenges for a complete analysis. Similar
difficulties have already been observed in the evaluation of other uniformity
invariants for this general case $\gamma\neq 0$. For example, a research[2]
for the boomerang uniformity of $Inv\circ(1,\gamma)$ with $\gamma\neq 0$
required 8 pages to be addressed in its preprint version; however, this was
ultimately omitted from the journal publication version. Furthermore, to the
best of our knowledge, there are no known results concerning the
$c$-differential uniformity or $c$-boomerang uniformity for this general case.

The rest of this paper are organized as follows. In Section 2, we give some
preliminaries. In Section 3, we investigate the second-order zero differential
uniformity of $Inv\circ(0,1)$. In Section 4, we study the second-order zero
differential uniformity of $Inv\circ(1,\gamma)$ when $\gamma\neq 0$. Finally,
we give the concluding remark in Section 5.

##  2 Preliminaries

In the following, we give some properties of the FBCT when $p=2$.

######  Proposition 2.1.

([1]) Let $f$ be a function on $\mathbb{F}_{2^{n}}$ and
$a,b\in\mathbb{F}_{2^{n}}$. Then,

  1. 1.

$\nabla_{f}(a,b)=\nabla_{f}(b,a)$.

  2. 2.

$\nabla_{f}(a,b)=2^{n}$ when $ab(a+b)=0$.

  3. 3.

$\nabla_{f}(a,b)\equiv 0\pmod{4}$.

  4. 4.

$\nabla_{f}(a,b)=\nabla(a,a+b)$.

  5. 5.

$f$ is APN if and only if $\nabla_{f}(a,b)=0$ for all
$a,b\in\mathbb{F}_{2^{n}}$ with $ab(a+b)\neq 0$.

By the third term of the above proposition, we see that $\nabla_{f}(a,b)\equiv
0\pmod{4}$ and hence $\omega_{i}=0$ for $i\not\equiv 0\pmod{4}$. The last term
of the above proposition means that $f$ is APN if and only if $f$ is second-
order zero differentially $0$-uniform. Hence, in studies of the second-order
zero differential uniformity, only functions that are not APN have been
considered, and the minimal possible second-order zero differential uniformity
is $4$, in this case. In this paper, we completely characterize second-order
zero differentially $4$-uniform swapped inverse functions.

Next, we give some properties of the FBCT when $p$ is an odd prime. Note that
some of the above proposition does not hold in this case.

######  Proposition 2.2.

([11]) Let $p$ be an odd prime and $f$ be a function on $\mathbb{F}_{p^{n}}$
and $a,b\in\mathbb{F}_{p^{n}}$. Then,

  1. 1.

$\nabla_{f}(a,b)=\nabla_{f}(b,a)$.

  2. 2.

$\nabla_{f}(a,b)=p^{n}$ when $ab=0$.

  3. 3.

$f$ is PN if and only if $\nabla_{f}(a,b)=0$ for all
$a,b\in\mathbb{F}_{p^{n}}$ with $ab\neq 0$.

Similarly with the case of characteristic $2$, only functions that are not PN
have been considered in this research. Unlike the binary case, there are
second-order zero differentially $1$-uniform functions[5, 11]. In this paper
we show that the second-zero differential uniformity of $Inv\circ(0,1)$ is at
most $4$ for odd characteristic case.

We can see that if $x$ is a solution of (1), then $x+a$ is a solution of
$f(x-a+b)-f(x-a)-f(x+b)+f(x)=0$, and $x+b$ is a solution of
$f(x+a-b)-f(x+a)-f(x-b)+f(x)=0$. So, (especially in odd characteristic) we
have the following.

$\nabla_{f}(a,b)=\nabla_{f}(-a,b)=\nabla_{f}(a,-b)=\nabla_{f}(-a,-b).$ (2)

Next we recall that the FBCT of $Inv$.

######  Proposition 2.3.

([4]) Let $p=2$. Then,

$\nabla_{Inv}(a,b)=\begin{cases}0,&\text{ if }ab(a+b)\neq
0,\frac{a}{b}\not\in\mathbb{F}_{4}\setminus\mathbb{F}_{2},\\\ 4,&\text{ if
}ab(a+b)\neq 0,\frac{a}{b}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2},\\\
2^{n},&\text{ if }ab(a+b)=0.\end{cases}$

Table 2 describes $\nabla_{Inv}(a,b)$ when $p$ is odd, which was studied in
[11]. In [11], the authors state that

$\nabla_{Inv}(a,b)=\begin{cases}1,&\text{ if }ab\neq 0,a\neq b,\\\ 3,&\text{
if }ab\neq 0,(a^{2}+ab+b^{2})(a^{2}-ab+b^{2})=a^{4}+a^{2}b^{2}+b^{4}=0,\\\
p^{n},&\text{ otherwise,}\end{cases}$ (3)

for the case $3\mid(p^{n}-1)$. But, the cases $\nabla_{Inv}(a,b)=1$ and
$\nabla_{Inv}(a,b)=3$ in (3) do not totally cover all nontrivial cases that
$ab\neq 0$, leading to a potential misunderstanding that $\nabla_{Inv}=p^{n}$.
Hence, in our opinion, this case should be clarified as shown in Table 2.

$\nabla_{Inv}(a,b)$ | 1 | 3 | $p^{n}$
---|---|---|---
$p=3$ |  $ab\neq 0$, $a^{2}\neq b^{2}$ |  $ab\neq 0$, $a^{2}=b^{2}$ | $ab=0$
$3\mid(p^{n}-1)$ |  $ab\neq 0$, $a^{4}+a^{2}b^{2}+b^{4}\neq 0$ |  $ab\neq 0$, $a^{4}+a^{2}b^{2}+b^{4}=0$ | $ab=0$
$p>3$ and $3\nmid(p^{n}-1)$ | $ab\neq 0$ | - | $ab=0$
Table 2: The value $\nabla_{Inv}(a,b)$ when $p$ is odd [11]

The following is the well-known lemma for the number of solutions of quadratic
equations on $\mathbb{F}_{2^{n}}$, which is useful for our results.

######  Lemma 2.4.

([12]) Let $a\in\mathbb{F}_{2^{n}}^{*}$ and $b,c\in\mathbb{F}_{2^{n}}$.

  1. 1.

$ax^{2}+bx+c=0$ has one solution in $\mathbb{F}_{2^{n}}$ if and only if $b=0$.

  2. 2.

$ax^{2}+bx+c=0$ has two solutions in $\mathbb{F}_{2^{n}}$ if and only if
$\operatorname{Tr}\left(\frac{ac}{b^{2}}\right)=0$.

  3. 3.

$ax^{2}+bx+c=0$ has no solution in $\mathbb{F}_{2^{n}}$ if and only if $b\neq
0$ and $\operatorname{Tr}\left(\frac{ac}{b^{2}}\right)=1$.

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

###  3.1 The second-order zero differential spectrum in even characteristic

In this subsection, we study the second-order zero differential spectrum of
$Inv\circ(0,1)$. In next theorem, we investigate the FBCT of
$f(x)=Inv\circ(0,1)$.

######  Theorem 3.1.

Let $p=2$ and $f(x)=Inv\circ(0,1)$. Then,

$\nabla_{f}(a,b)=\begin{cases}2^{n},&\text{ if }ab(a+b)=0,\\\ 8,&\text{ if
}a\neq b,a^{3}+a+1=b^{3}+b+1=0,\\\ 4,&\text{ if }ab(a+b)\neq 0\text{ and
}[a,b\in\mathbb{F}_{4}^{*}\text{ or }(a,b)\in S],\\\ 0,&\text{
otherwise.}\end{cases}$

where
$S=\\{(a,b)\in\mathbb{F}_{2^{n}}^{*}\times\mathbb{F}_{2^{n}}^{*}:a^{2}+b^{2}+ab\in\\{1,ab(a+b)\\}\\}\setminus\\{(a,b)\in\mathbb{F}_{2^{3}}^{*}\times\mathbb{F}_{2^{3}}^{*}:a^{3}+a+1=b^{3}+b+1=0\\}$.

###### Proof.

If $a=1$, then $b\neq 0,1$ and (1) is equivalent to
$f(x)+f(x+1)+f(x+b)+f(x+b+1)=0$. If $x\in\\{0,1,b,b+1\\}$, then
$f(x)+f(x+1)+f(x+b)+f(x+b+1)=0$ is equivalent to $b^{2}+b+1=0$, that is,
$b\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$. If $x\not\in\\{0,1,b,b+1\\}$,
then $f(x)+f(x+1)+f(x+b)+f(x+b+1)=0$ is equivalent to $b^{2}+b=0$ , a
contradiction. So, $\nabla_{f}(1,b)=4$ if and only if
$b\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$, and $\nabla_{f}(1,b)=0$
otherwise. By Proposition 2.1, we can see that
$\nabla(1,a)=\nabla_{f}(a,1)=\nabla(a,a+1)=4$ if and only if
$a\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$, and
$\nabla_{f}(a,1)=\nabla(a,a+1)=0$ otherwise. To summarize, when
$1\in\\{a,b,a+b\\}$,

$\nabla_{f}(a,b)=\begin{cases}4,&\text{ if }a,b\in\mathbb{F}_{4}^{*},\\\
0,&\text{ otherwise.}\end{cases}$

Now we assume that $ab(a+b)\neq 0$ and $1\not\in\\{a,b,a+b\\}$. If
$x\in\\{0,a,b,a+b\\}$, then (1) is equivalent to

$a^{2}b+ab^{2}+a^{2}+b^{2}+ab=0.$ (4)

If $x\in\\{1,a+1,b+1,a+b+1\\}$, then (1) is equivalent to

$a^{2}+b^{2}+ab=1.$ (5)

If $x\not\in\\{0,a,b,a+b\\}\cup\\{1,a+1,b+1,a+b+1\\}$, then (1) is equivalent
to $x^{2^{n}-2}+(x+a)^{2^{n}-2}+(x+b)^{2^{n}-2}+(x+a+b)^{2^{n}-2}=0$, which is
equivalent to $ab(a+b)=0$, a contradiction. Hence, (1) has at most $8$
solutions $0,a,b,a+b,1,a+1,b+1,a+b+1$.

Next we show that $\nabla_{f}(a,b)=8$ if and only if $a^{3}+a+1=b^{3}+b+1=0$.
Assume that $\nabla_{f}(a,b)=8$. Then, we have (4) and (5) hold. Adding (4)
and (5), we have

$a^{2}b+ba^{2}=1.$ (6)

Dividing $a^{2}$ on the both sides of (5) and dividing $a^{3}$ on the both
sides of (6), we have

$\left(\displaystyle\frac{b}{a}\right)^{2}+\displaystyle\frac{b}{a}=1+\left(\displaystyle\frac{1}{a}\right)^{2}\text{
and
}\left(\displaystyle\frac{b}{a}\right)^{2}+\displaystyle\frac{b}{a}=\left(\displaystyle\frac{1}{a}\right)^{3}$

respectively, and hence adding the two above equations implies that
$a^{3}+a+1=0$. By a symmetric method that dividing $b^{2}$ to (5) and dividing
$b^{3}$ to (6), we also have $b^{3}+b+1=0$.

Conversely, we assume that $a^{3}+a+1=b^{3}+b+1=0$ with $a\neq b$, and it is
enough to show that (4) and (5) hold. We denote $\alpha$ by another solution
of $x^{3}+x+1=0$ so that $\alpha=a+b$ and $\frac{1}{\alpha}=ab$. Then we have

$\displaystyle a^{2}b+ba^{2}+a^{2}+b^{2}+ab$
$\displaystyle=ab(a+b)+(a+b)^{2}+ab=1+\alpha^{2}+\frac{1}{\alpha}=\frac{\alpha^{3}+\alpha+1}{\alpha}=0,$
$\displaystyle a^{2}+b^{2}+ab$
$\displaystyle=(a+b)^{2}+ab=\alpha^{2}+\frac{1}{\alpha}=\frac{\alpha^{3}+1}{\alpha}=\frac{\alpha}{\alpha}=1,$

which completes the proof. ∎

Next, we study the second-order zero differential spectrum of
$f=Inv\circ(0,1)$. Since there are three elements $x\in\mathbb{F}_{2^{n}}$
such that $x^{3}+x+1=0$ if and only if $3\mid n$, we have the following :

$\omega_{8}=\begin{cases}6,&\text{ if }3\mid n,\\\ 0,&\text{ if }3\nmid
n.\end{cases}$ (7)

We denote

$\displaystyle S_{1}$
$\displaystyle=\\{(a,b)\in\mathbb{F}_{2^{n}}^{*}\times\mathbb{F}_{2^{n}}^{*}:\eqref{Inv01_char_even_proof_eq1}\text{
holds with }a\neq b\\}$ $\displaystyle S_{2}$
$\displaystyle=\\{(a,b)\in\mathbb{F}_{2^{n}}^{*}\times\mathbb{F}_{2^{n}}^{*}:\eqref{Inv01_char_even_proof_eq2}\text{
holds with }a\neq b\\}$ $\displaystyle S_{3}$
$\displaystyle=\\{(a,b)\in\mathbb{F}_{2^{n}}^{*}\times\mathbb{F}_{2^{n}}^{*}:(a,b)\in\mathbb{F}_{4}^{*}\times\mathbb{F}_{4}^{*},a\neq
b\\}$

Then, by Theorem 3.1, we have $\omega_{8}=\\#(S_{1}\cap S_{2})$ and

$\omega_{4}=\\#[(S_{1}\cup S_{2})-(S_{1}\cap
S_{2})]+\\#S_{3}=\\#S_{1}+\\#S_{2}+\\#S_{3}-2\omega_{8},$ (8)

because it can be easily checked that $S_{3}\cap(S_{1}\cup S_{2})=\emptyset$.
Note that (4) is equivalent to $(a+1)b^{2}+a(a+1)b+a^{2}=0$. So, for fixed
$a\in\mathbb{F}_{2^{n}}\setminus\\{0,1\\}$, there are two elements
$b\in\mathbb{F}_{2^{n}}$ satisfying (4) if and only if
$\operatorname{Tr}\left(\frac{1}{a+1}\right)=0$, by Lemma 2.4. Since
$\\{\frac{1}{a+1}:a\in\mathbb{F}_{2^{n}}\setminus\\{0,1\\}\\}=\mathbb{F}_{2^{n}}\setminus\\{0,1\\}$,
we have

$\\#S_{1}=\begin{cases}2^{n}-2,&\text{if }2\nmid n,\\\ 2^{n}-4,&\text{if
}2\mid n.\end{cases}$

Similarly, (5) is equivalent to $b^{2}+ab+a^{2}+1=0$. So, for fixed
$a\in\mathbb{F}_{2^{n}}^{*}$, there are two elements $b\in\mathbb{F}_{2^{n}}$
satisfying (5) if and only if $\operatorname{Tr}\left(\frac{1}{a}+1\right)=0$,
by Lemma 2.4. $(a,b)=(1,1)$ is a solution of (5) with $ab(a+b)=0$ and then we
have $\nabla_{f}(1,1)=2^{n}$, so the case $a=1$ need to be excluded. We have
$\\{\frac{1}{a}+1:a\in\mathbb{F}_{2^{n}}\setminus\\{0,1\\}\\}=\mathbb{F}_{2^{n}}\setminus\\{0,1\\}$
and hence $\\#S_{2}=\\#S_{1}$. Furthermore, $\\#S_{3}=6$ when $n$ is even, and
$\\#S_{3}=0$ when $n$ is odd. Therefore, by (8), we have

$\omega_{4}=\begin{cases}2^{n+1}-14,&\text{ if }2\mid n,3\mid n,\\\
2^{n+1}-2,&\text{ if }2\mid n,3\nmid n,\\\ 2^{n+1}-16,&\text{ if }2\nmid
n,3\mid n,\\\ 2^{n+1}-4,&\text{ if }2\nmid n,3\nmid n.\end{cases}$ (9)

Observe that
$\\#\\{(a,b)\in\mathbb{F}_{2^{n}}\times\mathbb{F}_{2^{n}}:ab(a+b)\neq
0\\}=(2^{n}-2)(2^{n}-1)$. By Theorem 3.1, we have

$\omega_{0}=(2^{n}-2)(2^{n}-1)-\omega_{4}-\omega_{8}=\begin{cases}2^{2n}-5\cdot
2^{n}+10,&\text{ if }2\mid n,3\mid n,\\\ 2^{2n}-5\cdot 2^{n}+4,&\text{ if
}2\mid n,3\nmid n,\\\ 2^{2n}-5\cdot 2^{n}+12,&\text{ if }2\nmid n,3\mid n,\\\
2^{2n}-5\cdot 2^{n}+6,&\text{ if }2\nmid n,3\nmid n.\end{cases}$

From (7) and (9), we have the following corollary on the second-order zero
differential uniformity of $Inv\circ(0,1)$.

######  Corollary 3.2.

Let $p=2$ and $f(x)=Inv\circ(0,1)$. Then,

$\nabla_{f}=\begin{cases}4,&\text{ if }3\nmid n,\\\ 8,&\text{ if }3\mid
n.\end{cases}$

###  3.2 The second-order zero differential uniformity in odd characteristic

In next theorem, we study the second-order zero differential uniformity of
$Inv\circ(0,1)$ in odd characteristic.

######  Theorem 3.3.

Let $p$ be an odd prime and $f(x)=Inv\circ(0,1)$ on $\mathbb{F}_{p^{n}}$.
Then, $\nabla_{f}=4$ if $p=29$ or $p=37$, and $\nabla_{f}\leq 3$ otherwise.
Furthermore, $\nabla_{f}(a,b)=4$ if and only if one of the following
conditions holds.

  1. 1.

$p=29$ and $(a,b)=(\pm 2,\pm 12)$ or $(a,b)=(\pm 12,\pm 2)$.

  2. 2.

$p=37$ and $(a,b)=(\pm 2,\pm 6)$ or $(a,b)=(\pm 6,\pm 2)$.

###### Proof.

We first consider the case that $a,b=\pm 1$. If $a=1$ then (1) is equivalent
to

$f(x+b+1)-f(x+1)-f(x+b)+f(x)=0.$ (10)

If $a=b=1$, then (10) is equivalent to

$f(x+2)-2f(x+1)+f(x)=0.$ (11)

  * •

If $x=0$, then (11) is equivalent to $2^{-1}=-1$, which implies $p=3$.

  * •

If $x=-1$, then (11) is equivalent to $-3=0$, which implies $p=3$.

  * •

If $x=-2$, then (11) is equivalent to $2^{-1}=3$, which implies $p=5$.

  * •

If $x=1$, then (11) holds when $p=3$, and we have $3=1$, which is a
contradiction, when $p>3$.

  * •

If $x\not\in\\{0,\pm 1,-2\\}$, then (14) is equivalent to $2=0$, which is a
contradiction.

Hence, we have

$\nabla_{f}(1,1)=\begin{cases}3,&\text{ if }p=3,\\\ 1,&\text{ if }p=5,\\\
0,&\text{ otherwise.}\end{cases}$ (12)

By (2), we also have $\nabla_{f}(1,-1)=\nabla_{f}(1,1)$ is expressed as (12).
Now we assume that $b\neq\pm 1$.

$\begin{array}[]{ll}x=0\ :\eqref{podd_a1_eqn}\ \Leftrightarrow\
b^{2}+b-1=0,&x=1\ :\eqref{podd_a1_eqn}\ \Leftrightarrow\ b^{2}+3b+4=0,\\\
x=-1\ :\eqref{podd_a1_eqn}\ \Leftrightarrow\ 2b^{2}-2b+1=0,&x=-b\
:\eqref{podd_a1_eqn}\ \Leftrightarrow\ b^{2}-b-1=0,\\\ x=-b+1\
:\eqref{podd_a1_eqn}\ \Leftrightarrow\ b^{2}-3b+4=0,&x=-b-1\
:\eqref{podd_a1_eqn}\ \Leftrightarrow\ 2b^{2}+2b+1=0.\end{array}$

If $x\not\in\\{0,\pm 1,-b,-b\pm 1\\}$, then (10) is equivalent to
$b(2x+b+1)=0$, and hence we have that $x=-\frac{b+1}{2}$ is a solution of
(10).

We consider the cases that (10) has at least two solutions in $\\{0,\pm
1,-b,-b\pm 1\\}$. We can see that $x=\alpha$ and $x=-b+\alpha$ cannot be
solutions of (10) simultaneously, where $\alpha\in\\{0,\pm 1\\}$. So, we do
not consider those cases below.

  * •

If $x=0$ and $x=1$ are solutions of (10), then we have
$b^{2}+b-1=b^{2}+3b+4=0$ which implies $b=-\frac{5}{2}$. If $b=-\frac{5}{2}$,
then $b^{2}+b-1=b^{2}+3b+4=\frac{11}{4}$. So, we have $p=11$, and
$b=-\frac{5}{2}=3$.

  * •

If $x=0$ and $x=-1$ are solutions of (10), then we have
$b^{2}+b-1=2b^{2}-2b+1=0$ which implies $b=\frac{3}{4}$. If $b=\frac{3}{4}$,
then $b^{2}+b-1=\frac{5}{16}$ and $2b^{2}-2b+1=\frac{5}{8}$. So, we have
$p=5$, and $b=\frac{3}{4}=2$.

  * •

If $x=0$ and $x=-b+1$ are solutions of (10), then we have
$b^{2}+b-1=b^{2}-3b+4=0$ which implies $b=\frac{5}{4}$. If $b=\frac{5}{4}$,
then $b^{2}+b-1=b^{2}-3b+4=\frac{29}{16}$. So, we have $p=29$, and
$b=\frac{5}{4}=-6$.

  * •

If $x=0$ and $x=-b-1$ are solutions of (10), then we have
$b^{2}+b-1=2b^{2}+2b+1=0$ which implies $3=0$ and hence we have $p=3$. Note
that $b\in\mathbb{F}_{3^{n}}$ with $b^{2}+b-1=0$ exists if and only if $n$ is
even.

  * •

If $x=1$ and $x=-1$ are solutions of (10), then we have
$b^{2}+3b+4=2b^{2}-2b+1=0$ which implies $b=-\frac{7}{8}$. If
$b=-\frac{7}{8}$, then $b^{2}+3b+4=\frac{137}{64}$ and
$2b^{2}-2b+1=\frac{137}{29}$. So, we have $p=137$, and $b=-\frac{7}{8}=119$.

  * •

If $x=1$ and $x=-b$ are solutions of (10), then we have
$b^{2}+3b+4=b^{2}-b-1=0$ which implies $b=-\frac{5}{4}$. If $b=-\frac{5}{4}$,
then $b^{2}+3b+4=b^{2}-b-1=\frac{29}{16}$. So, we have $p=29$, and
$b=-\frac{5}{4}=6$.

  * •

If $x=1$ and $x=-b-1$ are solutions of (10), then we have
$b^{2}+3b+4=2b^{2}+2b+1=0$ which implies $b=-\frac{7}{4}$. If
$b=-\frac{7}{4}$, then $2b^{2}+2b+1=\frac{29}{8}$ and
$b^{2}+3b+4=\frac{29}{16}$. So, we have $p=29$, and $b=-\frac{7}{4}=20$.

  * •

If $x=-1$ and $x=-b$ are solutions of (10), then we have
$2b^{2}-2b+1=b^{2}-b-1=0$ which implies $3=0$ and hence we have $p=3$. Note
that $b\in\mathbb{F}_{3^{n}}$ with $b^{2}-b-1=0$ exists if and only if $n$ is
even.

  * •

If $x=-1$ and $x=-b+1$ are solutions of (10), then we have
$2b^{2}-2b+1=b^{2}-3b+4=0$ which implies $b=\frac{7}{4}$. If $b=\frac{7}{4}$,
then $2b^{2}-2b+1=\frac{29}{8}$ and $b^{2}-3b+4=\frac{29}{16}$. So, we have
$p=29$, and $b=\frac{7}{4}=9$.

  * •

If $x=-b$ and $x=-b+1$ are solutions of (10), then we have
$b^{2}-b-1=b^{2}-3b+4=0$ which implies $b=\frac{5}{2}$. If $b=\frac{3}{4}$,
then $b^{2}-b-1=b^{2}-3b+4=\frac{11}{4}$. So, we have $p=11$, and
$b=\frac{5}{2}=8$.

  * •

If $x=-b$ and $x=-b-1$ are solutions of (10), then we have
$b^{2}-b-1=2b^{2}+2b+1=0$ which implies $b=-\frac{3}{4}$. If $b=-\frac{3}{4}$,
then $b^{2}-b-1=\frac{5}{16}$ and $2b^{2}+2b+1=\frac{5}{8}$. So, we have
$p=5$, and $b=-\frac{3}{4}=3$.

  * •

If $x=-b+1$ and $x=-b-1$ are solutions of (10), then we have
$b^{2}-3b+4=2b^{2}+2b+1=0$ which implies $b=\frac{7}{8}$. If $b=\frac{7}{8}$,
then $b^{2}-3b+4=\frac{137}{64}$ and $2b^{2}+2b+1=\frac{137}{29}$. So, we have
$p=137$, and $b=\frac{7}{8}=18$.

There are no common case in above, and hence (10) has at most two solutions in
$\\{0,\pm 1,-b,-b\pm 1\\}$ and at most one solution in
$\mathbb{F}_{p^{n}}\setminus\\{0,\pm 1,-b,-b\pm 1\\}$. Therefore, we have
$\nabla(1,b)\leq 3$ for all $b\in\mathbb{F}_{p^{n}}$.

Now we assume that $a,b\not\in\\{0,\pm 1\\}$. Next we consider the case that
$a\pm b\in\\{0,\pm 1\\}$.
(i) If $a+b=0$, then (1) is equivalent to

$2f(x)-f(x+a)-f(x-a)=0$ (13)

If $x=0$, $x=1$ and $x=-1$, then (13) is equivalent to $2=0$,
$(a+1)^{-1}=(a-1)^{-1}$ and $\frac{a^{2}}{(a-1)(a+1)}=0$, which are
contradictions. Moreover,

$\begin{array}[]{ll}x=-a\ :\eqref{podd_ab0_eqn}\ \Leftrightarrow\
a=-\frac{3}{2},&x=-a+1\ :\eqref{podd_ab0_eqn}\ \Leftrightarrow\ a=3^{-1},\\\
x=a\ :\eqref{podd_ab0_eqn}\ \Leftrightarrow\ a=\frac{3}{2},&x=a+1\
:\eqref{podd_ab0_eqn}\ \Leftrightarrow\ a=-3^{-1}.\end{array}$

If $x\not\in\\{0,\pm 1,\pm a,\pm a+1\\}$ then
$0,1\not\in\\{x,x+1,x+a,x+1-a\\}$, and hence (13) is equivalent to
$\frac{a^{2}}{x(x+a)(x-a)}=0$, which contradicts to the assumption.

(13) has two solutions $x=-a+1$ and $x=a$ when $3^{-1}=\frac{3}{2}$ or
equivalently $9=2$, which implies $p=7$, and we have $a=3^{-1}=5$. Similarly,
when $p=7$ and $a=-3^{-1}=-\frac{3}{2}=2$, (13) has two solutions $x=-a$ and
$x=a+1$. (13) has two solutions $x=-a$ and $x=-a+1$ when $3^{-1}=-\frac{3}{2}$
or equivalently $-9=2$, which implies $p=11$, and we have $a=3^{-1}=4$.
Similarly, when $p=11$ and$a=-3^{-1}=\frac{3}{2}=7$, (13) has two solutions
$x=-a$ and $x=-a+1$. Therefore, we have

$\nabla(a,-a)=\begin{cases}2,&\text{ if
}(p,a)\in\\{(7,2),(7,5),(11,4),(11,7)\\},\\\ 1,&\text{ if
}(p,a)\not\in\\{(7,2),(7,5),(11,4),(11,7)\\}\text{ and }a\in\\{\pm
3^{-1},\pm\frac{3}{2}\\},\\\ 0,&\text{otherwise.}\end{cases}$

(ii) If $a+b=1$, then (1) is equivalent to

$f(x+1)-f(x+a)-f(x+1-a)+f(x)=0$ (14)

Then, we have

$\begin{array}[]{ll}x=0\ :\eqref{podd_ab1_eqn}\ \Leftrightarrow\
a^{2}-a+1=0,&x=1\ :\eqref{podd_ab1_eqn}\ \Leftrightarrow\ a^{2}-a+4=0,\\\
x=-a\ :\eqref{podd_ab1_eqn}\ \Leftrightarrow\ 2a^{3}-2a+1=0,&x=-a+1\
:\eqref{podd_ab1_eqn}\ \Leftrightarrow\ 3a^{2}-7a+4=(3a-4)(a-1)=0\
\Rightarrow\ a=\frac{4}{3},\\\ x=a\ :\eqref{podd_ab1_eqn}\ \Leftrightarrow\
a=-3^{-1},&x=a-1\ :\eqref{podd_ab1_eqn}\ \Leftrightarrow\
2a^{3}-6a^{2}+4a-1=0.\end{array}$

If $x=-1$, then (14) is equivalent to $1=0$, which is a contradiction. If
$x\not\in\\{0,\pm 1,\pm a,\pm(a-1)\\}$ then $0,1\not\in\\{x,x+1,x+a,x+1-a\\}$,
and hence (14) is equivalent to $a(a-1)(2x+1)=0$, so (14) has a solution
$x=-\frac{1}{2}$.

We consider the cases that (14) has at least two solutions in $\\{0,\pm 1,\pm
a,\pm(a-1)\\}$.

  * •

If $x=0$ and $x=1$ are solutions of (14), then we have $p=3$. But, if $p=3$,
then $0=a^{2}-a+1=(a+1)^{2}$ implies $a=-1$, which contradicts to
$a\not\in\\{0,\pm 1\\}$.

  * •

If $x=0$ and $x=-a$ are solutions of (14), then we have $a^{2}-a+1=0$ and
$2a^{3}-2a+1=0$. So, $0=2a^{3}-2a+1=(a^{2}-a+1)(2a+2)-2a-1=-2a-1$ and hence we
have $a=-\frac{1}{2}$. Then $a^{2}-a+1=2a^{3}-2a+1=\frac{7}{4}$ and hence we
have $p=7$ and then $a=-\frac{1}{2}=3$.

  * •

If $x=0$ and $x=-a+1$ are solutions of (14), then $a=\frac{4}{3}$ and
$0=a^{2}-a+1=\frac{13}{9}$. So, we have $p=13$, and then $a=\frac{4}{3}=10$.

  * •

If $x=0$ and $x=a$ are solutions of (14), then $a=-\frac{1}{3}$ and
$0=a^{2}-a+1=\frac{13}{9}$. So, we have $p=13$, and then $a=-\frac{1}{3}=4$.

  * •

If $x=0$ and $x=a-1$ are solutions of (14), then we have $a^{2}-a+1=0$ and
$2a^{3}-6a^{2}+4a-1=0$. So,
$0=2a^{3}-6a^{2}+4a-1=(a^{2}-a+1)(2a-4)-2a+3=-2a+3$ and hence we have
$a=\frac{3}{2}$. Then $a^{2}-a+1=\frac{7}{4}$ and
$2a^{3}-6a^{2}+4a-1=-\frac{7}{4}$, and hence we have $p=7$ and
$a=\frac{3}{2}=5$.

  * •

If $x=1$ and $x=-a$ are solutions of (14), then we have $a^{2}-a+4=0$ and
$2a^{3}-2a+1=0$. So, $0=2a^{3}-2a+1=(a^{2}-a+4)(2a+2)-8a-7=-8a-7$ and hence we
have $a=-\frac{7}{8}$. Then $a^{2}-a+4=\frac{361}{64}$ and
$2a^{3}-2a+1=\frac{361}{256}$, so we have $p=19$ and $a=-\frac{7}{8}=11$.

  * •

If $x=1$ and $x=-a+1$ are solutions of (14), then we have $a=\frac{4}{3}$ and
$0=a^{2}-a+4=\frac{40}{9}$. So, we have $p=5$, and then $a=\frac{4}{3}=3$.

  * •

If $x=1$ and $x=a$ are solutions of (14), then we have $a^{2}-a+4=0$ and
$a=-\frac{1}{3}$ and $0=a^{2}-a+4=\frac{40}{9}$. So, we have $p=5$, and then
$a=-\frac{1}{3}=3$.

  * •

If $x=1$ and $x=a-1$ are solutions of (14), then we have $a^{2}-a+4=0$ and
$2a^{3}-6a^{2}+4a-1=0$. So,
$0=2a^{3}-6a^{2}+4a-1=(a^{2}-a+4)(2a-4)-8a+15=-8a+15$ and hence we have
$a=\frac{15}{8}$. Then $a^{2}-a+4=\frac{361}{64}$ and
$2a^{3}-6a^{2}+4a-1=-\frac{361}{256}$, so we have $p=19$ and
$a=\frac{15}{8}=9$.

  * •

If $x=-a$ and $x=-a+1$ are solutions of (14), then we have $a=\frac{4}{3}$ and
$0=2a^{3}-2a+1=\frac{83}{27}$. So, we have $p=83$, and then
$a=\frac{4}{3}=29$.

  * •

If $x=-a$ and $x=a$ are solutions of (14), then we have $a=-\frac{1}{3}$ and
$0=2a^{3}-2a+1=\frac{43}{27}$. So, we have $p=43$, and then
$a=-\frac{4}{3}=14$.

  * •

If $x=-a$ and $x=a-1$ are solutions of (14), then we have $2a^{3}-2a+1=0$ and
$2a^{3}-6a^{2}+4a-1=0$. Then, we have
$0=(2a^{3}-2a+1)(1-a)+(2a^{3}-6a^{2}+4a-1)(1+a)-2a^{3}$, which contradicts to
$a\neq 0$.

  * •

If $x=-a+1$ and $x=a$ are solutions of (14), then we have $a=\frac{4}{3}$ and
$a=-\frac{1}{3}$. Then, we have $p=5$ and $a=-\frac{1}{3}=3$.

  * •

If $x=-a+1$ and $x=a-1$ are solutions of (14), then we have $a=\frac{4}{3}$
and $0=2a^{3}-6a^{2}+4a-1=-\frac{43}{27}$. Then, we have $p=43$ and
$a=\frac{4}{3}=30$.

  * •

If $x=a$ and $x=a-1$ are solutions of (14), then we have $a=-\frac{1}{3}$ and
$0=2a^{3}-6a^{2}+4a-1=-\frac{83}{27}$. Then, we have $p=83$ and
$a=-\frac{1}{3}=55$.

In above considered cases, the only candidate that (14) has more than two
solutions in $\\{0,\pm 1,\pm a,\pm(a-1)\\}$ is when $p=5$ and $a=3$. In this
case, $x=1$, $x=-a+1$ and $x=a$ are solutions of (14), but we have $-a+1=3=a$,
and hence (14) has only two solutions in $\\{0,\pm 1,\pm a,\pm(a-1)\\}$.
Therefore, (14) has at most two solutions in $\\{0,\pm 1,\pm a,\pm(a-1)\\}$
and at most one solution $\mathbb{F}_{p^{n}}\setminus\\{0,\pm 1,\pm
a,\pm(a-1)\\}$, and hence $\nabla_{f}(a,1-a)\leq 3$ for all
$a\in\mathbb{F}_{p^{n}}$.

The other cases with $a\pm b\in\\{0,\pm 1\\}$ are covered by the above cases
(i) and (ii).

  * •

If $a+b=-1$, then $-a-b=1$. By (2) and (ii), we have
$\nabla_{f}(a,b)=\nabla_{f}(-a,-b)\leq 3$.

  * •

If $a-b=0$, then $b=a$. By (2) and (i), we have
$\nabla_{f}(a,a)=\nabla_{f}(a,-a)\leq 2$.

  * •

If $a-b=1$, then $b=a+1$. By (2) and the above case that $a+b=-1$, we have
$\nabla_{f}(a,a+1)=\nabla_{f}(a,-a-1)\leq 3$.

  * •

If $a-b=-1$, then $b=a-1$. By (2) and (ii), we have
$\nabla_{f}(a,a-1)=\nabla_{f}(a,-a+1)\leq 3$.

Next we consider that $a,b,a\pm b\not\in\\{0,\pm 1\\}$.

$\displaystyle x=0$ $\displaystyle:\eqref{FBU_eqn}\ \Leftrightarrow\
a^{2}+ab+b^{2}=ab(a+b).$ (15) $\displaystyle x=1$
$\displaystyle:\eqref{FBU_eqn}\ \Leftrightarrow\ a^{2}+ab+b^{2}=-2(a+b)-1.$
(16) $\displaystyle x=-a$ $\displaystyle:\eqref{FBU_eqn}\ \Leftrightarrow\
a^{2}-ab+b^{2}=ab(a-b).$ (17) $\displaystyle x=-a+1$
$\displaystyle:\eqref{FBU_eqn}\ \Leftrightarrow\ a^{2}-ab+b^{2}=2(a-b)-1.$
(18) $\displaystyle x=-b$ $\displaystyle:\eqref{FBU_eqn}\ \Leftrightarrow\
a^{2}-ab+b^{2}=-ab(a-b).$ (19) $\displaystyle x=-b+1$
$\displaystyle:\eqref{FBU_eqn}\ \Leftrightarrow\ a^{2}-ab+b^{2}=-2(a-b)-1.$
(20) $\displaystyle x=-a-b$ $\displaystyle:\eqref{FBU_eqn}\ \Leftrightarrow\
a^{2}+ab+b^{2}=-ab(a+b).$ (21) $\displaystyle x=-a-b+1$
$\displaystyle:\eqref{FBU_eqn}\ \Leftrightarrow\ a^{2}+ab+b^{2}=2(a+b)-1.$
(22)

Any two of equations among (15), (17), (19), and (21) cannot hold
simultaneously. For example, if $x=0$ and $x=-a$ are two solutions of (1),
then we have $ab(1-b)=0$ from (15) and (17), which is a contradiction to the
assumption of the case that $a,b\not\in\\{0,\pm 1\\}$. Hence (1) has at most
one solution in $\\{0,-a,-b,-a-b\\}$. Hence, we obtain that (1) cannot have
two solutions in $\\{0,-a,-b,-a-b\\}$.
We consider any two of (16), (18), (20) and (22) hold simultaneously.

  * •

If (16) and (18) holds, then we have $b=-2$ and then we have $a^{2}=-1$.

  * •

If (16) and (20) holds, then we have $a=-2$ and then we have $b^{2}=-1$.

  * •

If (16) and (22) holds, then we have $a+b=0$, which contradicts to
$a+b\not\in\\{0,\pm 1\\}$.

  * •

If (18) and (20) holds, then we have $a-b=0$, which contradicts to
$a-b\not\in\\{0,\pm 1\\}$.

  * •

If (18) and (22) holds, then we have $a=2$ and then we have $b^{2}=-1$.

  * •

If (18) and (22) holds, then we have $b=2$ and then we have $a^{2}=-1$.

We can see that (1) has at most two solutions in $\\{1,-a+1,-b+1,-a-b+1\\}$,
because we have $a,b=\pm 2$ and hence $a+b=0$ or $a-b=0$, which contradict to
the assumption that $a,b,a\pm b\not\in\\{0,\pm 1\\}$. We substitute the above
cases to (15), (17), (19), and (21). We only consider the cases $a=\pm 2$ and
$b^{2}=-1$, because the cases with $b=\pm 2$ and $a^{2}=-1$ are similar.

  * •

We substitute $a=2$, $b^{2}=-1$ to (15), then we have $b=\frac{5}{2}$. Then
$b^{2}=\frac{25}{4}=-1$ implies $p=29$, and then $b=\frac{5}{2}=17=-12$.

  * •

We substitute $a=-2$, $b^{2}=-1$ to (15), then we have $b=\frac{1}{6}$. Then
$b^{2}=\frac{1}{36}=-1$ implies $p=37$, and then $b=\frac{1}{6}=3-6$.

Similarly, we obtain the followings:

$\displaystyle a=2,b^{2}=-1\text{ in }\eqref{podd x=-a eqn}$ $\displaystyle\
\Rightarrow\ p=37\text{ and }b=\frac{1}{6}=-6.$ $\displaystyle
a=-2,b^{2}=-1\text{ in }\eqref{podd x=-a eqn}$ $\displaystyle\ \Rightarrow\
p=29\text{ and }b=\frac{5}{2}=-12.$ $\displaystyle a=2,b^{2}=-1\text{ in
}\eqref{podd x=-b eqn}$ $\displaystyle\ \Rightarrow\ p=29\text{ and
}b=-\frac{5}{2}=12.$ $\displaystyle a=-2,b^{2}=-1\text{ in }\eqref{podd x=-b
eqn}$ $\displaystyle\ \Rightarrow\ p=37\text{ and }b=-\frac{1}{6}=6.$
$\displaystyle a=2,b^{2}=-1\text{ in }\eqref{podd x=-a-b eqn}$ $\displaystyle\
\Rightarrow\ p=37\text{ and }b=-\frac{1}{6}=6.$ $\displaystyle
a=-2,b^{2}=-1\text{ in }\eqref{podd x=-a-b eqn}$ $\displaystyle\ \Rightarrow\
p=29\text{ and }b=-\frac{5}{2}=12.$

To summarize, (1) has three solutions in
$\\{0,1,-a,-a+1,-b,-b+1,-a-b,-a-b+1\\}$ if $(p,a,b)\in\\{(29,\pm 2,\pm
12),(37,\pm 2,\pm 6)\\}$. Similarly, for the cases that $b=\pm 2$ and
$a^{2}=-1$, we can see that (1) has three solutions in
$\\{0,1,-a,-a+1,-b,-b+1,-a-b,-a-b+1\\}$ if $(p,a,b)\in\\{(29,\pm 12,\pm
2),(37,\pm 6,\pm 2)\\}$.

If $x\not\in\\{0,1,-a,-a+1,-b,-b+1,-(a+b),-(a+b)+1\\}$, then
$0,1\not\in\\{x,x+a,x+b,x+a+b\\}$, and hence (1) is equivalent to
$ab(2x+a+b)=0$, hence $x=-\frac{a+b}{2}$ is a solution of (1). We can easily
check that $-\frac{a+b}{2}\not\in\\{0,1,-a,-a+1,-b,-b+1,-(a+b),-(a+b)+1\\}$
for all

$(p,a,b)\in\\{(29,\pm 2,\pm 12),(29,\pm 12,\pm 2),(37,\pm 2,\pm 6),(37,\pm
6,\pm 2)\\},$ (23)

and hence we have $\nabla_{f}(a,b)=4$ if and only if (23) holds, which
completes the proof. ∎

######  Remark 3.4.

By (12) and Theorem 3.3, we have $\nabla_{f}=3$ when $p=3$.

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

###  4.1 The second-order zero differential spectrum in even characteristic

######  Theorem 4.1.

Let $p=2$ and $f(x)=Inv\circ(1,\gamma)$ where
$\gamma\in\mathbb{F}_{2^{n}}\setminus\\{0,1\\}$. If $ab(a+b)\neq 0$, then we
have

$\nabla_{f}(a,b)=\begin{cases}8,&\text{if
}1\in\\{a,b,a+b\\},(\\{a,b\\}\setminus\\{1\\})\subset S_{\gamma,1},\\\
&\gamma\in\\{a,b,a+b\\},(\\{a,b\\}\setminus\\{\gamma\\})\subset
S_{\gamma,2},\\\ &\frac{b}{a}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}\text{
and
}a^{3}\in\\{\gamma+1,\gamma^{3}+\gamma^{2}\\},\\{a,b,a+b\\}\cap\\{1,\gamma\\}=\emptyset\\\
4,&\text{if
}\gamma\in\mathbb{F}_{4}\setminus\mathbb{F}_{2},a,b\in\mathbb{F}_{4}^{*},a\neq
b,\\\ &1\in\\{a,b,a+b\\},(\\{a,b\\}\setminus\\{1\\})\subset S_{\gamma,3}\cup
S_{\gamma,4}\\\
&\gamma\in\\{a,b,a+b\\},(\\{a,b\\}\setminus\\{\gamma\\})\subset
S_{\gamma,5}\cup S_{\gamma,6}\\\ &\frac{b}{a}\not\in\mathbb{F}_{4},(a,b)\in
S_{\gamma,7}\cup S_{\gamma,8},\\{a,b,a+b\\}\cap\\{1,\gamma\\}=\emptyset,\\\
&\frac{b}{a}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2},(a,b)\not\in
S_{\gamma,7}\cup S_{\gamma,8},\\{a,b,a+b\\}\cap\\{1,\gamma\\}=\emptyset,\\\
0,&\text{otherwise.}\end{cases}$

where

$\begin{array}[]{ll}S_{\gamma,1}=\\{c\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}:\gamma=c(c+1)\\},&S_{\gamma,2}=\\{c\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}:\gamma=c^{2}/(c+1)\\},\\\
S_{\gamma,3}=\\{c\in\mathbb{F}_{2^{n}}\setminus\mathbb{F}_{8}:\gamma=c(c+1)\\},&S_{\gamma,4}=\\{c\in\mathbb{F}_{2^{n}}\setminus\mathbb{F}_{8}:\gamma^{3}+\gamma^{2}+(c^{2}+c+1)\gamma=1\\},\\\
S_{\gamma,5}=\\{c\in\mathbb{F}_{2^{n}}\setminus\mathbb{F}_{8}:\gamma=c^{2}/(c+1)\\},&S_{\gamma,6}=\\{c\in\mathbb{F}_{2^{n}}\setminus\mathbb{F}_{8}:\gamma^{3}+\gamma^{2}+(c+1)\gamma+c^{2}=1\\},\\\
S_{\gamma,7}=\\{(a,b)\in\mathbb{F}_{2^{n}}^{*}\times\mathbb{F}_{2^{n}}^{*}:G_{\gamma,a,b}=\gamma+1\\},&S_{\gamma,8}=\\{(a,b)\in\mathbb{F}_{2^{n}}^{*}\times\mathbb{F}_{2^{n}}^{*}:G_{\gamma,a,b}=\gamma^{2}(\gamma+1)\\}\\\
\end{array}$

and $G_{\gamma,a,b}=ab(a+b)+(a^{2}+ab+b^{2})(\gamma+1)$.

###### Proof.

Our proof is divided into three cases, $1\in\\{a,b,a+b\\}$,
$\gamma\in\\{a,b,a+b\\}$ and $1,\gamma\not\in\\{a,b,a+b\\}$.

(Case 1) First, we consider the case $1\in\\{a,b,a+b\\}$. Let $a=1$ and $b\neq
0,1$. Then, (1) is equivalent to

$f(x)+f(x+1)+f(x+b)+f(x+b+1)=0.$ (24)

If $b\in\\{\gamma,\gamma+1\\}$, then (24) is equivalent to
$f(x)+f(x+1)+f(x+\gamma)+f(x+\gamma+1)=0$. If $x\in\\{0,1,\gamma,\gamma+1\\}$,
then $f(x)+f(x+1)+f(x+\gamma)+f(x+\gamma+1)=0$ is equivalent to
$\gamma^{2}+\gamma+1=0$. If $x\not\in\\{0,1,\gamma,\gamma+1\\}$, then
$f(x)+f(x+1)+f(x+\gamma)+f(x+\gamma+1)=0$ is equivalent to
$\gamma^{2}+\gamma=0$, which contradicts to the assumption that
$\gamma\not\in\\{0,1\\}$.
Assume that $b\not\in\\{\gamma,\gamma+1\\}$. If $x\in\\{0,1,b,b+1\\}$, then
(24) is equivalent to $\gamma=b(b+1)$. If
$x\in\\{\gamma,\gamma+1,\gamma+b,\gamma+b+1\\}$, then (24) is equivalent to
$\gamma(\gamma^{3}+\gamma^{2}+(b^{2}+b+1)\gamma+1)=0$. If
$x\not\in\\{0,1,b,b+1\\}\cup\\{\gamma,\gamma+1,\gamma+b,\gamma+b+1\\}$, (24)
is equivalent to $b(b+1)=0$, which contradicts to $b\neq 0,1$. (24) has 8
solutions $\\{0,1,b,b+1\\}\cup\\{\gamma,\gamma+1,\gamma+b,\gamma+b+1\\}$ if
and only if $\gamma=b(b+1)$ and $\gamma^{3}+\gamma^{2}+(b^{2}+b+1)\gamma+1=0$
hold, which implies that $b^{6}+b^{5}+b^{4}+b^{3}+b^{2}+b+1=0$ which is
equivalent to $b\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}$. So, we have the
following:

$\nabla_{f}(1,b)=\begin{cases}8,&\text{ if }\gamma=b(b+1)\text{ where
}b\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}\\\ 4,&\text{ if
}\gamma=b(b+1)\text{ or }\gamma^{3}+\gamma^{2}+(b^{2}+b+1)\gamma=1\text{ where
}b\not\in\mathbb{F}_{8}\\\ &\text{ or
}b,\gamma\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}\\\ 0,&\text{
otherwise.}\end{cases}$

Note that the cases $b=1$ and $a+b=1$ can be similarly analyzed with the case
$a=1$, and we summarize the cases $1\in\\{a,b,a+b\\}$ as follows :

$\nabla_{f}(a,b)=\begin{cases}8,&\text{ if }(\\{a,b\\}\setminus\\{1\\})\subset
S_{\gamma,1},\\\ 4,&\text{ if }(\\{a,b\\}\setminus\\{1\\})\subset
S_{\gamma,3}\cup S_{\gamma,4}\\\ &\text{ or
}b,\gamma\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}\\\ 0,&\text{
otherwise.}\end{cases}$

(Case 2) Next, we consider the case $\gamma\in\\{a,b,a+b\\}$. Let $a=\gamma$
and $b\neq 0,\gamma$. Then, (1) is equivalent to

$f(x)+f(x+\gamma)+f(x+b)+f(x+b+\gamma)=0.$ (25)

If $b\in\\{1,\gamma+1\\}$, then (25) is equivalent to
$f(x)+f(x+1)+f(x+\gamma)+f(x+\gamma+1)=0$. Similarly with the case $a=1$,
$f(x)+f(x+1)+f(x+\gamma)+f(x+\gamma+1)=0$ has 4 solutions $0,1,\gamma$ and
$\gamma+1$ if $\gamma\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$, and no
solution otherwise.
Assume that $b\not\in\\{1,\gamma+1\\}$. If $x\in\\{0,\gamma,b,b+\gamma\\}$,
then (25) is equivalent to $(b+1)\gamma+b^{2}=0$. If
$x\in\\{1,\gamma+1,b+1,b+\gamma+1\\}$, then (25) is equivalent to
$\gamma^{3}+\gamma^{2}+(b+1)\gamma+b^{2}+1=0$. If
$x\not\in\\{0,\gamma,b,b+\gamma\\}\cup\\{1,\gamma+1,b+1,b+\gamma+1\\}$, then
(25) is equivalent to $b\gamma(\gamma+b)=0$, which contradicts to the
assumption that $b\neq 0,\gamma$. (25) has 8 solutions
$\\{0,\gamma,b,b+\gamma\\}\cup\\{1,\gamma+1,b+1,b+\gamma+1\\}$ if and only if
$\gamma=b^{2}/(b+1)$ and $\gamma^{3}+\gamma^{2}+(b+1)\gamma+b^{2}+1=0$ which
implies that $b^{6}+b^{5}+b^{4}+b^{3}+b^{2}+b+1=0$ which is equivalent to
$b\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}$. Similarly with (Case 1), the
cases $b=\gamma$ and $a+b=\gamma$ can be similarly analyzed with the case
$a=\gamma$. So, we have the following:

$\nabla_{f}(\gamma,b)=\begin{cases}8,&\text{ if
}(\\{a,b\\}\setminus\\{\gamma\\})\subset S_{\gamma,2},\\\ 4,&\text{ if
}(\\{a,b\\}\setminus\\{\gamma\\})\subset S_{\gamma,5}\cup S_{\gamma,6}\\\
&\text{ or }b,\gamma\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}\\\ 0,&\text{
otherwise.}\end{cases}$

(Case 3) Now we assume that $1,\gamma\not\in\\{a,b,a+b\\}$.

$\displaystyle x\in\\{0,a,b,a+b\\}$ $\displaystyle:\eqref{FBU_eqn}\
\Leftrightarrow\ a^{2}+ab+b^{2}=0.$ (26) $\displaystyle
x\in\\{1,a+1,b+1,a+b+1\\}$ $\displaystyle:\eqref{FBU_eqn}\ \Leftrightarrow\
(a^{2}+ab+b^{2}+1)\gamma+1=a^{2}b+ab^{2}+a^{2}+ab+b^{2}.$ (27) $\displaystyle
x\in\\{\gamma,a+\gamma,b+\gamma,a+b+\gamma\\}$ $\displaystyle:\eqref{FBU_eqn}\
\Leftrightarrow\
\gamma^{3}+\gamma^{2}+(a^{2}+ab+b^{2})\gamma=a^{2}b+ab^{2}+a^{2}+ab+b^{2}.$
(28)

If
$x\not\in\\{0,a,b,a+b\\}\cup\\{1,a+1,b+1,a+b+1\\}\cup\\{\gamma,a+\gamma,b+\gamma,a+b+\gamma\\}$,
then (1) is equivalent to $ab(a+b)=0$, which implies trivial cases. So, (1)
has at most 12 solutions in
$\\{0,a,b,a+b\\}\cup\\{1,a+1,b+1,a+b+1\\}\cup\\{\gamma,a+\gamma,b+\gamma,a+b+\gamma\\}$.
Now we analyze that (1) has at least 8 solutions.
(i) Substituting (26) to (27), we have $\gamma+1=a^{2}b+ab^{2}$. By (26), we
have $\frac{b}{a}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$ and hence we obtain
$\gamma+1=a^{3}=b^{3}$.
(ii) Substituting (26) to (28), we have $\gamma^{3}+\gamma^{2}=a^{2}b+ab^{2}$.
Similarly with (i) we obtain
$\frac{b}{a}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$ and hence
$\gamma^{3}+\gamma^{2}=a^{3}=b^{3}$.
(iii) Adding (27) and (28), we have $(\gamma+1)^{3}=0$, which contradicts to
the assumption that $\gamma\neq 1$. Hence (27) and (28) cannot hold
simultaneously.

By (iii), (1) cannot have $12$ solutions, and hence
$\nabla_{f}(a,b)\in\\{0,4,8\\}$. By (i) and (ii), we have $\nabla_{f}(a,b)=8$
if and only if $\frac{b}{a}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$ and
$a^{3}=b^{3}=\gamma+1$ or $a^{3}=b^{3}=\gamma^{3}+\gamma^{2}$ with
$1,\gamma\not\in\\{a,b,a+b\\}$. Moreover, we have $\nabla_{f}(a,b)=4$ if only
one of (26), (27) and (28) holds. Hence, $\nabla_{f}(a,b)=4$ if and only if
$\frac{b}{a}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$ and both (27) and (28)
do not hold, or $\frac{b}{a}\not\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$ and
one of (27) and (28) holds.

Therefore, under the assumption that $1,\gamma\not\in\\{a,b,a+b\\}$, we have

$\nabla_{f}(a,b)=\begin{cases}8,&\text{ if
}\frac{b}{a}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2},a^{3}\in\\{\gamma+1,\gamma^{3}+\gamma^{2}\\},\\\
4,&\text{ if }\frac{b}{a}\not\in\mathbb{F}_{4},(a,b)\in S_{\gamma,7}\cup
S_{\gamma,8},\\\ &\text{ or
}\frac{b}{a}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2},(a,b)\not\in
S_{\gamma,7}\cup S_{\gamma,8},\\\ 0,&\text{ otherwise.}\end{cases}$

We complete the proof. ∎

Next, we further investigate the second-order zero differential spectrum of
$f=Inv\circ(1,\gamma)$. We denote

$\begin{array}[]{ll}A_{4}=\\{1\\}\cup S_{\gamma,1},&S_{4}=\\{(a,b)\in
A_{4}\times A_{4}:a\neq b\\},\\\ A_{5}=\\{\gamma\\}\cup
S_{\gamma,2},&S_{5}=\\{(a,b)\in A_{5}\times A_{5}:a\neq b\\}.\end{array}$

By Theorem 4.1, we have $\nabla_{f}(a,b)=8$ if $(a,b)\in S_{4}\cup S_{5}$ when
$\gamma\in\mathbb{F}_{8}$. By Lemma 2.4, $x^{2}+x+\gamma=0$ has two solutions
in $\mathbb{F}_{8}\setminus\mathbb{F}_{2}$ if and only if
$\operatorname{Tr}(\gamma)=0$, and $x^{2}+\gamma x+\gamma=0$ has two solutions
in $\mathbb{F}_{8}\setminus\mathbb{F}_{2}$ if and only if
$\operatorname{Tr}\left(\frac{1}{\gamma}\right)=0$. When
$\gamma\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}$, we have
$\operatorname{Tr}\left(\frac{1}{\gamma}\right)=\operatorname{Tr}(\gamma^{3})$
and

$\operatorname{Tr}\left(\gamma+\frac{1}{\gamma}\right)=\operatorname{Tr}(\gamma+\gamma^{3})=(\gamma+\gamma^{2}+\gamma^{2^{2}})+(\gamma^{3}+\gamma^{3\cdot
2}+\gamma^{3\cdot
2^{2}})=\gamma+\gamma^{2}+\gamma^{3}+\gamma^{4}+\gamma^{5}+\gamma^{6}=1.$ (29)

So, we have $\\#A_{4}=3$ and $\\#A_{5}=0$, or $\\#A_{4}=0$ and $\\#A_{5}=3$,
and hence $\\#(S_{4}\cup S_{5})=6$.

When $2\mid n$, we have $\nabla_{f}(a,b)=8$ if $(a,b)\in S_{6}\cup S_{7}$
where

$\displaystyle A_{6}=\\{a\in\mathbb{F}_{2^{n}}^{*}:a^{3}=\gamma+1\\},\ $
$\displaystyle S_{6}=\\{(a,b)\in A_{6}\times A_{6}:a\neq b,\\},$
$\displaystyle
A_{7}=\\{a\in\mathbb{F}_{2^{n}}^{*}:a^{3}=\gamma^{3}+\gamma^{2}\\},\ $
$\displaystyle S_{7}=\\{(a,b)\in A_{7}\times A_{7}:a\neq b,\\}.$

We consider the cases that $(a,b)\in S_{6}\cup S_{7}$ and
$\\{a,b,a+b\\}\cap\\{1,\gamma\\}\neq\emptyset$. We assume that
$a\in\\{1,\gamma\\}$, and the cases $b,a+b\in\\{1,\gamma\\}$ are similar. If
$a=1$, then we have $\gamma^{3}+\gamma^{2}=1$, since $\gamma\neq 0$. If
$a=\gamma$, then $\gamma+1=\gamma^{3}$, , since $\gamma\neq 0$. Note that
$\gamma\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}$ if and only if
$\gamma^{3}+\gamma+1=0$ or $\gamma^{3}+\gamma^{2}+1=0$. Therefore, we have
$\omega_{8}=12$ when $6\mid n$ and
$\gamma\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}$.

If $\gamma\not\in\mathbb{F}_{8}$, then we have $\nabla_{f}(a,b)=8$ for all
$(a,b)\in S_{6}\cup S_{7}$. And we also have $\\#(S_{6}\cup
S_{7})=6\cdot\\#(C_{n}\cap\\{\gamma+1,\gamma^{3}+\gamma^{2}\\})$, where

$C_{n}=\\{\alpha^{3}:\alpha\in\mathbb{F}_{2^{n}}\setminus\mathbb{F}_{4}\\}$
(30)

is the set of all cubic elements in $\mathbb{F}_{2^{n}}\setminus\\{0,1\\}$.

Therefore, we have the following

$\omega_{8}=\begin{cases}12,&\text{ if }6\mid
n,\gamma\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}\\\ 6,&\text{ if }3\mid
n,2\nmid n,\gamma\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}\\\
6\cdot\\#(C_{n}\cap\\{\gamma+1,\gamma^{3}+\gamma^{2}\\}),&\text{ if }2\mid
n,\gamma\not\in\mathbb{F}_{8},\\\ 0,&\text{ otherwise,}\end{cases}$ (31)

where $C_{n}$ is given in (30). In particular, we have the following corollary
from (31).

######  Corollary 4.2.

Let $p=2$ and $f(x)=Inv\circ(1,\gamma)$ where
$\gamma\in\mathbb{F}_{2^{n}}\setminus\mathbb{F}_{2}$. Then,

$\nabla_{f}=\begin{cases}8,&\text{ if }2\mid
n,\\{\gamma^{3}+\gamma^{2},\gamma+1\\}\cap C_{n}\neq\emptyset,\\\ &\ \ \ \
3\mid n,\gamma\in\mathbb{F}_{8}\setminus\mathbb{F}_{2},\\\ 4,&\text{
otherwise,}\end{cases}$

where $C_{n}$ is given in (30). In particular, we have $\nabla_{f}=4$ for all
$\gamma\in\mathbb{F}_{2^{n}}$, if $n$ is odd with $3\nmid n$.

Next we consider to compute $\omega_{4}$.

(Case 1) When $n$ is even, there are 6 pairs $(a,b)$ such that
$a,b\in\mathbb{F}_{4}^{*}$, $a\neq b$ for
$\gamma\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$.

(Case 2) We consider to compute the number of pairs $(a,b)$ such that

$\displaystyle 1\in\\{a,b,a+b\\},$ $\displaystyle\
(\\{a,b\\}\setminus\\{1\\})\subset S_{\gamma,3}\cup S_{\gamma,4},\text{ or }$
$\displaystyle\gamma\in\\{a,b,a+b\\},$ $\displaystyle\
(\\{a,b\\}\setminus\\{\gamma\\})\subset S_{\gamma,5}\cup S_{\gamma,6}.$

We denote

$\displaystyle A_{8}$ $\displaystyle=\\{1\\}\cup S_{\gamma,3},\
A_{9}=\\{1\\}\cup S_{\gamma,4},\ A_{10}=\\{\gamma\\}\cup S_{\gamma,5},\
A_{11}=\\{\gamma\\}\cup S_{\gamma,6},$ $\displaystyle S_{k}$
$\displaystyle=\\{(a,b)\in A_{k}\times A_{k}:a\neq b\\},\ k=8,9,10,11.$

By Theorem 4.1, we have $\nabla_{f}(a,b)=4$ if $(a,b)\in S_{8}\cup S_{9}\cup
S_{10}\cup S_{11}$. By Lemma 2.4,

$\begin{array}[]{ll}\\#S_{\gamma,3}=2\ \Leftrightarrow\
\operatorname{Tr}(\gamma)=0,&\ \\#S_{\gamma,4}=2\ \Leftrightarrow\
\operatorname{Tr}(1/\gamma)=\operatorname{Tr}(1),\\\ \\#S_{\gamma,5}=2\
\Leftrightarrow\ \operatorname{Tr}(1/\gamma)=0,&\ \\#S_{\gamma,6}=2\
\Leftrightarrow\ \operatorname{Tr}(\gamma)=\operatorname{Tr}(1),\end{array}$

assuming $\gamma\in\mathbb{F}_{2^{n}}\setminus\mathbb{F}_{8}$. If $n$ is even,
we have

$\\#(S_{8}\cup S_{11})=\begin{cases}12,&\text{ if
}\operatorname{Tr}(\gamma)=0,\\\ 0,&\text{ if
}\operatorname{Tr}(\gamma)=1,\end{cases}\ \ \ \\#(S_{9}\cup
S_{10})=\begin{cases}12,&\text{ if }\operatorname{Tr}(1/\gamma)=0,\\\
0,&\text{ if }\operatorname{Tr}(1/\gamma)=1.\end{cases}$

If $n$ is odd, then we have $\\#(S_{8}\cup S_{11})=\\#(S_{9}\cup S_{10})=6$.
Note that $\\#(S_{8}\cup S_{9}\cup S_{10}\cup S_{11})=0$ when
$\gamma\in\mathbb{F}_{8}$.

(Case 3) We consider to compute the number of pairs $(a,b)$ such that

$\frac{b}{a}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2},(a,b)\not\in
S_{\gamma,7}\cup S_{\gamma,8},\\{a,b,a+b\\}\cap\\{1,\gamma\\}=\emptyset.$

If $\frac{b}{a}\in\mathbb{F}_{4}\setminus\mathbb{F}_{2}$ then (27) is
equivalent to $a^{3}=b^{3}=\gamma+1$, and (28) is equivalent to
$a^{3}=b^{3}=\gamma^{2}(\gamma+1)$. So, there are
$6\cdot\\#(C_{n}\setminus\\{\gamma^{3},\gamma+1,\gamma^{3}+\gamma^{2}\\})=6[(2^{n}-1)/3-1]-\\#(C_{n}\cap\\{\gamma^{3},\gamma+1,\gamma^{3}+\gamma^{2}\\}=2^{n+1}-8-\\#(C_{n}\cap\\{\gamma^{3},\gamma+1,\gamma^{3}+\gamma^{2}\\}$
pairs $(a,b)$ with $a^{3}=b^{3}\neq\gamma+1,\gamma^{2}(\gamma+1)$ when $n$ is
even.

Therefore, we have

$\displaystyle\omega_{4}=$ $\displaystyle\\#((S_{\gamma,7}\cup
S_{\gamma,8})\setminus
S)-\\#(C_{n}\cap\\{\gamma^{3},\gamma+1,\gamma^{3}+\gamma^{2}\\})$
$\displaystyle+$ $\displaystyle\begin{cases}2^{n+1}+22,&\text{ if }2\mid
n,\operatorname{Tr}(\gamma)=\operatorname{Tr}(1/\gamma)=0,\gamma\not\in\mathbb{F}_{8}\\\
2^{n+1}+10,&\text{ if }2\mid
n,\operatorname{Tr}(\gamma+1/\gamma)=1,\gamma\not\in\mathbb{F}_{8}\\\
2^{n+1}-2,&\text{ if }2\mid
n,[\operatorname{Tr}(\gamma)=\operatorname{Tr}(1/\gamma)=1,\gamma\not\in\mathbb{F}_{8}]\text{
or }\gamma\in\mathbb{F}_{8}\setminus\mathbb{F}_{2},\end{cases}$

when $n$ is even, and

$\omega_{4}=\\#((S_{\gamma,7}\cup S_{\gamma,8})\setminus
S)+\begin{cases}12,&\text{ if }\gamma\not\in\mathbb{F}_{8},\\\ 0,&\text{ if
}\gamma\in\mathbb{F}_{8},\\\ \end{cases}$

when $n$ is odd, where

$S=\\{(a,b)\in\mathbb{F}_{2^{n}}^{*}\times\mathbb{F}_{2^{n}}^{*}:\\{1,\gamma\\}\cap\\{a,b,a+b\\}\neq\emptyset\\}.$

In particular, when $n$ is odd, we have

$\begin{cases}\omega_{8}=6,\ \ \omega_{4}=\\#((S_{\gamma,7}\cup
S_{\gamma,8})\setminus S),&\text{ if }3\mid
n,\gamma\in\mathbb{F}_{8}\setminus\mathbb{F}_{2},\\\ \omega_{8}=0,\ \
\omega_{4}=\\#((S_{\gamma,7}\cup S_{\gamma,8})\setminus S)+12,&\text{
otherwise.}\end{cases}$ (32)

We desire to further investigate $\omega_{4}$, when $n$ is odd. We consider to
compute the number of pairs $(a,b)$ such that
$\frac{b}{a}\not\in\mathbb{F}_{4},(a,b)\in S_{\gamma,7}\cup
S_{\gamma,8},\\{a,b,a+b\\}\cap\\{1,\gamma\\}=\emptyset.$ We rewrite (27) as

$b^{2}(a+\gamma+1)+ab(a+\gamma+1)+(a+1)^{2}(\gamma+1)=0.$ (33)

By Lemma 2.4, there are two $b\in\mathbb{F}_{2^{n}}$ satisfying (33) if and
only if

$\displaystyle 0$
$\displaystyle=\operatorname{Tr}\left(\displaystyle\frac{(a+1)^{2}(\gamma+1)}{a^{2}(a+\gamma+1)}\right)=\operatorname{Tr}\left(\displaystyle\frac{(a+1)^{2}}{a^{2}}\right)+\operatorname{Tr}\left(\displaystyle\frac{a(a+1)^{2}}{a^{2}(a+\gamma+1)}\right)$
$\displaystyle=\operatorname{Tr}\left(\displaystyle\frac{a+1}{a}\right)+\operatorname{Tr}\left(\displaystyle\frac{a^{2}+1}{a(a+\gamma+1)}\right)=\operatorname{Tr}\left(\displaystyle\frac{(a+1)(a+\gamma+1)+a^{2}+1}{a(a+\gamma+1)}\right)$
$\displaystyle=\operatorname{Tr}\left(\displaystyle\frac{\gamma(a+1)}{a(a+\gamma+1)}\right)=\operatorname{Tr}\left(\displaystyle\frac{\gamma}{a+\gamma+1}\right)+\operatorname{Tr}\left(\displaystyle\frac{\gamma}{a(a+\gamma+1)}\right).$

Hence, there are two $b\in\mathbb{F}_{2^{n}}$ satisfying (27) if and only if

$\operatorname{Tr}\left(\displaystyle\frac{\gamma}{a+\gamma+1}\right)=\operatorname{Tr}\left(\displaystyle\frac{\gamma}{a(a+\gamma+1)}\right).$
(34)

Similarly, there are two $b\in\mathbb{F}_{2^{n}}$ satisfying (28) if and only
if

$\operatorname{Tr}\left(\displaystyle\frac{1}{a+\gamma+1}\right)=\operatorname{Tr}\left(\displaystyle\frac{\gamma}{a(a+\gamma+1)}\right).$
(35)

If we set

$S_{\gamma,9}=\left\\{a\in\mathbb{F}_{2^{n}}^{*}:\eqref{Sr7_trace}\text{
holds},a\neq\gamma+1\right\\}\ \text{ and }\
S_{\gamma,10}=\left\\{a\in\mathbb{F}_{2^{n}}^{*}:\eqref{Sr8_trace}\text{
holds},a\neq\gamma+1\right\\},$

then for every $a\in\mathbb{F}_{2^{n}}\setminus\\{0,1,\gamma,\gamma+1\\}$,

$\\#\\{b\in\mathbb{F}_{2^{n}}:(a,b)\in S_{\gamma,7}\cup
S_{\gamma,8}\\}=\begin{cases}4,&\text{ if }a\in S_{\gamma,9}\cap
S_{\gamma,10},\\\ 2,&\text{ if }a\in(S_{\gamma,9}\setminus
S_{\gamma,10})\cup(S_{\gamma,10}\setminus S_{\gamma,9}),\\\ 0,&\text{ if
}a\not\in S_{\gamma,9}\cup S_{\gamma,10}.\end{cases}$

Since the right hand sides of (34) and (35) are identical,
$a\in(S_{\gamma,9}\setminus S_{\gamma,10})\cup(S_{\gamma,10}\setminus
S_{\gamma,9})$ is equivalent to
$\operatorname{Tr}\left(\frac{\gamma+1}{a+\gamma+1}\right)=1$, and then we
have $2$ pairs $(a,b)$ in $S_{\gamma,7}\cup S_{\gamma,8}$ in this case. Since
the field trace map is balanced, the number of
$a\in\mathbb{F}_{2^{n}}^{*}\setminus\\{\gamma+1\\}$ satisfying
$\operatorname{Tr}\left(\frac{\gamma+1}{a+\gamma+1}\right)=1$ is $2^{n-1}-1$,
when $n$ is odd.

######  Remark 4.3.

If $\operatorname{Tr}\left(\frac{\gamma+1}{a+\gamma+1}\right)=0$ or
equivalently $\operatorname{Tr}\left(\frac{a}{a+\gamma+1}\right)=1$, then
$a\in S_{\gamma,9}\cap S_{\gamma,10}$ or $a\not\in S_{\gamma,9}\cup
S_{\gamma,10}$ holds. By SageMath experiment, $\\#(S_{\gamma,9}\cap
S_{\gamma,10})=2^{n-2}-1$ for any
$\gamma\in\mathbb{F}_{2^{n}}\setminus\\{0,1\\}$, when $1<n\leq 17$ is odd. We
conjecture that this holds for every odd $n>1$.

From now on, we assume that the conjecture in Remark 4.3 is true. Then we have

$\displaystyle\\#((S_{\gamma,7}\cup S_{\gamma,8})\setminus S)$
$\displaystyle=4\cdot\\#(S_{\gamma,9}\cap
S_{\gamma,10})+2\cdot\\#\left((S_{\gamma,9}\setminus
S_{\gamma,10})\cup(S_{\gamma,10}\setminus
S_{\gamma,9})\right)-\\#(S\cap(S_{\gamma,7}\cup S_{\gamma,8}))$
$\displaystyle=4(2^{n-2}-1)+2(2^{n-1}-1)-\\#(S\cap(S_{\gamma,7}\cup
S_{\gamma,8}))$ $\displaystyle=2^{n+1}-6-\\#(S\cap(S_{\gamma,7}\cup
S_{\gamma,8})).$ (36)

It remains to compute $\\#(S\cap(S_{\gamma,7}\cup S_{\gamma,8}))$, and hence
we consider to compute the number of pairs
$(a,b)\in\mathbb{F}_{2^{n}}^{*}\times\mathbb{F}_{2^{n}}^{*}$ with
$\\{1,\gamma\\}\cap\\{a,b,a+b\\}\neq\emptyset$ such that (34) or (35). If
$a=1$, then (34) holds. But, $a=1$ in (27) implies that $b=0$ or $b=1$, a
contradiction. And, $a=\gamma$ in (34) is equivalent to
$\operatorname{Tr}(\gamma)=\operatorname{Tr}(1)=1$. Hence we have 4 pairs of
$(1,b)\in(S_{\gamma,7}\cup S_{\gamma,8}))$ when $\operatorname{Tr}(\gamma)=1$,
and 2 pairs otherwise. If $a=\gamma$, then (35) holds. But, $a=\gamma$ in (28)
implies that $b=0$ or $b=\gamma$, a contradiction. And, $a=1$ in (35) is
equivalent to
$\operatorname{Tr}\left(\frac{1}{\gamma}\right)=\operatorname{Tr}(1)=1$. Hence
we have 4 pairs of $(\gamma,b)\in(S_{\gamma,7}\cup S_{\gamma,8}))$ when
$\operatorname{Tr}\left(\frac{1}{\gamma}\right)=1$, and 2 pairs otherwise.

Applying $b=1$ in (27) implies that $a=0$ or $a=1$, which are already
considered in above. Applying $b=\gamma$ in (27) implies that $a^{2}+\gamma
a+(\gamma+1)^{3}=0$. By Lemma 2.4, $a^{2}+\gamma a+(\gamma+1)^{3}=0$ has two
solutions if and only if
$\operatorname{Tr}\left(\frac{(\gamma+1)^{3}}{\gamma^{2}}\right)=\operatorname{Tr}(\gamma+1)=0$
or equivalently $\operatorname{Tr}(\gamma)=1$. Since (27) and (28) have no
common solution, (34) holds and (35) does not hold for each solution $a$ of
$a^{2}+\gamma a+(\gamma+1)^{3}=0$. Since if $b=\gamma$ is a solution of (27)
then another solution of (27) is $b=a+\gamma$, so we have 2 pairs of
$(a,\gamma)\in S_{\gamma,7}\cup S_{\gamma,8}$ and 2 pairs of $(a,a+\gamma)\in
S_{\gamma,7}\cup S_{\gamma,8}$, when $\operatorname{Tr}(\gamma)=1$. Similarly,
$b=\gamma$ in (28) implies already considered case in above, and $b=1$ in (28)
implies $\gamma a^{2}+\gamma a+(\gamma+1)^{3}=0$, which has two solutions if
and only if
$\operatorname{Tr}\left(\frac{(\gamma+1)^{3}}{\gamma}\right)=\operatorname{Tr}\left(1+\frac{1}{\gamma}\right)=0$
or equivalently $\operatorname{Tr}\left(\frac{1}{\gamma}\right)=1$. (35) holds
and (34) does not hold for each solution $a$ of $\gamma a^{2}+\gamma
a+(\gamma+1)^{3}=0$. Hence, we have 2 pairs of $(a,1)\in S_{\gamma,7}\cup
S_{\gamma,8}$ and 2 pairs of $(a,a+1)\in S_{\gamma,7}\cup S_{\gamma,8}$, when
$\operatorname{Tr}\left(\frac{1}{\gamma}\right)=1$. To summarize, we have

$\\#(S\cap(S_{\gamma,7}\cup S_{\gamma,8}))=\begin{cases}4\cdot 0+2\cdot
2=4,&\text{ if
}\operatorname{Tr}(\gamma)=\operatorname{Tr}\left(\frac{1}{\gamma}\right)=0,\\\
4\cdot 4+2\cdot 0=16,&\text{ if
}\operatorname{Tr}(\gamma)=\operatorname{Tr}\left(\frac{1}{\gamma}\right)=1,\\\
4\cdot 2+2\cdot 1=10,&\text{ if
}\operatorname{Tr}\left(\gamma+\frac{1}{\gamma}\right)=1.\end{cases}$

So, if $3\mid n$ and $\gamma\in\mathbb{F}_{8}\setminus\mathbb{F}_{2}$, then we
have $\nabla_{f}=8$ and

$\omega_{8}=6,\ \ \omega_{4}=2^{n+1}-16,\ \
\omega_{0}=(2^{n}-2)(2^{n}-1)-\omega_{4}-\omega_{8}=2^{2n}-5\cdot 2^{n}+12.$

applying (29) on (36). Otherwise, we have $\nabla_{f}=4$, and applying (32)
and (36) we have

$\omega_{4}=2^{n+1}+6-\\#(S\cap(S_{\gamma,7}\cup
S_{\gamma,8}))=\begin{cases}2^{n+1}+2,&\text{ if
}\operatorname{Tr}(\gamma)=\operatorname{Tr}\left(\frac{1}{\gamma}\right)=0,\\\
2^{n+1}-10,&\text{ if
}\operatorname{Tr}(\gamma)=\operatorname{Tr}\left(\frac{1}{\gamma}\right)=1,\\\
2^{n+1}-4,&\text{ if
}\operatorname{Tr}\left(\gamma+\frac{1}{\gamma}\right)=1,\end{cases}$
$\omega_{0}=(2^{n}-1)(2^{n}-2)-\omega_{4}=\begin{cases}2^{2n}-5\cdot
2^{n},&\text{ if
}\operatorname{Tr}(\gamma)=\operatorname{Tr}\left(\frac{1}{\gamma}\right)=0,\\\
2^{2n}-5\cdot 2^{n}+12,&\text{ if
}\operatorname{Tr}(\gamma)=\operatorname{Tr}\left(\frac{1}{\gamma}\right)=1,\\\
2^{2n}-5\cdot 2^{n}+6,&\text{ if
}\operatorname{Tr}\left(\gamma+\frac{1}{\gamma}\right)=1.\end{cases}$

We confirm that the above holds for all
$\gamma\in\mathbb{F}_{2^{n}}\setminus\mathbb{F}_{2}$ via SageMath experiments,
when $1<n\leq 11$ is odd.

###  4.2 Some experimental results on the second-order zero differential
uniformity in odd characteristic

In this subsection, we consider the second-order zero differential uniformity
of $f=Inv\circ(1,\gamma)$ when $p$ is odd, by using similar idea with Section
3.2. However, unlike in Section 3.2 where $\gamma=0$ and $p=2$, $\gamma$ is
not fixed and $p$ is an arbitrary odd prime in this section, so we have to
deal with a much wider range of cases, which can be a very tedious process.
However, since the inverse function has low second-order zero differential
uniformity and $Inv\circ(1,\gamma)$ differs from the inverse function by only
two points, we can conjecture that $Inv\circ(1,\gamma)$ also has the low
second-order zero differential uniformity. We introduce several experimental
results using SageMath. When $p^{n}<1000$ with $p>3$ we have the followings

  * •

$\nabla_{f}=7$ when $p^{n}=11^{2}$ and $\gamma=-1$,

  * •

$\nabla_{f}=6$ when $p=5$, $\gamma^{4}=-1$ or $p=97$

  * •

otherwise, $2\leq\nabla_{f}\leq 5$.

On the other hand, when $p=3$, we have $\nabla_{f}\in\\{3,6,9\\}$. These
results can be partially explained by investigating $\nabla_{f}(a,a)$ for
$a\in\mathbb{F}_{3^{n}}^{*}$. Let $b=a$. Then (1) is equivalent to

$f(x)+f(x+a)+f(x-a)=0$ (37)

When $a=1$,

  * •

If $x\in\mathbb{F}_{3}$, then (37) is equivalent to $\gamma=1$, a
contradiction.

  * •

If $x\in\\{\gamma,\gamma\pm 1\\}$, then (37) is equivalent to
$\gamma^{2}-\gamma=1$, such $\gamma$ exists only if $n$ is even.

  * •

If $x\not\in\\{0,\pm 1,\gamma,\gamma\pm 1\\}$, then (37) is equivalent to
$-a^{2}=0$, a contradiction.

Therefore, $\nabla_{f}(1,1)=3$ when $\gamma^{2}-\gamma=1$, and
$\nabla_{f}(1,1)=0$ otherwise.

When $a=\gamma$,

  * •

If $x\in\\{1,1\pm\gamma\\}$, then (37) is equivalent to $\gamma^{2}+\gamma=1$,
such $\gamma$ exists only if $n$ is even.

  * •

If $x\in\\{0,\pm\gamma\\}$, then (37) is equivalent to $\gamma=1$, a
contradiction.

  * •

If $x\not\in\\{0,1,\pm\gamma,1\pm\gamma\\}$, then (37) is equivalent to
$-\gamma^{2}=0$, a contradiction.

Therefore, $\nabla_{f}(\gamma,\gamma)=3$ when $\gamma^{2}+\gamma=1$, and
$\nabla_{f}(\gamma,\gamma)=0$ otherwise.

If $a\neq 1,\gamma$, then we have

  * •

$x=0,\pm a$ are solutions of (37).

  * •

$x=1,\pm a+1$ are solutions of (37) if $\gamma=1-a^{2}$.

  * •

$x=\gamma,\pm a+\gamma$ are solutions of (37) if $\gamma(\gamma-1)=a^{2}$.

  * •

$x\not\in\\{0,1,\gamma,\pm a,1\pm a,\gamma\pm a\\}$, then (37) is equivalent
to $-1=0$, a contradiction.

We add $\gamma=1-a^{2}$ and $\gamma(\gamma-1)=a^{2}$, then we obtain
$\gamma^{2}=1$. Since $\gamma\neq 1$, we have $\gamma=-1$, and
$a^{2}=b^{2}=-1$. For $a\in\mathbb{F}_{3^{n}}$ with $a^{2}=-1$ to exist, $n$
need to be even. Therefore, we have $\nabla_{f}(a,a)=9$, if $2\mid
n,\gamma=-1,a^{2}=-1$.

Furthermore, we have $\nabla_{f}(a,a)=6$, if $a^{2}=1-\gamma$ or
$a^{2}=\gamma(\gamma-1)$, and
$1-\gamma,\gamma(\gamma-1)\not\in\\{1,\gamma^{2}\\}$. By the way,
$1-\gamma,\gamma(\gamma-1)\in\\{1,\gamma^{2}\\}$ implies that
$\gamma^{2}\pm\gamma=1$, since $\gamma\neq 0$. If
$\gamma^{2}+\gamma=1\Leftrightarrow 1-\gamma=\gamma^{2}=a^{2}$ or
$\gamma^{2}-\gamma=1=a^{2}$, then we already have $\nabla_{f}(a,a)=3$ in these
cases. Therefore, we have $\nabla_{f}(a,a)=6$ if
$a\in\\{1-\gamma,\gamma(\gamma-1)\\}\cap Q_{n}\neq\emptyset$ and
$\gamma^{2}\pm\gamma\neq 1$, where

$Q_{n}=\\{a^{2}\in\mathbb{F}_{3^{n}}:a\in\mathbb{F}_{3^{n}}\\}\setminus\mathbb{F}_{3}$

is the set of all squares in $\mathbb{F}_{3^{n}}\setminus\mathbb{F}_{3}$. To
summarize, we have

$\nabla_{f}(a,a)=\begin{cases}9,&\text{ if }2\mid n,\gamma=-1,a^{2}=-1\\\
6,&\text{ if }a^{2}\in\\{1-\gamma,\gamma(\gamma-1)\\}\cap Q_{n}\text{ and
}\gamma^{2}\pm\gamma\neq 1,\\\ 3,&\text{ otherwise.}\end{cases}$

Furthermore, note that we have the following :

$\nabla_{f}=\begin{cases}9,&\text{ if }2\mid n,\gamma=-1,\\\ 6,&\text{ if
}\\{1-\gamma,\gamma(\gamma-1)\\}\cap Q_{n}\neq\emptyset\text{ and
}\gamma^{2}\pm\gamma\neq 1,\\\ 3,&\text{ otherwise,}\end{cases}$

from SageMath experiments for $2\leq n\leq 8$. We conjecture that the above
holds for every $n\geq 2$.

##  5 Conclusion

In this paper, we studied the second-order zero differential uniformity of the
swapped inverse function $Inv\circ(1,\gamma)$ where
$\gamma\in\mathbb{F}_{p^{n}}\setminus\\{1\\}$. Specifically,

  * •

We studied the FBCT of $Inv\circ(0,1)$ and completely computed the second-
order zero differential spectrum of $Inv\circ(0,1)$ defined on
$\mathbb{F}_{2^{n}}$.

  * •

We showed that the second-order zero differential uniformity of
$Inv\circ(0,1)$ defined on $\mathbb{F}_{p^{n}}$ is at most $4$, when $p$ is an
odd prime.

  * •

We studied the FBCT of $Inv\circ(1,\gamma)$ defined on $\mathbb{F}_{2^{n}}$,
where $\gamma\in\mathbb{F}_{2^{n}}\setminus\\{0,1\\}$. Moreover, we further
studied the second-order zero differential spectrum of $Inv\circ(1,\gamma)$,
especially when $n$ is odd.

  * •

Due to the extensive range of cases, rather than conducting a detailed
investigation into the second-order zero differential uniformity of
$Inv\circ(1,\gamma)$ when $p$ is odd, we presented several experimental
results for some cases of $\gamma$ and $p$. Furthermore, from experimental
results, we suggested a conjecture on the second-order zero differential
uniformity of $Inv\circ(1,\gamma)$ when $p=3$.

In summary, the swapped inverse functions, known to have good cryptographic
properties, also have low second-order zero differential uniformity. In
particular, our paper is the first to characterize classes of non-power
functions with the second-order zero differential uniformity equal to $4$ over
binary fields.

Finding the second-order zero differential uniformity of functions with good
cryptographic properties has not been widely studied so far. Also, when
restricted to non-power functions, the results are very limited. Therefore, we
believe that this topic is worth further study, and we plan to explore the
second-order zero differential uniformity of other non-power functions.

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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