# Algebraic area enumeration for lattice paths

Stéphane Ouvry∗ Alexios P. Polychronakos†

###### Abstract

We give a summary of recent progress on the algebraic area enumeration of
closed paths on planar lattices. Several connections are made with quantum
mechanics and statistical mechanics. Explicit combinatorial formulae are
proposed which rely on sums labelled by the multicompositions of the length of
the paths.

* LPTMS, CNRS, Université Paris-Saclay, 91405 Orsay Cedex, France
<EMAIL_ADDRESS>
$\dagger$ Physics Department, the City College of New York, NY 10031, USA and
The Graduate Center of CUNY, New York, NY 10016, USA
<EMAIL_ADDRESS>
The seminal problem of the algebraic area enumeration of paths on planar
lattices of various kinds has been around for a long time. It is well known
that this purely combinatorial problem can be equivalently reformulated in the
realm of Hofstadter-like quantum mechanics models. Recently [1] it has been
given a boost in the form of an explicit enumeration formula which in turn
could be reinterpreted [2] in terms of statistical mechanics models with
exclusion statistics, again a purely quantum concept. It is a striking fact
that an enumeration quest regarding classical random paths should be in the
end so intimately connected to quantum physics, this in so many ways.

In this note we give a summary of this recent progress starting with the
original algebraic area enumeration problem for closed paths on a square
lattice and then enlarging the perspective to other kind of lattices and paths
via the statistical mechanics reinterpretation. So the first question we
address is : among the ${{\bf n}\choose{\bf n}/2}^{2}$ closed $\bf n$-steps
paths that one can draw on a square lattice starting from and returning to a
given point –note that ${\bf n}$ is then necessarily even ${\bf n}=2n$–, how
many of them enclose a given algebraic area $A$?

The algebraic area enclosed by a path is weighted by its winding numbers: if
the path moves around a region in counterclockwise (positive) direction, its
area counts as positive, otherwise negative; if the path winds around more
than once, the area is counted with multiplicity. These regions inside the
path are called winding sectors. In Figure 1 we give an explicit example of
what is meant by algebraic area for a closed path of length ${\bf n}=36$. We
see inside the path various winding sectors with winding numbers +2,+1,0,-1,-1
and various numbers of lattice cells per winding sectors: respectively
2,14,1,1,1. The $0$-winding number inside the path arises from a superposition
of a $+1$ and a $-1$ winding, $+1-1=0$. It does not contribute to the
algebraic area. Taking into account the non $0$-winding sectors we end up with
an algebraic area $A={2}\times{2}+{1}\times{14}+{(-1)}\times{2}=16$.

Figure 1: A closed path of length ${\bf n}=36$ starting from and returning to
the same bullet red point with its various winding sectors $m=+2,+1,0,-1,-1$.
Note the double arrow on the horizontal link which indicates that the path has
moved twice on this link, here in the same left direction.

Quite generally, calling $S_{m}$ the arithmetic area of the $m$-winding
sectors inside a path (i.e. the total number of lattice cells it encloses with
winding number $m$, where $m$ can be positive or negative) the algebraic area
is

$A=\sum_{m=-\infty}^{\infty}mS_{m}$

to be distinguished from the arithmetic area $\sum_{m=-\infty}^{\infty}S_{m}$.

Winding sectors for continuous Brownian curves as well as for discrete lattice
paths have been the subject of studies for a long time. In this respect we
note in the last few years some advances in [3] where an explicit formula for
the expected area $\langle S_{m}\rangle$ of the $m$-winding sectors inside
square lattice paths is proposed, to the exception of the $0$-winding sector,
for the simple reason that the latter is difficult to distinguish from the
outside –i.e., $0$-winding again– sector, which is of infinite size. Taking
the continuous limit allows to recover the results previously obtained in [4]
for Brownian curves. One notes that for Brownian curves the expected area
$\langle S_{0}\rangle$ of the $0$-winding sectors is also known by other means
thanks to the SLE machinery [5]. However it remains an open problem for
discrete lattice paths.

Counting on the square lattice the number of closed paths of length $\bf n$
enclosing an algebraic area $A$ amounts in a most straightforward way to
introducing two lattice hopping operators $u$ and $v$ respectively in the
right and up directions and declaring that they do not commute

$v\;u=\mathrm{Q}\;u\;v$

Clearly selecting the $u$ and $v$ independent part in

$\big{(}u+u^{-1}+v+v^{-1}\big{)}^{\bf n}={\sum_{A}C_{\bf
n}(A)\;\mathrm{Q}^{A}}+\ldots$ (1)

provides the number $C_{\bf n}(A)$ which counts the paths enclosing an
algebraic area $A$. For example it is not difficult to check that
$\big{(}u+u^{-1}+v+v^{-1}\big{)}^{4}=28+4\mathrm{Q}+4\mathrm{Q}^{-1}+\ldots$
indicating that among the ${4\choose 2}^{2}=36$ closed paths making $4$ steps
$C_{4}(0)=28$ enclose an algebraic area $A=0$ and $C_{4}(1)=C_{4}(-1)=4$
enclose an algebraic area $A=\pm 1$.

Now it is immediate to see that, provided $\mathrm{Q}$ is rewritten as
$\mathrm{Q}=e^{i2\pi\Phi/\Phi_{o}}$ where $\Phi$ is the flux of an external
magnetic field through the unit lattice cell and $\Phi_{o}$ the flux quantum,
the expression

$H=u+u^{-1}+v+v^{-1}$

can be interpreted as a Hamiltonian modelling a quantum particle hopping on a
square lattice and coupled to a perpendicular magnetic field. This famous
model is known under the name Hofstadter model [6].

Going a step further, a simplification arises when the flux is rational
$\mathrm{Q}=e^{i2\pi p/q}$ with $p,q$ two coprime integers: in this case
determining the Hofstadter spectrum narrows down to computing the eigenvalues
$E_{1},E_{2},\ldots,E_{q}$ of the finite $q\times q$ Hamiltonian matrix

$H_{q}=\begin{pmatrix}\mathrm{Q}e^{ik_{y}}+\mathrm{Q}^{-1}e^{-ik_{y}}&e^{ik_{x}}&0&\cdots&0&e^{-ik_{x}}\\\
e^{-ik_{x}}&\mathrm{Q}^{2}e^{ik_{y}}+\mathrm{Q}^{-2}{e^{-ik_{y}}}&\quad
e^{ik_{x}}\quad&\cdots&0&0\\\ 0&e^{-ik_{x}}&()&\cdots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\ 0&0&0&\cdots&()&e^{ik_{x}}\\\
e^{ik_{x}}&0&0&\cdots&\quad
e^{-ik_{x}}\quad&\mathrm{Q}^{q}e^{ik_{y}}+\mathrm{Q}^{-q}{e^{-ik_{y}}}\\\
\end{pmatrix}$

The $n$-th quantum trace follows as

$\textbf{Tr}\;H_{q}^{\bf
n}=\frac{1}{q}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{dk_{x}}{2\pi}\frac{dk_{y}}{2\pi}\sum_{i=1}^{q}E_{i}^{\bf
n}$ (2)

where one has summed over the $q$ eigenvalues and integrated over the two
continuous quantum numbers $k_{x}$ and $k_{y}$ while enforcing the proper
normalization $\textbf{Tr}\;I_{q}=1$, where $I_{q}$ is the $q\times q$
identity matrix.

Selecting as in (1) the $u,v$ independent part of
$\big{(}u+u^{-1}+v+v^{-1}\big{)}^{\bf n}$ translates in the quantum world to
computing the trace of $H_{q}^{{\bf n}}$ so that in view of the algebraic area
enumeration one should write

$\textbf{ Tr}\;H_{q}^{{\bf n}}=\sum_{A}C_{\bf n}(A)\;\mathrm{Q}^{A}$ (3)

As a consequence the integer $q$ has to be understood as a priori
undetermined, i.e., as a free parameter allowing for the $\mathrm{Q}^{A}$
basis –where $A=0,\pm 1,\pm 2,\ldots$ stands for the possible values of the
algebraic area– on which the enumeration (3) can take place. On the other hand
keeping a particular value of $q$ would amount to counting the algebraic area
modulo $q$.

All the machinery of quantum mechanics is now at our disposal. It is known
that the determinant of the secular matrix $I_{q}-zH_{q}$ simplifies to

$\det(I_{q}-zH_{q})=\sum_{n=0}^{\lfloor
q/2\rfloor}(-1)^{n}Z(n)z^{2n}-2\big{(}\cos(qk_{x})+\cos(qk_{y})\big{)}z^{q}$

where the $Z(n)$’s are independent of $k_{x}$ and $k_{y}$ and by convention
$Z(0)=1$. Even more, Kreft [7] was able to rewrite them in a closed form as
trigonometric multiple nested sums

${Z(n)=\sum_{k_{1}=1}^{q-2n+2}\sum_{k_{2}=1}^{k_{1}}\cdots\sum_{k_{n}=1}^{k_{n-1}}s_{k_{1}+2n-2}s_{k_{2}+2n-4}\cdots
s_{k_{n-1}+2}s_{k_{n}}}$ (4)

where

$s_{k}=4\sin^{2}(\pi kp/q)$ (5)

From the knowledge of $Z(n)$ in (4) the algebraic area enumeration can
proceed. We give here a summary of the procedure, more details can be found in
[1, 2]. First introduce the $b(n)$’s via

$\log\left(\sum_{n=0}^{\lfloor
q/2\rfloor}Z(n)z^{n}\right)=\sum_{n=1}^{\infty}b(n)z^{n}$ (6)

It is not difficult to see that $b(n)$ selects the part of $Z(n)$ which is
obtained by rewriting it as a linear combination of trigonometric single sums
plus other terms which are products of such single sums, which are then
ignored. In other words the $b(n)$’s end up being proportional to $q$ –i.e.,
the right scaling for the trace of a $q\times q$ matrix– whereas the $Z(n)$’s
also contain terms proportional to $q^{2},\ldots,q^{n}$. This rewriting is
encoded in the coefficients $c(l_{1},l_{2},\ldots,l_{j})$ labeled by the
compositions $l_{1},l_{2},\ldots,l_{j}$ of $n$ (meaning the ordered partitions
of $n$: there are $2^{n-1}$ compositions of $n$, for example
$3=3,2+1,1+2,1+1+1$) so that $b(n)$ is expressed as

$b(n)=(-1)^{n+1}\hskip-8.5359pt{\sum_{\begin{subarray}{c}l_{1},l_{2},\ldots,l_{j}\\\
{\rm composition}\;{\rm
of}\;n\end{subarray}}\hskip-11.38092pt{c(l_{1},l_{2},\ldots,l_{j})}}{\sum_{k=1}^{q-j+1}s^{l_{j}}_{k+j-1}\cdots
s^{l_{2}}_{k+1}{s}^{l_{1}}_{k}}$ (7)

with 111The expression for the $c(l_{1},l_{2},\ldots,l_{j})$’s is closely
related to Dyck paths counting (up to a factor $l_{1}$), see [8] p. 516. We
plan to further elaborate on this connection in a forthcoming publication.

$c(l_{1},l_{2},\ldots,l_{j})=\frac{{l_{1}+l_{2}\choose
l_{1}}}{l_{1}+l_{2}}\;\;l_{2}\frac{{l_{2}+l_{3}\choose
l_{2}}}{l_{2}+l_{3}}\;\cdots\;\;l_{{j}-1}\frac{{l_{{j}-1}+l_{j}\choose
l_{{j}-1}}}{l_{{j}-1}+l_{j}}$

As announced, solely trigonometric single sums appear in (7).

Now one can use the identity $\log\;\det(I-zM)={{\text{Tr}}}\;\log(I-zM)$
valid for any matrix $M$ where Tr here means the usual matrix trace. After
some manipulations one reaches, not surprisingly, that the quantum trace (2)
is proportional to $b(n)$

$\textbf{ Tr}\;H_{q}^{{\bf{n}}=2n}=2n(-1)^{n+1}{1\over q}b(n)$

It follows that, using (7), it can be rewritten as composition-dependent
trigonometric single sums ${\sum_{k=1}^{q-j+1}s^{l_{j}}_{k+j-1}\cdots
s^{l_{2}}_{k+1}{s}^{l_{1}}_{k}}$ weighted by the combinatorial coefficient
$c(l_{1},l_{2},\ldots,l_{j})$ and summed over all compositions
$l_{1},l_{2},\ldots,l_{j}$ of the integer $n={\bf n}/2$

$\textbf{ Tr}\;H_{q}^{{\bf
n}=2n}=2n{\sum_{\begin{subarray}{c}l_{1},l_{2},\ldots,l_{j}\\\ {\rm
composition}\;{\rm
of}\;n\end{subarray}}\hskip-11.38092pt{c(l_{1},l_{2},\ldots,l_{j})}}{1\over
q}{\sum_{k=1}^{q-j+1}s^{l_{j}}_{k+j-1}\cdots s^{l_{2}}_{k+1}{s}^{l_{1}}_{k}}$

The trigonometric single sums remain to be computed, which can also be done,
still keeping in mind that, as said earlier, $q$ is an a priori undetermined
free parameter. Finally one extracts from (3) the desired number of closed
paths of length $\bf n$ enclosing a given algebraic area $A$ as

$\displaystyle C_{\bf{n}}(A)=$ $\displaystyle
2n{\sum_{l_{1},l_{2},\ldots,l_{j}\;\atop\text{ composition}\;\text{
of}\;n}{\frac{{l_{1}+l_{2}\choose
l_{1}}}{l_{1}+l_{2}}\;\;l_{2}\frac{{l_{2}+l_{3}\choose
l_{2}}}{l_{2}+l_{3}}\;\ldots\;\;l_{{j}-1}\frac{{l_{{j}-1}+l_{j}\choose
l_{{j}-1}}}{l_{{j}-1}+l_{j}}}}$
$\displaystyle{\sum_{k_{3}=0}^{2l_{3}}\sum_{k_{4}=0}^{2l_{4}}\ldots\sum_{k_{j}=0}^{2l_{j}}\prod_{i=3}^{j}{2l_{i}\choose
k_{i}}}{{2l_{1}\choose l_{1}+A+\sum_{i=3}^{j}(i-2)(k_{i}-l_{i})}{2l_{2}\choose
l_{2}-A-\sum_{i=3}^{j}(i-1)(k_{i}-l_{i})}}$ (8)

This formula grows quickly in complexity since the number of compositions on
which one has to sum increases like $2^{n}$ with the number of steps of the
paths. Still it has the benefit of being explicit.We leave as an open problem
to the interested reader to prove that in the continuous limit where the
elementary lattice size $a\to 0$, the number of steps ${\bf n}\to\infty$ with
the scaling ${\bf n}a^{2}=2t$,

${{\bf n}\;C_{\bf{n}}(A/a^{2})\over{{\bf n}\choose{\bf
n}/2}^{2}}\to{\pi}{1\over\cosh^{2}(\pi A/t)}$

i.e., one recovers Levy’s law for the distribution of the algebraic area
enclosed by Brownian curves after a time $t$ (the convergence has been checked
numerically to improve with increasing ${\bf n}$).

Why in (4) and (6) the particular choice of the notations $Z_{n}$ and $b_{n}$?
In statistical mechanics $Z(n)$ usually refers to an $n$-body partition
function and $b(n)$ to its associated $n$-th cluster coefficient. Let us
interpret $s_{k}$ in (5) as $s_{k}=e^{-\beta\epsilon_{k}}$ ($\beta$ is the
inverse temperature), i.e., as a spectral function for a quantum 1-body
spectrum $\epsilon_{k}$ labeled by an integer $k$. The structure of $Z(n)$ in
(4) with the $+2$ shifts in the spectral function arguments then precisely
corresponds to an $n$-body partition function for a gas of particles with
exclusion statistics $g=2$ (no two particles can occupy two adjacent quantum
states) and 1-body spectrum $\epsilon_{k}$. Exclusion statistics is a purely
quantum (again) concept which describes the statistical mechanical properties
of identical particles. Usual particles are either Bosons ($g=0$) or Fermions
($g=1$). Here for square lattice paths one goes beyond Fermi exclusion with
$g=2$. In general for $g$-exclusion the $n$-body partition function (4) would
become

${Z(n)=\sum_{k_{1}=1}^{q-gn+g}\sum_{k_{2}=1}^{k_{1}}\cdots\sum_{k_{n}=1}^{k_{n-1}}s_{k_{1}+gn-g}s_{k_{2}+gn-2g}\cdots
s_{k_{n-1}+g}s_{k_{n}}}$

with a shift in the arguments of the spectral function which is $g$ instead of
$2$. In line with (6, 7) the associated $n$-th cluster coefficient would end
up rewriting as

$b(n)=(-1)^{n+1}\hskip-8.5359pt{\sum_{\begin{subarray}{c}l_{1},l_{2},\ldots,l_{j}\\\
{\rm g-composition}\;{\rm
of}\;n\end{subarray}}\hskip-11.38092pt{c_{g}(l_{1},l_{2},\ldots,l_{j})}}{\sum_{k=1}^{q-j+1}s^{l_{j}}_{k+j-1}\cdots
s^{l_{2}}_{k+1}{s}^{l_{1}}_{k}}$ (9)

where

${c_{g}(l_{1},l_{2},\ldots,l_{j})}={{{(l_{1}+\dots+l_{g-1}-1)!\over
l_{1}!\cdots l_{g-1}!}~{}\prod_{i=1}^{j-g+1}{l_{i}+\dots+l_{i+g-1}-1\choose
l_{i+g-1}}}}$

In (9) one sums over all $g$-compositions of the integer $n$ obtained by
inserting at will inside the usual compositions (i.e., the $2$-compositions)
no more than $g-2$ zeroes in succession. For example for $n=3$ and $g=3$ one
has $9$ such $3$-compositions
$n=3=2+1=1+2=1+1+1=2+0+1=1+0+2=1+0+1+1=1+1+0+1=1+0+1+0+1$. For general $g$
there are $g^{n-1}$ such $g$-compositions of the integer $n$ (see [9] for an
analysis of these extended compositions, also called multicompositions).

One has reached the conclusion that the square lattice paths algebraic area
enumeration is described by a quantum gas of particles with statistical
exclusion $g=2$. To see this more explicitly on the Hofstadter Hamiltonian
itself let us perform on the hopping lattice operators $u$ and $v$ the modular
transformation

$u\to-u\;v~{},~{}~{}~{}v\to v$

to get the new Hamiltonian

$H=-u\;v-v^{-1}\;u^{-1}+v+v^{-1}$ (10)

still describing the same paths but on the deformed lattice of Figure 2.

Figure 2: The deformed square lattice after the modular transformation.

The secular matrix corresponding to the Hamiltonian (10) is

$I_{q}-zH_{q}=\begin{pmatrix}{1}&-{(1-\mathrm{Q})}z&0&\cdots&0&-{(1-{{1\over\mathrm{Q}^{q}}})}z\\\
-{(1-{1\over{\mathrm{Q}}})}z&{1}&-{(1-\mathrm{Q}^{2})}z&\cdots&0&0\\\
0&-{(1-{1\over\mathrm{Q}^{2}})}z&{1}&\cdots&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&{1}&-{(1-{\mathrm{Q}^{q-1}})}z\\\
-{(1-\mathrm{Q}^{q})}z&0&0&\cdots&-{(1-{1\over\mathrm{Q}^{q-1}})}z&{1}\\\
\end{pmatrix}$ (11)

where one has set $k_{x}=k_{y}=0$ for simplicity. The Hofstadter spectral
function (5) is recovered as

$s_{k}=(1-\mathrm{Q}^{k})(1-{1\over\mathrm{Q}^{k}})$

(11) is a particular case of the more general class of secular matrices

$I_{q}-zH_{q}=\begin{pmatrix}{1}&-{f(1)}z&{0}&\cdots&0&-{g(q)}z\\\
-{g(1)}z&{1}&-{f(2)}z&\cdots&0&0\\\ 0&-{g(2)}z&{1}&\cdots&0&0\\\
\vdots&\vdots&\vdots&{\ddots}&\vdots&\vdots\\\ 0&0&0&\cdots&{1}&-{f(q-1)}z\\\
-{f(q)}z&0&0&\cdots&-{g(q-1)}z&{1}\\\ \end{pmatrix}$ (12)

and associated spectral functions

${s_{k}=g(k)f(k)}$

which become the building blocks of the $Z(n)$’s in (4) (up to spurious
umklapp terms which would disappear if either $f(q)$ or $g(q)$ vanish).

In a natural way (12) becomes in the $g=3$ case

$I_{q}-zH_{q}=\begin{pmatrix}{1}&-{f(1)}z&0&0&\cdots&0&-{g(q-1)}z&{0}\\\
{0}&{1}&-{f(2)}z&0&\cdots&0&0&-{g(q)}z\\\
-{g(1)}z&{0}&{1}&-{f(3)}z&\cdots&0&0&0\\\ 0&-{g(2)}z&{0}&{1}&\cdots&0&0&0\\\
\vdots&\vdots&\vdots&\vdots&{\ddots}&\vdots&\vdots&\vdots\\\
0&0&0&0&\cdots&{1}&-{f(q-2)}z&0\\\ 0&0&0&0&\cdots&{0}&{1}&-{f(q-1)}z\\\
-{f(\mathrm{Q})}z&0&0&0&\cdots&-{g(q-2)}z&{0}&{1}\\\ \end{pmatrix}$ (13)

with below the unity main diagonal an empty sub-diagonal made only of $0$’s
which is the manifestation of the stronger $g=3$ exclusion. The spectral
function then follows as

${s_{k}=g(k)f(k)f(k+1)}$

For $g$-exclusion the generalization of (13) amounts to a Hamiltonian of the
form

$H=F(u)v+v^{{1-g}}G(u)$ (14)

with spectral parameters

$F(\mathrm{Q}^{k})=f(k)\quad\quad G(\mathrm{Q}^{k})=g(k)$

spectral function

${s_{k}=g(k)f(k)f(k+1)\ldots f(k+g-2)}$

and a secular matrix with now $g-2$ empty sub-diagonals below the main
diagonal (here $q$ is always understood to be larger than $g$). Clearly the
Hofstadter Hamiltonian (10), which rewrites as $H=(1-u)v+v^{{1-2}}(1-u^{-1})$,
is a particular case of (14) with $g=2$ and $F(u)=1-u$, $G(u)=1-u^{-1}$.

Let us illustrate this mechanism in the case of $g=3$ exclusion with the
specific example of chiral paths on a triangular lattice (Kreweras type
paths). The three chiral hopping operators $U,V$ and
$W=\mathrm{Q}U^{-1}V^{-1}$ described in Figure 3 are such that

$VU=\mathrm{Q}^{2}UV$

The triangular lattice Hamiltonian is, in a self-explanatory form,

$H=U+V+W$

To bring it to the exclusion form (14) one chooses the representation
$U=-i\,u\,v$ and $V=i\,u^{-1}\,v$ in which case $H$ rewrites as

$H=i(-u+u^{-1})v+v^{-2}$

It is indeed an Hamiltonian of the type (14) for $g=3$ exclusion,
$F(u)=i(-u+u^{-1})$, $G(u)=1$ and with spectral parameters

$f(k)=-i(\mathrm{Q}^{k}-{1\over\mathrm{Q}^{k}})\quad g(k)=1$

spectral function

$s_{k}=g(k)f(k)f(k+1)=4\sin(2\pi pk/q)\sin\big{(}2\pi p(k+1)/q\big{)}$ (15)

and secular matrix

$I_{q}-zH_{q}=\begin{pmatrix}{1}&i(\mathrm{Q}-{1\over\mathrm{Q}})z&0&0&\cdots&0&-z&{0}\\\
{0}&{1}&i(\mathrm{Q}^{2}-{1\over\mathrm{Q}^{2}})z&0&\cdots&0&0&-z\\\
-z&{0}&{1}&i(\mathrm{Q}^{3}-{1\over\mathrm{Q}^{3}})z&\cdots&0&0&0\\\
0&-z&{0}&{1}&\cdots&0&0&0\\\
\vdots&\vdots&\vdots&\vdots&{\ddots}&\vdots&\vdots&\vdots\\\
0&0&0&0&\cdots&{1}&i(\mathrm{Q}^{q-2}-{1\over\mathrm{Q}^{q-2}})z&0\\\
0&0&0&0&\cdots&{0}&{1}&i(\mathrm{Q}^{q-1}-{1\over\mathrm{Q}^{q-1}})z\\\
i(\mathrm{Q}^{q}-{1\over\mathrm{Q}^{q}})z&0&0&0&\cdots&-z&{0}&{1}\\\
\end{pmatrix}$

which is indeed of the type (13) with a vanishing bottom-left entry. Note that
the non Hermiticity of the triangular Hamiltonian and thus of the secular
matrix is a consequence of the fact that the chiral paths carry an orientation
on the lattice.

The triangular algebraic area enumeration follows with an expression similar
to (8) provided that the trigonometric single sums appearing in (9) pertaining
to the triangular spectral function (15) can be computed [2] and that the sum
is made on all $3$-compositions of the length of the triangular paths
considered.

In conclusion we have shown how various tools available in quantum and
statistical physics allowed for an explicit algebraic area enumeration of
closed paths on planar lattices. The enumeration formulae rely on an explicit
sum over compositions whose number grows quickly with the length of the path.
It would be certainly rewarding to rewrite this sum with a smaller number of
terms. Trivially by symmetry one can restrict to mirror-free compositions
weighted twice except for the palindromic ones. We leave this issue as well as
other questions of interest to the lattice path combinatorics community.

Figure 3: The three hopping operators $U,V$ and $W$ on the triangular lattice.
Figure 4: Examples of closed chiral paths on the triangular lattice.

## References

  * [1] S. Ouvry and S. Wu, ”The algebraic area of closed lattice random walks”, J. Phys. A: Math. Theor. 52 (2019) 255201.
  * [2] S. Ouvry and A. P. Polychronakos, ”Exclusion statistics and lattice random walks”, NPB 948 (2019) 114731; ”Lattice walk area combinatorics, some remarkable trigonometric sums and Apéry-like numbers”, NPB Volume 960 (2020) 115174; see also L. Gan, S. Ouvry and A. P. Polychronakos, ”Algebraic area enumeration of random walks on the honeycomb lattice”, arXiv:2107.10851, Phys. Rev. E (to be published).
  * [3] T. Budd, ”Winding of simple walks on the square lattice”, preprint, arXiv:1709.04042 and talk given at the Lattice Paths, Combinatorics and Interactions CIRM Luminy conference (2021).
  * [4] A. Comtet, J. Desbois and S. Ouvry, ”Winding of planar Brownian curves”, Journal of Physics A : Math. Gen. 23 (1990) 3563.
  * [5] C. Garban and J. A. Trujillo Ferreras, ”The expected area of the filled planar Brownian loop is $\pi/5$”, Commun. Math. Phys. 264, 3 (2006) 797–810.
  * [6] D. Hofstadter, ”Energy levels and wave functions of Bloch electrons in rational and irrational magnetic fields”, Phys. Rev. B 14 (1976) 2239.
  * [7] C. Kreft, https://www-sfb288.math.tu-berlin.de/Publications/preprint-list/51/100, SFB 288 Preprint No. 89, TU-Berlin (1993).
  * [8] C. Krattenthaler, ”Permutations with restricted patterns and Dyck paths”, Advances in Applied Mathematics 27 (2001) 510–530.
  * [9] B. Hopkins and S. Ouvry,”Combinatorics of multicompositions, Proceedings of the Combinatorial and Additive Number Theory conference, CUNY Graduate Center NY NY, Springer (2020).


2021

[1]Haode Yan

1]School of Mathematics, Southwest Jiaotong University, Chengdu, 610031, China

# On the $c$-differential spectrum of power functions over finite fields

Kun Zhang<EMAIL_ADDRESS><EMAIL_ADDRESS>[

###### Abstract

Recently, a new concept called multiplicative differential was introduced by
Ellingsen et al. Inspired by this pioneering work, power functions with low
$c$-differential uniformity were constructed. Wang et al. defined the
$c$-differential spectrum of a power function wang2021several . In this paper,
we present some properties of the $c$-differential spectrum of a power
function. Then we apply these properties to investigate the $c$-differential
spectra of some power functions. A new class of AP$c$N function is also
obtained.

###### keywords:

$c$-differential uniformity, Almost perfect $c$-nonlinear, $c$-differential
spectrum

###### pacs:

[

MSC Classification]11T71, 94A60

##  1 Introduction

Differential cryptanalysis 1991Differential ; 1993DifferentialBS  is the first
statistical attack to decipher iterative block ciphers. Substitution boxes
(S-boxes for short) play an crucial role in the field of symmetric block
ciphers, which can be seen as cryptographic functions over finite fields. Let
${\mathbb{F}_{q}}$ be the finite fields with $q$ elements. For a function
$F(x)$ from ${\mathbb{F}}_{q}$ to itself, the main tools to handle $F$
regarding the differential attack are the Difference Distribution Table (DDT
for short) and the differential uniformity $\Delta_{F}$, introduced by Nyberg
NK1993  in 1994. For any $a,b\in{\mathbb{F}}_{2^{n}}$, the DDT entry at point
$(a,b)$, denoted by $\delta_{F}(a,b)$, is defined as

${\delta_{F}}\left({a,b}\right)=\lvert\\{x\in{\mathbb{F}}_{q}:~{}F(x+a)-F(x)=b\\}\rvert,$

where $\lvert S\rvert$ denotes the cardinality of a set $S$. The differential
uniformity of the function $F(x)$, denoted by ${\Delta_{F}}$, is defined as

${\Delta_{F}}=\max\limits_{a\in{\mathbb{F}}_{q}^{*}}\max\limits_{b\in{\mathbb{F}}_{q}}\delta_{F}(a,b),$

where ${\mathbb{F}}_{q}^{*}={\mathbb{F}}_{q}\setminus\\{0\\}$. When $F$ is
used as an S-box inside a cryptosystem, the smaller the value $\Delta_{F}$ is,
the better $F$ to the resistance against differential attack. When
$\Delta_{F}=1$, $F$ is called perfect nonlinear (PN for short) function. When
$\Delta_{F}=2$, $F$ is called almost perfect nonlinear (APN for short)
function. Note that PN functions over even characteristic finite fields do not
exist. PN and APN functions play an important role in both theory and
applications. The recent progress can be found in CM1997 ; DNiho ; DO1968 ;
DWelch ; HRS ; HS ; ZW2009 ; ZKW2009  and their references. Power functions
with low differential uniformity serve as good candidates for the design of
S-boxes not only because of their strong resistance to differential attacks
but also for the usually low implementation cost in hardware. When $F(x)$ is a
power function, i.e., $F(x)=x^{d}$ for an integer $d$, one can easily see that
$\delta_{F}(a,b)=\delta_{F}(1,{b/{a^{d}}})$ for all $a\in{\mathbb{F}}_{q}^{*}$
and $b\in{\mathbb{F}}_{q}$. That is to say, the differential properties of
$F(x)$ is completely determined by the values of $\delta_{F}(1,b)$ as $b$ runs
through ${\mathbb{F}}_{q}$. Therefore, Blondeau, Canteaut and Charpin first
defined the difference spectrum of a power function in BCC . Denote

$\omega_{i}=\lvert\left\\{b\in{\mathbb{F}}_{q}:\delta_{F}(1,b)=i\right\\}\rvert,\,\,0\leq
i\leq\Delta_{F},$

where $\Delta_{F}$ is the differential uniformity of $F$. The differential
spectrum of $F$ is defined as the multi-set

$DS_{F}=\left\\{\omega_{i}>0~{}:~{}0\leq i\leq\Delta_{F}\right\\}.$

The distribution of DDT of a power function can be deduced via its
differential spectrum. Moreover, the differential spectrum is also an
important notion for estimating its resistance against variants of
differential cryptanalysis (BCC , bracken2010highly , charpin2014sparse ).
However, it is difficult to completely determine the differential spectrum of
a power function. For a known results on differential spectrum of power
functions, the readers are referred to BCC ; BCC2 ; BP ; XYY ; Li ;
Dobbertin2001 ; 2020XZL ; CHNC ; Yan ; LRF ; XY2017 ; YXLHXL . The
distribution of DDT of a power function can be deduced via its differential
spectrum.

In 2002BN , the authors used a new type of differential, namely multiplicative
differential, that is quite useful from a practical perpective for ciphers
that utilize modular multiplication as a primitive operation. It is an
extension of differential cryptanalysis, and it cryptanalyzes some existing
ciphers (like a variant of the well-known IDEA cipher). The authors argue that
one should look at other types of differential for a cryptographic function
$F$, not only the usual $(F(x+a),F(x))$ but $(F(x+a),cF(x))$. Moreover, they
first introduced the $c$-Differential Distribution Table ($c$DDT for short) .
For a function $F$ from ${\mathbb{F}}_{{q}}$ to itself and
$c\in{\mathbb{F}}_{{q}}$, the entry at point $(a,b)$ of the $c$DDT, denoted by
${}_{c}\delta_{F}(a,b)$, is defined as

${}_{c}\delta_{F}(a,b)=\\#\\{x\in{\mathbb{F}}_{q}:F(x+a)-cF(x)=b\\}.$

The corresponding $c$-differential uniformity is defined as follows.

######  Definition 1.

(2020CEP ) Let ${\mathbb{F}}_{q}$ denote the finite field with $q$ elements,
where $q$ is a prime power. For a function
$F:{{\mathbb{F}}_{q}}\to{{\mathbb{F}}_{q}}$, and $a,b,c\in{\mathbb{F}}_{q}$,
we call

${}_{c}{\Delta_{F}}=\max\left\\{{{}_{c}{\delta_{F}}\left(a,b\right):a,b\in{{\mathbb{F}}_{{q}}},\mathrm{and}~{}a\neq
0~{}\mathrm{if}~{}c=1}\right\\}$

the $c$-differential uniformity of $F$.

If ${}_{c}{\Delta_{F}}=\delta$, then we say that $F$ is differentially
$(c,\delta)$-uniform. Similarly, the smaller the value ${}_{c}\Delta_{F}$ is,
the better $F$ to the resistance against multiplicative differential attack.
If the $c$-differential uniformity of $F$ equals $1$, then $F$ is called a
perfect $c$-nonlinear (P$c$N) function. P$c$N function over odd characteristic
finite fields are also called $c$-planar functions. If the $c$-differential
uniformity of $F$ is $2$, then $F$ is called an almost perfect $c$-nonlinear
(AP$c$N) function. It is easy to conclude, for $c=1$ and $a\neq 0$, the
$c$-differential uniformity becomes the usual differential uniformity, and the
P$c$N and AP$c$N functions become PN and APN functions respectively. It is
know that APN functions over finite fields of even characteristic have lowest
differential uniformity. However, for the $c$-differential uniformity, there
exist P$c$N functions over even characteristic finite fields. There are a few
functions with low $c$-differential uniformity reported. The readers can refer
to wang2021several ; 2021SomeZha ; tu2021class ; yan2021on1 ; 2021SYZ ;
2020CEP ; 2020MD ; hasan2021c ; WLZ ; BC .

Similarly, when $F(x)$ is a power function, one easily sees that
${}_{c}{\delta_{F}(a,b)}$=${}_{c}{\delta_{F}(1,{b/{a^{d}}})}$ for all
$a\in{\mathbb{F}}_{{q}}^{*}$ and $b\in{\mathbb{F}}_{q}$. That is to say, the
$c$-differential spectrum of $F(x)$ is completely determined by the values of
${}_{c}\delta_{F}(1,b)$ as $b$ runs through ${\mathbb{F}}_{q}$. Therefore, the
$c$-differential spectrum of a power function can be defined as follows.

######  Definition 2.

(wang2021several ) Let $F\left(x\right)={x^{d}}$ be a power function over
${\mathbb{F}}_{q}$ with $c$-differential uniformity ${}_{c}\Delta_{F}$. Denote
by
${}_{c}{\omega_{i}}=\\#\\{{b\in{\mathbb{F}}_{q}:{{}_{c}{\delta_{F}}(1,b)=i}.}\\}$
for each $0\leq i\leq{}_{c}{\Delta_{F}}$. The $c$-differential spectrum of $F$
is defined to be the multi-set

$\mathbb{S}=\\{{{}_{c}{\omega_{i}}:{0\leq
i\leq{}_{c}{\Delta_{F}}}~{}\mathrm{and}{}~{}_{c}{\omega_{i}}>0}\\}.$

When $c=1$, the $c$-differential spectrum becomes the usual differential
spectrum. As far as we know, only two classes of power functions have known
$c$-differential spectra. In wang2021several , the authors computed the
$c$-differential spectrum of Gold function. The $c$-differential spectrum of
$x^{d}$ over ${\mathbb{F}}_{2^{4m}}$ was determined in tu2021class , where
$d=2^{3m}+2^{2m}+2^{m}-1$. We summarized the known results in Table 1 as well
as the results obtained in this paper. The rest of this paper is organized as
follows. Section II introduces some notation and useful lemmas. In Section
III, we present the properties of $c$-differential spectrum. In Section IV, we
calculate the $c$-differential spectrum of several power functions with low
$c$-differential uniformity. In Section V, we construct an infinite family of
AP$c$N power mappings over ${\mathbb{F}}_{{5^{n}}}$, and determine its
$c$-differential spectrum. Section VI concludes this paper.

Table 1: Power functions $F(x)=x^{d}$ over ${\mathbb{F}}_{p^{n}}$ with known
$c$-differential spectrum

$p$ | $d$ | Condition | ${}_{c}\Delta_{F}$ | Ref.
---|---|---|---|---
2 | $2^{3m}+2^{2m}+2^{m}-1$ |  $n=4m$, $0,1\neq c\in{\mathbb{F}}_{2^{n}}$, $c^{1+2^{2m}}=1$ | $2$ | tu2021class
odd | $p^{k}+1$ |  $1\neq c\in{\mathbb{F}}_{p^{\gcd(n,k)}}$ and $\frac{n}{\gcd(n,k)}$ is odd, or $c\notin{\mathbb{F}}_{p^{\gcd(n,k)}}$, $n$ is even and $k=\frac{n}{2}$ | $2$ | wang2021several
2 | $2^{n}-2$ |  $c\neq 0$, ${\mathrm{Tr}}_{n}(c)={\mathrm{Tr}}_{n}(c^{-1})=1$ | $2$ | Thm 7
2 | $2^{n}-2$ |  $c\neq 0$, ${\mathrm{Tr}}_{n}(c)=0$ or ${\mathrm{Tr}}_{n}(c^{-1})=0$ | $3$ | Thm 7
odd | $p^{n}-2$ |  $c=4,4^{-1}$, or $\chi(c^{2}-4c)=-1$ and $\chi(1-4c)=-1$ | 2 | Thm 8
odd | $p^{n}-2$ |  $c\neq 0,4,4^{-1}$, $\chi(c^{2}-4c)=1$ or $\chi(1-4c)=1$ | 3 | Thm 8
3 | $\frac{3^{n}+3}{2}$ |  $c=-1$, $n$ even | $2$ | Thm 9
3 | $3^{n}-3$ |  $c=-1$, $n=0({\bmod 4})$ | 6 | Thm 10
3 | $3^{n}-3$ |  $c=-1$, $n\neq 0({\bmod 4})$ | 4 | Thm 10
odd | $\frac{p^{k}+1}{2}$ |  $c=-1$, $\gcd(n,k)=1$, $k$ odd | $\frac{p+1}{2}$ | Thm 11, 12
5 | $\frac{p^{n}-3}{2}$ | $c=-1$ | 2 | Thm 16
|  |  |  |

00footnotetext:  ${\mathrm{Tr}}_{n}\left(\cdot\right)$ denotes the absolute
trace mapping from ${\mathbb{F}}_{{2^{n}}}$ to
${\mathbb{F}}_{2}$.00footnotetext: $\chi\left(\cdot\right)$ denotes the
quadratic multiplicative character on ${\mathbb{F}}_{{p^{n}}}^{*}$.

##  2 Preliminaries

In this section, we first fix some notation and list some facts which will be
used in this paper unless otherwise stated.

  * •

Let ${\mathbb{F}}_{q}$ denote the finite field with $q$ elements.

  * •

${\mathbb{F}}_{{q}}^{*}={\mathbb{F}}_{{q}}\setminus\\{0\\}$,
${\mathbb{F}}_{{q}}^{\\#}={{\rm{{\mathbb{F}}}}_{{q}}}\backslash\\{0,-1\\}$.

  * •

$\Delta_{c}(x)=(x+1)^{d}-cx^{d}$, where $c\in{\mathbb{F}}_{{q}}$.

  * •

${\delta_{c}}\left(b\right)=\\#\left\\{{x\in{{\mathbb{F}}_{{q}}}:{\Delta_{c}}\left(x\right)=b}\right\\}$.

  * •

$\chi$ denotes the quadratic multiplicative character on ${\mathbb{F}}_{{q}}$,
i.e.,

$\chi(x)=\left\\{\begin{aligned}
1,~{}~{}&\mathrm{if}~{}{x}~{}\mathrm{a~{}square~{}},\\\
0,~{}~{}&\mathrm{if}~{}x=0,\\\
-1,~{}~{}&\mathrm{if}~{}{x}~{}\mathrm{a~{}nonsquare~{}}.\end{aligned}\right.$

It is well-known that $\sum_{x\in{\mathbb{F}}_{q}}\chi(x)=0$.

  * •

${S_{i,j}}:=\left\\{{x\in{\mathbb{F}}_{{q}}^{\\#}:\chi(x)=i,\chi(x+1)=j}\right\\}$,
where $i,j\in\\{\pm 1\\}$.

  * •

${S_{1,1}}\cup{S_{-1,-1}}\cup{S_{1,-1}}\cup{S_{-1,1}}={\mathbb{F}}_{{q}}^{\\#}$.

Secondly, we introduce some lemmas which will be used in the sequel. To
determine the greatest common divisor of integers, the following lemma plays
an important role in the rest of this paper.

######  Lemma 1.

let $p,k,n$ be integers greater than or equal to $1$. Then

$\gcd\left({{p^{k}}+1,{p^{n}}-1}\right)=\left\\{\begin{aligned}
{}\frac{{{2^{\gcd\left({2k,n}\right)}}-1}}{{{2^{\gcd\left({k,n}\right)}}-1}},~{}~{}~{}&\mathrm{if}~{}{p}=2,\\\
2,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}&\mathrm{if}~{}\frac{n}{{\gcd(n,k)}}~{}\mathrm{is}~{}\mathrm{odd},\\\
{p^{\gcd\left({k,n}\right)}}+1,~{}~{}~{}~{}&\mathrm{if}~{}\frac{n}{{\gcd\left({n,k}\right)}}~{}\mathrm{is}~{}\mathrm{even}.\end{aligned}\right.$

The following lemma can be used in solving equations over finite fields.

######  Lemma 2.

FF  Let $n$ be a positive integer. We have:

(i) The equation $x^{2}+ax+b=0$, with $a,b\in{\mathbb{F}}_{2^{n}}$, $a\neq 0$,
has two solutions in ${\mathbb{F}}_{2^{n}}$ if
${\mathrm{Tr}}_{n}(\frac{b}{a^{2}})=0$, and zero solutions otherwise.

(ii) The equation $x^{2}+ax+b=0$, with $a,b\in{\mathbb{F}}_{p^{n}}$, $p$ odd,
has (two, respectively, one) solutions in ${\mathbb{F}}_{p^{n}}$ if and only
if the discriminant $a^{2}-4b$ is a (nonzero, respectively, zero) square in
${\mathbb{F}}_{p^{n}}$.

At last, we introduce the following result on the quadratic multiplicative
character sums.

######  Lemma 3.

(FF, , Theorem 5.48) Let $f(x)=a_{2}x^{2}+a_{1}x+a_{0}\in{\mathbb{F}}_{q}[x]$
with $q$ odd and $a_{2}\neq 0$. Put $d=a^{2}_{1}-4a_{0}a_{2}$ and let $\chi$
be the quadratic character of ${\mathbb{F}}_{q}$. Then

$\displaystyle\sum_{x\in{\mathbb{F}}_{q}}\chi(f(x))=\left\\{\begin{array}[]{lllll}-\chi(a_{2}),~{}~{}~{}~{}~{}~{}~{}\mathrm{if}~{}d\neq
0.\\\ (q-1)\chi(a_{2}),~{}\mathrm{if}~{}d=0.\end{array}\right.\ \ $

##  3 The properties of the $c$-differential spectrum

The usual differential spectrum of a power function satisfies several
identities (see BCC ). It is natural to wonder how it behaves with respect to
the $c$-differential spectrum of a power function. In this section, we give
the following identities of the $c$-differential spectrum.

######  Theorem 4.

Let $F\left(x\right)={x^{d}}$ be a power function over ${\mathbb{F}}_{q}$ with
$c$-differential uniformity ${}_{c}{\Delta_{F}}$ for some $1\neq
c\in{\mathbb{F}}_{q}$. Recall that
${{}_{c}\omega_{i}}=\\#\left\\{{b\in{\mathbb{F}}_{q}:{{}_{c}{\Delta_{F}}\left(1,b\right)=i}.}\right\\}$
for each $0\leq i\leq{}_{c}{\Delta_{F}}$. We have

$\sum_{i=0}^{{}_{c}{\Delta_{F}}}{}_{c}\omega_{i}=\sum_{i=0}^{{}_{c}{\Delta_{F}}}i\cdot{}_{c}\omega_{i}=q.$
(2)

Moreover, we have

$\sum\limits_{i=0}^{{}_{c}{\Delta_{F}}}{{i^{2}}\cdot{}_{c}{\omega_{i}}}=\frac{{{\rm{}}{{}_{c}N_{4}}-1}}{{q}-1}-\gcd(d,q-1),$
(3)

where

${}_{c}N_{4}=\\#\left\\{{\left({{x_{1}},{x_{2}},{x_{3}},{x_{4}}}\right)\in({\mathbb{F}}_{q})^{4}:{\Bigg{\\{}\begin{array}[]{l}{x_{1}}-{x_{2}}+{x_{3}}-{x_{4}}=0\\\
x_{1}^{d}-cx_{2}^{d}+cx_{3}^{d}-x_{4}^{d}=0\end{array}}}\right\\}.$ (4)

Proof: According to the definition of ${}_{c}\omega_{i}$, we have

$\sum\limits_{i=0}^{{}_{c}{\Delta_{F}}}{{{}_{c}\omega_{i}}}=\sum\limits_{i=0}^{{}_{c}{\Delta_{F}}}{\\#\left\\{{b\in{\mathbb{F}}_{q}:{{}_{c}{\Delta_{F}}\left(1,b\right)=i}}\right\\}}=\sum\limits_{b\in{\mathbb{F}}_{q}}1=q.$

and

$\displaystyle\sum\limits_{i=0}^{{}_{c}{\Delta_{F}}}{i\cdot{{}_{c}\omega_{i}}}$
$\displaystyle=\sum\limits_{i=0}^{{}_{c}{\Delta_{F}}}{i\cdot\\#\left\\{{b\in{\mathbb{F}}_{q}:{{}_{c}{\Delta_{F}}(1,b)=i}}\right\\}}$
$\displaystyle=\sum_{b\in{\mathbb{F}}_{q}}\\#\\{x\in{\mathbb{F}}_{q}:(x+1)^{d}-cx^{d}=b\\}$
$\displaystyle=\sum\limits_{x\in{\mathbb{F}}_{q}}1$ $\displaystyle=q.$

Next we prove the last statement. For $c\neq 1$, we define

${}_{c}n({\alpha,\beta})=\\#\Bigg{\\{}{({{x},{y}})\in({\mathbb{F}}_{q})^{2}:{\bigg{\\{}\begin{array}[]{l}{x}-{y}=\alpha\\\
x^{d}-cy^{d}=\beta\end{array}}}\Bigg{\\}}.$

It is obvious that

${{}_{c}N_{4}}=\sum\limits_{\alpha,\beta\in{{\mathbb{F}}_{q}}}{{(_{c}n}({\alpha,\beta}))^{2}}.$

Moreover, one can easily check that ${}_{c}n(0,0)=1$. For $\beta\neq 0$,
${}_{c}n(0,\beta)$ is equal to the number of solutions $y\in{\mathbb{F}}_{q}$
of the equation $y^{d}=\frac{\beta}{1-c}$. Let $\gamma$ be a primitive element
of ${\mathbb{F}}_{q}^{*}$ and $\frac{\beta}{1-c}=\gamma^{k}$ for some integer
$k$. Let $e=\gcd(d,q-1)$. Then ${}_{c}n(0,\beta)=e$ if $e\mid k$ and
${}_{c}n(0,\beta)=0$ otherwise. For $\alpha,\beta\neq 0$, we have
${}_{c}n(\alpha,\beta)={}_{c}n(1,\frac{\beta}{\alpha^{d}})$. We immediately
obtain the following.

$\displaystyle{}{{}_{c}N_{4}}$
$\displaystyle=\sum\limits_{\alpha,\beta\in{{\mathbb{F}}_{q}}}{{(_{c}n}({\alpha,\beta}))^{2}}$
$\displaystyle={({}_{c}n}({0,0}))^{2}+\sum\limits_{\beta\in{\mathbb{F}}_{q}^{*}}({{}_{c}n}({0,\beta}))^{2}+\sum\limits_{\alpha\in{\mathbb{F}}_{q}^{*}}{{\sum\limits_{\beta\in{{\mathbb{F}}_{q}}}{({{}_{c}n}\left({\alpha,\beta}\right))^{2}}}}$
$\displaystyle=1+\sum_{1\leq k\leq q-1,e\mid
k}e^{2}+\sum\limits_{\alpha\in{\mathbb{F}}^{*}_{q}}{{\sum\limits_{\beta\in{{\mathbb{F}}_{q}}}{({{}_{c}n}({1,\frac{\beta}{{{\alpha^{d}}}}}))^{2}}}}$
$\displaystyle=1+\frac{{q}-1}{e}{{e}^{2}}+\sum\limits_{\alpha\in{\mathbb{F}}_{q}^{*}}{{\sum\limits_{b\in{{\mathbb{F}}_{q}}}{({{}_{c}n}({1,b}))^{2}}}}$
$\displaystyle=1+\big{(}q-1\big{)}{e}+\sum\limits_{\alpha\in{\mathbb{F}}_{q}^{*}}{{\sum\limits_{b\in{{\mathbb{F}}_{q}}}{{{\big{(}{{\delta_{c}}(b)}\big{)}}^{2}}}}}$
$\displaystyle=1+\big{(}q-1\big{)}{{{e}}}+(q-1)\sum\limits_{i=0}^{{}_{c}{\Delta_{F}}}{{i^{2}}\cdot{{}_{c}\omega_{i}}}.$

The fourth identity holds since for fixed $a\neq 0$, when $\beta$ runs through
${\mathbb{F}}_{q}$, so does $\frac{\beta}{\alpha^{d}}$. We complete the proof.

##  4 The $c$-differential spectrum of some power functions

In this section, we investigate the $c$-differential spectra of some power
functions with low $c$-differential uniformity.

###  4.1 The $c$-differential spectrum of the inverse function

Since there has been quite a bit of effort to investigate the inverse function
over ${\mathbb{F}}_{2^{n}}$ as it is relevant in Rijndael and Advance
Encryption Standard, it is natural to wonder how it behaves with respect to
the $c$-differential uniformity. In 2020CEP , the $c$-differential uniformity
of the inverse function $x\in{\mathbb{F}}_{p^{n}},x\mapsto x^{p^{n}-2}$ has
been studied. We have the following lemmas.

######  Lemma 5 (2020CEP ).

Let $n$ be a positive integer, $1\neq c\in{\mathbb{F}}_{2^{n}}$ and
$F:{{\mathbb{F}}_{{2^{n}}}}\to{{\mathbb{F}}_{{2^{n}}}}$ be the inverse
function defined by $F(x)=x^{2^{n}-2}$. We have:

(i) If $c=0$, then $F$ is P$c$N ;

(ii) If $c\neq 0$ and ${\mathrm{Tr}}_{n}(c)={\mathrm{Tr}}_{n}(\frac{1}{c})=1$,
the $c$-differential uniformity of $F$ is $2$;

(iii) If $c\neq 0$ and ${\mathrm{Tr}}_{n}(c)=0$ or
${\mathrm{Tr}}_{n}(\frac{1}{c})=0$, the $c$-differential uniformity of $F$ is
$3$.

######  Lemma 6 (2020CEP ).

Let $p$ be an odd prime. $n\geq 1$ be a positive integer, $1\neq
c\in{\mathbb{F}}_{p^{n}}$ and
$F:{{\mathbb{F}}_{{p^{n}}}}\to{{\mathbb{F}}_{{p^{n}}}}$ be the inverse
function defined by $F(x)=x^{p^{n}-2}$. We have:

(i) If $c=0$, then $F$ is P$c$N .

(ii) If $c\neq 0,4,4^{-1}$, $\chi(c^{2}-4c)=1$, or $\chi(1-4c)=1$, the
$c$-differential uniformity of $F$ is $3$.

(iii) If $c=4,4^{-1}$, the $c$-differential uniformity of $F$ is $2$.

(iv) If $c\neq 0$, $\chi(c^{2}-4c)=\chi(1-4c)=-1$, the $c$-differential
uniformity of $F$ is $2$.

The $c$-differential spectrum of the inverse function is given in the
following theorem.

######  Theorem 7.

Let $F(x)=x^{2^{n}-2}$ be a power function over ${\mathbb{F}}_{{2^{n}}}$. When
$0,1\neq c\in{\mathbb{F}}_{2^{n}}$, the inverse function is AP$c$N with
$c$-differential spectrum

$\mathbb{S}=\\{{}_{c}\omega_{0}=2^{n-1}-2,{}_{c}\omega_{1}=4,{}_{c}\omega_{2}=2^{n-1}-2\\}$

if ${\mathrm{Tr}}_{n}(c)={\mathrm{Tr}}_{n}(\frac{1}{c})=1$. Moreover, the
inverse function is differentially $(c,3)$-uniform with $c$-differential
spectrum

$\mathbb{S}=\\{{}_{c}\omega_{0}=2^{n-1}-1,{}_{c}\omega_{1}=3,{}_{c}\omega_{2}=2^{n-1}-3,{}_{c}\omega_{3}=1\\}$

if ${\mathrm{Tr}}_{n}(c)=1$, ${\mathrm{Tr}}_{n}(\frac{1}{c})=0$ or
${\mathrm{Tr}}_{n}(c)=0$, ${\mathrm{Tr}}_{n}(\frac{1}{c})=1$, and is
differentially $(c,3)$-uniform with $c$-differential spectrum

$\mathbb{S}=\\{{}_{c}\omega_{0}=2^{n-1},{}_{c}\omega_{1}=2,{}_{c}\omega_{2}=2^{n-1}-4,{}_{c}\omega_{3}=2\\}$

if ${\mathrm{Tr}}_{n}(c)={\mathrm{Tr}}_{n}(\frac{1}{c})=0$.

Proof: For $b\in{\mathbb{F}}_{{2^{n}}}$, we consider the $c$-differential
equation

$(x+1)^{2^{n}-2}+cx^{2^{n}-2}=b.$ (5)

In order to calculate the $c$-differential spectrum, we first determine the
values of ${}_{c}\omega_{1}$ and ${}_{c}\omega_{3}$.

Case 1: $b=0$. (5) has a unique solution $x=\frac{c}{c+1}$.

Case 2: $b=1$. Then $x=0$ is a solution of (5), and $x=1$ is not a solution of
(5) since $c\neq 1$. Assume that $x\neq 0,1$, by multiplying $x(x+1)$ on both
sides of (5), we get $x^{2}+cx+c=0$. By Lemma 2 (i), this equation has two
solutions iff ${\mathrm{Tr}}_{n}(\frac{1}{c})=0$. Consequently,
$\delta_{c}(1)=3$ if ${\mathrm{Tr}}_{n}(\frac{1}{c})=0$, and $\delta_{c}(1)=1$
if ${\mathrm{Tr}}_{n}(\frac{1}{c})=1$.

Case 3: $b=c$. In this case, $x=1$ is a solution of (5), while $x=0$ is not a
solution. Next we assume that $x\neq 0,1$. Similarly, we get
$x^{2}+c^{-1}x+1=0$. By Lemma 2 (i), this equation has two solutions iff
${\mathrm{Tr}}_{n}(c)=0$. Consequently, $\delta_{c}(c)=3$ if
${\mathrm{Tr}}_{n}(c)=0$, and $\delta_{c}(c)=1$ if ${\mathrm{Tr}}_{n}(c)=1$.

Case 4: $b\neq 0,1,c$. In this case, $x\neq 0,1$. Multiplying $x(x+1)$ on both
sides of (5), we obtain

$bx^{2}+(1+b+c)x+c=0.$ (6)

When $b=1+c$, (6) has unique solution $x=(\frac{c}{c+1})^{2^{n-1}}$. By Lemma
2 (i), (6) has 0 or 2 solutions if $1+b+c\neq 0$. We have $\delta_{c}(1+c)=1$,
and $\delta_{c}(b)=0$ or $2$ when $b\neq 0,1,c,1+c$.

Summarizing the above, we obtain that

${}_{c}\omega_{1}=\left\\{\begin{aligned}
{}4,~{}~{}~{}&\mathrm{if}~{}{\mathrm{Tr}}_{n}(c)={\mathrm{Tr}}_{n}(\frac{1}{c})=1,\\\
3,~{}~{}~{}&\mathrm{if}~{}{\mathrm{Tr}}_{n}(c)=1,{\mathrm{Tr}}_{n}(\frac{1}{c})=0~{}\mathrm{or}~{}{\mathrm{Tr}}_{n}(c)=0,{\mathrm{Tr}}_{n}(\frac{1}{c})=1,\\\
2,~{}~{}~{}&\mathrm{if}~{}{\mathrm{Tr}}_{n}(c)={\mathrm{Tr}}_{n}(\frac{1}{c})=0,\end{aligned}\right.$

and

${}_{c}\omega_{3}=\left\\{\begin{aligned}
{}0,~{}~{}~{}&\mathrm{if}~{}{\mathrm{Tr}}_{n}(c)={\mathrm{Tr}}_{n}(\frac{1}{c})=1,\\\
1,~{}~{}~{}&\mathrm{if}~{}{\mathrm{Tr}}_{n}(c)=1,{\mathrm{Tr}}_{n}(\frac{1}{c})=0~{}\mathrm{or}~{}{\mathrm{Tr}}_{n}(c)=0,{\mathrm{Tr}}_{n}(\frac{1}{c})=1,\\\
2,~{}~{}~{}&\mathrm{if}~{}{\mathrm{Tr}}_{n}(c)={\mathrm{Tr}}_{n}(\frac{1}{c})=0.\end{aligned}\right.$

By solving the equation system (2), the desired result follows.

Next, we calculate the $c$-differential spectrum of the inverse function over
${\mathbb{F}}_{p^{n}}$ when $p$ is odd. We have the following results.

######  Theorem 8.

Let $p$ be any odd prime. $F(x)=x^{p^{n}-2}$ be a power function over
${\mathbb{F}}_{p^{n}}$. When $0,1,4,4^{-1}\neq c\in{\mathbb{F}}_{p^{n}}$, the
$c$-differential spectrum of it is given as the following six cases:

(i)
$\mathbb{S}=\\{{}_{c}\omega_{0}=\frac{p^{n}-3}{2},{}_{c}\omega_{1}=3,{}_{c}\omega_{2}=\frac{p^{n}-3}{2}\\}$
if $\chi(c^{2}-4c)=\chi(1-4c)=-1$ and $\chi(c)=-1$;

(ii)
$\mathbb{S}=\\{{}_{c}\omega_{0}=\frac{p^{n}-5}{2},{}_{c}\omega_{1}=5,{}_{c}\omega_{2}=\frac{p^{n}-5}{2}\\}$
if $\chi(c^{2}-4c)=\chi(1-4c)=-1$ and $\chi(c)=1$;

(iii)
$\mathbb{S}=\\{{}_{c}\omega_{0}=\frac{p^{n}-1}{2},{}_{c}\omega_{1}=2,{}_{c}\omega_{2}=\frac{p^{n}-5}{2},{}_{c}\omega_{3}=1\\}$
if $\chi(c^{2}-4c)\chi(1-4c)=-1$ and $\chi(c)=-1$;

(iv)
$\mathbb{S}=\\{{}_{c}\omega_{0}=\frac{p^{n}-3}{2},{}_{c}\omega_{1}=4,{}_{c}\omega_{2}=\frac{p^{n}-7}{2},{}_{c}\omega_{3}=1\\}$
if $\chi(c^{2}-4c)\chi(1-4c)=-1$ and $\chi(c)=1$;

(v)
$\mathbb{S}=\\{{}_{c}\omega_{0}=\frac{p^{n}+1}{2},{}_{c}\omega_{1}=1,{}_{c}\omega_{2}=\frac{p^{n}-7}{2},{}_{c}\omega_{3}=2\\}$
if $\chi(c^{2}-4c)=\chi(1-4c)=1$ and $\chi(c)=-1$;

(vi)
$\mathbb{S}=\\{{}_{c}\omega_{0}=\frac{p^{n}-1}{2},{}_{c}\omega_{1}=3,{}_{c}\omega_{2}=\frac{p^{n}-9}{2},{}_{c}\omega_{3}=2\\}$
if $\chi(c^{2}-4c)=\chi(1-4c)=1$ and $\chi(c)=1$.

Proof: For $b\in{\mathbb{F}}_{{p^{n}}}$, we consider the $c$-diffferential
equation

$(x+1)^{p^{n}-2}-cx^{p^{n}-2}=b.$ (7)

Similarly, we determine the values of ${}_{c}\omega_{1}$ and
${}_{c}\omega_{3}$.

Case 1:  $b=0$. It can be easily seen that $x=\frac{c}{1-c}$ is the unique
solution of (7), i.e. $\delta_{c}(0)=1$.

Case 2:  $b=1$. Then $x=0$ is a solution of (7), and $x=-1$ is not a solution
of (7). Assume that $x\neq 0,-1$, by multiplying $x(x+1)$ on both sides of
(7), we get $x^{2}+cx+c=0$. The discriminant of this quadratic equation is
$c^{2}-4c$, which is not zero since $c\neq 0,4$. By Lemma 2(ii), this equation
has two solutions if $\chi(c^{2}-4c)=1$, and no solution if
$\chi(c^{2}-4c)=-1$. We obtain $\delta_{c}(1)=3$ if $\chi(c^{2}-4c)=1$,
$\delta_{c}(1)=1$ if $\chi(c^{2}-4c)=-1$.

Case 3:  $b=c$. Then $x=-1$ is a solution of (7), while $x=0$ is not a
solution. Next we assume that $x\neq 0,-1$. By multiplying $x(x+1)$ on both
sides of (7), we get $cx^{2}+(2c-1)x+c=0$. The discriminant of this quadratic
equation is $1-4c$, which is not zero since $c\neq 0,4^{-1}$. By Lemma 2(ii),
this equation has two solutions if $\chi(1-4c)=1$, and no solution if
$\chi(1-4c)=-1$. We obtain $\delta_{c}(c)=1$ if $\chi(1-4c)=-1$ and
$\delta_{c}(c)=3$ if $\chi(1-4c)=1$.

Case 4:  Let $b\neq 0,1,c$. Then $x\neq 0,1$. By multiplying $x(x+1)$ on both
sides of (7), we obtain

$bx^{2}+(b+c-1)x+c=0.$ (8)

The equation (8) has a unique solution if and only if the discriminant
$(b+c-1)^{2}-4bc=0$. That is,

$b^{2}-2(c+1)b+(c-1)^{2}=0.$

The above equation on $b$ is quadratic, and the discriminant is $16c$. We
assert that, if $\chi(c)=1$, there exist two $b$’s such that (8) has one
solution, and if $\chi(c)=-1$, such $b$ do not exist.

By discussions as above, we summarize the values of ${}_{c}{\omega_{1}}$ and
${}_{c}{\omega_{3}}$ in the following table. By solving the equation system
(2), the desired result follows.

Table 2: The values of ${}_{c}{\omega_{1}}$ and ${}_{c}{\omega_{3}}$ ${}_{c}{\omega_{1}}$ | ${}_{c}{\omega_{3}}$ | $\chi(c^{2}-4c)$ | $\chi(1-4c)$ | $\chi(c)$
---|---|---|---|---
$3$ | $2$ | $1$ | $1$ | $1$
$1$ | $2$ | $1$ | $1$ | $-1$
$2$ | $1$ | $1$ | $-1$ | $-1$
$4$ | $1$ | $1$ | $-1$ | $1$
$4$ | $1$ | $-1$ | $1$ | $1$
$2$ | $1$ | $-1$ | $1$ | $-1$
$3$ | $0$ | $-1$ | $-1$ | $-1$
$5$ | $0$ | $-1$ | $-1$ | $1$
|  |  |  |

###  4.2 The $(-1)$-differential spectra of some power functions

The case that $c=-1$ is a special case. Sometimes the $(-1)$-differential
uniformity of a power function is lower than its $c$-differential uniformity
for other $c$. In this subsection, we compute the $(-1)$-differential spectra
of some power functions. We begin this subsection with a simple result.

In 2021SYZ , the authors reported that the power function
$x^{\frac{3^{n}+3}{2}}$ over ${\mathbb{F}}_{3^{n}}$ is AP$c$N if $n$ is even.
We have the following theorem.

######  Theorem 9.

Let $G\left(x\right)={x^{\frac{{{3^{n}}+3}}{2}}}$ on ${\mathbb{F}}_{3^{n}}$,
$n\geq 2$. The $(-1)$-differential spectrum of $G(x)$ is

$\mathbb{S}=\left\\{{{{}_{-1}\omega_{0}}=\frac{{{3^{n}}-1}}{2},{{}_{-1}\omega_{1}}=1,{{}_{-1}\omega_{2}}=\frac{{{3^{n}}-1}}{2}}\right\\}.$

Proof: To determine the $(-1)$-differential spectrum of
$x^{\frac{{{3^{n}}+3}}{2}}$, we consider the equation

$\Delta_{-1}(x)={\left({x+1}\right)^{\frac{{{3^{n}}+3}}{2}}}+{x^{\frac{{{3^{n}}+3}}{2}}}=b$
(9)

for $b\in{\mathbb{F}}_{{3^{n}}}$. Note that $x$ is a solution of (9) if and
only if $-x-1$ is a solution of (9). When $x=-x-1$, then $x=1$ and the
corresponding $b=-1$ satisfies that $\delta_{-1}(-1)$ is odd. Then
$\delta_{-1}(-1)=1$ since $F(x)$ is AP$c$N. We get ${}_{-1}\omega_{1}=1$. Then
by solving the equation (2), we complete the proof.

It was shown that $x^{3^{n}-3}$ is an APN power mapping of
${\mathbb{F}}_{3^{n}}$ when $n>1$ is odd HRS , is differentially $4$-uniform
when $n=2\left({\bmod 4}\right)$ 2020XZL , and is differentially $5$-uniform
when $n=0\left({\bmod 4}\right)$ 2020XZL . Moreover, 2021SYZ  showed that this
power function has low $(-1)$-differential uniformity. In the following
theorem, we determine the $(-1)$-differential spectrum of $x^{3^{n}-3}$ over
${\mathbb{F}}_{3^{n}}$.

######  Theorem 10.

Let $H\left(x\right)={x^{3^{n}-3}}$ be a power function over
${{\mathbb{F}}_{{3^{n}}}}$. When $n=0\left({\bmod 4}\right)$, $H$ is
differentially $(-1,6)$-uniform with $(-1)$-differential spectrum

$\begin{gathered}\mathbb{S}=\left\\{{{{}_{-1}\omega_{0}}=\frac{{5\cdot{3^{n}}-3}}{8},{{}_{-1}\omega_{1}}=1,{{}_{-1}\omega_{2}}=\frac{{{3^{n}}+3}}{4},{{}_{-1}\omega_{4}}=\frac{{{3^{n}}-17}}{8},{{}_{-1}\omega_{6}}=1}\right\\}.\hfill\\\
\end{gathered}$

When $n=2\left({\bmod 4}\right)$, $H$ is differentially $(-1,4)$-uniform with
$(-1)$-differential spectrum

$\mathbb{S}=\left\\{{{{}_{-1}\omega_{0}}=\frac{{5\cdot{3^{n}}-13}}{8},{{}_{-1}\omega_{1}}=1,{{}_{-1}\omega_{2}}=\frac{{{3^{n}}+7}}{4},{{}_{-1}\omega_{4}}=\frac{{{3^{n}}-9}}{8}}\right\\}.$

When $n=1,3\left({\bmod 4}\right)$, $H$ is differentially $(-1,4)$-uniform
with $(-1)$-differential spectrum

$\mathbb{S}=\left\\{{{{}_{-1}\omega_{0}}=\frac{{5\cdot{3^{n}}-7}}{8},{{}_{-1}\omega_{1}}=1,{{}_{-1}\omega_{2}}=\frac{{{3^{n}}+1}}{4},{{}_{-1}\omega_{4}}=\frac{{{3^{n}}-3}}{8}{\rm{}}}\right\\}.$

Proof: To determine the $(-1)$-differential spectrum of this power function,
we consider the equation

$\Delta_{-1}(x)=(x+1)^{3^{n}-3}+x^{3^{n}-3}=b$ (10)

for $b\in{\mathbb{F}}_{{3^{n}}}$. Firstly we consider $b=1$. Note that $x=0$
and $x=-1$ are both solutions of (10). For $x\neq 0,-1$, we multiply both
sides of (10) by $x^{2}(x+1)^{2}$ and obtain

$x^{4}-x^{3}-x^{2}+x-1=0.$ (11)

Let $y=x-1$, then $y\neq-1,1$, and (11) becomes

$y^{4}-y^{2}-1=0.$ (12)

Next we prove that the solutions of (12) are all in
${\mathbb{F}}_{3^{4}}\setminus{\mathbb{F}}_{3^{2}}$. Obviously, the solutions
of (12) are in ${\mathbb{F}}_{3^{4}}$. If $y\in{\mathbb{F}}_{3^{2}}$ is a
solution of (12), then $y\neq 0$ and $y^{4}=\pm 1$. If $y^{4}=1$
(respectively, $y^{4}=-1$), we can obtain $y^{2}=0$ by (12) (respectively,
$y^{2}=1$), which is a contradiction. By the discussion as above, we conclude
that $\delta_{-1}(1)=6$ when $n=0\left({\bmod 4}\right)$ and
$\delta_{-1}(1)=2$ otherwise.

For $b\neq 1$, we know that $x=0$ and $x=-1$ are not solutions of (10). When
$b=0$, the equation (10) becomes

$x^{2}+x-1=0.$ (13)

It can be similarly proved that the solutions of (13) are in
${\mathbb{F}}_{3^{2}}\setminus{\mathbb{F}}_{3}$. We conclude that
$\delta_{-1}(0)=2$ if $n$ is even, and $\delta_{-1}(0)=0$ if $n$ is odd.

If $b\neq 0,1$, then $x\neq 0,1$. The differential equation (10) can be
deduced as

$bx^{4}-bx^{3}+(b+1)x^{2}+x-1=0.$ (14)

Let $y=x-1$, then (14) becomes

$by^{4}+(b+1)y^{2}+b+1=0.$ (15)

We consider the solutions of equation (15). Let $z=y^{2}$, then we have

$bz^{2}+(b+1)z+b+1=0.$ (16)

Note that (16) is a quadratic equation on $z$, and the discriminant of this
quadratic equation is $\Delta=b+1$. We discuss in the following three cases.

Case (i) $b=-1$. We get $z=0$, then $y=0$ and $x=1$ consequently. So
$\delta_{-1}(-1)=1$.

Case (ii) $\chi\left({b+1}\right)=1$. Herein, $\chi$ denotes the quadratic
character on ${\mathbb{F}}_{3^{n}}$. We know that (16) has two distinct
nonzero solutions, namely $z_{1}$ and $z_{2}$. Now we need to determine
whether $y^{2}=z_{i}$ ($i=1,2$) has solutions over ${\mathbb{F}}_{3^{n}}$. If
$\chi\left({z_{1}}\right)=\chi\left({z_{2}}\right)=1$, then (15) has four
solutions. If $\chi\left({{z_{1}}{z_{2}}}\right)=-1$, (15) has two solutions.
If $\chi\left({z_{1}}\right)=\chi\left({z_{2}}\right)=-1$, (15) has no
solution.

Case (iii) $\chi\left({b+1}\right)=-1$. Obviously, (15) has no solution.

Summarizing the discussion as above, (10) cannot have $3$ or $5$ solutions. We
have, ${}_{-1}\omega_{3}=_{-1}\omega_{5}=0$. Moreover, ${}_{-1}\omega_{1}=1$,
${}_{-1}\omega_{6}=1$ if $n=0\left({\bmod 4}\right)$ and ${}_{-1}\omega_{6}=0$
otherwise. By (2), we know that ${}_{-1}\omega_{0}$, ${}_{-1}\omega_{2}$ and
${}_{-1}\omega_{4}$ satisfy

$\displaystyle\left\\{\begin{array}[]{ll}_{-1}{\omega_{0}}{+_{-1}}{\omega_{2}}{+_{-1}}{\omega_{4}}&={3^{n}}-1{-_{-1}}{\omega_{6}},\\\
{2_{-1}}{\omega_{2}}+{4_{-1}}{\omega_{4}}&={3^{n}}-1-{6_{-1}}{\omega_{6}}.\end{array}\right.$
(19)

Denote by
$N=\\#\\{b\in{\mathbb{F}}_{3^{n}}\setminus{\mathbb{F}}_{3}~{}:~{}\chi(b+1)=1,\chi(b)=-1\\}$.
Then ${}_{-1}\omega_{2}=N+1$ when $n=0,1,3\left({\bmod 4}\right)$,
${}_{-1}\omega_{2}=N+2$ when $n=2\left({\bmod 4}\right)$. By Lemma 3, we have

$\displaystyle N$
$\displaystyle=\frac{1}{4}\sum\limits_{b\in{\mathbb{F}}_{3^{n}}\setminus{\mathbb{F}}_{3}}{(\chi(b+1)+1)}{({1-\chi(b)})}$
$\displaystyle=\frac{1}{4}(\sum\limits_{b\in{\mathbb{F}}_{3^{n}}}{(\chi(b+1)+1)}{({1-\chi(b)})-3+\chi(-1))}$
$\displaystyle=\frac{1}{4}(-\sum\limits_{b\in{\mathbb{F}}_{3^{n}}}\chi(b(b+1))+\sum\limits_{b\in{\mathbb{F}}_{3^{n}}}\chi(b+1)-\sum\limits_{b\in{\mathbb{F}}_{3^{n}}}\chi(b)+\sum\limits_{b\in{\mathbb{F}}_{3^{n}}}1-3+\chi(-1))$
$\displaystyle=\frac{3^{n}-2+\chi(-1)}{4}.$

Then by solving (19), the desired result follows.

In 2021SYZ , the authors studied the $(-1)$-differential uniformity of the
power mapping $x^{\frac{p^{k}+1}{2}}$, where $p$ is an odd prime. In the
following two theorems, we determine the $(-1)$-differential spectrum of
$x^{\frac{p^{k}+1}{2}}$ with some conditions.

######  Theorem 11.

Let $F(x)=x^{\frac{p^{k}+1}{2}}$ be a power function over
${\mathbb{F}}_{p^{n}}$, where $p\equiv 1\left({\bmod~{}4}\right)$,
$\gcd(n,k)=1$, and $\frac{2n}{\gcd(2n,k)}$ is even. The $(-1)$-differential
spectrum of $F(x)$ is given by

$\mathbb{S}=\left\\{{{}_{-1}{\omega_{0}}=\frac{{({p^{n}}+1)(p-1)}}{2(p+1)},{}_{-1}{\omega_{1}}=\frac{{{p^{n}}-3}}{2},{}_{-1}{\omega_{\frac{{p+3}}{4}}}=2,{}_{-1}{\omega_{\frac{{p+1}}{2}}}=\frac{{{p^{n}}-p}}{{p+1}}}\right\\}$

when $n$ is odd, and is given by

$\mathbb{S}=\left\\{{{}_{-1}{\omega_{0}}=\frac{{(p^{n}-1)(p-1)}}{2(p+1)},{}_{-1}{\omega_{1}}=\frac{{{p^{n}}-1}}{2},{}_{-1}{\omega_{\frac{{p+3}}{4}}}=2,{}_{-1}{\omega_{\frac{{p+1}}{2}}}=\frac{{{p^{n}}-p-2}}{{p+1}}}\right\\}$

when $n$ is even.

Proof: Since $\frac{2n}{\gcd(2n,k)}$ is even and $\gcd(n,k)=1$, then $k$ is
odd. We consider the $(-1)$-differential function
$\Delta_{-1}\left(x\right)={\left({x+1}\right)^{d}}+{x^{d}}$ on
${\mathbb{F}}_{p^{n}}$. There exist $\alpha,\beta\in{\mathbb{F}}_{p^{2n}}$
such that $x+1={\alpha^{2}}$, $x={\beta^{2}}$. Let
$\alpha-\beta=\theta\in{\mathbb{F}}_{{p^{2n}}}^{*}$, then
$\alpha+\beta={\theta^{-1}}$, $\alpha=\frac{1}{2}(\theta+{\theta^{-1}})$,
$\beta=-\frac{1}{2}(\theta-{\theta^{-1}})$, and
$x=\frac{1}{4}{\left({\theta-{\theta^{-1}}}\right)^{2}}$. We can obtain that
$\theta^{2(p^{n}+1)}=1$ or $\theta^{2(p^{n}-1)}=1$ since
$x\in{\mathbb{F}}_{p^{n}}$. Note that $x=0$ if and only if $\theta^{2}=1$, and
$x=-1$ if and only if $\theta^{2}=-1$. For $\theta^{2}\neq\pm 1$, we can check
that $\pm\theta,\pm\theta^{-1}$ are pairwise distinct. We mention that for
$x\neq 0,-1$, each $x$ corresponds to four distinct $\theta$’s
$(\pm\theta,\pm\theta^{-1})$. Moreover, we have

$\displaystyle\Delta_{-1}\left(x\right)$
$\displaystyle={\alpha^{{p^{k}}+1}}+{\beta^{{p^{k}}+1}}$
$\displaystyle=\frac{1}{4}{\left({\theta+{\theta^{-1}}}\right)^{{p^{k}}+1}}+\frac{1}{4}{\left({\theta-{\theta^{-1}}}\right)^{{p^{k}}+1}}$
$\displaystyle=\frac{1}{2}({\theta^{{p^{k}}+1}}+{\theta^{-{p^{k}}-1}}).$

Although we can choose different $\theta$,
$\Delta_{-1}\left(x\right)=\frac{1}{2}({\theta^{{p^{k}}+1}}+{\theta^{-{p^{k}}-1}})$
always holds, no matter which $\theta$ is chosen. To determine the
$(-1)$-differential spectrum of this function, we need to investigate the
image of the $\Delta_{-1}(x)$. Let $\mathrm{Im}(\Delta_{-1})\mid_{S}$ denote
the image set of the differential function $\Delta_{-1}(x)$ restricted on some
set $S$, i.e.,
$\mathrm{Im}(\Delta_{-1})\mid_{S}=\\{\Delta_{-1}(x):x=\frac{1}{4}(\theta-\theta^{-1})^{2},\theta\in
S\\}$. We define

${S_{1}}=\left\\{{\theta:{{\theta^{2\left({{p^{n}}+1}\right)}}=1}}\right\\},{S_{2}}=\left\\{{\theta:{{\theta^{2\left({{p^{n}}-1}\right)}}=1}}\right\\}.$

Then we have

${S_{1}}=\left\\{{{\gamma^{i}}:{i=\frac{{{p^{n}}-1}}{2}j,0\leq j\leq
2{p^{n}}+1}}\right\\},{S_{2}}=\left\\{{{\gamma^{i}}:{i=\frac{{{p^{n}}+1}}{2}j,0\leq
j\leq 2{p^{n}}-3}}\right\\},$

where $\gamma$ is a generator of ${\mathbb{F}}_{{p^{2n}}}^{*}$. Define

${C_{i}}=\left\\{{{\theta^{{p^{k}}+1}}:{\theta\in{S_{i}}}}\right\\},i=1,2.$

Then,
${C_{1}}=\left\langle{{\gamma^{\gcd(\frac{{({p^{n}}-1)({p^{k}}+1)}}{2},{p^{2n}}-1)}}}\right\rangle$,
${C_{2}}=\left\langle{{\gamma^{\gcd(\frac{{({p^{n}}+1)({p^{k}}+1)}}{2},{p^{2n}}-1)}}}\right\rangle$
and $C_{1}\cap C_{2}=\\{\pm 1\\}$. Let $\phi$ be a mapping defined on
$C_{1}\cup C_{2}$ with $\phi(u)=\frac{1}{2}(u+u^{-1})$. It is easy to see that
$\phi$ is a $2$-to-$1$ mapping on $(C_{1}\cup C_{2})\setminus\\{\pm 1\\}$, and

$\mathrm{Im}(\phi)\mid_{C_{1}\setminus\\{\pm
1\\}}\cap\mathrm{Im}(\phi)\mid_{C_{2}\setminus\\{\pm 1\\}}=\emptyset.$

We consider the following two cases.

Case 1: $n$ is odd. In this case we have
$\gcd(\frac{{({p^{n}}-1)({p^{k}}+1)}}{2},p^{2n}-1)=\frac{(p^{n}-1)(p+1)}{2}$
and $\gcd(\frac{{({p^{n}}+1)({p^{k}}+1)}}{2},p^{2n}-1)=p^{n}+1$. Then
${C_{1}}=\left\langle{\gamma^{\frac{(p^{n}-1)(p+1)}{2}}}\right\rangle$,
${C_{2}}=\left\langle{\gamma^{p^{n}+1}}\right\rangle$. Then the function
$\theta^{p^{k}+1}$ is $(p+1)$-to-1 on $S_{1}$ and $2$-to-$1$ on $S_{2}$.

For any $b\in\mathrm{Im}(\phi)\mid_{C_{1}\setminus\\{\pm 1\\}}$, there are two
$u$’s in $C_{1}\setminus\\{\pm 1\\}$ such that $\phi(u)=b$. Each $u$
corresponds $(p+1)$ $\theta$’s in $S_{1}$. Hence the $(-1)$-differential
equation $\Delta_{-1}(x)=b$ has $\frac{p+1}{2}$ solutions. The number of such
$b$ is $\frac{p^{n}+1}{p+1}-1$ since $\\#\\{{C_{1}\setminus\\{\pm
1\\}}\\}=2(\frac{p^{n}+1}{p+1}-1)$. For
$b\in\mathrm{Im}(\phi)\mid_{C_{1}\setminus\\{\pm 1\\}}$, we can discuss it
similarly.

For $b=1$ (respectively, $b=-1$), there is a unique $u=1$ (respectively,
$u=-1$) such that $\phi(u)=b$. Each $u$ corresponds $(p+1)$ $\theta$’s in
$S_{1}$ and two $\theta$’s in $S_{2}$, i.e.
$\theta\in\\{\theta^{p+1}=1:\theta\in S_{1}\\}\cup\\{\theta^{2}=1:\theta\in
S_{2}\\}$ (respectively, $\theta\in\\{\theta^{p+1}=-1:\theta\in
S_{1}\\}\cup\\{\theta^{2}=-1:\theta\in S_{2}\\}$). Note that four of them
satisfy $\theta^{2}=1$ (respectively, $\theta^{2}=-1$), the other $(p-1)$
$\theta$’s do not satisfy $\theta^{2}=\pm 1$ since $p\equiv
1\left({\bmod~{}4}\right)$. Based on the corresponding relation of $x$ and
$\theta$, we conclude that $\Delta_{-1}(x)=1$ has $\frac{p-1}{4}+1$ solutions.
Summarizing the discussions as above, we have the following table.

Table 3:  Set $\mathbb{S}$ | $\\#~{}\mathbb{S}$ | $\\#\\{{x\in{{\mathbb{F}}_{{p^{n}}}}:{{\Delta_{-1}}(x)=b}.}\\},b\in\mathbb{S}$
---|---|---
$\mathrm{Im}(\phi)\mid_{C_{1}\setminus\\{\pm 1\\}}$ | $\frac{p^{n}-p}{p+1}$ | $\frac{p+1}{2}$
$\mathrm{Im}(\phi)\mid_{C_{2}\setminus\\{\pm 1\\}}$ | $\frac{p^{n}-3}{2}$ | $1$
$\\{1\\}$ | $1$ | $\frac{p+3}{4}$
$\\{-1\\}$ | $1$ | $\frac{p+3}{4}$
${\mathbb{F}}_{{p^{n}}}\setminus$ the above | $\frac{{({p^{n}}+1)(p-1)}}{2(p+1)}$ | $0$
|  |

Case 2: $n$ is even. In this case we have
$\gcd(\frac{(p^{n}-1)(p^{k}+1)}{2},p^{2n}-1)=p^{n}-1$ and
$\gcd(\frac{(p^{n}+1)(p^{k}+1)}{2},p^{2n}-1)=\frac{(p^{n}+1)(p+1)}{2}$, then
${C_{1}}=\left\langle{\gamma^{p^{n}-1}}\right\rangle$,
${C_{2}}=\left\langle{\gamma^{\frac{(p^{n}+1)(p+1)}{2}}}\right\rangle$. Then
the function $\theta^{p^{k}+1}$ is $2$-to-$1$ on $S_{1}$ and $(p+1)$-to-$1$ on
$S_{2}$. By a similar discussion, we obtain the following table. We finish the
proof.

Table 4:  Set $\mathbb{S}$ | $\\#~{}\mathbb{S}$ | $\\#\\{{x\in{{\mathbb{F}}_{{p^{n}}}}:{{\Delta_{-1}}(x)=b}.}\\},b\in\mathbb{S}$
---|---|---
$\mathrm{Im}(\phi)\mid_{C_{1}\setminus\\{\pm 1\\}}$ | $\frac{p^{n}-1}{2}$ | $1$
$\mathrm{Im}(\phi)\mid_{C_{2}\setminus\\{\pm 1\\}}$ | $\frac{p^{n}-p-2}{p+1}$ | $\frac{p+1}{2}$
$\\{1\\}$ | $1$ | $\frac{p+3}{4}$
$\\{-1\\}$ | $1$ | $\frac{p+3}{4}$
${\mathbb{F}}_{{p^{n}}}\setminus$ the above | $\frac{{({p^{n}}-1)(p-1)}}{2(p+1)}$ | $0$
|  |

######  Theorem 12.

Let $F(x)=x^{\frac{p^{k}+1}{2}}$ be a power function over
${\mathbb{F}}_{p^{n}}$, where $p>7$, $p\equiv 3\left({\bmod~{}4}\right)$,
$\gcd(n,k)=1$, and $\frac{2n}{\gcd(2n,k)}$ is even. If $n$ is odd, the
$(-1)$-differential spectrum of $F(x)$ is

$\begin{gathered}\mathbb{S}=\left\\{{{}_{-1}{\omega_{0}}=\frac{{{p^{n}}(3p-1)-(p+5)}}{{4(p+1)}}{,_{-1}}{\omega_{2}}=\frac{{{p^{n}}-3}}{4}{,_{-1}}{\omega_{\frac{{p+1}}{4}}}=1,}\right.\hfill\\\
~{}~{}~{}~{}~{}~{}{\text{
}}\left.{{}_{-1}{\omega_{\frac{{p+5}}{4}}}=1{,_{-1}}{\omega_{\frac{{p+1}}{2}}}=\frac{{{p^{n}}-p}}{{p+1}}}\right\\}.\hfill\\\
\end{gathered}$

If $n$ is even, the $(-1)$-differential spectrum of $F(x)$ is

$\begin{gathered}S=\left\\{{{}_{-1}{\omega_{0}}=\frac{{({p^{n}}-1)(3p-1)}}{{4(p+1)}}{,_{-1}}{\omega_{2}}=\frac{{{p^{n}}-1}}{4}{,_{-1}}{\omega_{\frac{{p+1}}{4}}}=1,}\right.\hfill\\\
~{}~{}~{}~{}~{}~{}{\text{
}}\left.{{}_{-1}{\omega_{\frac{{p+5}}{4}}}=1{,_{-1}}{\omega_{\frac{{p+1}}{2}}}=\frac{{{p^{n}}-p-2}}{{p+1}}{\text{
}}}\right\\}.\hfill\\\ \end{gathered}$

The proof is similar to that of Theorem 11 and we omit it.

##  5 A new class of AP$c$N power permutations and their $c$-differential
spectra

Very recently, the usual differential properties of the power permutation
$F(x)=x^{\frac{5^{n}-3}{2}}$ over ${\mathbb{F}}_{{5^{n}}}$ were studied in
YHDLCM . In this section, we prove that $F(x)$ is AP$c$N when $c=-1$. The
$(-1)$-differential spectrum of $F(x)$ is also given. First, we investigate
the $(-1)$-differential uniformity of $x^{\frac{5^{n}-3}{2}}$. We introduce
the following lemma. The proof is very similar to that of Lemma $3$ in YHDLCM
and we omit it.

######  Lemma 13.

Let $b\in{{\mathbb{F}}_{{5^{n}}}}\backslash\left\\{{\pm 1}\right\\}$ be a
nonzero nonsquare element. If both of the two quadratic equations
$x^{2}+x-b^{-1}=0$ and $y^{2}+y+b^{-1}=0$ have solutions in
${\mathbb{F}}_{{5^{n}}}$, then the solution $z\in{\mathbb{F}}_{{5^{n}}}$ of
the quadratic equation

$z^{2}+(1-2b^{-1})z-b^{-1}=0$

satisfies $\chi(z(z+1))=-1$.

Based on the above lemma, we have the following theorem.

######  Theorem 14.

Denote by ${}_{-1}\Delta_{F}$ the $(-1)$-differential uniformity of
$F(x)=x^{d}$ over ${\mathbb{F}}_{{5^{n}}}$, where $d=\frac{5^{n}-3}{2}$. We
have ${}_{-1}\Delta_{F}=2$.

Proof: For $b\in{\mathbb{F}}_{{5^{n}}}$, we consider the $(-1)$-differential
equation

$(x+1)^{d}+x^{d}=b$ (20)

over ${\mathbb{F}}_{{5^{n}}}$. It is obvious that when $b=0$, (20) has a
unique solution $x=-\frac{1}{2}$ since $d$ is odd and $\gcd(d,5^{n}-1)=1$. In
the following, we may assume that $b\neq 0$. Recall that $\chi$ denotes the
quadratic multiplicative character of ${\mathbb{F}}_{{5^{n}}}$. For all
$x\in{\mathbb{F}}_{{p^{n}}}^{\\#}$, (20) becomes

$\chi(x+1)(x+1)^{-1}+\chi(x)x^{-1}=b.$ (21)

Depending on the values of $\chi(x)$ and $\chi(x+1)$, we have four quadratic
equations. By solving these four equations, we obtain Table 5.

Table 5: Simplification of (21) in four cases Case | $\chi\left(x\right)$ | $\chi\left(x+1\right)$ | Equation | $x_{1},x_{2}$ | $x_{1}x_{2}$
---|---|---|---|---|---
$\rm I$ | 1 | 1 | $x^{2}+(1-2b^{-1})x-b^{-1}=0$ | $\frac{-(1-2b^{-1})\pm\sqrt{1-b^{-2}}}{2}$ | $-b^{-1}$
$\rm II$ | 1 | -1 | $x^{2}+x-b^{-1}=0$ | $\frac{-1\pm\sqrt{1-b^{-1}}}{2}$ | $-b^{-1}$
$\rm III$ | -1 | 1 | $x^{2}+x+b^{-1}=0$ | $\frac{-1\pm\sqrt{1+b^{-1}}}{2}$ | $-b^{-1}$
$\rm IV$ | $-1$ | $-1$ | $x^{2}+(1+2b^{-1})x+b^{-1}=0$ | $\frac{-(1+2b^{-1})\pm\sqrt{1-b^{-2}}}{2}$ | $b^{-1}$
|  |  |  |  |

Next we discuss the possible solutions of (21) for a given $b\neq\pm 1$. Note
that $\chi(-1)=1$, if $b$ is a square element, then
$\chi(-b^{-1})=\chi(b^{-1})=1$. In both Cases $\rm II$ and $\rm III$, if $x$
is a solution, then $\chi(x(x+1))=-1$. So (21) has no solution in $S_{1,-1}$
and $S_{-1,1}$. Now we consider Cases $\rm I$ and $\rm IV$. Let
$x_{1,2}=\frac{-(1-2b^{-1})\pm\sqrt{1-b^{-2}}}{2}$ and
$x_{3,4}=\frac{-(1+2b^{-1})\pm\sqrt{1-b^{-2}}}{2}$ be solutions of
$x^{2}+(1-2b^{-1})x-b^{-1}=0$ and $x^{2}+(1+2b^{-1})x+b^{-1}=0$, respectively.
Then $x_{3}=-(x_{2}+1)$ and $x_{4}=-(x_{1}+1)$. We assert that the number of
solutions of (21) in Cases I and IV is at most two since
$\chi(x_{1}+1)=\chi(x_{2}+1)=1$ but $\chi(x_{3})=\chi(x_{4})=-1$. Then (21)
has at most two solutions when $b$ is a square element.

If $b$ is a nonsquare element, then $\chi(-b^{-1})=\chi(b^{-1})=-1$. We also
consider the solutions of (21) in each case. In Case $\rm I$, the product of
two solutions of equation $x^{2}+(1-2b^{-1})x-b^{-1}=0$ is $-b^{-1}$, which is
a nonsquare element, this means (21) has at most one solution in $S_{1,1}$.
Similarly, we can prove that (21) has at most one solution in $S_{-1,-1}$. Now
we consider Case $\rm II$. Let $x_{5}$ and $-x_{5}-1$ be the two solutions of
the quadratic equation $x^{2}+x-b^{-1}=0$. It is easy to check that $x_{5}\in
S_{1,-1}$ if and only if $-x_{5}-1\in S_{1,-1}$. This implies that (21) has at
most one solution in $S_{1,-1}$. Similarly, we can prove that (21) has at most
one solution in $S_{-1,1}$. If (21) has solutions in Cases $\rm I$ and $\rm
III$ simultaneously, then the discriminates of quadratic euqations
$x^{2}+(1-2b^{-1})x-b^{-1}=0$ and $x^{2}+x+b^{-1}=0$ are both square elements,
i.e., $\chi(1-b^{-2})=\chi(1+b^{-1})=1$. Then $\chi(1-b^{-1})=1$, which
implies that the equation $x^{2}+x+b^{-1}=0$ has two solutions in
${\mathbb{F}}_{{5^{n}}}$. This contradicts Lemma 13, since the solution $x$ in
Case $\rm I$ satisfies $\chi(x(x+1))=1$. Then $(\ref{p=5simplification})$
cannot have solutions in Cases $\rm I$ and $\rm III$ simultaneously.
Similarly, we can prove that (21) has no solution in Cases $\rm IV$ and $\rm
III$ simultaneously. Since Case $\rm II$ has solutions if and only if Case
$\rm III$ has solutions, we conclude that for
$b\in{{\mathbb{F}}_{{5^{n}}}}\backslash\left\\{{\pm 1}\right\\}$ (20) has
solutions in at most two cases, i.e., $\delta_{-1}(b)\leq 2$.

For $b=1$, one can easy check that $x=0$ is a solution of (20). Then we
consider (21) in the four cases. It is obvious that Cases $\rm II$ and $\rm
III$ has no solution. In Case $\rm I$, (21) becomes $x^{2}-x-1=0$. This
quadratic equation has a unqiue solution $x=-2$. We know that $\chi(-2)=-1$
for odd $n$ and $\chi(-2)=1$ for even $n$. Then Case $\rm I$ has one solution
when $n$ is even and no solution when $n$ is odd. In Case $\rm IV$, (21)
becomes $x^{2}-2x+1=0$. This quadratic equation has a unique solution $x=1$.
Therefore (21) has no solution in $S_{-1,-1}$ when $b=1$. We conclude that
$\delta_{-1}(1)=2$ for even $n$ and $\delta_{-1}(1)=1$ for odd $n$. For
$b=-1$, we can similarly obtain that $\delta_{-1}(-1)=2$ for even $n$ and
$\delta_{-1}(-1)=1$ for odd $n$. The proof is finished.

Next we will determine the $(-1)$-differential spectrum of
$F(x)=x^{\frac{5^{n}-3}{2}}$. It is sufficient to determine ${}_{-1}N_{4}$,
which denotes the number of solutions in $({\mathbb{F}}_{{5^{n}}})^{4}$ of the
equation system (4) when $c=-1$. Moreover, we need the value of the character
sum

${\Gamma_{5,n}}=\sum_{x\in{\mathbb{F}}_{5^{n}}}\chi{(x(x-1)(x+1))},$

where $\chi$ is the quadratic multiplicative character over
${\mathbb{F}}_{{5^{n}}}$. It was proved in YHDLCM  that

${\Gamma_{5,n}}={\left({-1}\right)^{n+1}}\sum\limits_{k=0}^{\left\lfloor{\frac{n}{2}}\right\rfloor}{{{\left({-1}\right)}^{k}}\binom{n}{2k}{2^{2k+1}}},$
(22)

which is always an integer. We have the following lemma on the value of
${}_{-1}N_{4}$, which is the number of the solutions
$(x_{1},x_{2},x_{3},x_{4})\in({\mathbb{F}}_{{5^{n}}})^{4}$ of the equation
system

$\Bigg{\\{}\begin{array}[]{l}{x_{1}}-{x_{2}}+{x_{3}}-{x_{4}}=0\\\
x_{1}^{d}+x_{2}^{d}-x_{3}^{d}-x_{4}^{d}=0\end{array}.$ (23)

######  Lemma 15.

We have

${{}_{-1}N_{4}}=\left\\{\begin{array}[]{l}1+4\left({{5^{n}}-1}\right)+({5^{n}}-1)\left({-\frac{1}{4}{\Gamma_{5,n}}+\frac{{7\cdot{5^{n}}-17}}{4}}\right),~{}\mathrm{if}~{}n~{}\mathrm{is}~{}\mathrm{even},\\\
1+2\left({{5^{n}}-1}\right)+({5^{n}}-1)\left({-\frac{1}{4}{\Gamma_{5,n}}+\frac{{7\cdot{5^{n}}-17}}{4}}\right),~{}\mathrm{if}~{}n~{}\mathrm{is}~{}\mathrm{odd},\end{array}\right.$

where $\Gamma_{5,n}$ was given in (22).

Proof: Since $d$ is odd, the number of solutions
$(x_{1},x_{2},x_{3},x_{4})\in({\mathbb{F}}_{{5^{n}}})^{4}$ of the equation
system

$\left\\{\begin{array}[]{l}{x_{1}}+{x_{2}}+{x_{3}}+{x_{4}}=0\\\
{x^{d}_{1}}-{x^{d}_{2}}-{x^{d}_{3}}+{x^{d}_{4}}=0\end{array}\right.$ (24)

is also ${}_{-1}N_{4}$. For a solution $(x_{1},x_{2},x_{3},x_{4})$ of (24),
first we consider that there exists $x_{i}=0$ for some $0\leq i\leq 4$. It is
easy to see that $(0,0,0,0)$ is a solution of (24), and (24) has no solution
containing only three zeros. If there are only two zeros in
$(x_{1},x_{2},x_{3},x_{4})$, one can get that $(x,0,0,-x)$ and $(0,x,-x,0)$
are solutions of (24), where $x\in{\mathbb{F}}_{{5^{n}}}^{*}$. That is, (24)
has $2(5^{n}-1)$ solutions containing only two zeros. We consider that there
is only one zero in $(x_{1},x_{2},x_{3},x_{4})$. If $x_{4}=0$, then
$x_{1},x_{2}$ and $x_{3}$ are nonzero and they satisfy

$\left\\{\begin{array}[]{l}{x_{1}}+{x_{2}}+{x_{3}}=0\\\
x^{d}_{1}-x^{d}_{2}-x^{d}_{3}=0.\end{array}\right.$

Let $y_{i}=\frac{x_{i}}{x_{3}}$ for $i=1,2$, we have $y_{1}+y_{2}+1=0$ and
${y^{d}_{1}}-{y^{d}_{2}}-1=0$ with $y_{1},y_{2}\neq 0$. Then $y_{2}=-y_{1}-1$
and $(y_{1}+1)^{d}+y^{d}_{1}=1$. By the proof of Theorem 14, we know that
equation $(y_{1}+1)^{d}+y^{d}_{1}=1$ has $\delta_{-1}(1)-1$ nonzero solutions
in ${\mathbb{F}}_{{5^{n}}}$. Similarly, we can determine the number of
solutions of (24) with only $x_{i}=0$, $i=1,2$ and $3$. Then (24) has
$4(5^{n}-1)(\delta_{-1}(1)-1)$ solutions containing only one zero. We conclude
that (24) has $1+(5^{n}-1)(4\delta_{-1}(1)-2)$ solutions containing zeros.

Next we consider $x_{i}\neq 0$ for $1\leq i\leq 4$. Let
$y_{i}=\frac{x_{i}}{x_{4}}$ for $i=1,2,3$. Then $y_{i}\neq 0$ and satisfy

$\left\\{\begin{array}[]{l}{y_{1}}+{y_{2}}+{y_{3}}+1=0\\\
{y^{d}_{1}}-{y^{d}_{2}}-{y^{d}_{3}}+1=0.\end{array}\right.$ (25)

We denote by ${}_{-1}n_{4}$ the number of solutions of (25). It can be seen
that

${}_{-1}N_{4}=1+(5^{n}-1)(4\delta_{-1}(1)-2)+(5^{n}-1){}_{-1}n_{4}.$ (26)

Recall that ${y^{d}_{i}}=\chi(y_{i})y^{-1}_{i}$, we discuss (25) in the
following cases.

Case 1. $y_{1},y_{2}$ and $y_{3}$ are all square elements. Then (25) becomes

$\left\\{\begin{array}[]{l}{y_{1}}+{y_{2}}+{y_{3}}+1=0\\\
{y^{-1}_{1}}-{y^{-1}_{2}}-{y^{-1}_{3}}+1=0.\end{array}\right.$ (27)

We denote by $n_{(1,1,1)}$ the number of solution of (27). Next, we determine
$n_{(1,1,1)}$. If $y_{1}=-1$, then $y_{2}=-y_{3}$, $(-1,y_{2},-y_{2})$ is a
solution of (27) when $y_{2}$ is a square element. If $y_{1}\neq-1$, then
$y_{2}y_{3}=(y_{2}+y_{3})({{y^{-1}_{2}}+{y^{-1}_{3}}})^{-1}=-(y_{1}+1)({{y^{-1}_{1}}+1})^{-1}=-y_{1}$.
Combining with the first equation of (27), we obtain
$y_{3}=\frac{y_{2}+1}{y_{2}-1}$ since $y_{2}\neq 1$. Consequently,
$y_{1}=-\frac{y_{2}(y_{2}+1)}{y_{2}-1}$. If $(y_{1},y_{2},y_{3})$ is a desired
solution, then $\chi(\frac{y_{2}+1}{y_{2}-1})=\chi({y^{2}_{2}}-1)=1$ and
$\chi(y_{2})=1$. The number of such $y_{2}$ is

$\begin{array}[]{l}\frac{1}{4}\sum\limits_{{y_{2}}\neq 0,\pm
1}{({\chi({{y_{2}}})+1})({\chi({{y^{2}_{2}}-1})+1})}\\\
=\frac{1}{4}({5^{n}}-6+{\sum\limits_{{y_{2}}\in{{\mathbb{F}}_{{5^{n}}}}}{\chi({y^{3}_{2}-{y_{2}}})}+\sum\limits_{{y_{2}}\in{{\mathbb{F}}_{{5^{n}}}}}{\chi({{y_{2}}})}+\sum\limits_{{y_{2}}\in{{\mathbb{F}}_{{5^{n}}}}}{\chi({y_{2}^{2}-1})}})\\\
=\frac{1}{4}({{\Gamma_{5,n}}+{5^{n}}-7}).\end{array}$

Note that there may exist $y_{2}$ with $\chi(y_{2})=\chi({y^{2}_{2}}-1)=1$,
such that $y_{1}=-\frac{y_{2}(y_{2}+1)}{y_{2}-1}=-1$, i.e., $y_{2}=\pm 2$. Two
solutions $(-1,2,-2)$ and $(-1,-2,2)$ should be subtracted, this only occurs
when $n$ is even. We obtain

${n_{(1,1,1)}}=\left\\{\begin{array}[]{l}\frac{{3\cdot{5^{n}}-9}}{4}+\frac{1}{4}{\Gamma_{5,n}},~{}~{}\mathrm{if}~{}n~{}\mathrm{is}~{}\mathrm{odd}.\\\
\frac{{3\cdot{5^{n}}-17}}{4}+\frac{1}{4}{\Gamma_{5,n}},~{}\mathrm{if}~{}n~{}\mathrm{is}~{}\mathrm{even}.\end{array}\right.$

Case 2. Two of $y_{1},y_{2}$ and $y_{3}$ are square element, the other one is
a nonsquare element. We first assume that $\chi(y_{1})=\chi(y_{2})=1$ and
$\chi(y_{3})=-1$. Then (25) becomes

$\left\\{\begin{array}[]{l}{y_{1}}+{y_{2}}+{y_{3}}+1=0\\\
{y^{-1}_{1}}-{y^{-1}_{2}}+{y^{-1}_{3}}+1=0.\end{array}\right.$ (28)

We denote by $n_{(1,1,-1)}$ the number of solution of (28). Next, we determine
$n_{(1,1,-1)}$. Note that (28) has no solution when $y_{2}=\pm 1$. If
$y_{2}\neq\pm 1$, we can obtain $y_{1}+y_{3}\neq 0$, moreover,
$y_{1}y_{3}=(y_{1}+y_{3})({{y^{-1}_{1}}+{y^{-1}_{3}}})^{-1}=\left({1+{y_{2}}}\right)\frac{{{y_{2}}}}{{{y_{2}}-1}}=\frac{y_{2}(y_{2}-1)}{y_{2}-1}$.
It is mentioned that $\chi(\frac{y_{2}+1}{y_{2}-1})=-1$ since
$\chi(y_{1})=\chi(y_{2})=1$ and $\chi(y_{3})=-1$. Moreover, $y_{1}$ satisfies
the following quadratic equation

${y^{2}_{1}}+(y_{2}+1)y_{1}+\frac{y_{2}(y_{2}+1)}{y_{2}-1}=0.$ (29)

The discriminate of (29) is
$\Delta=(y_{2}+1)^{2}+\frac{y_{2}(y_{2}+1)}{y_{2}-1}=\frac{(y_{2}+1)(y_{2}-2)^{2}}{y_{2}-1}$.
If (29) has solutions in ${\mathbb{F}}_{{5^{n}}}$, then $y_{2}$ must be $2$
since $\chi(\frac{y_{2}+1}{y_{2}-1})=-1$. Consequently, $y_{1}=y_{3}=1$. Note
that $\chi(y_{3})=-1$, we conclude that $n_{(1,1,-1)}=0$. Next we assume that
$\chi(y_{1})=\chi(y_{3})=1$ and $\chi(y_{2})=-1$. Then (25) becomes

$\left\\{\begin{array}[]{l}{y_{1}}+{y_{2}}+{y_{3}}+1=0\\\
{y^{-1}_{1}}+{y^{-1}_{2}}-{y^{-1}_{3}}+1=0.\end{array}\right.$ (30)

We denote by $n_{(1,-1,1)}$ the number of solution of (30). Similarly, we can
get that $n_{(1,-1,1)}=0$. Next, we assume that $\chi(y_{2})=\chi(y_{3})=1$
and $\chi(y_{1})=-1$. Then (25) becomes

$\left\\{\begin{array}[]{l}{y_{1}}+{y_{2}}+{y_{3}}+1=0\\\
-{y^{-1}_{1}}-{y^{-1}_{2}}-{y^{-1}_{3}}+1=0.\end{array}\right.$ (31)

We denote by $n_{(-1,1,1)}$ the number of solution of (31). Let
$z_{1}=\frac{y_{2}}{y_{3}}$, $z_{2}=\frac{y_{1}}{y_{3}}$ and
$z_{3}=\frac{1}{y_{3}}$. Then $\chi(z_{1})=\chi(z_{3})=1$ and
$\chi(z_{2})=-1$. From (31) we obtain

$\left\\{\begin{array}[]{l}{z_{1}}+{z_{2}}+{z_{3}}+1=0\\\
{z^{-1}_{1}}+{z^{-1}_{2}}-{z^{-1}_{3}}+1=0.\end{array}\right.$ (32)

The number of solutions of (32) is $n_{(1,-1,1)}$, i.e.,
$n_{(-1,1,1)}=n_{(1,-1,1)}=0$.

Case 3. Two of $y_{1},y_{2}$ and $y_{3}$ are nonsquare element, the other one
is a square element. We first assume that $\chi(y_{1})=1$ and
$\chi(y_{2})=\chi(y_{3})=-1$. Then (25) becomes

$\left\\{\begin{array}[]{l}{y_{1}}+{y_{2}}+{y_{3}}+1=0\\\
{y^{-1}_{1}}+{y^{-1}_{2}}+{y^{-1}_{3}}+1=0.\end{array}\right.$ (33)

We denote by $n_{(1,-1,-1)}$ the number of solution of (33). Next, we
determine $n_{(1,-1,-1)}$. Note that $y_{3}\neq-1$, since $\chi(y_{3})=-1$.
Then $y_{1}+y_{2}\neq 0$, moreover,
$y_{1}y_{2}=(y_{1}+y_{2})({y^{-1}_{1}}+{y^{-1}_{2}})^{-1}=(-y_{3}-1)({-y^{-1}_{3}}-1)^{-1}=y_{3}$.
Then we obtain $(y_{1}+1)(y_{2}+1)=0$ from the first equation of (33). If (33)
has solutions in ${\mathbb{F}}_{{5^{n}}}$, then $y_{1}$ must be $-1$ since
$\chi(y_{2})=-1$. Equation system (33) has solutions with types
$(-1,y_{2},-y_{2})$, where $\chi(y_{2})=-1$. Then we have
$n_{(1,-1,-1)}=\frac{5^{n}-1}{2}$.

Then we assume that $\chi(y_{2})=1$ and $\chi(y_{1})=\chi(y_{3})=-1$. Then
(25) becomes

$\left\\{\begin{array}[]{l}{y_{1}}+{y_{2}}+{y_{3}}+1=0\\\
-{y^{-1}_{1}}-{y^{-1}_{2}}+{y^{-1}_{3}}+1=0.\end{array}\right.$ (34)

We denote by $n_{(-1,1,-1)}$ the number of solution of (34). Next, we
determine $n_{(-1,1,-1)}$. It is similar to Case $1$. We obtain
$y_{1}y_{2}=-y_{3}$ and $y_{2}=\frac{y_{1}+1}{y_{1}-1}$ since $y_{1}\neq\pm
1$. Consequently, $y_{3}=-\frac{y_{1}(y_{1}+1)}{y_{1}-1}$. If
$(y_{1},y_{2},y_{3})$ is a desired solution, then
$\chi(\frac{y_{1}+1}{y_{1}-1})=\chi({y^{2}_{1}}-1)=1$ and $\chi(y_{1})=-1$.
The number of such $y_{1}$ is

$\begin{array}[]{l}-\frac{1}{4}\sum\limits_{{y_{1}}\neq 0,\pm
1}{({\chi({{y_{1}}})-1})({\chi({{y^{2}_{1}}-1})+1})}\\\
=-\frac{1}{4}({\sum\limits_{{y_{1}}\in{{\mathbb{F}}_{{5^{n}}}}}{\chi({y_{1}^{3}-{y_{1}}})}+\sum\limits_{{y_{1}}\in{{\mathbb{F}}_{{5^{n}}}}}{\chi({{y_{1}}})}-\sum\limits_{{y_{1}}\in{{\mathbb{F}}_{{5^{n}}}}}{\chi({y_{1}^{2}-1})-{5^{n}}+2}})\\\
=-\frac{1}{4}({{\Gamma_{5,n}}-{5^{n}}+3}).\end{array}$

We obtain $n_{(-1,1,-1)}=-\frac{1}{4}{\Gamma_{5,n}}+\frac{{{5^{n}}-3}}{4}$.
Then we assume that $\chi(y_{3})=1$ and $\chi(y_{1})=\chi(y_{2})=-1$. Then
(25) becomes

$\left\\{\begin{array}[]{l}{y_{1}}+{y_{2}}+{y_{3}}+1=0\\\
{y^{-1}_{1}}-{y^{-1}_{2}}+{y^{-1}_{3}}-1=0.\end{array}\right.$ (35)

We denote by $n_{(-1,-1,1)}$ the number of solution of (35). Similarly, we can
get that $n_{(-1,-1,1)}=-\frac{1}{4}{\Gamma_{5,n}}+\frac{{{5^{n}}-3}}{4}$.

Case 4. $y_{1},y_{2}$ and $y_{3}$ are all nonsquare elements, i.e.,
$\chi(y_{1})=\chi(y_{2})=\chi(y_{3})=-1$. Then (25) becomes

$\left\\{\begin{array}[]{l}{y_{1}}+{y_{2}}+{y_{3}}+1=0\\\
-{y^{-1}_{1}}+{y^{-1}_{2}}+{y^{-1}_{3}}+1=0.\end{array}\right.$ (36)

We denote by $n_{(-1,-1,-1)}$ the number of solution of (36). Next, we
determine $n_{(-1,-1,-1)}$. Note that $y_{1}\neq-1$, since $\chi(y_{1})=-1$.
It is similar to Case $2$. Since $\chi(y_{1})=\chi(y_{2})=\chi(y_{3})=-1$, we
can get that (36) has no solution in ${\mathbb{F}}_{{5^{n}}}$.

By discussions as above, ${{}_{-1}n_{4}}=\sum\limits_{{i,j,k}\in\\{\pm
1\\}}{{n_{\left({i,j,k}\right)}}}$ and then ${}_{-1}N_{{}_{4}}$ follows by
(26) and the value of $\delta_{-1}(1)$.

Then we can determine the $(-1)$-differential spectrum of
$F(x)=x^{\frac{5^{n}-3}{2}}$.

######  Theorem 16.

The power function $F(x)=x^{\frac{5^{n}-3}{2}}$ over ${\mathbb{F}}_{{5^{n}}}$
when $c=-1$ is AP$c$N with $(-1)$-differential spectrum

$\begin{gathered}\mathbb{S}=\left\\{{{}_{-1}{\omega_{0}}=-\frac{1}{8}{\Gamma_{5,n}}+\frac{{3\cdot{5^{n}}-5}}{8}{,_{-1}}{\omega_{1}}=\frac{1}{4}{\Gamma_{5,n}}+\frac{{{5^{n}}+5}}{4},}\right.\hfill\\\
~{}~{}~{}~{}~{}~{}{\text{
}}\left.{{}_{-1}{\omega_{2}}=-\frac{1}{8}{\Gamma_{5,n}}+\frac{{3\cdot{5^{n}}-5}}{8}{\text{
}}}\right\\}.\hfill\\\ \end{gathered}$

when $n$ is even, and with $(-1)$-differential spectrum

$\begin{gathered}\mathbb{S}=\left\\{{{}_{-1}{\omega_{0}}=-\frac{1}{8}{\Gamma_{5,n}}+\frac{{3\cdot{5^{n}}-13}}{8}{,_{-1}}{\omega_{1}}=\frac{1}{4}{\Gamma_{5,n}}+\frac{{{5^{n}}+13}}{4}}\right.,\hfill\\\
~{}~{}~{}~{}~{}~{}{\text{
}}\left.{{}_{-1}{\omega_{2}}=-\frac{1}{8}{\Gamma_{5,n}}+\frac{{3\cdot{5^{n}}-13}}{8}{\text{
}}}\right\\}.\hfill\\\ \end{gathered}$

when $n$ is odd, where $\Gamma_{5,n}$ is determined in (22).

Proof: By (3) and Lemma 15, we obtian that the elements in the
$(-1)$-differential spectrum satisfy

$\sum\limits_{i=0}^{{}_{c}{\Delta_{F}}}{{i^{2}}\cdot{{}_{-1}\omega_{i}}}=\left\\{\begin{array}[]{l}-\frac{1}{4}{\Gamma_{5,n}}+\frac{{7\cdot{5^{n}}-5}}{4},~{}~{}\mathrm{if}~{}n~{}\mathrm{is}~{}\mathrm{even}.\\\
-\frac{1}{4}{\Gamma_{5,n}}+\frac{{7\cdot{5^{n}}-13}}{4},~{}\mathrm{if}~{}n~{}\mathrm{is}~{}\mathrm{odd}.\end{array}\right.$

Then by solving the equation (2) in Lemma 4. The proof is finished.

##  6 concluding remarks

In this paper, we mainly studied the $c$-differential spectra of power
functions over finite fields. Some basic properties of the $c$-differential
spectrum of a power function were given. The $c$-differential spectra of some
classes of power functions were determined. Moreover, we proposed a class of
AP$c$N function over ${\mathbb{F}}_{{5^{n}}}$ with $c=-1$. Our future work is
to find more power functions with low $c$-differential uniformity and to
determine their $c$-differential spectra. It is worth finding applications of
these functions in sequences, coding theory and combinatorial designs.

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