A note on the differential spectrum of the Ness-Helleseth function
Ketong Ren Maosheng Xiong Haode Yan

 	
	
	
	A note on the differential spectrum of the Ness-Helleseth function
	
    Ketong Ren1, Maosheng Xiong2 and Haode Yan 1

	1School of Mathematics, Southwest Jiaotong University, Chengdu, China.
 E-mail: mailto: rkt@my.swjtu.edu.cnrkt@my.swjtu.edu.cn, mailto: hdyan@swjtu.edu.cnhdyan@swjtu.edu.cn

	2Department of Mathematics, The Hong Kong University of Science and Technology, Hong Kong, China.
 E-mail: mamsxiong@ust.hk

	Corresponding author: Haode Yan

	
    
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§ ABSTRACT 
		Let n⩾3 be an odd integer and u an element in the finite field _3^n. The Ness-Helleseth function is the binomial f_u(x)=ux^d_1+x^d_2 over _3^n, where d_1=3^n-1/2-1 and d_2=3^n-2. In 2007, Ness and Helleseth showed that f_u is an APN function when χ(u+1)=χ(u-1)=χ(u), is differentially 3-uniform when χ(u+1)=χ(u-1)≠χ(u), and has differential uniformity at most 4 if χ(u+1)≠χ(u-1) and u∉_3. Here χ(·) denotes the quadratic character on _3^n. Recently, Xia et al. determined the differential uniformity of f_u for all u and computed the differential spectrum of f_u for u satisfying χ(u+1)=χ(u-1) or u∈_3. The remaining problem is the differential spectrum of f_u with χ(u+1)≠χ(u-1) and u∉_3. In this paper, we fill in the gap. By studying differential equations arising from the Ness-Helleseth function f_u more carefully, we express the differential spectrum of f_u for such u in terms of two quadratic character sums. This complements the previous work of Xia et al. 
	
	
	Keywords: cryptographic function; differential uniformity; differential spectrum; character sum
	
	Mathematics Subject Classification: 11T06, 94A60.
	


§ INTRODUCTION

Substitution boxes (S-boxes for short) are crucial in symmetric block ciphers. Cryptographic functions used in S-boxes can be considered as functions defined over finite fields. Let _q be the finite field with q elements, where q is a prime power (i.e. q=p^n and n is a positive integer). We denote by _q^*:=_q ∖{0} the multiplicative cyclic subgroup of _q. Any function F: _q→_q can be uniquely represented as a univariate polynomial of degree less than q. For a cryptographic function F, the main tools to study F regarding the differential attack <cit.> are the difference distribution table (DDT for short) and the differential uniformity introduced by Nyberg <cit.> in 1994. The DDT entry at point (a,b) for any a,b∈_q, denoted by δ_F(a,b), is defined as

    δ_F(a,b)=|{x∈_q|𝔻_aF(x)=b}|,

where 𝔻_a F(x)=F(x+a)-F(x) is the derivative function of F at the element a. The differential uniformity of F, denoted by Δ_F, is defined as

    Δ_F=max{δ_F(a,b)|a∈_q^*, b∈_q}.

Generally speaking, the smaller the value of Δ_F, the stronger the resistance of F used in S-boxes against the differential attack. A cryptographic function F is called differentially k-uniform if Δ_F=k. Particularly when Δ_F=1, F is called a planar function <cit.> or a perfect nonlinear (abbreviated as PN) function <cit.>. When Δ_F=2, F is called an almost perfect nonlinear (abbreviated as APN) function <cit.>, which is of the lowest possible differential uniformity over _2^n as in such finite fields, no PN functions exist. It has been of great research interest to find new functions with low differential uniformity. Readers may refer to <cit.>, <cit.>, <cit.>, <cit.>, <cit.>, <cit.>, <cit.>, <cit.>, <cit.>, <cit.> and references therein for some of the new development.

To investigate further differential properties of nonlinear functions, the concept of differential spectrum was devised as a refinement of differential uniformity <cit.>. 

Let F be a function from _p^n to _p^n with differential uniformity k, and

    ω_i = |{(a,b) ∈_p^n^* ×_p^n | δ_F(a,b) = i}|, 0 ⩽ i ⩽ k.

The differential spectrum of F is defined as the ordered sequence
 
    𝕊 =[ω_0,ω_1,...,ω_k].


According to the Definition <ref>, we have the following two identities

    ∑_i=0^kω_i=(p^n-1)p^n,∑_i=0^kiω_i=(p^n-1)p^n.

The differential spectrum of a cryptographic function, compared with the differential uniformity, provides much more detailed information. In particular, the value distribution of the DDT is given directly by the differential spectrum. Differential spectrum has many applications such as in sequences <cit.>, <cit.>, coding theory <cit.>, <cit.>, combinatorial design <cit.> etc. However, to determine the differential spectrum of a cryptographic function is usually a difficult problem. There are two variables a and b to consider in each ω_i. When F is a power function, i.e., F(x)=x^d for some positive integer d, since δ_F(a,b)=δ_F(1,b/a^d), the problem of the value distribution of {δ_F(a,b)|b∈_q} is the same as that of  {δ_F(1,b)|b∈_q}, so in this case two variables a and b degenerate into one variable b and the problem becomes much easier. Power functions with known differential spectra are summarized in Table <ref>.

	


For a polynomial that is not a power function, the investigation of its differential spectrum is much more difficult. There are very few such functions whose differential spectra were known <cit.>, <cit.>. The main focus of this paper is the Ness-Helleseth function. Let n be a positive odd integer, d_1=3^n-1/2-1, d_2=3^n-2 and u∈_3^n. The Ness-Helleseth function, denoted as f_u(x), is a binomial over _3^n defined as

    f_u(x)=ux^d_1+x^d_2.

To describe the differential properties of the Ness-Helleseth function f_u(x) which obviously depend on u, we define certain sets of u as in <cit.> 

    {[       𝒰_0={u∈_3^n | χ(u+1)≠χ(u-1)},;       𝒰_1={u∈_3^n | χ(u+1)=χ(u-1)},; 𝒰_10={u∈_3^n | χ(u+1)=χ(u-1)≠χ(u)},; 𝒰_11={u∈_3^n | χ(u+1)=χ(u-1)=χ(u)}. ].

Here χ denotes the quadratic character on _3^n^*. It is easy to see that 𝒰_0∩𝒰_1=∅, 𝒰_10∩𝒰_11=∅ and 𝒰_10∪𝒰_11=𝒰_1.

In 2007, Ness and Helleseth showed that (see <cit.>)


  1). f_u is an APN function when u∈𝒰_11; 


  2). f_u is differentially 3-uniform when u∈𝒰_10;  


  3). f_u has differential uniformity at most 4 if u∈𝒰_0∖_3. 

Moreover, Ness and Helleseth observed by numerical computation that in 1), the constraint imposed on u, namely u∈𝒰_11, appears to be necessary for f_u to be an APN function. 

In a recent paper <cit.>, Xia et al. conducted a further investigation into the differential properties of the Ness-Helleseth function f_u. They determined the differential uniformity of f_u for all u ∈_3^n (see <cit.>), hence confirming, in particular, that f_u is indeed APN if and only if u∈𝒰_11. Moreover, for the cases of 1) and 2), they also computed the differential spectrum of f_u explicitly in terms of a quadratic character sum T(u) (see <cit.>). However, for u∈𝒰_0∖_3, while it was shown that f_u has differential uniformity 4, the differential spectrum of f_u remains open. The purpose of this paper is to fill in this gap, that is, in this paper, we will compute the differential spectrum of f_u explicitly for any u∈𝒰_0∖_3 and similar to <cit.>, the result will be expressed in terms of quadratic character sums depending on u. 

Let us make a comparison of the methods used in this paper and in <cit.>. We first remark that for u∈𝒰_0∖_3, to determine the differential uniformity of f_u is already a quite difficult problem, as was shown in <cit.>, the final result involved 32 different quadratic character sums, about one-half of which can not be evaluated easily (see <cit.>). Instead, the authors applied Weil's bound on many of these character sums over finite fields to conclude that the differential uniformity of f_u is 4. While our paper is based on <cit.> and can be considered as a refinement, since we are dealing with the differential spectrum which is a much more difficult problem, it is conceivable that the techniques involved in this paper would be even more complicated. This is indeed the case, as will be seen in the proofs later on. In particular, we have found many relations among these 32 character sums, some of which are quite technical and surprising, that help up in computing the differential spectrum. 




This paper is organized as follows. Section <ref> presents certain quadratic character sums that are essential for the computation of the differential spectrum. In Section <ref>, the necessary and sufficient conditions of the differential equation to have i  (i=0,1,2,3,4) solutions are given. The differential spectrum of f_u is investigated in Section <ref>. Section <ref> concludes this paper.






§ ON QUADRATIC CHARACTER SUMS

In this section, we will introduce some results on the quadratic character sum over finite fields. Let χ(·) be the quadratic character of _p^n (p is an odd prime), which is defined as

    χ(x)=
    {[        1, x _p^n^*,;       -1, x _p^n^*,;        0,      x=0. ].

Let _p^n[x] be the polynomial ring over _p^n. We consider the character sum of the form

    ∑_a∈_p^nχ(f(a))

with f∈_p^n[x]. The case of (f) =1 is trivial, and for (f) =2, the following explicit formula was established in <cit.>.
<cit.>
	Let f(x)=a_2x^2+a_1x+a_0∈_q[x] with q odd and a_2≠ 0. Put d=a_1^2-4a_0a_2 and let χ(·) be the quadratic character of _q. Then
	
    ∑_a∈_qχ(f(a))=
    	{[     -χ(a_2),     if d≠ 0,; (q-1)χ(a_2),      if d=0.;              ].




The character sum plays an important role in determining the differential spectrum of the Ness-Helleseth function. Let p=3 and n be an odd integer. For any fixed u ∈𝒰_0∖_3, define g_i∈_3^n[x] (i∈{1,2,3,4,5}) as follows: 

    {[                                                      g_1(x)=-(u+1)x,;                                                     g_2(x)=x(x-1-u),;                                                     g_3(x)=x(x-1+u),;                       g_4(x)=x^2-x+u^2=(x+1+√(1-u^2))(x+1-√(1-u^2)),; g_5(x)=-φ(u)(x+u^2/φ(u))=-φ(u)(x+1-√(1-u^2)), where φ(u)=1+√(1-u^2).;                                                                      ].

Herein and hereafter, for a square x ∈_3^n^*, we denote by √(x) the square root of x in _3^n such that χ(√(x))=1. Since n is odd, χ(-1)=-1, this √(x) is uniquely determined by x. For u ∈𝒰_0∖_3, the element 1-u^2 is always a square in _3^n, so √(1-u^2) is well defined, and we have χ((u+1)φ(u))=χ(-(u+1+√(1-u^2))^2)=-1. Additionally, let A={0,1± u,-1±√(1-u^2)}. It is easy to note that the set A contains all the zeros of g_i(x), i=1,2,3,4,5. The values of χ(g_i(x)) on A are displayed in the Table <ref>.
	
	
For any u ∈𝒰_0\_3, the following character sums were meticulously computed in <cit.>:	
<cit.>
Let u ∈𝒰_0\_3, we have 

	
  * ∑_z∈_3^nχ(g_1(z)g_2(z))=-1, 
	
  * ∑_z∈_3^nχ(g_1(z)g_3(z))=-1, 
	
  * ∑_z∈_3^nχ(g_1(z)g_5(z))=1, 
	
  * ∑_z∈_3^nχ(g_2(z)g_3(z))=-2.
	
  * ∑_z∈_3^nχ(g_1(z)g_2(z)g_5(z))=2, 
	
  * ∑_z∈_3^nχ(g_1(z)g_3(z)g_5(z))=2.




In what follows, we give a series of lemmas on quadratic character sums involving g_i(x). The first three lemmas can be proved directly.
When u ∈𝒰_0\_3, we have
	
    ∑_z∈_3^nχ(g_4(z)g_5(z))=-χ(φ(u)).


We have,
	
    ∑_z∈_3^nχ(g_4(z)g_5(z))   =-χ(φ(u))∑_z∈_3^nχ((z+1-√(1-u^2))^2(z+1+√(1-u^2))) 
       =-χ(φ(u))∑_z∈_3^n,z≠√(1-u^2)-1χ(z+1+√(1-u^2)) 
       =-χ(φ(u))χ(√(1-u^2))
    		=-χ(φ(u)).


When u ∈𝒰_0\_3, we have
	
    ∑_z∈_3^nχ(g_1(z)g_4(z)g_5(z))=1+χ(φ(u)).



 We have,
	
    ∑_z∈_3^nχ(g_1(z)g_4(z)g_5(z))   =χ(u+1)χ(φ(u))∑_z∈_3^nχ( z(z+1+√(1-u^2))(z+1-√(1-u^2))^2) 
       =-∑_z∈_3^n,z≠√(1-u^2)-1χ( z(z+1+√(1-u^2))) 
       =-(-1-χ((√(1-u^2)-1)(-√(1-u^2))))
       =1-χ(√(1-u^2)-1)
    		=1+χ(φ(u)).



When u ∈𝒰_0\_3, we have
	
    ∑_z∈_3^nχ(g_1(z)g_3(z)g_4(z)g_5(z))=2-χ((√(1-u^2)+1+u)),

	and
	
    ∑_z∈_3^nχ(g_1(z)g_2(z)g_4(z)g_5(z))=2-χ((√(1-u^2)+1-u)).


We only prove the first identity, as the proof of the second is very similar. 
	
    ∑_z∈_3^nχ(g_1(z)g_3(z)g_4(z)g_5(z))=   χ((u+1)φ(u))∑_z∈_3^nχ((z-1+u)(z+1+√(1-u^2))z^2(z+1-√(1-u^2))^2)
    
    		=   -∑_z∈_3^n^*,z≠√(1-u^2)-1χ((z-1+u)(z+1+√(1-u^2)))
    
    		=   -∑_z∈_3^nχ((z-1+u)(z+1+√(1-u^2)))
       +χ((u-1)φ(u))+χ((√(1-u^2)+1+u)(-√(1-u^2)))
    
    		=   1-χ((u+1)φ(u))-χ((√(1-u^2)+1+u))
    
    		=   2-χ((√(1-u^2)+1+u)).




When u ∈𝒰_0\_3, we have
	
    ∑_z∈_3^nχ(g_2(z)g_3(z)g_4(z))=-2.



We have,
	
    ∑_z∈_3^nχ(g_2(z)g_3(z)g_4(z))    =∑_z∈_3^nχ(z^2(z-1+u)(z-1-u)(z^2-z+u^2)) 
       =∑_z∈_3^n^*χ((z-1+u)(z-1-u)(z^2-z+u^2)) 
       =∑_z∈_3^n^*, z≠ 1± uχ(z^2-z+u^2/z^2+z+1-u^2) 
       =∑_z∈_3^n, z≠ 1 ± uχ(z^2-z+u^2/z^2+z+1-u^2)-χ(u^2/1-u^2).

	Let t=z^2-z+u^2/z^2+z+1-u^2. Then (t-1)z^2+(t+1)z+t(1-u^2)-u^2=0. We know that t=1 if and only if z=1+u^2. When t≠ 1, the discriminant of the quadratic equation on z is Δ_t=(t+1)^2-(t-1)(t(1-u^2)-u^2)=u^2t^2+1-u^2. The number of z with a fixed t≠ 1 is 1+χ(Δ_t). Hence, we have
	
    ∑_z∈_3^nχ(g_2(z)g_3(z)g_4(z))    =∑_z∈_3^n, z≠ 1 ± uχ(z^2-z+u^2/z^2+z+1-u^2)-χ(u^2/1-u^2) 
       =∑_t∈_3^n, t≠ 1χ(t)(1+χ(Δ_t))-χ(u^2/1-u^2) 
       =∑_t∈_3^nχ(t)(1+χ(Δ_t))-χ(1)-χ(u^2/1-u^2) 
       =∑_t∈_3^nχ(t)+∑_t∈_3^nχ(t(u^2t^2+1-u^2))-χ(1)-χ(u^2/1-u^2).

	Note that ∑_t∈_3^nχ(t(u^2t^2+1-u^2))=∑_t∈_3^nχ(-t(u^2t^2+1-u^2))=-∑_t∈_3^nχ(t(u^2t^2+1-u^2)), then ∑_t∈_3^nχ(t(u^2t^2+1-u^2))=0. This with χ(u^2/1-u^2)=1 leads to ∑_z∈_3^nχ(g_2(z)g_3(z)g_4(z))=-2.


When u ∈𝒰_0\_3, we have
	
    ∑_z∈_3^nχ(g_1(z)g_4(z))+∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z))=0.


First,
	
    ∑_z∈_3^nχ(g_1(z)g_4(z))=-χ(u+1)∑_z∈_3^nχ(z(z^2-z+u^2)),

	and
	
    ∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z))    =-χ(u+1)∑_z∈_3^nχ(z^3(z-(u+1))(z+(u-1)))
       =-χ(u+1)∑_z∈_3^n^*χ(z(z^2+z+1-u^2)).

	Note that
	 
    ∑_z∈_3^n^*χ(z(z^2+z+1-u^2))=-∑_z∈_3^n^*χ(z(z^2-z+1-u^2))=-∑_z∈_3^n^*χ(z^2-z+1-u^2/z).

	Let z^2-z+1-u^2/z=t. Then t satisfies the quadratic equation 
    z^2-(t+1)z+1-u^2=0.
 Clearly, z=0 is not the solution of this quadratic equation for any t∈_3^n since u∉_3. For each t, the number of solutions of z is 1+χ(Δ_t), where Δ_t=(t+1)^2-(1-u^2)=t^2-t+u^2. Hence
	 
    ∑_z∈_3^n^*χ(z^2-z+1-u^2/z)=∑_t∈_3^nχ(t)(1+χ(Δ_t))=∑_t∈_3^nχ(tΔ_t)=∑_t∈_3^nχ(t(t^2-t+u^2)).

	The desired result follows.


When u ∈𝒰_0\_3, we have
	
    ∑_z∈_3^nχ(g_2(z)g_4(z))+∑_z∈_3^nχ(g_1(z)g_2(z)g_4(z))=-2,

	and
		
    ∑_z∈_3^nχ(g_3(z)g_4(z))+∑_z∈_3^nχ(g_1(z)g_3(z)g_4(z))=-2.


We only prove the first identity. The proof of the second one is similar, so we omit it.
	Note that
	
    ∑_z∈_3^nχ(g_2(z)g_4(z))=∑_z∈_3^nχ(z(z-1-u)(z^2-z+u^2)),

	and
	 
    ∑_z∈_3^nχ(g_1(z)g_2(z)g_4(z))=-χ(u+1)∑_z∈_3^nχ(z^2(z-1-u)(z^2-z+u^2)).

	We have
	
    ∑_z∈_3^nχ(g_1(z)g_2(z)g_4(z))    =-χ(u+1)∑_z∈_3^n^*χ((z-1-u)(z^2-z+u^2)) 
       =-χ(u+1)∑_z∈_3^n, z -1-uχ(z((z+1+u)^2-(z+1+u)+u^2)) 
       =-χ(u+1)∑_z∈_3^n, z -1-uχ(z(z^2-(u-1)z-(u^2-u))) 
       =-χ(u+1)(∑_z∈_3^nχ(z(z^2-(u-1)z-(u^2-u))-χ(-(u+1)u^2))) 
       =-1-χ(u+1)∑_z∈_3^nχ(z(z^2-(u-1)z-(u^2-u))),

	and 	
	 
    ∑_z∈_3^nχ(g_2(z)g_4(z))=∑_z∈_3^nχ(z(z-1-u)(z^2-z+u^2))=∑_z∈_3^n^*, z u+1χ(z^2-z+u^2/z(z-1-u)).

	Let z^2-z+u^2/z(z-1-u)=t. Then t satisfies
	
    (t-1)z^2+(1-(u+1)t)z-u^2=0.

	We know that t=1 if and only if z=-u. When t≠ 1, the discriminant of the quadratic equation on z is Δ_t=(1-(u+1)t)^2+(t-1)u^2=(u+1)^2t^2+(u-1)^2t+1-u^2. The number of z with a fixed t≠ 1 is 1+χ(Δ_t). Hence, we have
	
    ∑_z∈_3^nχ(g_2(z)g_4(z))   =∑_z∈_3^n^*, z u+1χ(z^2-z+u^2/z(z-1-u)) 
       =1+∑_t∈_3^n, t 1χ(t)(1+χ(Δ_t)) 
       =1+∑_t∈_3^nχ(t)(1+χ(Δ_t))-(1+χ(u^2)) 
       =∑_t∈_3^nχ(t)(1+χ(Δ_t))-1.

	Note that ∑_t∈_3^nχ(tΔ_t)=∑_t∈_3^nχ(t((u+1)^2t^2+(u-1)^2t+1-u^2))=∑_t∈_3^n^*χ((u+1)^2t^2+(u-1)^2t+1-u^2/t). Let v=(u+1)^2t^2+(u-1)^2t+1-u^2/t. Then
	
    (u+1)^2t^2+((u-1)^2-v)t+(1-u^2)=0,

	which is a quadratic equation on t. Δ_v=((u-1)^2-v)^2-(u+1)^2(1-u^2)=v^2+(u-1)^2v-(u-1)u^3.
	Then
	
    ∑_t∈_3^nχ(tΔ_t)   =∑_v∈_3^nχ(v(1+χ(Δ_v)))
       =∑_v∈_3^nχ(v(v^2+(u-1)^2v-(u-1)u^3)) 
       =∑_w∈_3^nχ((u-1)^2w((u-1)^4w^2+(u-1)^4w-(u-1)u^3)) 
       =∑_w∈_3^nχ(w(w^2+w-u^3/(u-1)^3)) 
       =∑_w∈_3^nχ(w^1/3((w^1/3)^2+w^1/3-u^3/(u-1)^3)) 
       =∑_w∈_3^nχ(w(w^2+w-u/u-1)) 
       =∑_w∈_3^nχ(w/u-1((w/u-1)^2+w/u-1-u/u-1)) 
       =χ(u-1)∑_w∈_3^nχ(w(w^2+(u-1)w-(u^2-u))) 
       =-χ(u-1)∑_w∈_3^nχ(w(w^2-(u-1)w-(u^2-u))).

	We conclude that
	 
    ∑_z∈_3^nχ(g_1(z)g_2(z)g_4(z))=-1-χ(u+1)∑_z∈_3^nχ(z(z^2-(u-1)z-(u^2-u))),

	and
	 
    ∑_z∈_3^nχ(g_2(z)g_4(z))=-1-χ(u-1)∑_w∈_3^nχ(w(w^2-(u-1)w-(u^2-u))).

	Then we have
	
    ∑_z∈_3^nχ(g_2(z)g_4(z))+∑_z∈_3^nχ(g_1(z)g_2(z)g_4(z))=-2.


When u ∈𝒰_0\_3, we have
	
    ∑_z∈_3^nχ(g_2(z)g_3(z)g_5(z))+∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z)g_5(z))=2.



Note that
	
    ∑_z∈_3^nχ(g_2(z)g_3(z)g_5(z))
       =-χ(φ(u))∑_z∈_3^nχ(z^2(z-1-u)(z-1+u)(z+1-√(1-u^2))) 
       =-χ(φ(u))∑_z∈_3^n^*χ((z-1-u)(z-1+u)(z+1-√(1-u^2))) 
       =-χ(φ(u))(-χ((-1-u)(-1+u)(1-√(1-u^2)))+∑_z∈_3^nχ((z-1-u)(z-1+u)(z+1-√(1-u^2))))
       =1-χ(φ(u))∑_z∈_3^nχ((z-1-u)(z-1+u)(z+1-√(1-u^2))),

		and
	
    ∑_z∈_3^nχ((z-1-u)(z-1+u)(z+1-√(1-u^2))) 
    
    		=   ∑_z∈_3^nχ((z+√(1-u^2)+1-u)(z+√(1-u^2)+1+u)z) 
    
    		=   ∑_z∈_3^nχ(z(z^2-(√(1-u^2)+1)z+(u^2-1-√(1-u^2)))) 
    
    		 =   χ(√(1-u^2)+1)∑_z∈_3^nχ(z(z^2-z+-u^2+1-√(1-u^2)/u^2)).

	Then
	 
    ∑_z∈_3^nχ(g_2(z)g_3(z)g_5(z))=1-∑_z∈_3^nχ(z(z^2-z+-u^2+1-√(1-u^2)/u^2)).
			 
	Moreover,
	
    ∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z)g_5(z)) 
    
    		 =   χ(u+1)χ(φ(u))∑_z∈_3^nχ(z^3(z-1-u)(z-1+u)(z+1-√(1-u^2))) 
    
    		=   -∑_z∈_3^nχ(z(z-1-u)(z-1+u)(z+1-√(1-u^2))) 
    
    		 =   -∑_t∈_3^n^*χ(1/t(1/t-1-u)(1/t-1+u)(1/t+1-√(1-u^2))) 
    
    		=   -∑_t∈_3^n^*χ((1-(1+u)t)(1-(1-u)t)(1+(1-√(1-u^2))t)) 
    
    		=   1-∑_t∈_3^nχ((1-(1+u)t)(1-(1-u)t)(1+(1-√(1-u^2))t)) 
    
    		 =   1-χ((1+u)(1-u)(1-√(1-u^2)))∑_t∈_3^nχ((1/1+u-t)(1/1-u-t)(1/1-√(1-u^2)+t)) 
    
    		 =   1-χ(1-√(1-u^2))∑_t∈_3^nχ((t-1/1+u)(t-1/1-u)(t+1/1-√(1-u^2))) 
    
    		 =   1-χ(1-√(1-u^2))∑_t∈_3^nχ((t-1/1-√(1-u^2)-1/1+u)(t-1/1-√(1-u^2)-1/1-u)t) 
    
    		 =   1-χ(1-√(1-u^2))∑_t∈_3^nχ(t(t^2+1+(1-u^2)√(1-u^2)/u^2(1-u^2)t-u^4+u^2+1+√(1-u^2)/u^4(1-u^2))) 
    
    		=   1-χ(1-√(1-u^2))χ(1+(1-u^2)√(1-u^2)/u^2(1-u^2))
       ·∑_t∈_3^nχ(t(t^2+t-u^4+u^2+1+√(1-u^2)/u^4(1-u^2)/(1+(1-u^2)√(1-u^2)/u^2(1-u^2))^2))
    
    		 =   1-∑_t∈_3^nχ(t(t^2+t-u^4+u^2+1+√(1-u^2)/u^4(1-u^2)/(1+(1-u^2)√(1-u^2)/u^2(1-u^2))^2))
    
    		=   1-∑_t∈_3^nχ(t(t^2+t+-(u^2-1)^3-(√(1-u^2))^3/u^6))
    
    		=   1-∑_t∈_3^nχ(t(t^2+t+-u^2+1-√(1-u^2)/u^2))
    
    		=   1+∑_t∈_3^nχ(t(t^2-t+-u^2+1-√(1-u^2)/u^2)).

	The desired result follows.



When u ∈𝒰_0\_3, we have
	
    ∑_z∈_3^nχ(g_2(z)g_3(z)g_4(z)g_5(z))+∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z)g_4(z)g_5(z))=2.



We have
	
    ∑_z∈_3^nχ(g_2(z)g_3(z)g_4(z)g_5(z)) 
    
    		 =   -χ(φ(u))∑_z∈_3^nχ(z^2(z-1-u)(z-1+u)(z+1+√(1-u^2))(z+1-√(1-u^2))^2) 
    
    		=   -χ(φ(u))∑_z∈_3^n, z 0, z√(1-u^2)-1χ((z-1-u)(z-1+u)(z+1+√(1-u^2))) 
    
    		 =   -χ(φ(u))(-χ((-1-u)(-1+u)(1+√(1-u^2)))-χ((√(1-u^2)+1-u)(√(1-u^2)+1+u)(-√(1-u^2))) 
       +∑_z∈_3^nχ((z-1-u)(z-1+u)(z+1+√(1-u^2))))
    
    		=   2-χ(φ(u))∑_z∈_3^nχ((z-1-u)(z-1+u)(z+1+√(1-u^2))).

	Note that
	
    ∑_z∈_3^nχ((z-1-u)(z-1+u)(z+1+√(1-u^2))) 
    
    		=   ∑_z∈_3^nχ(z(z+1-√(1-u^2)-u)(z+1-√(1-u^2)+u))
    
    		=   ∑_z∈_3^nχ(z(z^2+(√(1-u^2)-1)z+(u^2-1+√(1-u^2))))
    
    		 =   χ(√(1-u^2)-1)∑_z∈_3^nχ(z(z^2+z+1-u^2+√(1-u^2)/u^2)).

	
	Then
	 
    ∑_z∈_3^nχ(g_2(z)g_3(z)g_4(z)g_5(z))=2+∑_z∈_3^nχ(z(z^2+z+1-u^2+√(1-u^2)/u^2)).

	Moreover,
	
    ∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z)g_4(z)g_5(z))
       =χ(φ(u))χ(u+1)∑_z∈_3^nχ(z^3(z-1-u)(z-1+u)(z+1+√(1-u^2))(z+1-√(1-u^2))^2)
       =-∑_z∈_3^n,z√(1-u^2)-1χ(z(z-1-u)(z-1+u)(z+1+√(1-u^2))) 
       =-(-χ((√(1-u^2)-1)(√(1-u^2)+1-u)(√(1-u^2)+1+u)(-√(1-u^2))) 
         +∑_z∈_3^nχ(z(z-1-u)(z-1+u)(z+1+√(1-u^2)))) 
       =-1-∑_z∈_3^nχ(z(z-1-u)(z-1+u)(z+1+√(1-u^2))).

	
	Note that
	
    ∑_z∈_3^nχ(z(z-1-u)(z-1+u)(z+1+√(1-u^2))) 
    
    		 =   ∑_t∈_3^n^*χ(1/t(1/t-1-u)(1/t-1+u)(1/t+1+√(1-u^2)))
    
    		=   ∑_t∈_3^n^*χ((1-(1+u)t)(1-(1-u)t)(1+(1+√(1-u^2))t))
    
    		=   -1+∑_t∈_3^nχ((1-(1+u)t)(1-(1-u)t)(1+(1+√(1-u^2))t))
    
    		 =   -1+χ(1+u)χ(1-u)χ(1+√(1-u^2))∑_t∈_3^nχ((t-1/1+u)(t-1/1-u)(t+1/1+√(1-u^2)))
    
    		 =   -1+χ(1+√(1-u^2))∑_t∈_3^nχ(t(t-1/1+√(1-u^2)-1/1+u)(t-1/1+√(1-u^2)-1/1-u))
    
    		 =   -1+χ(1+√(1-u^2))∑_t∈_3^nχ(t(t^2+1-(1-u^2)√(1-u^2)/u^2(1-u^2)t+-u^4-u^2-1+√(1-u^2)/u^4(1-u^2)))
    
    		=   -1+χ(1+√(1-u^2))χ(1-(1-u^2)√(1-u^2)/u^2(1-u^2))
       ·∑_t∈_3^nχ(t(t^2+t+-u^4-u^2-1+√(1-u^2)/u^4(1-u^2)/(1-(1-u^2)√(1-u^2)/u^2(1-u^2))^2))
    
    		=   -1+∑_t∈_3^nχ(t(t^2+t+(√(1-u^2)/1-√(1-u^2))^3))
    
    		=   -1+∑_t∈_3^nχ(t(t^2+t+√(1-u^2)/1-√(1-u^2)))
    
    		=   -1+∑_t∈_3^nχ(t(t^2+t+1-u^2+√(1-u^2)/u^2)).

	Then
	 
    ∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z)g_4(z)g_5(z))=-∑_t∈_3^nχ(t(t^2+t+1-u^2+√(1-u^2)/u^2)).

	The desired result follows.



When u ∈𝒰_0\_3, we have
	
    ∑_z∈_3^nχ(g_2(z)g_5(z))+∑_z∈_3^nχ(g_3(z)g_4(z)g_5(z))=χ(√(1-u^2)+1+u),

	and
		
    ∑_z∈_3^nχ(g_3(z)g_5(z))+∑_z∈_3^nχ(g_2(z)g_4(z)g_5(z))=χ(√(1-u^2)+1-u).


 We only prove the first identity, as the proof of the second is very similar. 
	
    ∑_z∈_3^nχ(g_2(z)g_5(z))=-χ(φ(u))∑_z∈_3^nχ(z(z-1-u)(z+1-√(1-u^2))).

	Note that
	
    ∑_z∈_3^nχ(z(z-1-u)(z+1-√(1-u^2)))
    
    		=   ∑_z∈_3^nχ(z(z^2-(u+√(1-u^2))z-(u+1)(1-√(1-u^2))))
    
    		=   ∑_z∈_3^n^*χ(z^2-(u+√(1-u^2))z-(u+1)(1-√(1-u^2))/z).

	Let t=z^2-(u+√(1-u^2))z-(u+1)(1-√(1-u^2))/z. Then
	
    z^2-(t+u+√(1-u^2))z-(u+1)(1-√(1-u^2))=0,

	and Δ_t=(t+u+√(1-u^2))^2+(u+1)(1-√(1-u^2))=t^2-(u+√(1-u^2))t+(u-1)(1+√(1-u^2)). 

	We have
	
    ∑_z∈_3^nχ(z(z-1-u)(z+1-√(1-u^2)))
    
    		=   ∑_t∈_3^nχ(t)(1+χ(Δ_t))
    
    		=   ∑_t∈_3^nχ(t(t^2-(u+√(1-u^2))t+(u-1)(1+√(1-u^2))))
    
    		=   -∑_t∈_3^nχ(t(t^2+(u+√(1-u^2))t+(u-1)(1+√(1-u^2)))),

	then

    ∑_z∈_3^nχ(g_2(z)g_5(z))=χ(φ(u))∑_t∈_3^nχ(t(t^2+(u+√(1-u^2))t+(u-1)(1+√(1-u^2)))).

	On the other hand,
	
    ∑_z∈_3^nχ(g_3(z)g_4(z)g_5(z))
       =-χ(φ(u))∑_z∈_3^nχ(z(z-1+u)(z+1+√(1-u^2))(z+1-√(1-u^2))^2)
       =-χ(φ(u))∑_z∈_3^n,z≠√(1-u^2)-1χ(z(z-1+u)(z+1+√(1-u^2)))
       =-χ(φ(u))(-χ((√(1-u^2)-1)(√(1-u^2)+1+u)(-√(1-u^2)))
         +∑_z∈_3^nχ(z(z-1+u)(z+1+√(1-u^2))))
       =χ(√(1-u^2)+1+u)-χ(φ(u))∑_z∈_3^nχ(z(z^2+(u+√(1-u^2))z+(u-1)(1+√(1-u^2)))).

	Hence, the first identity ensues. 






§ ON THE NUMBER OF SOLUTIONS OF THE DIFFERENTIAL EQUATION OF F_U

Let n be a positive odd integer, u∈_3^n. Recall that the Ness-Helleseth function is defined as

    f_u(x)=ux^d_1+x^d_2,

where d_1=3^n-1/2-1 and d_2=3^n-2. To determine the differential spectrum of f_u(x), attention should be given to the differential equation

    𝔻_af_u(x)=u(x+a)^3^n-1/2-1+(x+a)^3^n-2-ux^3^n-1/2-1-x^3^n-2=b,

where (a,b)∈_3^n^*×_3^n. This equation was studied in <cit.>. For the sake of completeness, we give some details here. 

We denote by N(a,b), N_1(a,b) and N_2(a,b) the numbers of solutions of (<ref>) in the sets _3^n, {0,-a} and _3^n\{0,-a} respectively. Then 
    N(a,b)=N_1(a,b)+N_2(a,b).

The following lemma is given in  <cit.>.
<cit.>
	The value of N_1(a,b) is determined as follows:
	
    N_1(a,b)=
    	{[                            2,            if b=a^-1 and u=0,;                            1, if b=a^-1(1± uχ(a)) and u  0,;                            0,                    otherwise. ].

	When b = 0, the value of N_2(a,b) is given as
	
    N_2(a,0)=
    	{[        3^n-3/4,     if u∈{± 1},;              0, if u ∈𝒰_0\{±1}.;                 ].


What needs to be calculated is N_2(a,b) for b∈_3^n^*. When x∉{0, -a}, the differential equation is equivalent to

    u(x+a)^3^n-1/2x+x-ux^3^n-1/2(x+a)-(x+a)=bx(x+a),

which can be simplified as

    bx^2+(ba-u(τ_a-τ_0))x+a(uτ_0+1)=0,

where τ_a=χ(x+a) and τ_0=χ(x).
The discussion of the solutions of the quadratic equation above when b≠0 has been clarified by Helleseth in <cit.> and results are listed in Table <ref>, in which x_1 and x_2 denote the two solutions of the quadratic equations in each case.



Drawing upon the information in Table <ref>, the subsequent pivotal results have been unveiled. Note that the term a desired solution refers to a solution of a certain quadratic equation in any case in Table <ref> that indeed satisfies the corresponding condition on (τ_a,τ_0). In the rest of this paper, we always assume that u ∈𝒰_0 \_3={u∈_3^n∖_3 | χ(u+1)≠χ(u-1)}, then χ(1-u^2)=1 and (a,b) ∈_3^n^* ×_3^n^*. 
For the sake of brevity and clarity, for such fixed u and (a,b), we denote by N_I (respectively N_II, N_III and N_IV) the number of desired solutions in Case I (respectively, Case II, Case III and Case IV). Consequently, N_2(a,b)=N_I+N_II+N_III+N_IV. We discuss the values of N_I, N_II, N_III and N_IV as follows. It was proved in <cit.> that N_I≤ 1 and N_IV≤ 1. Moreover, the following proposition was proposed.
(<cit.>) We have,
	
		
  * N_I=1 if and only if 
    χ(1-u+1/ab)=1 and χ(u+1/ab)=-1.

		
  * N_IV=1 if and only if 
    χ(1+u-1/ab)=1 and χ(u+1/ab)=-1.

	

Since χ(u+1/ab)≠ 0, the following corollary can be deduced immediately.
 We have,
	
		
  * N_I=0 if one of the subsequent three disjoint conditions is met:
		
			
  * χ(1-u+1/ab)=0.
			
  * χ(1-u+1/ab)=-1.
			
  * χ(1-u+1/ab)=1 and χ(u+1/ab)=1.
		
		
  * N_IV=0 if one of the subsequent three disjoint conditions is met:
		
			
  * χ(1+u-1/ab)=0.
			
  * χ(1+u-1/ab)=-1.
			
  * χ(1+u-1/ab)=1 and χ(u+1/ab)=1.
		
	



As has been demonstrated in <cit.>, if x is a solution of the quadratic equation in Case II, then -(x+a) is a solution of the quadratic equation in Case III, and vice versa. Besides, x and -(x+a) cannot be desired solutions simultaneously. Therefore, it can be concluded that N_II+N_III≤ 2. More specifically, the following proposition showed the sufficient and necessary condition of  N_II+N_III=2.
(<cit.>)We have,
	 N_II+N_III=2 if and only if
	
    {[      χ(u^2+a^2b^2-ab)=1,; χ(-u^2-ab-ab√(1-u^2))=1.;                          ].


Next, we specifically consider the case when N_II+N_III=1. We have the following proposition.
We have,
	N_II+N_III=1 if and only if
		
    {[ χ(u^2+a^2b^2-ab)=0,;    χ(a^2b^2-u^2)=1.;                     ].


 The sufficiency is obvious. We only prove the necessity. When χ(u^2+a^2b^2-ab)=1, the quadratic equation in Case II has two solutions, namely x_1 and x_2. Then the solutions of the quadratic equation in Case III are -x_1-a and -x_2-a. Note that x_1 (x_2, respectively) is a desired solution if and only if -x_2-a (-x_1-a, respectively) is a desired solution. Then N_II+N_III≠ 1 when  χ(u^2+a^2b^2-ab)=1. Moreover, when  χ(u^2+a^2b^2-ab)=-1, N_II+N_III=0. We conclude that if N_II+N_III=1, then χ(u^2+a^2b^2-ab)=0.
	
When χ(u^2+a^2b^2-ab)=0, let x_0 be the unique solution of the quadratic equation in Case II, then
x_0=u+ab/b. Moreover, the unique solution of the quadratic equation in Case III is x'_0=-u-ab/b. If x_0 is a desired solution, then χ(x_0)=χ(u+ab/b)=-1 and χ(x_0+a)=χ(u-ab/b)=1. If x'_0 is a desired solution, then χ(x'_0)=χ(-u-ab/b)=1 and χ(x'_0+a)=χ(-u+ab/b)=-1. Obviously, x_0 and x'_0 cannot be desired solutions simultaneously. If N_II+N_III=1, then χ(u+ab/b)χ(u-ab/b)=-1, i.e., χ(u^2-a^2b^2)=-1. The proof is completed.

Note that χ(-u^2-ab-ab√(1-u^2))=0 implies that χ(u^2+a^2b^2-ab)=0. Moreover, χ(u^2+a^2b^2-ab)=0 and χ(a^2b^2-u^2)=0 cannot hold simultaneously for  u ∈𝒰_0 \_3. Then we have the following corollary.
 N_II+N_III=0 if one of the subsequent three disjoint conditions is met:
	
		
  * χ(u^2+a^2b^2-ab)=-1.
		
  * χ(u^2+a^2b^2-ab)=0,χ(a^2b^2-u^2)=-1.
		
  * χ(u^2+a^2b^2-ab)=1 and χ(-u^2-ab-ab√(1-u^2))=-1.
	


When χ(u^2+a^2b^2-ab)=0, then ab=-1±√(1-u^2), which implies that χ(u+1/ab)=-χ((u+1)φ(u))=1. Then we conclude that N_I=N_IV=0 when N_II+N_III=1. Moreover, by propositions and corollaries demonstrated previously in this section, the discussion on the value of N_2(a,b)=N_I+N_II+N_III+N_IV is finished. Recall that N(a,b)=N_1(a,b)+N_2(a,b). By Lemma <ref>, For u∈𝒰_0 \_3, N_1(a,b)=1 or 0. When N_1(a,b)=1, then ab=1± u, the conditions in Proposition <ref> cannot hold, hence N_I=N_IV=0. Moreover, for u∈𝒰_0 \_3, if ab=1± u, then u^2+a^2b^2-ab≠ 0. We conclude that N_II+N_III≠1 when N_1(a,b)=1. We summarize the above discussion in the following Table <ref>.


	

By Table <ref>, we obtain the following sufficient and necessary conditions about the numbers of solutions of the differential equation (<ref>). We mention that the sufficient and necessary condition for (<ref>) to have 4 solutions was given in <cit.>.

<cit.>
	When u ∈𝒰_0 \_3, the differential equation 𝔻_af_u(x)=b has four solutions if and only if (a,b) satisfies the following conditions
	
    {[            χ(u+1/ab)=-1,;           χ(1-u+1/ab)=1,;           χ(1-u-1/ab)=1,;      χ(u^2+a^2b^2-ab)=1,; χ(-u^2-ab-ab√(1-u^2))=1.;                          ].

	


	When u ∈𝒰_0 \_3, the differential equation 𝔻_af_u(x)=b of the function f_u(x) has three solutions if and only if (a,b) satisfies one of the following conditions
	
		
  * ab=1± u,χ(u^2+a^2b^2-ab)=1, χ(-u^2-ab-ab√(1-u^2))=1.
		
  * χ(1-u+1/ab)=1, χ(a(u+1)/b)=-1, χ(1+u-1/ab)=-1, χ(u^2+a^2b^2-ab)=1, χ(-u^2-ab-ab√(1-u^2))=1.
		
  * χ(1-u+1/ab)=-1, χ(a(u+1)/b)=-1, χ(1+u-1/ab)=1, χ(u^2+a^2b^2-ab)=1, χ(-u^2-ab-ab√(1-u^2))=1.
	
	


	When u ∈𝒰_0 \_3, the differential equation 𝔻_af_u(x)=b of the function f_u(x) has two solutions if and only if (a,b) satisfies one of the following conditions
	
		
  * χ(1-u+1/ab)=1, χ(a(u+1)/b)=-1, χ(1+u-1/ab)=1, χ(u^2+a^2b^2-ab)=-1.
		
  * χ(1-u+1/ab)=1, χ(a(u+1)/b)=-1, χ(1+u-1/ab)=1, χ(u^2+a^2b^2-ab)=1,χ(-u^2-ab-ab√(1-u^2))=-1.
		
  * χ(1-u+1/ab)=-1,χ(1+u-1/ab)=-1,χ(u^2+a^2b^2-ab)=1,χ(-u^2-ab-ab√(1-u^2))=1.
		
  * χ(1-u+1/ab)=-1,χ(1+u-1/ab)=1,χ(a(u+1)/b)=1,χ(u^2+a^2b^2-ab)=1,χ(-u^2-ab-ab√(1-u^2))=1.
		
  * χ(1-u+1/ab)=1,χ(a(u+1)/b)=1,χ(1+u-1/ab)=-1,χ(u^2+a^2b^2-ab)=1,χ(-u^2-ab-ab√(1-u^2))=1.
		
  * χ(1-u+1/ab)=1,χ(a(u+1)/b)=1,χ(1+u-1/ab)=1,χ(u^2+a^2b^2-ab)=1,χ(-u^2-ab-ab√(1-u^2))=1.
	
	


	When u ∈𝒰_0 \_3, the differential equation 𝔻_af_u(x)=b of the function f_u(x) has one solution if and only if (a,b) satisfies one of the following conditions
	
		
  * ab=1± u,χ(u^2+a^2b^2-ab)=-1.
		
  * ab=1± u,χ(u^2+a^2b^2-ab)=1,χ(-u^2-ab-ab√(1-u^2))=-1.
		
  * χ(u^2+a^2b^2-ab)=0,χ(a^2b^2-u^2)=1.
		
  * χ(1-u+1/ab)=1, χ(a(u+1)/b)=-1, χ(1+u-1/ab)=-1, χ(u^2+a^2b^2-ab)=-1.
		
  * χ(1-u+1/ab)=-1, χ(a(u+1)/b)=-1, χ(1+u-1/ab)=1, χ(u^2+a^2b^2-ab)=-1.
		
  * χ(1-u+1/ab)=1, χ(a(u+1)/b)=-1, χ(1+u-1/ab)=-1, χ(u^2+a^2b^2-ab)=1, χ(-u^2-ab-ab√(1-u^2))=-1.
		
  * χ(1-u+1/ab)=-1, χ(a(u+1)/b)=-1, χ(1+u-1/ab)=1, χ(u^2+a^2b^2-ab)=1, χ(-u^2-ab-ab√(1-u^2))=-1.
	
	


	When u ∈𝒰_0 \_3, the differential equation 𝔻_af_u(x)=b of the function f_u(x) has no solution if and only if (a,b) satisfies one of the following conditions
	
		
  * b=0.
		
  * χ(u^2+a^2b^2-ab)=0,χ(a^2b^2-u^2)=-1.
		
  * χ(1-u+1/ab)=-1, χ(1+u-1/ab)=-1, χ(u^2+a^2b^2-ab)=-1.
		
  * χ(1-u+1/ab)=-1, χ(1+u-1/ab)=1,χ(a(u+1)/b)=1,χ(u^2+a^2b^2-ab)=-1.
		
  * χ(1-u+1/ab)=1, χ(a(u+1)/b)=1, χ(1+u-1/ab)=-1, χ(u^2+a^2b^2-ab)=-1.
		
  * χ(1-u+1/ab)=1, χ(a(u+1)/b)=1, χ(1+u-1/ab)=1, χ(u^2+a^2b^2-ab)=-1.
		
  * χ(1-u+1/ab)=-1, χ(1+u-1/ab)=-1, χ(u^2+a^2b^2-ab)=1, χ(-u^2-ab-ab√(1-u^2))=-1.
		
  * χ(1-u+1/ab)=-1, χ(1+u-1/ab)=1,χ(a(u+1)/b)=1,χ(u^2+a^2b^2-ab)=1, χ(-u^2-ab-ab√(1-u^2))=-1.
		
  * χ(1-u+1/ab)=1, χ(a(u+1)/b)=1, χ(1+u-1/ab)=-1, χ(u^2+a^2b^2-ab)=1,χ(-u^2-ab-ab√(1-u^2))=-1.
		
  * χ(1-u+1/ab)=1, χ(a(u+1)/b)=1, χ(1+u-1/ab)=1, χ(u^2+a^2b^2-ab)=1,χ(-u^2-ab-ab√(1-u^2))=-1.
	
	





§ THE DIFFERENTIAL SPECTRUM OF F_U WHEN Χ(U+1)≠Χ(U-1)

Recall that ω_i=|{(a,b)∈_p^n^*×_p^n|δ_F(a,b)=i}|,0⩽ i⩽Δ_F, where δ_F(a,b) denotes the number of solutions to the differential equation 𝔻_aF=b. We are ready to investigate the differential spectrum of f_u.
As a prerequisite, we define two quadratic character sums, namely Γ_3 and Γ_4, as enumerated below.

    Γ_3    =∑_z∈_3^nχ(g_1(z)g_4(z))=-χ(u+1)∑_z∈_3^nχ(z^3-z^2+u^2z). 
    Γ_4    =∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z)g_4(z))=-χ(u+1)∑_z∈_3^nχ(z^5-(u^2+1)z^2+(u^2-u^4)z).

These two character sums will be used in the differential spectrum of f_u. The main result of this paper is given as follows.

	Let n≥3 be an odd integer and f_u(x)=ux^d_1+x^d_2 be the Ness-Helleseth function over _3^n with d_1=3^n-1/2-1 and d_2=3^n-2. Then, when u ∈𝒰_0∖_3, the differential spectrum of f_u is given by
	
    [
    			ω_0   =(3^n-1)(-1+ε+1/32(5·3^n+1-17-Γ_4)),
    ω_1   =(3^n-1)(3-ε+1/16(3^n+1+3+2Γ_3+Γ_4)),
    ω_2   =(3^n-1)(-ε+1/4(3^n-7-Γ_3)),
    ω_3   =(3^n-1)(ε+1/16(3^n+1+2Γ_3-Γ_4)), 
    ω_4   =(3^n-1)/32(3^n+1+Γ_4)
    		],

	where
	
    ε   =
    	{[                                          1, χ(u)=χ(u+1),χ((u+1)√(1-u^2)+(u-1)^2)=-1,or,;                                                χ(u)=χ(u-1),χ((1-u)√(1-u^2)+(u+1)^2)=-1;;                                          0,                                  otherwise.;                                             ].

	



The proof of Theorem <ref> will be divided into five parts, where in each part ω_i (for i∈{0,1,2,3,4}) will be calculated.


  * Proof of ω_4. The sufficient and necessary condition for (<ref>) to have 4 solutions was shown in Proposition <ref>. Let ab=z. For each z∈_3^n^*, there are 3^n-1 pairs of (a,b) such that ab=z. Further we have, 
    ω_4=(3^n-1)n_4,
 where n_4 denotes the number of z satisfying the following system.
	
    {[ χ(g_1(z))=1,; χ(g_2(z))=1,; χ(g_3(z))=1,; χ(g_4(z))=1,; χ(g_5(z))=1,;              ].
 where g_i(i=1,2,3,4,5) are defined previously. Then by character sum n_4 can be expressed as
	
    n_4=1/32∑_z∈_3^n\ A(1+χ(g_1(z)))(1+χ(g_2(z)))(1+χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z))).

	By Table <ref>,
	 
    ∑_z∈ A(1+χ(g_1(z)))(1+χ(g_2(z)))(1+χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z)))=0.

	By the lemmas in Section II, it follows that
	
    n_4   =1/32∑_z∈_3^n(1+χ(g_1(z)))(1+χ(g_2(z)))(1+χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z))) 
       =1/32(3^n+1+∑_z∈_3^n(g_1(z)g_2(z)g_3(z)g_4(z))-χ(u+1)χ(φ(u)-1)-χ(φ(u))χ(φ(u)-1)) 
       =1/32(3^n+1+Γ_4).

	The last identity holds since χ(u+1)χ(φ(u))=-1 and χ(u+1)+χ(φ(u))=0. Then the value of ω_4 follows.

  * Proof of ω_3.
The sufficient and necessary condition for (<ref>) to have 3 solutions was shown in Proposition <ref>. Let ab=z. For each z∈_3^n^*, there are 3^n-1 pairs of (a,b) such that ab=z. Further we have 
    ω_3=(3^n-1)(n_3,1+n_3,2+n_3,3),
 where the definitions of n_3,1 , n_3,2 and n_3,3 will be detailed below. 

Let n_3,1 denote the number of z satisfying
	
    {[      z=1± u,; χ(g_4(z))=1,; χ(g_5(z))=1.;              ].

Then we get
	
    n_3,1=
    	{[                                          1, χ(u)=χ(u+1),χ((u+1)√(1-u^2)+(u-1)^2)=-1,or,;                                                χ(u)=χ(u-1),χ((1-u)√(1-u^2)+(u+1)^2)=-1;;                                          0,                                  otherwise.;                                             ].

Let n_3,2, n_3,3 denote the number of z satisfying the following two equation systems respectively:
	
    {[  χ(g_1(z))=1,;  χ(g_2(z))=1,; χ(g_3(z))=-1,;  χ(g_4(z))=1,;  χ(g_5(z))=1,;               ].     {[  χ(g_1(z))=1,; χ(g_2(z))=-1,;  χ(g_3(z))=1,;  χ(g_4(z))=1,;  χ(g_5(z))=1,;               ].
 where g_i(i=1,2,3,4,5) are defined previously. Then by character sum, n_3,1 can be expressed as
	
    32n_3,2=∑_z∈_3^n\ A(1+χ(g_1(z)))(1+χ(g_2(z)))(1-χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z))).

	By Table <ref>,  
    ∑_z∈ A(1+χ(g_1(z)))(1+χ(g_2(z)))(1-χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z)))=0.

	It follows that
	
    32n_3,2=∑_z∈_3^n(1+χ(g_1(z)))(1+χ(g_2(z)))(1-χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z))).

	Similarly, it can be concluded that
	
    32n_3,3=∑_z∈_3^n(1+χ(g_1(z)))(1-χ(g_2(z)))(1+χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z))).

	By utilizing the lemmas presented in Section II, the following sum can be derived
	
    n_3,1+n_3,2+n_3,3
    
    	=   ε+1/16[3^n+1+2∑_z∈_3^nχ(g_1(z)g_4(z))-∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z)g_4(z))
       -χ(φ(u))χ(φ(u)-1)-χ(u+1)χ(φ(u)-1)] 
    
    	 =   ε+1/16[3^n+1+2Γ_3-Γ_4-χ(φ(u))χ(φ(u)-1)-χ(u+1)χ(φ(u)-1)] 
    
    	=   ε+1/16(3^n+1+2Γ_3-Γ_4),

	where ϵ was defined in (<ref>).
	


 
 

 




  * Proof of ω_2.
	The sufficient and necessary condition for (<ref>) to have 2 solutions was shown in Proposition <ref>.  Let ab=z. For each z∈_3^n^*, there are 3^n-1 pairs of (a,b) such that ab=z. Further we have 
    ω_2=(3^n-1)(n_2,1+n_2,2+n_2,3+n_2,4+n_2,5+n_2,6),
 where n_2,1,n_2,2,n_2,3,n_2,4,n_2,5,n_2,6 denote the number of z satisfying the following six equation systems respectively:
	
    {[  χ(g_1(z))=1,;  χ(g_2(z))=1,;  χ(g_3(z))=1,; χ(g_4(z))=-1,;               ].     {[ χ(g_2(z))=-1,; χ(g_3(z))=-1,;  χ(g_4(z))=1,;  χ(g_5(z))=1,;               ].     {[  χ(g_1(z))=1,;  χ(g_2(z))=1,;  χ(g_3(z))=1,;  χ(g_4(z))=1,; χ(g_5(z))=-1,;               ].

    {[ χ(g_1(z))=-1,; χ(g_2(z))=-1,;  χ(g_3(z))=1,;  χ(g_4(z))=1,;  χ(g_5(z))=1,;               ].     {[ χ(g_1(z))=-1,;  χ(g_2(z))=1,; χ(g_3(z))=-1,;  χ(g_4(z))=1,;  χ(g_5(z))=1,;               ].     {[ χ(g_1(z))=-1,;  χ(g_2(z))=1,;  χ(g_3(z))=1,;  χ(g_4(z))=1,;  χ(g_5(z))=1,;               ].
 where g_i(i=1,2,3,4,5) are defined previously.
	Then by character sum, n_21 can be expressed as
	
    16n_2,1=∑_z∈_3^n\ A(1+χ(g_1(z)))(1+χ(g_2(z)))(1+χ(g_3(z)))(1-χ(g_4(z))).

	By Table <ref>, 
    ∑_z∈ A(1+χ(g_1(z)))(1+χ(g_2(z)))(1+χ(g_3(z)))(1-χ(g_4(z)))=0.

	It follows that
	
    16n_2,1=∑_z∈_3^n(1+χ(g_1(z)))(1+χ(g_2(z)))(1+χ(g_3(z)))(1-χ(g_4(z))).

	Similarly, it can be concluded that
	
    16n_2,2   =∑_z∈_3^n\ {1± u,-1±√(1-u^2)}(1-χ(g_2(z)))(1-χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z))).
    
    	32n_2,3   =∑_z∈_3^n(1+χ(g_1(z)))(1+χ(g_2(z)))(1+χ(g_3(z)))(1+χ(g_4(z)))(1-χ(g_5(z)))-4.
    
    	32n_2,4   =∑_z∈_3^n\ A(1-χ(g_1(z)))(1-χ(g_2(z)))(1+χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z))).
    
    	32n_2,5   =∑_z∈_3^n\ A(1-χ(g_1(z)))(1+χ(g_2(z)))(1-χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z))).
    
    	32n_2,6   =∑_z∈_3^n\ A(1-χ(g_1(z)))(1+χ(g_2(z)))(1+χ(g_3(z)))(1+χ(g_4(z)))(1+χ(g_5(z))).

	By utilizing the lemmas presented in Section II, the following sum can be derived
	
    n_2,1+n_2,2+n_2,3+n_2,4+n_2,5+n_2,6
    
    	=   1/8[2·3^n-14-2∑_z∈_3^nχ(g_1(z)g_4(z))-χ(φ(u))χ(φ(u)-1)+χ(u+1)χ(φ(u)-1) 
       -2(1+χ(u-u^2))(1-χ((u+1)(√(1-u^2))+(u-1)^2))-2χ(u^2-1-√(1-u^2)) 
       -2(1-χ(u^2+u))(1-χ((1-u)(√(1-u^2))+(u+1)^2))] 
    
    	 =   1/8[2·3^n-14-2Γ_3-χ(φ(u))χ(φ(u)-1)+χ(u+1)χ(φ(u)-1) 
       -2(1+χ(u-u^2))(1-χ((u+1)(√(1-u^2))+(u-1)^2))-2χ(u^2-1-√(1-u^2)) 
       -2(1-χ(u^2+u))(1-χ((1-u)(√(1-u^2))+(u+1)^2))] 
    
    	 =   1/4(3^n-7-Γ_3+χ(u+1)-(1-χ(u^2+u))(1-χ((1-u)(√(1-u^2))+(u+1)^2))
       -(1+χ(u-u^2))(1-χ((u+1)(√(1-u^2))+(u-1)^2))-χ(u^2-1-√(1-u^2)))
    
    	=   1/4(3^n-7-4ε-Γ_3+χ(u+1)-χ(u^2-1-√(1-u^2)))
    
    	=   -ε+1/4(3^n-7-Γ_3),

	
	where ε has been defined in (<ref>). The last identity holds since
 χ(u^2-1-√(1-u^2))=χ(√(1-u^2)(-√(1-u^2)-1))=-χ(φ(u)) and χ(u+1)χ(φ(u))=-1. 

  * Proof of ω_1.
	The sufficient and necessary condition for (<ref>) to have 1 solution was shown in Proposition <ref>. Let ab=z. For each z∈_3^n^*, there are 3^n-1 pairs of (a,b) such that ab=z. Further we have 
    ω_1=(3^n-1)(n_1,1+n_1,2+n_1,3+n_1,4+n_1,5+n_1,6+n_1,7),
 where the definitions of n_1,1, n_1,2, n_1,3, n_1,4, n_1,5, n_1,6 and n_1,7 will be detailed below. 

	Let n_1,1, n_1,2, n_1,3 denote the number of z satisfying the following two equation systems respectively:
	
    {[       z=1± u,; χ(g_4(z))=-1,;               ].     {[       z=1± u,;  χ(g_4(z))=1,; χ(g_5(z))=-1,;               ].     {[ χ(g_4(z))=0,; χ(g_5(z))=1.;              ].

Then we get

    n_1,1   =
    	{[                          1, χ(u)=χ(u-1) or χ(u)=χ(u+1);;                          0,                  otherwise.;                             ].
    
    	n_1,2   =
    	{[                                        1, χ(u)=χ(u+1),χ((u+1)√(1-u^2)+(u-1)^2)=1,or;                                             χ(u)=χ(u-1),χ((1-u)√(1-u^2)+(u+1)^2)=1;;                                        0,                                otherwise.;                                           ].
    
    	n_1,3   =
    	{[                                          1, χ(u^2-1+√(1-u^2))=1 or χ(u^2-1-√(1-u^2))=1;;                                          0,                                  otherwise.;                                             ].

	Note that either of χ(u)=χ(u-1) or χ(u)=χ(u+1) must hold since χ(u-1)≠χ(u+1). Similarly, χ(u^2-1+√(1-u^2))≠χ(u^2-1-√(1-u^2)) since (u^2-1+√(1-u^2))(u^2-1-√(1-u^2))=u^2(u^2-1), which is a nonsquare. It follows that either of χ(u^2-1+√(1-u^2))=1 or χ(u^2-1-√(1-u^2))=1 must hold since χ(u^2-1+√(1-u^2))≠(u^2-1-√(1-u^2)) and neither of them could be 0. Then we can conclude that n_1,1=1 and n_1,3=1. 

	Let n_1,4, n_1,5, n_1,6, n_1,7 denote the number of z satisfying the following four equation systems respectively:
	
    {[  χ(g_1(z))=1,;  χ(g_2(z))=1,; χ(g_3(z))=-1,; χ(g_4(z))=-1,;               ].     {[  χ(g_1(z))=1,; χ(g_2(z))=-1,;  χ(g_3(z))=1,; χ(g_4(z))=-1,;               ].     {[  χ(g_1(z))=1,;  χ(g_2(z))=1,; χ(g_3(z))=-1,;  χ(g_4(z))=1,; χ(g_5(z))=-1,;               ].      {[  χ(g_1(z))=1,; χ(g_2(z))=-1,;  χ(g_3(z))=1,;  χ(g_4(z))=1,; χ(g_5(z))=-1,;               ].
 where g_i(i=1,2,3,4,5) are defined previously.
	Then by character sum, n_1,3 can be expressed as
	
    16n_1,4=∑_z∈_3^n\ A(1+χ(g_1(z)))(1+χ(g_2(z)))(1-χ(g_3(z)))(1-χ(g_4(z))).

	By Table <ref>, 
    ∑_z∈ A(1+χ(g_1(z)))(1+χ(g_2(z)))(1-χ(g_3(z)))(1-χ(g_4(z)))=0.

	It follows that
	
    16n_1,4=∑_z∈_3^n(1+χ(g_1(z)))(1+χ(g_2(z)))(1-χ(g_3(z)))(1-χ(g_4(z))).

	Similarly, it can be concluded that
	
    16n_1,5   =∑_z∈_3^n(1+χ(g_1(z)))(1-χ(g_2(z)))(1+χ(g_3(z)))(1-χ(g_4(z))).
    
    	 32n_1,6   =∑_z∈_3^n(1+χ(g_1(z)))(1+χ(g_2(z)))(1-χ(g_3(z)))(1+χ(g_4(z)))(1-χ(g_5(z)))-4.
    
    	 32n_1,7   =∑_z∈_3^n(1+χ(g_1(z)))(1-χ(g_2(z)))(1+χ(g_3(z)))(1+χ(g_4(z)))(1-χ(g_5(z)))-4.

	By utilizing the lemmas presented in Section II, the following sum can be derived
	
    n_1,1+n_1,2+n_1,3+n_1,4+n_1,5+n_1,6+n_1,7
    
    	 =   2+(1-ε)+1/16[3·3^n+3+2∑_z∈_3^nχ(g_1(z)g_4(z))+∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z)g_4(z))
        +χ(φ(u))χ(φ(u)-1)+χ(u+1)χ(φ(u)-1)]
    
    	 =   3-ε+1/16[3·3^n+3+2Γ_3+Γ_4+χ(φ(u))χ(φ(u)-1)+χ(u+1)χ(φ(u)-1)] 
    
    	=   3-ε+1/16(3^n+1+3+2Γ_3+Γ_4),

	where ε has been defined in (<ref>).

  * Proof of ω_0.
	The sufficient and necessary condition for (<ref>) to have no solution was shown in Proposition <ref>. Let ab=z. For each z∈_3^n^*, there are 3^n-1 pairs of (a,b) such that ab=z. Further we have 
    ω_0=(3^n-1)(n_0,1+n_0,2+n_0,3+n_0,4+n_0,5+n_0,6+n_0,7+n_0,8+n_0,9+n_0,10),
 where n_0,1=1 for the condition z=0 and the definitions of n_0,2, n_0,3, n_0,4, n_0,5, n_0,6, n_0,7, n_0,8, n_0,9 and n_0,10 will be detailed below. 

	Let n_0,2 denote the number of z satisfying
	
    {[   χ(g_4(z))=0,; χ(z^2-u^2)=-1,;                ].

	then
	
    n_0,2=
    	{[                                            1, χ(u^2-1+√(1-u^2))=-1 or χ(u^2-1-√(1-u^2))=-1;;                                            0,                                    otherwise.;                                               ].

	Note that χ(u^2-1+√(1-u^2))≠χ(u^2-1-√(1-u^2)) since (u^2-1+√(1-u^2))(u^2-1-√(1-u^2))=u^2(u^2-1), which is a nonsquare. It follows that either of χ(u^2-1+√(1-u^2))=-1 or χ(u^2-1-√(1-u^2))=-1 must hold. Then we can conclude that n_0,2=1. 

	Let n_0,3, n_0,4, n_0,5, n_0,6, n_0,7, n_0,8, n_0,9, n_0,10 denote the number of z satisfying the following eight equation systems respectively:
	
    {[ χ(g_2(z))=-1,; χ(g_3(z))=-1,; χ(g_4(z))=-1,;               ].     {[ χ(g_1(z))=-1,; χ(g_2(z))=-1,;  χ(g_3(z))=1,; χ(g_4(z))=-1,;               ].     {[ χ(g_1(z))=-1,;  χ(g_2(z))=1,; χ(g_3(z))=-1,; χ(g_4(z))=-1,;               ].     {[ χ(g_1(z))=-1,;  χ(g_2(z))=1,;  χ(g_3(z))=1,; χ(g_4(z))=-1,;               ].

    {[ χ(g_2(z))=-1,; χ(g_3(z))=-1,;  χ(g_4(z))=1,; χ(g_5(z))=-1,;               ].     {[ χ(g_1(z))=-1,; χ(g_2(z))=-1,;  χ(g_3(z))=1,;  χ(g_4(z))=1,; χ(g_5(z))=-1,;               ].     {[ χ(g_1(z))=-1,;  χ(g_2(z))=1,; χ(g_3(z))=-1,;  χ(g_4(z))=1,; χ(g_5(z))=-1,;               ].     {[ χ(g_1(z))=-1,;  χ(g_2(z))=1,;  χ(g_3(z))=1,;  χ(g_4(z))=1,; χ(g_5(z))=-1,;               ].
 where g_i(i=1,2,3,4,5) are defined previously.
	Then by character sum, n_0,3 can be expressed as
	
    8n_0,3=∑_z∈_3^n\ A(1-χ(g_2(z)))(1-χ(g_3))(1-χ(g_4(z))).

	Similarly, it can be concluded that
	
    16n_0,4   =∑_z∈_3^n\ A(1-χ(g_1(z)))(1-χ(g_2(z)))(1+χ(g_3(z)))(1-χ(g_4(z))).
    
    	16n_0,5   =∑_z∈_3^n\ A(1-χ(g_1(z)))(1+χ(g_2(z)))(1-χ(g_3(z)))(1-χ(g_4(z))).
    
    	16n_0,6   =∑_z∈_3^n\ A(1-χ(g_1(z)))(1+χ(g_2(z)))(1+χ(g_3(z)))(1-χ(g_4(z))).
    
    	16n_0,7   =∑_z∈_3^n\ A(1-χ(g_2(z)))(1-χ(g_3(z)))(1+χ(g_4(z)))(1-χ(g_5(z))).
    
    	32n_0,8   =∑_z∈_3^n\ A(1-χ(g_1(z)))(1-χ(g_2(z)))(1+χ(g_3(z)))(1+χ(g_4(z)))(1-χ(g_5(z))).
    
    	32n_0,9   =∑_z∈_3^n\ A(1-χ(g_1(z)))(1+χ(g_2(z)))(1-χ(g_3(z)))(1+χ(g_4(z)))(1-χ(g_5(z))).
    
    	32n_0,10   =∑_z∈_3^n\ A(1-χ(g_1(z)))(1+χ(g_2(z)))(1+χ(g_3(z)))(1+χ(g_4(z)))(1-χ(g_5(z))).

	By utilizing Table <ref> and the lemmas  presented in Section II, the following sum can be derived
	
    n_0,1+n_0,2+n_0,3+n_0,4+n_0,5+n_0,6+n_0,7+n_0,8+n_0,9+n_0,10
    
    	=   2+1/32[15·3^n-81-∑_z∈_3^nχ(g_1(z)g_2(z)g_3(z)g_4(z))+5χ(φ(u))χ(φ(u)-1)-3χ(u+1)χ(φ-1)
       +8χ(u^2-1-√(1-u^2))-8(1+χ(u-u^2))(1+χ((u+1)(√(1-u^2))+(u-1)^2)) 
       -8(1-χ(u^2+u))(1+χ((1-u)(√(1-u^2))+(u+1)^2))] 
    
    	 =   2+1/32[15·3^n-81-Γ_4+5χ(φ(u))χ(φ(u)-1)-3χ(u+1)χ(φ-1)
       +8χ(u^2-1-√(1-u^2))-8(1+χ(u-u^2))(1+χ((u+1)(√(1-u^2))+(u-1)^2)) 
       -8(1-χ(u^2+u))(1+χ((1-u)(√(1-u^2))+(u+1)^2))] 
    
    	 =   (2+1/32(15·3^n-81-Γ_4-8χ(u+1)-8(1-χ(u^2+u))(1+χ((1-u)(√(1-u^2))+(u+1)^2))
       +8χ(u^2-1-√(1-u^2))-8(1+χ(u-u^2))(1+χ((u+1)(√(1-u^2))+(u-1)^2))))
    
    	 =   (2+1/32(5·3^n+1-81-32(1-ε)-Γ_4-8χ(u+1)+8χ(u^2-1-√(1-u^2)))) 
    
    	 =   -1+ε+1/32(5·3^n+1-17-Γ_4),

	where ε has beem defined in (<ref>).

This completes the proof of Theorem <ref>.



	Recall that the elements ω_i (i=0,1,2,3,4) satisfy two identities  in (<ref>). Namely,
		
    {[ ω_0+ω_1+ω_2+ω_3+ω_4                   =         (3^n-1)3^n,;  ω_1+2ω_2+3ω_3+4ω_4                   =         (3^n-1)3^n. ].

	After the values of ω_4, ω_3 and ω_2 are determined, ω_1 and ω_0 can be deduced by solving the above system.


 Let p=3, n=3 and u=w^4, where w is a primitive element in _3^n^*. Then u∈𝒰_0\_3, ε=0, Γ_3=-4 and Γ_4=4. 
	By Theorem <ref>, the differential spectrum of f_u is 
		
    𝕊=[ω_0=286, ω_1=208, ω_2=156, ω_3=26, ω_4=26],

	which coincides with the result calculated directly by MAGMA.



 Let p=3, n=5 and u=w^210, where w is a primitive element in _3^n^*. Then u∈𝒰_0\_3, ε=1,Γ_3=-4 and Γ_4=12.
	By Theorem <ref>, the differential spectrum of f_u is 
   
    𝕊=[ω_0=27346, ω_1=11616, ω_2=14278, ω_3=3630, ω_4=1936].

   which coincides with the result calculated directly by MAGMA.



 Let p=3, n=7 and u=w, where w is a primitive element in _3^n^*. Then u∈𝒰_0\_3, ε=1,Γ_3=-28 and Γ_4=-12.
	By Theorem <ref>, the differential spectrum of f_u is 
   
    𝕊=[ω_0=2240650, ω_1=891888, ω_2=1204486, ω_3=295110, ω_4=148648].

   which coincides with the result calculated directly by MAGMA.






§ CONCLUDING REMARKS

In this paper, we conducted an in-depth investigation of the differential properties of the Ness-Helleseth function. For u∈𝒰_0∖_3, we expressed the differential spectrum in terms of quadratic character sums. This completed the work on the differential properties of the Ness-Helleseth function. Besides, we obtained a series of identities of character sums, which may be used in some other areas. It may be interesting to consider applications of the differential spectrum of the Ness-Helleseth function in other areas such as sequence design, coding theory and combinatorial design. Moreover, the study of the Ness-Helleseth function can be extended to p>3 <cit.>, <cit.>, and the investigation of the differential spectrum of such function will be undertaken in our further work.
	
	

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