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Research article

On distribution properties of cubic residues

  • Received: 09 April 2020 Accepted: 15 July 2020 Published: 23 July 2020
  • MSC : 11A15, 11L40

  • In this paper, we use the elementary methods, the properties of the Gauss sums and the estimate for character sums to study the calculating problems of a certain cubic residues modulo p, and give some interesting identities and asymptotic formulas for their counting functions.

    Citation: Hu Jiayuan, Chen Zhuoyu. On distribution properties of cubic residues[J]. AIMS Mathematics, 2020, 5(6): 6051-6060. doi: 10.3934/math.2020388

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  • In this paper, we use the elementary methods, the properties of the Gauss sums and the estimate for character sums to study the calculating problems of a certain cubic residues modulo p, and give some interesting identities and asymptotic formulas for their counting functions.


    Let p be an odd prime, and a be an integer with (a,p)=1. In order to study of the properties of quadratic residues modulo p, Legendre first introduced the characteristic function of the quadratic residues (ap), which was later called Legendre's symbol as follows:

    (ap)={1,  if a is a quadratic residue modulo p;1,  if a is a quadratic non-residue modulo p;0  if pa.

    The introduction of this symbol greatly promotes the research about the properties of the quadratic residues and non-residues modulo p. Therefore, a great number of scholars began to study related work and obtained a series of valuable research results. For example, if p be an odd prime with p=4k+1, then for any quadratic residue r and quadratic non-residue s modulo p, one has the identity (see [2]: Theorem 4–11)

    (12p1a=1(a+ra1p))2+(12p1a=1(a+sa1p))2=p,

    where a1 is the inverse of amodp. In detail, a1 satisfy the equation xa1modp.

    Many other papers related to power residues and non-residues modulo p can also be found in references [4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20], here we will not describe it in detail.

    Recently, Wang Tingting and Lv Xingxing [5] studied the distribution properties of a certain quadratic residues and non-residues modulo p, and proved the following two conclusions:

    Theorem A. For any prime p with p3mod4, one has the identities

    N(p,1)={18(p3),  if p3mod8;18(p7),  if p7mod8

    and

    N(p,1)={18(p3),  if p3mod8;18(p+1),  if p7mod8.

    where N(p,1) denotes the number of all integers 1ap1 such that a, a+a1 and aa1 are all quadratic residues modulo p, N(p,1) denotes the number of all integers 1ap1 such that a is a quadratic non-residue modulo p, a+a1 and aa1 are quadratic residues modulo p.

    Theorem B. For any prime p with p1mod4, one has the asymptotic formulas

    N(p,1)={18(p3)+E(p,1),  if p5mod8;18(p17)+E1(p,1),  if p1mod8

    and

    N(p,1)={18(p+3)+E(p,1),  if p5mod8;18(p+3)+E1(p,1),  if p1mod8

    where we have the estimates |E(p,1)|34p, |E1(p,1)|54p, |E(p,1)|34p and |E1(p,1)|54p.

    As two corollaries of these results, Wang Tingting and Lv Xingxing [5] solved two open problems proposed by professor Sun Zhiwei. That is, they proved:

    For any prime p101, there is at least one integer a, such that a, a+a1 and aa1 are all quadratic residues modulo p.

    For any prime p18, there is at least one quadratic non-residue amodp, such that a+a1 and aa1 are quadratic residues modulo p.

    In the field of the quadratic residues research, it is worth mentioning Z. H. Sun's important work [6]. We believe that Theorem A and B can also be derived from earlier results in [6]. All research works are very meaningful. In fact, the distribution of power residues plays an vital role in mathematics and cryptography, a number of important number theory and information security problems are closely related to it. A survey on this can be found in F. L. Ţiplea, S. Iftene, G. Teşeleanu, A. M. Nica [10]. Therefore, it is necessary to continue to study the distribution properties of the power residues.

    Inspired by the works in [5] and [6], we will naturally ask that what happens to the cubic residues modulo p?

    It is clear that if (p1,3)=1, then for any integer a with (a,p)=1, the congruence equation x3amodp has one solution. So it is a trivial situation in the condition of (p1,3)=1. But if p1mod3, then what happens?

    This paper focuses on such problems. Let p be a prime with p1mod3, and N(p) denotes the number of all integers 1<a<p1 such that a+a1 and aa1 are cubic residues modulo p. In this paper, using the elementary methods, the properties of the third-order characters and Gauss sums, and the estimate for character sums, we are going to give an exact identity and some asymptotic formulas for N(p). Our main results are summarized as following three conclusions:

    Theorem 1. Let p be an odd prime with p7mod12. If 2 is a cubic residue modp, then we have the identity

    N(p)=19(p+4d11),

    where d is uniquely determined by 4p=d2+27b2 and d1mod3.

    Theorem 2. Let p be an odd prime with p7mod12. If 2 is not a cubic residue modp, then we have the asymptotic formula

    N(p)=19(p5)+E(p),

    where we have the estimates |E(p)|23p.

    Theorem 3. Let p be an odd prime with p1mod12, then we have the asymptotic formula

    N(p)={19(p13)+E1(p),  if 2 is a cubic residue mod p;19(p1)+E1(p),  if 2 is not a cubic residue mod p.

    where we have the estimates |E1(p)|<269p.

    According to our theorems, we can deduce the following:

    Corollary. Let p>700 be a prime with p1mod3, then there exists at least one integer 1<a<p1 such that a+a1 and aa1 are cubic residues modp.

    To prove our main results, we need following several basic lemmas. For simplicity, there is no need to repeat some elementary knowledge of number theory and analytic number theory, which can be found in references [1,2,3].

    Lemma 1. Let p be a prime with p1mod3. Then for any third-order character λmodp, we have the identity

    τ3(λ)+τ3(¯λ)=dp,

    where τ(χ)=p1a=1χ(a) e(ap) denotes the classical Gauss sums, and e(y)=e2πiy, d is uniquely determined by 4p=d2+27b2 and d1mod3.

    Proof. This result is Theorem 1 in Zhang Wenpeng and Hu Jiayuan [21].

    Lemma 2. Let p be an odd prime with p7mod12. Then for any third-order character λmodp, we have the identity

    p1a=1λ(a+a1)¯λ(aa1)+p1a=1¯λ(a+a1)λ(aa1)=4.

    Proof. Note that p3mod4 and (1p)=1, so based on the properties of the Legendre's symbol and complete residue system modp, we obtain

    p1a=1λ(a+a1)¯λ(aa1)=p1a=1λ(a2+1)¯λ(a21)=p1a=1(1+(ap))λ(a+1)¯λ(a1)=p2a=1λ(a+2)¯λ(a)+p1a=1(ap)λ(a+1)¯λ(a1)=p1a=1λ(1+2a1)1p1a=1(ap)λ(a1)¯λ(a+1)=2p1a=1(ap)λ(a1)¯λ(a+1). (2.1)

    Similarly, we can also deduce that

    p1a=1¯λ(a+a1)λ(aa1)=2+p1a=1(ap)λ(a1)¯λ(a+1). (2.2)

    Combining (2.1) and (2.2) we have the identity

    p1a=1λ(a+a1)¯λ(aa1)+p1a=1¯λ(a+a1)λ(aa1)=4.

    This proves Lemma 2.

    Lemma 3. Let p be an odd prime with p1mod3. Then for any third-order character λmodp, we have the identity

    p1a=1(λ(aa1)+¯λ(aa1))=d+1p(λ(2)τ3(λ)+¯λ(2)τ3(¯λ)).

    Proof. Write χ2=(p). Note that the identities ¯λ=λ2, from the properties of the Gauss sums and the Legendre's symbol modp we have

    p1a=1λ(aa1)=p1a=1λ2(a)λ(a21)=p1a=1λ(a)λ(a1)+p1a=1(ap)λ(a)λ(a1)=1τ(¯λ)p1b=1¯λ(b)p1a=1λ(a)e(b(a1)p)+1τ(¯λ)p1b=1¯λ(b)p1a=1λ(a)(bp)e(b(a1)p)=τ2(λ)τ(¯λ)+τ(λχ2)τ(¯λ)p1b=1¯λ2(b)χ2(b)e(bp)=τ2(λ)τ(¯λ)+(1p)τ2(λχ2)τ(¯λ). (2.3)

    Now according to the properties of complete residue system modp, we have

    p1a=1λ(a21)=1+p1a=1λ(a2+2a)=1+p1a=1λ(a)λ(a+2)=1+1τ(¯λ)p1b=1¯λ(b)p1a=1λ(a)e(b(a+2)p)=1+¯λ(2)τ2(λ)τ(¯λ). (2.4)

    On the other hand, from the properties of Legendre's symbol modp we also have

    p1a=1λ(a21)=p1a=1λ(a1)+p1a=1χ2(a)λ(a1)=1+p1a=1λ(a)+1τ(¯λ)p1b=1¯λ(b)p1a=1χ2(a)e(b(a1)p)=1+τ(χ2)τ(¯λ)p1b=1¯λ(b)χ2(b)e(bp)=1+(1p)τ(χ2)τ(¯λχ2)τ(¯λ). (2.5)

    Since λ3(2)=1, so applying (2.4) and (2.5) we can deduce the identity

    τ2(λ)=(1p)λ(2)τ(χ2)τ(¯λχ2). (2.6)

    Note that τ2(χ2)=χ2(1)p and τ(λ)τ(¯λ)=τ(λ)¯τ(λ)=p, combining Lemma 1, (2.3) and (2.6) we can deduce the identity

    p1a=1(λ(aa1)+¯λ(aa1))=τ3(λ)p+τ3(¯λ)p+(1p)τ(λ)τ2(λχ2)p+(1p)τ(¯λ)τ2(¯λχ2)p=d+(1p)λ(2)τ(¯λ)τ4(λ)pτ2(χ2)+(1p)¯λ(2)τ(λ)τ4(¯λ)pτ2(χ2)=d+1p(λ(2)τ3(λ)+¯λ(2)τ3(¯λ)).

    This proves Lemma 3.

    Lemma 4. Let p be an odd prime with p1mod3. Then for any third-order character λmodp, we have the identity

    p1a=1(λ(a+a1)+¯λ(a+a1))=d+(1p)1p(λ(2)τ3(λ)+¯λ(2)τ3(¯λ)).

    Proof. It can be deduced from the same methods of proving Lemma 3.

    Lemma 5. Let p be an odd prime with p7mod12. Then for any third-order character λmodp, we have the identity

    p1a=1(λ(a2a2)+¯λ(a2a2))=d+1p(λ(2)τ3(λ)+¯λ(2)τ3(¯λ)).

    Proof. Note that the identity

    p1a=1(ap)λ(aa1)=p1a=1(ap)λ(a+a1)=p1a=1(ap)λ(aa1).

    From the properties of the Legendre's symbol modp we have

    p1a=1λ(a2a2)=p1a=1λ(aa1)+p1a=1(ap)λ(aa1)=p1a=1λ(aa1). (2.7)

    Applying (2.7) and Lemma 3 we may immediately get

    p1a=1(λ(a2a2)+¯λ(a2a2))=p1a=1(λ(aa1)+¯λ(aa1))=d+1p(λ(2)τ3(λ)+¯λ(2)τ3(¯λ)).

    This proves Lemma 5.

    In this section, we shall complete the proofs of our main results. First we prove Theorem 1. For any prime p with p7mod12, note that (1p)=1, so for all integers 1<a<p1, we have (a+a1,p)=1. From Lemma 2–5 we have

    N(p)=19p2a=2(1+λ(a+a1)+¯λ(a+a1))(1+λ(aa1)+¯λ(aa1))=19p1a=1(1+λ(a+a1)+¯λ(a+a1))(1+λ(aa1)+¯λ(aa1))29(1+λ(2)+¯λ(2))=p19+19p1a=1(λ(a+a1)¯λ(aa1)+¯λ(a+a1)λ(aa1))+19p1a=1(λ(a+a1)+λ(aa1)+¯λ(a+a1)+¯λ(aa1))+19p1a=1(λ(a2a2)+¯λ(a2a2))29(1+λ(2)+¯λ(2))=19(p+3d72λ(2)2¯λ(2))+19p(λ(2)τ3(λ)+¯λ(2)τ3(¯λ)). (3.1)

    If 2 is a cubic residue modp, then λ(2)=1, from (3.1) and Lemma 1 we have

    N(p)=19(p+4d11).

    This proves Theorem 1.

    If 2 is not a cubic residue modp, then 1+λ(2)+¯λ(2)=0, note that the estimate |τ(λ)|=p, from (3.1) and Lemma 1 we have

    N(p)=p72λ(2)2¯λ(2)9+(3+λ(2))τ3(λ)+(3+¯λ(2))τ3(¯λ)9p=p59+(2¯λ(2))τ3(λ)+(2λ(2))τ3(¯λ)9p=p59+E(p),

    where |E(p)|=19p|(2¯λ(2))τ3(λ)+(2λ(2))τ3(¯λ)|23p.

    This proves Theorem 2.

    If p1mod12, then there exist integer r such that r2=(r)21modp and r+r10modp. For this integer 1<r<p1, we also have λ(r)=¯λ2(r)=¯λ(r2)=¯λ(1)=1, so we have

    N(p)=19p2a=2(a2+1,p)=1(1+λ(a+a1)+¯λ(a+a1))(1+λ(aa1)+¯λ(aa1))=19p1a=1(1+λ(a+a1)+¯λ(a+a1))(1+λ(aa1)+¯λ(aa1))49(1+λ(2)+¯λ(2))=p19+19p1a=1(λ(a+a1)¯λ(aa1)+¯λ(a+a1)λ(aa1))+19p1a=1(λ(a+a1)+λ(aa1)+¯λ(a+a1)+¯λ(aa1))+19p1a=1(λ(a2a2)+¯λ(a2a2))49(1+λ(2)+¯λ(2)). (3.2)

    From the Weil's work [22] we have the estimates

    |p1a=1λ(a2a2)|=|p1a=1λ(a)λ(a41)|=|p1a=1λ(a5a)|5p (3.3)
    |p1a=1λ(a+a1)¯λ(aa1)|=|p1a=1λ(a2+1)¯λ(a21)|4p. (3.4)

    If λ(2)=1, then note that |d|2p, applying (3.2–3.4) and Lemma 1–5 we have asymptotic formula

    N(p)=19(p13)+E1(p), (3.5)

    where |E1(p)|269p.

    If λ(2)1, then applying (3.2–3.4) and Lemma 2–5 we have asymptotic formula

    N(p)=19(p1)+E1(p), (3.6)

    where |E1(p)|269p.

    Now Theorem 3 follows from asymptotic formulas (3.5) and (3.6).

    This completes the proofs of our all results.

    The main results of this paper are three theorems and a corollary. Theorem 1 obtained an exact formula for N(p) with p=12k+7 and 2 is a cubic residue modulo p. Theorem 2 and Theorem 3 established two asymptotic formulas for N(p) with p1mod3 and 2 is not a cubic residue modulo p. At the same time, we also give two sharp upper bound estimates for the error terms. As some applications, we also deduced following corollary:

    If p>700 is a prime with p1mod3, then there is at least one integer 1<a<p1 such that a+a1 and aa1 are cubic residues modulo p.

    The authors would like to thank the referee for their very helpful and detailed comments.

    This work is supported by the N. S. F. (2017MS0114) of Inner Mongolia, Talent introduction research Foundation of Hetao College (HYRC2019007) and the N. S. F. (11771351) of P. R. China.

    The authors declare that there are no conflicts of interest regarding the publication of this paper.



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