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The hybrid power mean of the generalized Gauss sums and the generalized two-term exponential sums

  • This article applied the properties of character sums, quadratic character, and classical Gauss sums to study the calculations of the hybrid power mean of the generalized Gauss sums and the generalized two-term exponential sums. It also provided exact formulas for calculating these hybrid power means.

    Citation: Xue Han, Tingting Wang. The hybrid power mean of the generalized Gauss sums and the generalized two-term exponential sums[J]. AIMS Mathematics, 2024, 9(2): 3722-3739. doi: 10.3934/math.2024183

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  • This article applied the properties of character sums, quadratic character, and classical Gauss sums to study the calculations of the hybrid power mean of the generalized Gauss sums and the generalized two-term exponential sums. It also provided exact formulas for calculating these hybrid power means.



    The concept of exponential sums was first introduced to solve the Waring problem. The estimation of exponential sums has consistently been a fundamental focus in analytic number theory. Throughout the history of research on exponential sums, researchers have observed that individual values of exponential sums show highly irregular behavior, while their higher power mean and hybrid power mean exhibit regular patterns. Investigating two-term exponential sums is vital for various areas, including integer factorization, prime factorization, and the study of number theory functions. Therefore, in order to promote the mutual advancement of analytic number theory and its related fields, it is essential to explore the higher power mean and hybrid power mean of two-term exponential sums.

    Let q>3 be a positive integer and m and n be integers satisfying (mn,q)=1. For any Dirichlet character χ, the generalized Gauss sums G(m,n,χ;q) and the generalized two-term exponential sums C(m,k,h,χ;q) are defined as follows:

    G(m,n,χ;q)=q1a=0χ(a)e(manq),C(m,k,h,χ;q)=q1a=1χ(a)e(mak+ahq),

    where e(y)=e2πiy, k1, n1, and h>1 are integers.

    Due to the importance of the generalized Gauss sum in analytic number theory, it has attracted the attention of many experts and resulted in some significant findings. For example, according to the results of A. Weil [1], an upper bound estimate can be obtained as |p1x=1χ(x)e(axn+bxp)|np, where p is an odd prime.

    Therefore, we can speculate that

    |p1a=1χ(a)e(ma2p)|2p.

    Zhang Wenpeng and Lv Xingxing [2] studied the following identity that holds when p is an odd prime satisfying p3mod4 and for any integer m satisfying (m,p)=1,

    χmodp|p1a=1χ(a)e(ma2p)|2|p1b=1χ(b+¯b)|2=(p1)(3p26p1).

    Goran Djanković et al. [3] discussed the following equation that holds when p is an odd prime and n is an integer satisfying (n,p)=1,

    χmodpp1m=0|p1a=1χ2(a)e(ma2p)|2|p1b=1χ(b)e(mb+n¯bp)|2={p(p1)(2p25p2(np)+5),if p1mod4,p(p1)(2p25p+5),if p3mod4,

    where (p) denotes the Legendre symbol and ¯b denotes the multiplicative inverse of bmodp.

    Li Xiaoxue and Wu Chengjing [4] discussed the following statement that holds for any integer m and n satisfying (mn,p)=1,

    χmodp|p1a=1χ2(a)e(ma2p)|2|p1b=1χ(b)e(nb+¯bp)|2=p(p1)2+p(p1)p1b=1e(2nb+2ˉbp)(p1)(p1b=1e(2nb+2ˉbp))2,

    where p is an odd prime satisfying p3mod4.

    Liu Xiaoge and Meng Yuanyuan [5] obtained the following result when p is an odd prime with p1mod12,

    p1m=1|p1a=1λ(a)e(ma4p)|2|p1b=1χ4(b)e(ma3p)|2=12p2(p1),

    for any cubic character λ and quartic character χ4 modulo p.

    Consulting references [6,7,8,9,10,11] reveals additional significant results related to two-term exponential sums; these findings will not be restated in this paper.

    The main focus of this paper is to study the hybrid power mean of the generalized Gauss sums G(m,n,χ;p) and the generalized two-term exponential sums C(m,k,h,χ;p) given by

    χmodpp1m=0|p1a=1λ(a)e(manp)|2l|p1b=1χ(b)e(bh+mbkp)|2, (1.1)

    where χmodp denotes the sum over all characters χ modulo p and λ=χ2 or λ=χ.

    This paper considers the computational problem posed by Eq (1.1), and it seems that no one has studied this specific topic before, at least not in the existing literature. Generally, summing over χmodp does not yield ideal results. However, through calculations, we can gain a more accurate understanding of the relationship between the hybrid power mean of the generalized Gauss sums and the generalized two-term exponential sums.

    In this paper, we utilize elementary and analytic methods, as well as the properties of character sums and classical quadratic Gauss sums, to investigate the computational problems presented in Eq (1.1). Here, we will discuss the hybrid power mean of several kinds of generalized two-term sums with generalized Gauss sums under the conditions of n=2,4;l=1,2;h=3,4;k=1,2. As a result, we derive several meaningful computational formulas. In other words, we can obtain the following:

    Theorem 1.1. When p is an odd prime satisfying (3,p1)=1 and χ is any character modulo p, we have

    χmodpp1m=0|p1a=1χ2(a)e(ma2p)|2|p1b=1χ(b)e(b3+mb2p)|2={p(p1)(2p25p+7),if p5mod24,p(p1)(2p25p+5),if p3,11mod24,p(p1)(2p25p+11),if p17mod24,p(p1)(2p25p+9),if p23mod24.

    Theorem 1.2. When p is an odd prime satisfying (3,p1)=1 and χ is any character modulo p, we can derive the following identity

    χmodpp1m=0|p1a=1χ2(a)e(ma2p)|2|p1b=1χ(b)e(b3+mbp)|2={p(p1)(2p25p2p+5),if p5mod12,p(p1)(2p25p+5),if p11mod12orp=3.

    Theorem 1.3. Let p be an odd prime satisfying (3,p1)=1 and χ is any character modulo p, and we can obtain

    χmodpp1m=0|p1a=1χ(a)e(ma2p)|2|p1b=1χ(b)e(b3+mb2p)|2=p(p1)(p25p+8).

    Theorem 1.4. Let p be an odd prime satisfying (3,p1)=1 and χ is any character modulo p, and we will obtain the identity

    χmodpp1m=0|p1a=1χ(a)e(ma2p)|2|p1b=1χ(b)e(b3+mbp)|2={p(p1)(p23p2τ(χ2)+4+χ4(3)τ(¯χ4)2+¯χ4(3)τ(χ4)2),ifp5mod12,p(p1)(p23p+4),ifp11mod12orp=3.

    Theorem 1.5. Let p be an odd prime satisfying p>5 and χ is any character modulo p, and we can conclude that

    χmodpp1m=0|p1a=1χ(a)e(ma4p)|2|p1b=1χ(b)e(mb4+bp)|2={p(p1)(p25p+8),if4p1,p(p1)(p29p+24),if4p1.

    Theorem 1.6. Let p be an odd prime satisfying (3,p1)=1 and χ is any character modulo p, and we can conclude that

    χmodpp1m=0|p1a=1χ(a)e(ma2p)|4|p1b=1χ(b)e(b3+mb2p)|2=p(p1)(3p318p2+27p+14+6(1p)).

    Theorem 1.7. When p>5 is an odd prime satisfying p3mod4 and χ is any character modulo p, we can conclude that

    χmodpp1m=0|p1a=1χ(a)e(ma4p)|4|p1b=1χ(b)e(mb4+bp)|2=p(p1)(3p318p2+27p+8).

    In this section, we will provide some important lemmas that are necessary for proving the theorems. In the following, we will apply the properties of characters modulo p and quadratic Gauss sums, and relevant content can be referenced from literature [12,13,14,15]. Therefore, we will not elaborate on it here. First, we introduce the relevant properties of quadratic character in [13]:

    χmodp(x(ax+b)p)=χmodp(ax+bp)=(ap),

    where (ab,p)=1.

    Next, we present the following lemmas:

    Lemma 2.1. Let p be an odd prime satisfying (3,p1)=1, then we can obtain the following equation:

    p1a=1p1b=1p1c=1a2+c2b2+1modpa2cb2modpc31modp1=2p2.

    Proof. Through applying the properties of exponential sums and residue systems, we can conclude that

    p1a=1p1b=1p1c=1a2+c2b2+1modpa2cb2modpc31modp1=p1a=1p1b=1p1c=1a2+c2b2+1modpa2cb2modpc1modp1=p1a=1p1b=1a2+1b2+1modpa2b2modp1=2p2.

    This result verifies Lemma 2.1.

    Lemma 2.2. Let p be an odd prime satisfying (3,p1)=1, and we will obtain the identity

    p1a=1p1b=1p1c=1a2+c2b2+1modpacbmodp1=4p8.

    Proof. By utilizing the solutions to congruence equations and the computational method described in Lemma 2.1, we can easily obtain the following result:

    p1a=1p1b=1p1c=1a2+c2b2+1modpacbmodp1=p1a=1p1c=1a2+c2a2c2+1modp1=p1a=1p1c=1(a21)(c21)0modp1=4(p1)4=4p8.

    This proves Lemma 2.2.

    Lemma 2.3. Let p be a prime satisfying p>5, and we will obtain the identity

    p1a=1p1b=1p1c=1a4+c4b4+1modpacbmodp1={4p8,if 4p1,8p24,if 4p1.

    Proof. By utilizing the solutions to congruence equations, we obtain the following result:

    p1a=1p1b=1p1c=1a4+c4b4+1modpacbmodp1=p1a=1p1c=1a4+c4a4c4+1modp1=p1a=1p1c=1(a41)(c41)0modp1={4p8,if 4p1,8p24,if 4p1.

    This proves Lemma 2.3.

    Lemma 2.4. Let p be an odd prime with (3,p1)=1, then we have

    p1a=1p1b=1p1c=1a2+c2b2+1modpa2cb2modp1={3p7,if p5mod24,3p5,if p3,11mod24,3p11,if p17mod24,3p9,if p23mod24.

    Proof. Through utilizing the properties of character sums and residue systems, we can derive the following precise calculation formula:

    p1a=1p1b=1p1c=1a2+c2b2+1modpa2cb2modp1=p1a=1p1c=1p1b=1a2+c2b+1modpa2cbmodp(1+χ2(b))=p1a=1p1c=1a2+c2a2c+1modp(1+χ2(a2c))=p1a=1p1c=1a2+c2a2c+1modp1+p1a=1p1c=1a2+c2a2c+1modpχ2(c)=p1a=1p1c=1(c1)(c+1a2)0modp1+p1a=1p1c=1(c1)(c+1a2)0modpχ2(c)=p1a=1p1c=1c10modp1+p1a=1p1c=1c+1a20modp1p1a=1(1+1a2)0modp1+p1a=1p1c=1c10modpχ2(c)+p1a=1p1c=1c+1a20modpχ2(c)p1a=1(1+1a2)0modp1=p1(1+χ2(2))+p1a=1p1c=1c+1a0modp(1+χ2(a))+p1(1+χ2(2))+p1a=1p1c=1c+1a0modp(1+χ2(a))χ2(c)=2(p1)2(1+χ2(2))+p1a=1p1c=1c+1a0modp1+p1a=1p1c=1c+1a0modpχ2(a)+p1a=1p1c=1c+1a0modpχ2(c)+p1a=1p1c=1c+1a0modpχ2(ac)=2(p1)2(1+χ2(2))+p211χ2(1)=3p82χ2(2)χ2(1)={3p7,if p5mod24,3p5,if p3,11mod24,3p11,if p17mod24,3p9,if p23mod24.

    This proves Lemma 2.4.

    Lemma 2.5. When p is an odd prime that satisfies (3,p1)=1, we can prove the following:

    p1a=1p1b=1p1c=1p1d=1a2dmodpb2cmodpcdmodpe(c3d3p)={p+32p,if p5mod12,p+3,if p11mod12orp=3.

    Proof. For p>3, by solving congruence equations and utilizing the properties of classical Gauss sums, we can obtain

    p1a=1p1b=1p1c=1p1d=1a2dmodpb2cmodpcdmodpe(c3d3p)=p1a=1p1b=1p1c=1p1d=1a2dmodpb2cmodpe(c3d3p)p1a=1p1b=1p1c=1p1d=1a2dmodpb2cmodpcdmodpe(c3d3p)=p1a=1p1b=1p1c=1p1d=1admodpbcmodp(1+χ2(a))(1+χ2(b))e(c3d3p)2(p1)=p1a=1p1b=1p1c=1p1d=1admodpbcmodpe(c3d3p)+p1a=1p1b=1p1c=1p1d=1admodpbcmodpχ2(a)e(c3d3p)+p1a=1p1b=1p1c=1p1d=1admodpbcmodpχ2(b)e(c3d3p)+p1a=1p1b=1p1c=1p1d=1admodpbcmodpχ2(ab)e(c3d3p)2(p1)=p1a=1p1b=1e(b3a3p)+p1a=1p1b=1χ2(a)e(b3a3p)+p1a=1p1b=1χ2(b)e(b3a3p)+p1a=1p1b=1χ2(ab)e(b3a3p)2(p1)=p1a=1p1b=1e(a3(b31)p)+τ(χ2)p1b=1χ2(b)χ2(b31)+τ(χ2)p1b=1χ2(b31)+pp1b=1b31modpχ2(b)p1b=1χ2(b)2(p1)=1+τ(χ2)p1b=1χ2(b)χ2(b31)+τ(χ2)p1b=1χ2(b31)+pp1b=1b31modpχ2(b)p1b=1χ2(b)2(p1)=p+1+τ(χ2)p1b=1(1+χ2(b))χ2(b31)2(p1)=p+1+τ(χ2)p1b=1(1+χ2(b))χ2(b1)2(p1)={p+32p,if p5mod12,p+3,if p11mod12.

    Specifically, when p=3, and we can calculate the sums:

    2a=12b=12c=12d=1a2dmod3b2cmod3cdmod3e(c3d33)=2a=12b=12c=12d=1a2dmod3b2cmod3e(c3d33)2a=12b=12c=12d=1a2dmod3b2cmod3cdmod3e(c3d33)=2a=12b=12c=12d=1a2dmod3b2cmod3e(c3d33)4=2a=12b=1a21mod3b21mod314=0,

    which equals p+3.

    This proves Lemma 2.5.

    Lemma 2.6. Let p be an odd prime that satisfies (3,p1)=1. We can prove the following:

    p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpacbdmodpd±cmodpe(c3d3p)={χ4(3)τ(¯χ4)2+¯χ4(3)τ(χ4)22τ(χ2)p+3,if p5mod12,p+3,if p11mod12orp=3.

    Proof. Since the case when p = 3 is trivial, we will not discuss it further here. By solving congruence equations and reducing the residue classes modulo p, we can deduce that a2b2+cd0modp and acbdmodp are equivalent to b2(ˉc2d21)dcmodp and abˉcdmodp, or b2c2¯(d+c)modp and abˉcdmodp. The number of solutions for this equation can be represented as 1+χ2(c2(¯c+d)) = 1+χ2(c+d), where, in the case of p5mod12, χ2=χ24, χ2() represents the Legendre symbol and χ4() is any fourth character modulo p. Therefore, we can prove

    p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpacbdmodpd±cmodpe(c3d3p)=p1a=1p1b=1p1c=1p1d=1b2(ˉc2d21)dcmodpabˉcdmodpd±cmodpe(c3d3p)=p1c=1p1d=1d±cmodp(1+χ2(c+d))e(c3d3p)=p1c=1p1d=1(1+χ2(c+d))e(c3d3p)p1c=1(1+χ2(2c))p1c=1e(2c3p)=p1c=1p1d=1e(c3d3p)+p1c=1p1d=1χ2(c+d)e(c3d3p)(p1)+1=1+p1c=1p1d=1χ2(c+d)e(c3d3p)(p1)+1=p1c=1p1d=1χ2(c+1)χ2(d)e(d3(c31)p)p+3=p1c=1p1d=1χ2(c+1)χ2(d)e(d(c31)p)p+3=τ(χ2)p1c=1χ2(c+1)χ2(c31)p+3=τ(χ2)p1c=0χ2(c+1)χ2(c31)χ2(1)τ(χ2)p+3=τ(χ2)p1c=0χ2(c+2)χ2(c3+3c2+3c)τ(χ2)p+3=τ(χ2)p1c=1χ2(2c+1)χ2(3c2+3c+1)τ(χ2)p+3=τ(χ2)p1c=0χ2(2c+1)χ2(3c2+3c+1)2τ(χ2)p+3=τ(χ2)p1c=1χ2(c)χ2(3c2+1)2τ(χ2)p+3=p1b=1χ2(b)e(bp)p1c=1χ2(c)e(3bc2p)2τ(χ2)p+3=p1b=1χ4(b2)e(bp)p1c=1χ4(c2)e(3bc2p)2τ(χ2)p+3=p1b=1χ4(b2)e(bp)p1c=1χ4(c)(1+χ2(c))e(3bcp)2τ(χ2)p+3=p1b=1χ4(b2)e(bp)p1c=1(χ4(c)+¯χ4(c))e(3bcp)2τ(χ2)p+3=p1b=1χ2(b)e(bp)p1c=1χ4(c)e(3bcp)+p1b=1χ2(b)e(bp)p1c=1¯χ4(c)e(3bcp)2τ(χ2)p+3=χ4(3)τ(¯χ4)2+¯χ4(3)τ(χ4)22τ(χ2)p+3,

    when p11mod12, and we have χ2(a)=χ2(a); thus,

    p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpacbdmodpd±cmodpe(c3d3p)=p1a=1p1b=1p1c=1p1d=1b2(ˉc2d21)dcmodpabˉcdmodpd±cmodpe(c3d3p)=p1c=1p1d=1d±cmodp(1+χ2(c+d))e(c3d3p)=p1c=1p1d=1χ2(c+1)χ2(d)e(d(c31)p)p+3=τ(χ2)p1c=1χ2(c+1)χ2(c31)p+3=τ(χ2)p1c=0χ2(c+1)χ2(c31)χ2(1)τ(χ2)p+3=τ(χ2)p1c=1χ2(c)χ2(3c2+1)+τ(χ2)p+3=p+3.

    This proves Lemma 2.6.

    Lemma 2.7. Let p be an odd prime, and we can establish the identity

    p1a=1p1b=1p1c=1p1d=1p1e=1a2+b2+c2d2+e2+1modpabcdemodp1=p3+6p219p146(1p).

    Proof. Through utilizing the properties of the reduced residue system modulo p, we know that if d and e traverse reduced residue system modulo p, then da and eb will also traverse this system. We first clarify the results based on Lemma 1 presented in article [16], then improve the results for two specific parts

    p1a=1p1b=1p1d=1p1e=1a2+b2+d2e2a2d2+b2e2+1modp1=p1a=1p1b=1p1d=1p1e=1a+b+dead+be+1modp(1+2χ2(d)+2χ2(a)+χ2(de)+χ2(ab))+p1a=1p1b=1p1d=1p1e=1a+b+dead+be+1modp(2χ2(ad)+2χ2(bd)+2χ2(ade)+2χ2(dab)+χ2(abde)=W1+2W2+2W3+W4+W5+2W6+2W7+2W8+2W9+W10, (2.1)

    where χ2() denotes the Legendre symbol modulo p.

    W2=p1a=1p1b=1p1d=1p1e=1a(d1)+b(e1)de1modpχ2(d)=p1a=1p1b=11+p1a=1p1b=1p1e=2b(e1)e1modp1+p1a=1p1b=1p1d=2a(d1)d1modpχ2(d)+p1a=1p1b=1p1d=2p1e=2a(d1)+b(e1)de1modpχ2(d)=(p1)2+(p1)(p2)+(p1)p1d=2χ2(d)+p1a=0p1b=1p1d=2p1e=2a+bde1modpχ2(d)p1b=1p1d=2p1e=2bde1modpχ2(d)=(p1)(2p4)(p1)(p2)+p3=p22p1. (2.2)
    W7=p1a=1p1b=1p1d=1p1e=1a(d1)+b(e1)de1modpχ2(bd)=p1a=1p1b=1χ2(b)+p1a=1p1b=1p1e=2b(e1)e1modpχ2(b)+p1a=1p1b=1p1d=2a(d1)d1modpχ2(bd)+p1a=1p1b=1p1d=2p1e=2a(d1)+b(e1)de1modpχ2(bd)=p1a=1p1b=1p1e=2b(e1)e1modpχ2(b)+p1a=1p1b=1p1d=2p1e=2a+b(e1)de1modpχ2(bd)=(p1)(p2)+(p3)=p22p1. (2.3)

    Consequently, by combining Eqs (2.1)–(2.3) and Lemma 1 of article [16], we can deduce the following result:

    p1a=1p1b=1p1c=1p1d=1p1e=1a2+b2+c2d2+e2+1modpabcdemodp1=p3+6p219p146(1p).

    Lemma 2.8. Let p be an odd prime that satisfies p3mod4, and we have the following identity:

    p1a=1p1b=1p1c=1p1d=1p1e=1a4+b4+c4d4+e4+1modpabcdemodp1=p3+6p219p8.

    Proof. When p satisfies p3mod4 and is an odd prime, then χ2(abde)=χ2(abde). By utilizing the properties of the reduced residue system modulo p and the solutions to congruence equations and combining them with the results presented in Lemma 2.7, we can obtain

    p1a=1p1b=1p1c=1p1d=1p1e=1a4+b4+c4d4+e4+1modpabcdemodp1=p1a=1p1b=1p1d=1p1e=1a4+b4+d4e4a4d4+b4e4+1modp1=p1a=1p1b=1p1d=1p1e=1a2+b2+d2e2a2d2+b2e2+1modp(1+χ2(a))(1+χ2(b))(1+χ2(d))(1+χ2(e))=p1a=1p1b=1p1d=1p1e=1a2+b2+d2e2a2d2+b2e2+1modp1=p3+6p219p8.

    We know the trigonometric identity:

    p1a=0e(map)={p,if p|m,0,if pm. (3.1)

    First, we give the proofs for Theorems 1.1–1.4. Let p be an odd prime and suppose (3,p1)=1, then by Lemmas 2.1 and 2.4, we can obtain

    χmodpp1m=0|p1a=1χ2(a)e(ma2p)|2|p1b=1χ(b)e(b3+mb2p)|2=p1a=1p1b=1p1c=1p1d=1e(c3d3p)χmodpχ(a2c)¯χ(b2d)p1m=0e(m(a2b2+c2d2)p)=p(p1)p1a=1p1b=1p1c=1p1d=1a2b2+c2d20modpa2cb2dmodpe(c3d3p)=p2(p1)p1a=1p1b=1p1c=1a2b2+c210modpa2cb2modpc31modp1p(p1)p1a=1p1b=1p1c=1a2b2+c210modpa2cb2modp1=p(p1)(2p22p)p(p1)(3p82χ2(2)χ2(1))={p(p1)(2p25p+7),if p5mod24,p(p1)(2p25p+5),if p3,11mod24,p(p1)(2p25p+11),if p17mod24,p(p1)(2p25p+9),if p23mod24,

    and by combining Lemma 2.5, we have

    χmodpp1m=0|p1a=1χ2(a)e(ma2p)|2|p1b=1χ(b)e(b3+mbp)|2=p1a=1p1b=1p1c=1p1d=1e(c3d3p)χmodpχ(a2c)¯χ(b2d)p1m=0e(m(a2b2+cd)p)=p(p1)p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpa2cb2dmodpe(c3d3p)=p(p1)p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpa2cb2dmodpcdmodpe(c3d3p)+p(p1)p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpa2cb2dmodpcdmodpe(c3d3p)=2p(p1)(p1)2+p(p1)p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpa2cb2dmodpcdmodpe(c3d3p)={p(p1)(2p25p2p+5),if p5mod12,p(p1)(2p25p+5),if p11mod12orp=3,

    therefore, according to Lemma 2.2, we can obtain the identity

    χmodpp1m=0|p1a=1χ(a)e(ma2p)|2|p1b=1χ(b)e(b3+mb2p)|2=p1a=1p1b=1p1c=1p1d=1e(c3d3p)χmodpχ(ac)¯χ(bd)p1m=0e(m(a2b2+c2d2)p)=p(p1)p1a=1p1b=1p1c=1p1d=1a2b2+c2d20modpacbdmodpe(c3d3p)=p2(p1)p1a=1p1b=1p1c=1a2b2+c210modpacbmodpc31modp1p(p1)p1a=1p1b=1p1c=1a2b2+c210modpacbmodp1=p(p1)(p2p)p(p1)(4p8)=p(p1)(p25p+8).

    We combine Lemma 2.6 to provide the proof of Theorem 1.4:

    χmodpp1m=0|p1a=1χ(a)e(ma2p)|2|p1b=1χ(b)e(b3+mbp)|2=p1a=1p1b=1p1c=1p1d=1e(c3d3p)χmodpχ(ac)¯χ(bd)p1m=0e(m(a2b2+cd)p)=p(p1)p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpacbdmodpe(c3d3p)=p(p1)p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpa2cb2dmodpd±cmodpe(c3d3p)+p(p1)p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpacbdmodpd±cmodpe(c3d3p)=p(p1)(p1)2+p(p1)p1a=1p1b=1p1c=1p1d=1a2b2+cd0modpacbdmodpd±cmodpe(c3d3p)={p(p1)(p23p2τ(χ2)+4+χ4(3)τ(¯χ4)2+¯χ4(3)τ(χ4)2),if p5mod12,p(p1)(p23p+4),if p11mod12orp=3.

    Here, according to Lemma 2.3, we can obtain

    χmodpp1m=0|p1a=1χ(a)e(ma4p)|2|p1b=1χ(b)e(mb4+bp)|2=p1a=1p1b=1p1c=1p1d=1e(cdp)χmodpχ(ac)¯χ(bd)p1m=0e(m(a4b4+c4d4)p)=p(p1)p1a=1p1b=1p1c=1p1d=1a4b4+c4d40modpacbdmodpe(cdp)=p2(p1)p1a=1p1b=1p1c=1a4b4+c410modpacbmodpc1modp1p(p1)p1a=1p1b=1p1c=1a4b4+c410modpacbmodp1={p(p1)(p25p+8),if 4p1,p(p1)(p29p+24),if 4p1.

    This proves Theorem 1.5.

    Now, combining Lemmas 2.2 and 2.7 with odd prime p and satisfying the condition (3,p1)=1, we can get the identity

    χmodpp1m=0|p1a=1χ(a)e(ma2p)|4|p1b=1χ(b)e(b3+mb2p)|2=p1a=1p1b=1p1c=1p1d=1p1e=1p1f=1e(e3f3p)χmodpχ(ace)¯χ(bdf)p1m=0e(m(a2+c2+e2b2d2f2)p)=p(p1)p1a=1p1b=1p1c=1p1d=1p1e=1p1f=1a2+c2+e2b2+d2+f2modpacebdmodpe(e3f3p)=p2(p1)p1a=1p1b=1p1c=1p1d=1p1e=1a2+c2+e2b2+d2+1modpacebdmodpe31modp1p(p1)p1a=1p1b=1p1c=1p1d=1p1e=1a2+c2+e2b2+d2+1modpacebdmodp1=p2(p1)2p1a=1p1b=1p1c=1a2+c2b2+1modpacbmodp1p(p1)p1a=1p1b=1p1c=1p1d=1p1e=1a2+c2+e2b2+d2+1modpacebdmodp1=p(p1)(4p312p2+8p)p(p1)(p3+6p219p146(1p))=p(p1)(3p318p2+27p+14+6(1p)).

    This proves Theorem 1.6.

    Finally, according to Lemmas 2.3 and 2.8, when p3mod4, we can prove Theorem 1.7:

    χmodpp1m=0|p1a=1χ(a)e(ma4p)|4|p1b=1χ(b)e(mb4+bp)|2=p1a=1p1b=1p1c=1p1d=1p1e=1p1f=1e(efp)χmodpχ(ace)¯χ(bdf)p1m=0e(m(a4+c4+e4b4d4f4)p)=p(p1)p1a=1p1b=1p1c=1p1d=1p1e=1p1f=1a4+c4+e4b4+d4+f4modpacebdmodpe(efp)=p2(p1)p1a=1p1b=1p1c=1p1d=1p1e=1a4+c4+e4b4+d4+1modpacebdmodpe1modp1p(p1)p1a=1p1b=1p1c=1p1d=1p1e=1a4+c4+e4b4+d4+1modpacebdmodp1=p2(p1)2p1a=1p1b=1p1c=1a4+c4b4+1modpacbmodp1p(p1)p1a=1p1b=1p1c=1p1d=1p1e=1a4+c4+e4b4+d4+1modpacebdmodp1=p2(p1)2p1a=1p1b=1p1c=1a2+c2b2+1modpacbmodp1p(p1)p1a=1p1b=1p1c=1p1d=1a2+c2+b2d2a2b2+c2d2+1modp1=p(p1)(4p312p2+8p)p(p1)(p3+6p219p8)=p(p1)(3p318p2+27p+8).

    This completes the proofs of our results.

    The main result of this paper was to investigate the computational problem involving the hybrid power mean of the generalized Gauss sums with the generalized two-term exponential sums. We also obtained exact computational formulas. The main results were obtained by using Lemma 2.7 to complete the proof. At this point, we let

    h(p)=p1a=1p1b=1p1c=1p1d=1p1e=1a2+b2+c2d2+e2+1modpabcdemodp1.

    Here, we provided an example to calculate the exact results for prime numbers p. The precise calculation results were summarized in Table 1. Additionally, our results also offered effective solutions to the problem of calculating the higher power mean of two-term exponential sums. We believe that these works will play a positive role in advancing the research on related problems.

    Table 1.  The calculation of h(p).
    p h(p) p h(p)
    101 h(101)=1089568 103 h(103)=1154416
    109 h(109)=1364227 107 h(107)=1291696
    113 h(113)=1517344 127 h(127)=2142736
    137 h(137)=2681344 131 h(131)=2348560
    149 h(149)=3438304 139 h(139)=2798868
    173 h(173)=5353984 151 h(151)=3576880
    181 h(181)=6122848 157 h(157)=4014796
    197 h(197)=7874464 163 h(163)=4487056
    229 h(229)=12319264 167 h(167)=4821616
    233 h(233)=12970624 191 h(191)=7183120

     | Show Table
    DownLoad: CSV

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This work is supported by the N. S. B. R. P. (2022JM-013) of Shaanxi Province.

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



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