In this paper, we use the elementary methods, the properties of Dirichlet character sums and the classical Gauss sums to study the estimation of the mean value of high-powers for a special character sum modulo a prime, and derive an exact computational formula. It can be conveniently programmed by the “Mathematica” software, by which we can get the exact results easily.
Quadratic characterthe classical gauss sumsthe mean value of high-powercomputational formulaIntroduction
Let p be an odd prime, the quadratic character modulo p is called the Legendre symbol, which is defined by
(ap)={1,ifaisaquadraticresiduemodulop;−1,ifaisaquadraticnon-residuemodulop;0,ifp∣a.
Many mathematicians have studied the properties of the Legendre symbol and obtained a series of important results (see [1–13]). Perhaps the most representative properties of the Legendre’s symbol are as follows:
Let p and q be two distinct odd primes, then one has the quadratic reciprocal formula (see [14]: Theorem 9.8 or [15]: Theorems 4–6)
(pq)⋅(qp)=(−1)(p−1)(q−1)4.
For any odd prime p with p≡1mod4, there exists two non-zero integers αp and βp such that (see [15]: Theorems 4–11)
p=αp2+βp2.
In fact, the integers αp and βp in Eq. (1) can be represented by the Jacobsthal sums ϕ2(r), which is (see [16]: Definition of the Jacobsthal sums)
ϕk(r)=∑a=1p−1(ap)(ak+rp)andαp=12ϕ2(1),βp=12ϕ2(s),where s is any quadratic non-residue modulo p.
Now we consider a sum A(r) be similar to βp. For any integers r with (r,p)=1 and k≥0, let A(r) and Sk(p) be defined as follows:
A(r)=1+∑a=1p−1(a2+ra¯p)andSk(p)=1p−1∑r=1p−1Ak(r).
In this paper, we give an exact computational formula for Sk(p) with p≡1mod6, and prove the following result:
Theorem. Let p be a prime with p≡1mod6, for any integer k, we have the identity
Sk(p)=13⋅[dk+(−d+9b2)k+(−d−9b2)k],where d and b are uniquely determined by 4p=d2+27b2, d≡1mod3 and b>0.
From this Theorem, we can immediately deduce the following four Corollaries:
Corollary 1. Let p be a prime with p≡1mod6, then we have
1p−1∑r=1p−111+∑a=1p−1(a2+ra¯p)=1p−1∑r=1p−111+∑a=1p−1(a4+rap)=pd⋅(3p−d2).
Corollary 2. Let p be a prime with p≡1mod6, we have
1p−1∑r=1p−11[1+∑a=1p−1(a2+ra¯p)]2=1p−1∑r=1p−11[1+∑a=1p−1(a4+rap)]2=3⋅p2d2⋅(3p−d2)2.
Corollary 3. Let p be a prime with p≡1mod6, then we have
1p−1∑r=1p−1[1+∑a=1p−1(a2+ra¯p)]4=1p−1∑r=1p−1[1+∑a=1p−1(a4+rap)]4=6⋅p2.
Corollary 4. Let p be a prime with p≡1mod6, we have
1p−1∑r=1p−1[1+∑a=1p−1(a2+ra¯p)]6=1p−1∑r=1p−1[1+∑a=1p−1(a4+rap)]6=18p3+d2⋅(d2−3p)2.
Some notes: In our Theorem, we only discuss the case p≡1mod6. If p≡5mod6, the result is trivial, see Proposition 6.1.2 in [16]. In this case, for any integer r with (r,p)=1, we have the identity
A(r)=1+∑a=1p−1(a2+ra¯p)=1+∑a=1p−1(a3p)(a3+rp)=1+∑a=1p−1(1+ra¯p)=∑a=0p−1(1+rap)=0.
Thus, for all prime p with p≡5mod6 and k≥1, we have Sk(p)=0.
In addition, our Theorem holds for all negative integers.
Obviously, the advantage of our work is that it can transfer a complex mathematical computational problem into a simple form suitable for computer programming. It means that for any fixed prime p with p≡1mod6 and integer k, the exact value of Sk(p) can be calculated by our Theorem and a simple computer program. In Section 4, we give an example to calculate the exact results of the prime number p within 200 satisfying conditions p≡1mod6 and d≡1mod3. The exact results of calculation are summarised in Table 1.
The calculation of Sk(p)
p
d
b
k
Sk(p)
S1(7)=0,S2(7)=14,
S3(7)=−20,S4(7)=294,
7
1
1
1, 2, 3, 4, 5, 6, 7, 8
S5(7)=−700,S6(7)=6574
S7(7)=−20580,S8(7)=152054
S1(19)=0,S2(19)=38,
S3(19)=−56,S4(19)=2166,
19
7
1
1, 2, 3, 4, 5, 6, 7, 8
S5(19)=−5329,S6(19)=126598
S7(19)=−424536,S8(19)=7514006
S1(31)=0,S2(31)=62,
S3(31)=−308,S4(31)=5766,
31
4
2
1, 2, 3, 4, 5, 6, 7, 8
S5(31)=−47740,S6(31)=631102
S7(31)=−6215748,S8(31)=73396406
S1(61)=0,S2(62)=122,
S3(61)=−182,S4(61)=22326,
61
1
3
1, 2, 3, 4, 5, 6, 7, 8
S5(61)=−55510,S6(61)=4118782
S7(61)=−14221662,S8(61)=763839926
S1(73)=0,S2(73)=146,
S3(73)=−1190,S4(73)=31974,
73
7
3
1, 2, 3, 4, 5, 6, 7, 8
S5(73)=−434350,S6(73)=8418406
S7(73)=−133171710,S8(73)=2360507414
S1(97)=0,S2(97)=194,
S3(97)=1330,S4(97)=56454,
97
19
1
1, 2, 3, 4, 5, 6, 7, 8
S5(97)=645050,S6(97)=18197014
S7(97)=262793370,S8(97)=6153247574
S1(103)=0,S2(103)=206,
S3(103)=−1820,S4(103)=63654,
103
13
3
1, 2, 3, 4, 5, 6, 7, 8
S5(103)=−937300,S6(103)=22981486
S7(103)=−405475980,S8(103)=8087165174
S1(151)=0,S2(151)=302,
S3(151)=−1748,S4(151)=136806,
151
19
3
1, 2, 3, 4, 5, 6, 7, 8
S5(151)=−1319740,S6(151)=65028622
S7(151)=−836979108,S8(151)=31764871286
S1(163)=0,S2(163)=326,
S3(163)=3400,S4(163)=159414,
163
25
1
1, 2, 3, 4, 5, 6, 7, 8
S5(163)=2771000,S6(163)=89513446
S7(163)=1897026600,S8(163)=53193475094
S1(181)=0,S2(181)=362,
S3(181)=−3458,S4(181)=196566,
181
7
5
1, 2, 3, 4, 5, 6, 7, 8
S5(181)=−3129490,S6(181)=118693102
S7(181)=−2379038298,S8(181)=75272130806
Several Lemmas
In this section, we give some simple Lemmas, which are necessary in the proofs of our Theorem. In addition, we need some properties of the classical Gauss sums and character sums, which can be found in many number theory books, such as [14,15] or [17], and we will not repeat them. First, we have the following:
Lemma 1. Let p be a prime with p≡1mod3, for any third-order character λ modulo p, we have the identity
τ3(λ)+τ3(λ¯)=dp,where τ(χ)=∑a=1p−1χ(a)e(ap) denotes the classical Gauss sums with e(y)=e2πiy and i2=−1, d is the same as the one in the Theorem.
Proof. See references [18] or [19].
Lemma 2. Let p be an odd prime, for any non-principal character χ modulo p, we have the identity
τ(χ2)=χ2(2)τ(χ2)⋅τ(χ)⋅τ(χχ2),where χ2=(∗p) denotes the Legendre’s symbol modulo p.
Proof. From the properties of the classical Gauss sums we have
∑a=0p−1χ(a2−1)=∑a=0p−1χ((a+1)2−1)=∑a=1p−1χ(a)χ(a+2)=1τ(χ¯)∑b=1p−1χ¯(b)∑a=1p−1χ(a)e(b(a+2)p)=τ(χ)τ(χ¯)∑b=1p−1χ¯(b)χ¯(b)e(2bp)=τ(χ)τ(χ¯)∑b=1p−1χ¯2(b)e(2bp)=χ2(2)⋅τ(χ)⋅τ(χ¯2)τ(χ¯).
On the other hand, for any integer b with (b,p)=1, from the identity
∑a=0p−1e(ba2p)=1+∑a=1p−1(1+χ2(a))e(bap)=∑a=1p−1χ2(a)e(bap)=χ2(b)⋅τ(χ2)we also have
∑a=0p−1χ(a2−1)=1τ(χ¯)∑a=0p−1∑b=1p−1χ¯(b)e(b(a2−1)p)=1τ(χ¯)∑b=1p−1χ¯(b)e(−bp)∑a=0p−1e(ba2p)=τ(χ2)τ(χ¯)∑b=1p−1χ¯(b)χ2(b)e(−bp)=χ2(−1)χ¯(−1)τ(χ2)⋅τ(χ¯χ2)τ(χ¯).
From Eqs. (2) and (3) we have the identity
τ(χ¯2)=χ¯2(2)⋅χ2(−1)χ¯(−1)⋅τ(χ2)⋅τ(χ¯χ2)τ(χ)or
τ(χ2)=χ2(2)τ(χ2)⋅τ(χ)⋅τ(χχ2).
This proves Lemma 2.
Lemma 3. Let p be a prime p with p≡1mod6, then for any integer r with (r,p)=1 and three order character λ modulo p, we have the identity
∑a=1p−1(a2+ra¯p)=−1+1p⋅(λ¯(2r)⋅τ3(λ)+λ(2r)⋅τ3(λ¯)).
Proof. From the characteristic function of the cubic residue modulo p, we have
13⋅(1+λ(a)+λ¯(a))={1ifaisacubicresiduemodulop;0otherwise.
Applying Eq. (4) we have
∑a=1p−1(a2+ra¯p)=∑a=1p−1χ2(a¯3)χ2(a3+r)=∑a=1p−1χ2(a3)χ2(a3+r)=∑a=1p−1(1+λ(a)+λ¯(a))χ2(a)χ2(a+r)=∑a=1p−1χ2(1+ra¯)+∑a=1p−1λ(a)χ2(a)χ2(a+r)+∑a=1p−1λ¯(a)χ2(a)χ2(a+r)=−1+λ(r)∑a=1p−1λ(a)χ2(a)χ2(a+1)+λ¯(r)∑a=1p−1λ¯(a)χ2(a)χ2(a+1).
From the properties of the classical Gauss sums, we have
∑a=1p−1λ(a)χ2(a)χ2(a+1)=1τ(χ2)⋅∑b=1p−1χ2(b)∑a=1p−1λ(a)χ2(a)e(b(a+1)p)=1τ(χ2)⋅τ(λχ2)⋅τ(λ¯).
Taking χ=λ in Lemma 2, we have
τ(λ¯)=λ¯(2)τ(χ2)⋅τ(λ)⋅τ(λχ2).
Note that τ(λ)⋅τ(λ¯)=p, from Eqs. (6) and (7) we have
∑a=1p−1λ(a)χ2(a)χ2(a+1)=λ(2)p⋅τ3(λ¯).
Similarly, we also have
∑a=1p−1λ¯(a)χ2(a)χ2(a+1)=λ¯(2)p⋅τ3(λ).
Combining Eqs. (5), (8) and (9) we can deduce that
∑a=1p−1(a2+ra¯p)=−1+1p⋅(λ¯(2r)⋅τ3(λ)+λ(2r)⋅τ3(λ¯)).
This proves Lemma 3.
Lemma 4. Let p be any odd prime with p≡1mod6, then for any integers k≥3 and r with (r,p)=1, we have the third order recursive formula
Ak(r)=3p⋅Ak−2(r)+(d3−3dp)⋅Ak−3(r),where d is the same as defined in the Theorem.
Proof. Note that λ3=λ¯3=χ0, the principal character modulo p, from Lemma 1 and Lemma 3 we have
A3(r)=1p3⋅(λ¯(2r)⋅τ3(λ)+λ(2r)⋅τ3(λ¯))3=1p3⋅[τ9(λ)+τ9(λ¯)+3p3⋅(λ¯(2r)⋅τ3(λ)+λ(2r)⋅τ3(λ¯))]=1p3⋅[(τ3(λ)+τ3(λ¯))3−3p3(τ3(λ)+τ3(λ¯))+3p4⋅A(r)]=d3−3dp+3p⋅A(r).
Indeed, for any integer k≥3, from Eq. (10) we have the third order recursive formula
Ak(r)=Ak−3(r)⋅A3(r)=3p⋅Ak−2(r)+(d3−3dp)⋅Ak−3(r).
This proves Lemma 4.
Lemma 5. Let p be any odd prime with p≡1mod6, then we have
S0(p)=1,S1(p)=0,S2(p)=2p,S3(p)=d⋅(d2−3p)andSk(p)=3p⋅Sk−2(p)+(d3−3pd)⋅Sk−3(p)for allk≥4.
Proof. From the definition
Sk(p)=1p−1∑r=1p−1Ak(r)and the orthogonality of characters modulo p, we have
S0(p)=1,S1(p)=1p(p−1)∑r=1p−1(λ¯(2r)⋅τ3(λ)+λ(2r)⋅τ3(λ¯))=0,S2(p)=1p2(p−1)∑r=1p−1(λ¯(2r)⋅τ3(λ)+λ(2r)⋅τ3(λ¯))2=2p.
From Eq. (10) we also have
S3(p)=1p3(p−1)∑r=1p−1(λ¯(2r)⋅τ3(λ)+λ(2r)⋅τ3(λ¯))3=1p−1∑r=1p−1(d3−3dp+3p⋅A(r))=d⋅(d2−3p).
If k≥4, then from Lemma 4 we have
Sk(p)=3p⋅Sk−2(p)+(d3−3dp)⋅Sk−3(p).
Now Lemma 5 follows from Eqs. (11)–(14).
Proof of the Theorem
In this section, we complete the proof of our Theorem. It is clear that the characteristic equation of the third order linear recursive formula
Sk(p)=3p⋅Sk−2(p)+(d3−3dp)⋅Sk−3(p)is
x3−3px−(d3−3dp)=0.
Note that 4p=d2+27b2, from Eq. (16) we have
(x−d)(x+d+9b2)(x+d−9b2)=0.
It is clear that the three roots of Eq. (16) are x1=d, x2=−d+9d2 and x3=−d−9d2. Indeed, the general term of Eq. (15) is
Sk(p)=C1⋅dk+C2⋅(−d+9b2)k+C3⋅(−d−9b2)k,k≥0.
From Lemma 5 we have
{C1+C2+C3=1,C1⋅d+C2⋅(−d+9b2)+C3⋅(−d−9b2)=0,C1⋅d2+C2⋅(−d+9b2)2+C3⋅(−d−9b2)2=2p.
Solving the Eq. (18) we can get C1=C2=C3=13. From Eq. (17) we have
Sk(p)=13[dk+(−d+9b2)k+(−d−9b2)k],k≥0.
This proves our Theorem.
Obviously, using Lemma 4 we can also extend k in Lemma 5 to all negative integers, which leads to the Corollary 1 and the Corollary 2.
This completes the proofs of our all results.
Conclusion
In this paper, we give an exact computational formula for Sk(p) with p≡1mod6, which is, for any integer k, we have the identity
Sk(p)=13⋅[dk+(−d+9b2)k+(−d−9b2)k],where d and b are uniquely determined by 4p=d2+27b2, p≡1mod6 and b>0.
Meanwhile, the problems of calculating the mean value of high-powers of quadratic character sums modulo a prime are given.
In the end, we use the mathematical software “Mathematica” to program and calculate the exact values of S1(p) to S8(p) of the prime number p within 200 satisfying conditions p≡1mod6 and d≡1mod3, as shown in Table 1. Its application can also extend to Sk(p) that satisfies conditions p≡1mod6 and d≡1mod3 (where 4p=d2+27b2) for any k. See the Appendix A for this specific computer program.
The authors would like to thank the editor and referees for their suggestions and critical comments that substantially improve the presentation of this work.
Funding Statement: This work was supported by the N. S. F. (12126357) of China.
Conflicts of Interest: The authors declare that they have no conflicts of interest to report regarding the present study.
ReferencesAnkeny, N. C. (1952). The least quadratic non-residue. ,55,65–72. DOI 10.2307/1969420.Peralta, R. (1992). On the distribution of quadratic residues and non-residues modulo a prime number. ,58,433–440. DOI 10.1090/S0025-5718-1992-1106978-9.Sun, Z. H. (2002). Consecutive numbers with the same legendre symbol. ,130,2503–2507. DOI 10.1090/S0002-9939-02-06600-5.Hummel, P. (2003). On consecutive quadratic non-residues: A conjecture of Issai Schur. ,103(2),257–266. DOI 10.1016/j.jnt.2003.06.003.Garaev, M. Z. (2003). A note on the least quadratic non-residue of the integer-sequences. ,68(1),1–11. DOI 10.1017/S0004972700037369.Kohnen, W. (2008). An elementary proof in the theory of quadratic residues. ,45(2),273–275. DOI 10.4134/BKMS.2008.45.2.273.Yuk-Kam, L., Jie, W. (2008). On the least quadratic non-residue. ,4,423–435. DOI 10.1142/S1793042108001432.Schinzel, A. (2011). Primitive roots and quadratic non-residues. ,149,161–170. DOI 10.4064/aa149-2-5.Wright, S. (2013). Quadratic residues and non-residues in arithmetic progression. ,133(7),2398–2430. DOI 10.1016/j.jnt.2013.01.004.Dummit, D. S., Dummit, E. P., Kisilevsky, H. (2016). Characterizations of quadratic, cubic, and quartic residue matrices. ,168,167–179. DOI 10.1016/j.jnt.2016.04.014.Ţiplea, F. L., Iftene, S., Teşeleanu, G., Nica, A. M. (2020). On the distribution of quadratic residues and non-quadratic residues modulo composite integers and applications to cryptography. ,372,124993.Wang, T. T., Lv, X. X. (2020). The quadratic residues and some of their new distribution properties. ,12(3),421.Zhang, J. F., Meng, Y. Y. (2021). The mean values of character sums and their applications. ,9(4),318.Apostol, T. M. (1976). . New York: Springer-Verlag.Zhang, W. P., Li, H. L. (2013). . Xi’an, China: Shaanxi Normal University Press.Berndt, B. C., Evans, R. J., Williams, K. S. (1999). Gauss and jacobi sums. ,83(497),349–351.Narkiewicz, W. (1986). . Warszawa: Polish Scientifc Publishers.Zhang, W. P., Hu, J. Y. (2018). The number of solutions of the diagonal cubic congruence equationmodp. ,20(1),73–80.Berndt, B. C., Evans, R. J. (1981). The determination of gauss sums. ,5(2),107–128.Appendix A.