In this paper, we give some relations between Gauss sums of order 3. As application, we give the number of solutions of some cubic diagonal equations. These generalize the earlier results obtained by Hong-Zhu and solve the sign problem raised by Zhang-Zhang.
Citation: Wenxu Ge, Weiping Li, Tianze Wang. A remark for Gauss sums of order 3 and some applications for cubic congruence equations[J]. AIMS Mathematics, 2022, 7(6): 10671-10680. doi: 10.3934/math.2022595
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In this paper, we give some relations between Gauss sums of order 3. As application, we give the number of solutions of some cubic diagonal equations. These generalize the earlier results obtained by Hong-Zhu and solve the sign problem raised by Zhang-Zhang.
For a prime p≡1(mod3), let Fp be the finite field of residues (modp), let G be the multiplicative group of non-zero residues (modp) and let H be the subgroup of non-zero cubic residues (modp). For any a∈G, we defined the sums
S(a)=p−1∑k=0e(ak3/p) |
and
G(χ)=p−1∑k=1χ(k)e(k/p), |
where χ is a multiplicative character of order 3 over Fp and e(x)=e2πix in this paper. Both S(a) and G(χ) are called Gauss sums of order 3. Gauss sums is very important in the analytic number theory and related research filed. Many scholars studied its properties and obtained a series of interesting results (see [5,6,8,9,10,11,13]).
Let z∈G∖H. By a classical result of Gauss [4] (also see Theorem 4.1.2 of [1]), S(1),S(z) and S(z2) are three roots of the cubic equation
x3−3px−pc=0, |
where c is uniquely determined by
4p=c2+27d2, c≡1(mod3). | (1.1) |
However, how to determine which of the three roots corresponds to S(1) is still an open problem.
In this paper, for a fixed z∈G∖H, we find a relation between S(1),S(z) and S(z2).
Theorem 1.1. Let p≡1(mod3) and z∈G∖H. Then
S(1)=2√pcos(θp), S(z)=2√pcos(θp−sgn(d)23π), S(z2)=2√pcos(θp+sgn(d)23π), |
where θp=13arccos(−c2√p)+jp23π; jp is one of three values −1,0,1 and only dependent on p; c and d are uniquely determined by
4p=c2+27d2, c≡1(mod3), 9d≡c(2zp−13+1)(modp). | (1.2) |
Moreover, there is a unique multiplicative character χ of order 3 over Fp such that
χ(z)=−1+√3i2, G(χ)=√peisgn(d)θp. |
As application, we consider some congruence equations modp. For a1,a2,a3∈G, let M(a1,a2,a3) be the number of solutions of
a1x31+a2x32+a3x33≡0(modp), |
and let N(a1,a2,a3) be the number of solutions of
a1x31+a2x32≡a3(modp). |
In [2], Chowla, Cowles and Cowles showed that M(1,1,1)=p2+c(p−1). As pointed out in [3], the following is essentially included in the derivation of the cubic equation of periods by Gauss [4]: For a prime p≡1(mod3) and for z∈G∖H, then one has
M(1,1,z)=p2+12(p−1)(9d−c), |
where c and d are uniquely determined by (1.1) (except for the sign of d).
Chowla, Cowles and Cowles [3] determined the sign of d for the case 2∈G∖H as the following result shows.
Proposition 1.2. [3] Let a prime p≡1(mod3). If 2∈G∖H, then for any z∈G∖H, one has
M(1,1,z)=p2+12(p−1)(9d−c), |
where c and d are uniquely determined by (1.1) with
d≡c(mod4) for z≡2(modH) |
and
d≡−c(mod4) for z≡4(modH). |
Recently, Hong and Zhu [7] solve the Gauss sign problem. In fact, they gave the following result.
Proposition 1.3. [7] Let a prime p≡1(mod3) and z∈G∖H. Let g be a generator of the multiplicative group G. one has
M(1,1,z)=p2+12(p−1)(−c−δz(p)d), |
where c and d are uniquely determined by (1.1) with d>0 and
δz(p)=(−1)⟨indg(d)⟩3⋅sgn(Im(r1+3√3r2i)). |
Here r1 and r2 are uniquely determined by
4p=r21+27r22, r1≡1(mod3), 9r2≡(2gp−13+1)r1(modp). |
Indeed, their result need to use the generator of group G (that is the primitive root of module p). However, for a large prime p, it is not easy to find the primitive root of module p. In this paper, we consider M(a1,a2,a3), N(a1,a2,a3) and determine the sign of d immediately by the coefficients a1,a2 and a3. We have the following three more general results.
Theorem 1.4. Let a prime p≡1(mod3) and a1,a2,a3∈G.
(1) For the case a1a2a3∈H, M(a1,a2,a3)=p2+c(p−1);
(2) For the case a1a2a3∉H, M(a1,a2,a3)=p2+12(p−1)(9d−c),
where c and d are uniquely determined by
4p=c2+27d2, c≡1(mod3), 9d≡c(2(a1a2a3)p−13+1)(modp). | (1.3) |
Theorem 1.5. Let p≡1(mod3) and a1,a2,a3∈G.
(1) For the case a1a2a3∈H,
N(a1,a2,a3)={p−2+c,ifa1≡a2(modH);p+1+c,otherwise. |
(2) For the case a1a2a3∉H,
N(a1,a2,a3)={p−2+12(9d−c),ifa1≡a2(modH);p+1+12(9d−c),otherwise, |
where c and d are uniquely determined by (1.3).
Corollary 1.6. Let p≡1(mod3) and a1,a2,a3∈G. Then
M(a1,a2,a3)≡−c(a1a2a3)p−13(modp). |
In [14], H. Zhang and W. P. Zhang proposed the following open problem:
Can the number of solutions to the cubic congruence equation
x31+x32+x33+x34≡z(modp) | (1.4) |
be calculated when z∈G?
Let L(z) be the number of solutions of the above Eq (1.4). In [12], W. P. Zhang and J. Y. Hu proved that
L(z)={p3−6p−12p(5c±27d),ifz∈G∖H;p3−6p+5cp,ifz∈H. | (1.5) |
However, in [12], they also proposed an interesting open problem: How to determine the choice of sign in (1.5). In this paper, we solve the sign problem in (1.5), and get the following result.
Theorem 1.7. Let p be a prime number and p≡1(mod3), let z∈G∖H. Then
L(z)=p3−6p−12p(5c−27d), |
where c and d are uniquely determined by
4p=c2+27d2, c≡1(mod3), 9d≡c(2zp−13+1)(modp). |
Lemma 2.1 (Theorem 3.1.3 of [1]). Let p≡1(mod3) and χ be a multiplicative character of order 3 over Fp. Then
J(χ,χ)=c+3√3di2, |
where the Jacobi sum J(χ,χ)=∑p−1a=1χ(a)χ(1−a), c and d are uniquely determined by
4p=c2+27d2, c≡1(mod3), 9d≡c(2gp−13+1)(modp) |
with g being the generator of the multiplicative group G of non-zero residues (modp) such that χ(g)=−1+√3i2.
Lemma 2.2 (Lemma 4.1.1 of [1]). Let p≡1(mod3). Let g be a generator of the multiplicative group G of non-zero residues (modp) with χ(g)=−1+√3i2. Then
G3(χ)=pJ(χ,χ). |
Lemma 2.3. Let p≡1(mod3) and z∈G∖H. Then there is a unique multiplicative character χ of order 3 over Fp such that
χ(z)=−1+√3i2, G3(χ)=p⋅c+3√3di2, |
where c and d are uniquely determined by (1.2).
Proof. Let g′ be a generator of the group G. Note that z∈G∖H. So we have indg′z≡±1(mod3). If indg′z≡1(mod3), we take g=g′; If indg′z≡−1(mod3), we take g=(g′)−1. Hence g also is a generator of the group G and indgz≡1(mod3). Thus we have
zp−13≡(gindgz)p−13≡gp−13indgz≡gp−13(modp). |
We take the multiplicative character χ(⋅)=e(indg(⋅)3). Obviously, we have
χ(z)=e(indgz3)=e(13)=−1+√3i2=χ(g). |
Obviously, all of the multiplicative non-principal characters of order 3 over Fp are χ and ¯χ, ¯χ(z)=¯χ(z)=−1−√3i2. Thus χ is the unique multiplicative character of order 3 over Fp with χ(z)=−1+√3i2.
Note that G3(χ)=pJ(χ,χ) by Lemma 2.2. Finally, using the Lemma 2.1, one immediately arrive the Lemma 2.3 as required.
Lemma 2.4. Let χ be a multiplicative character of order 3. Then for any a∈G, we have
S(a)=¯χ(a)G(χ)+χ(a)G(¯χ). | (2.1) |
Proof. Let χ be any multiplicative character of order 3. Then we have
1+χ(k)+¯χ(k)={3,ifk∈H;0,ifk∈G∖H. |
Thus for any a∈G, we have
S(a)=p−1∑k=0e(ak3/p)=1+p−1∑k=1(1+χ(k)+¯χ(k))e(ak/p)=1+p−1∑k=1e(ak/p)+p−1∑k=1χ(k)e(ak/p)+p−1∑k=1¯χ(k)e(ak/p)=¯χ(a)p−1∑k=1χ(ak)e(ak/p)+χ(a)p−1∑k=1¯χ(ak)e(ak/p)=¯χ(a)G(χ)+χ(a)G(¯χ). |
In this section, we prove Theorem 1.1. First, by Lemma 2.3, there is a unique multiplicative character χ of order 3 such that
χ(z)=−1+√3i2, G3(χ)=p⋅c+3√3di2, |
where c and d are uniquely determined by (1.2). We can rewrite G3(χ) by argument, and get
G3(χ)=p32e3iθsgn(d), |
where θ=13arccos(−c2√p). Thus we have
G(χ)=√pei(sgn(d)θ+j23π)=√peisgn(d)(θ+sgn(d)j23π), |
where j is one of three values −1,0,1. Let jp=sgn(d)j. Thus we have
G(χ)=√peisgn(d)(θ+jp23π). |
Next, we will prove that jp does not depend on the sign of d. Note that G(¯χ)=χ(−1)¯G(χ)=√pe−isgn(d)(θ+jp23π). By Lemma 2.4, we have
S(1)=¯χ(1)G(χ)+χ(1)G(¯χ)=G(χ)+G(¯χ)=2√pcos[sgn(d)(θ+jp23π)]=2√pcos(θ+jp23π). |
Obviously, by the definition of S(1), the value of S(1) doesn't depend on the sign of d. Thus we have that jp does not depend on the sign of d.
Take θp=θ+jp23π. We have G(χ)=√peisgn(d)θp and S(1)=2√pcos(θp). By Lemma 2.4, we have
S(z)=¯χ(z)G(χ)+χ(z)G(¯χ)=−1−√3i2⋅√peisgn(d)θp+−1+√3i2⋅√pe−isgn(d)θp=√pei(sgn(d)θp−2π3)+√pe−i(sgn(d)θp−2π3)=2√pcos(sgn(d)θp−2π3)=2√pcos(θp−sgn(d)2π3). |
Similarly, we have
S(z2)=2√pcos(θp+sgn(d)23π). |
This completes the proof of the Theorem 1.1.
In this section, we prove Theorem 1.4, 1.5 and 1.7. First, we begin with the proof of Theorem 1.4.
Proof of Theorem 1.4. By the orthogonality of additive character, we have
M(a1,a2,a3)=1pp−1∑m=0p−1∑x1=0p−1∑x2=0p−1∑x3=0e(m(a1x31+a2x32+a3x33)p)=p2+1pp−1∑m=1S(ma1)S(ma2)S(ma3). |
Then by Lemma 2.4, for any multiplicative character χ of order 3, we have
M(a1,a2,a3)=p2+1pp−1∑m=1[3∏j=1(¯χ(maj)G(χ)+χ(maj)G(¯χ))]=p2+1pp−1∑m=1[¯χ(a1a2a3)G3(χ)+χ(a1a2a3)G3(¯χ)]+G(χ)(χ(¯a1¯a2a3)+χ(¯a1a2¯a3)+χ(a1¯a2¯a3))p−1∑m=1¯χ(m)+G(¯χ)(χ(¯a1a2a3)+χ(a1¯a2a3)+χ(a1a2¯a3)))p−1∑m=1χ(m)=p2+p−1p[¯χ(a1a2a3)G3(χ)+χ(a1a2a3)G3(¯χ)]. |
If a1a2a3∈H, thus we have χ(a1a2a3)=¯χ(a1a2a3)=1. Then by Lemma 2.3, we have
M(a1,a2,a3)=p2+p−1p(G3(χ)+G3(¯χ))=p2+(p−1)[c+3√3di2+c−3√3di2]=p2+c(p−1). |
If a1a2a3∈G∖H, then by Lemma 2.3, we can take multiplicative character χ of order 3 satisfying
χ(a1a2a3)=−1+√3i2, G3(χ)=p⋅c+3√3di2, |
where c and d are uniquely determined by (1.3). Thus we have
M(a1,a2,a3)=p2+(p−1)(−1−√3i2⋅c+3√3di2+−1+√3i2⋅c−3√3di2)=p2+12(p−1)(9d−c). |
This completes the proof of the Theorem 1.4.
Proof of Theorem 1.5. We have
M(a1,a2,a3)=p−1∑x1,x2,x3=0a1x31+a2x32+a3x33≡0(modp)1=p−1∑x3=1p−1∑x1,x2=0a1x31+a2x32+a3x33≡0(modp)1+p−1∑x1,x2=0a1x31+a2x32≡0(modp)1=p−1∑x3=1p−1∑x1,x2=0a1(−x1¯x3)3+a2(x2¯x3)3≡a3(modp)1+1+p−1∑x1=1p−1∑x2=1(−¯x1x2)3≡a1¯a2(modp)1=(p−1)p−1∑x1,x2=0a1x31+a2x32≡a3(modp)1+1+p−1∑x1=1p−1∑x=1x3≡a1¯a2(modp)1=(p−1)N(a1,a2,a3)+1+p−1∑x1=1p−1∑x=1x3≡a1¯a2(modp)1. |
If a1≡a2(modH), the number of solutions of the congruence equation x3≡a1¯a2(modp) is exactly 3. Thus we have
M(a1,a2,a3)=(p−1)N(a1,a2,a3)+1+3(p−1)=(p−1)N(a1,a2,a3)+3p−2. |
If a1≢a2(modH), the congruence equation x3≡a1¯a2(modp) has no solution. Thus we have
M(a1,a2,a3)=(p−1)N(a1,a2,a3)+1. |
Hence Theorem 1.5 immediately follows from Theorem 1.4.
Proof of Theorem 1.7. First, by Lemma 2.3, there is a unique multiplicative character χ of order 3 such that
χ(z)=−1+√3i2, G3(χ)=p⋅c+3√3di2, |
where c and d are uniquely determined by (1.2).
Note that χ(−1)=1. By the orthogonality of additive character and Lemma 2.3, we have
L(z)=1pp−1∑m=0p−1∑x1=0p−1∑x2=0p−1∑x3=0p−1∑x4=0e(m(x31+x32+x33+x34−z)p)=p3+1pp−1∑m=1S4(m)e(−mzp)=p3+1pp−1∑m=1[¯χ(m)G(χ)+χ(m)G(¯χ)]4e(−mzp)=p3−6p+1pp−1∑m=1[¯χ(m)G4(χ)+4pχ(m)G2(χ)+4p¯χ(m)G2(¯χ)+χ(m)G4(¯χ)]e(−mzp)=p3−6p+1pG4(χ)p−1∑m=1¯χ(m)e(−mzp)+1pG4(¯χ)p−1∑m=1χ(m)e(−mzp) +4G2(χ)p−1∑m=1χ(m)e(−mzp)+4G2(¯χ)p−1∑m=1¯χ(m)e(−mzp)=p3−6p+1pG4(χ)χ(−z)G(¯χ)+1pG4(¯χ)¯χ(−z)G(χ)+4¯χ(−z)G3(χ)+4χ(−z)G3(¯χ)=p3−6p+χ(z)G3(χ)+¯χ(z)G3(¯χ)+4¯χ(z)G3(χ)+4χ(z)G3(¯χ)=p3−6p+p⋅−1+√3i2⋅c+3√3di2+p⋅−1−√3i2⋅c−3√3di2 +4p⋅−1−√3i2⋅c+3√3di2+4p⋅−1+√3i2⋅c−3√3di2=p3−6p−12p(5c−27d). |
This completes the proof of the Theorem 1.7.
Example 4.1. We take F31:={¯0,¯1,⋯,¯30}. Consider the cubic equations x31+2x32+3x33≡0(mod31) and x31+2x32≡3(mod31).
If the integers c and d satisfying that 4⋅31=c2+27d2,c≡1(mod3),9d≡c(2×631−13+1)(mod31), then c=4,d=2. One can check that 231−13≡1(mod31) and 631−13≡25(mod31), so 6 is not a cubic element in F31 and 2 is a cubic element in F31. Thus 6∉H and 1≡2(modH).
It then follows from Theorems 1.4 and 1.5 that the numbers M(1,2,3) and N(1,2,3) of the cubic equations x31+2x32+3x33≡0(mod31) and x31+2x32≡3(mod31) are given by
M(1,2,3)=312+12(31−1)(9×2−4)=1171 |
and
N(1,2,3)=31−2+12(9×2−4)=36. |
We list the solutions of equation x31+2x32≡3(mod31) as belove:
(¯1,¯1);(¯1,¯5);(¯1,¯25);(¯5,¯1);(¯5,¯5);(¯5,¯25);(¯25,¯1);(¯25,¯5);(¯25,¯25);(¯6,¯4);(¯6,¯7);(¯6,¯20);(¯26,¯4);(¯26,¯7);(¯26,¯20);(¯30,¯4);(¯30,¯7);(¯30,¯20);(¯4,¯8);(¯4,¯9);(¯4,¯14);(¯7,¯8);(¯7,¯9);(¯7,¯14);(¯20,¯8);(¯20,¯9);(¯20,¯14);(¯16,¯17);(¯16,¯22);(¯16,¯23);(¯18,¯17);(¯18,¯22);(¯18,¯23);(¯28,¯17);(¯28,¯22);(¯28,¯23). |
The authors are partially supported by the National Natural Science Foundation of China (Grant No. 11871193, 12071132) and the Natural Science Foundation of Henan Province (No. 222300420493, 202300410031).
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1. | Wenxu Ge, Weiping Li, Tianze Wang, A note on some diagonal cubic equations over finite fields, 2024, 9, 2473-6988, 21656, 10.3934/math.20241053 |