1. Introduction
Let q≥3 be an integer and χ denotes a Dirichlet character modq. For any integers m and n with (mn,q)=1, we define the two-term exponential sum C(m,n,k;q) and two-term character sum N(m,n,χ;q) as follows:
(1) C(m,n,k;q)=∑a=1qe(mak+naq),N(m,n,k,χ;q)=∑a=1qχ(mak+na),
where e(x)=e2πix, χ denotes a nonprincipal Dirichlet character modq, and k is a fixed positive integer.
These sums play a very important role in the study of analytic number theory, so they caused many number theorists’ interest and favor. Some works related to C(m,n,k;q) can be found in [1–5]. For example, Cochrane and Zheng [1] show that
(2)|C(m,n,k;q)|≤kω(q)q1/2,
where ω(q) denotes the number of all distinct prime divisors of q.
On the other hand, the sums N(m,n,k,χ;q) are a special case of the general character sums of the polynomials
(3)∑a=N+1N+Mχ(f(a)),
where M and N are any positive integers and f(x) is a polynomial. If q=p is an odd prime, then Weil (see [6]) obtained the following important conclusion.
Let χ be a qth-order character modp; if f(x) is not a perfect qth power modp, then we have the estimate
(4)∑x=N+1N+Mχ(f(x))≪p1/2lnp,
where “≪” constant depends only on the degree of f(x). Some related results can also be found in [7–10].
Now we are concerned about whether there exists an asymptotic formula for the hybrid mean value
(5)∑m=1q-1|∑a=1q-1χ(mak+a)|2·|∑b=1q-1e(mbk+bq)|2.
In this paper, we will use the analytic method and the properties of character sums to study this problem and give a sharp asymptotic formula for (5) with q=p, an odd prime. That is, we will prove the following.
Theorem 1.
Let p be an odd prime, let χ be any nonprincipal even character modp, and let χ3≠χ0 be the principal character modp. Then we have the asymptotic formula
(6)∑m=1p-1|∑a=1p-1χ(ma3+a)|2·|∑b=1p-1e(mb3+bp)|2=2p3+E(p),
where E(p) satisfies the inequalities -12p2-2p≤E(p)≤4p2-2p.
From this theorem we may immediately deduce the following.
Corollary 2.
For any odd prime p and any nonprincipal even character χmodp with χ3≠χ0, one has
(7)∑m=1p-1|∑a=1p-1χ(ma3+a)|2·|∑b=1p-1e(mb3+bp)|2=2p3+O(p2).
In the theorem, we only consider the polynomial f(x)=mx3+x. For general polynomial f(x)=mxk+xh with k≥4 and 1≤h<k, whether there exists an asymptotic formula is complex problem for (5), it needs us to further study.
For general positive integer q≥4, whether there exists an asymptotic formula for (5) is also an interesting open problem.
2. Several Lemmas
To complete the proof of our theorem, we need the following several lemmas.
Lemma 1.
Let p be an odd prime and let χ be any nonprincipal even character modp. Then for any integer m with (m,p)=1, the identity
(8)∑a=1p-1χ(ma3+a)=τ(χ1)τ(χ1¯3)χ1¯(m)τ(χ¯) ×(1+(mp)τ(χ1χ2)τ(χ1¯3χ2)τ(χ1)τ(χ1¯3)),
where (*/p)=χ2 denotes the Legendre symbol and χ=χ12.
Proof.
Since χ(-1)=1, there exists one and only one character χ1modp such that χ=χ12. Thus, from the properties of Gauss sums we have
(9)∑a=1p-1χ(ma3+a)=1τ(χ¯)∑a=1p-1∑b=1p-1χ¯(b)e(b(ma3+a)p)=1τ(χ¯)∑a=1p-1∑b=1p-1χ¯(ba¯)e(ba¯(ma3+a)p)=1τ(χ¯)∑b=1p-1χ¯(b)e(bp)∑a=1p-1χ(a)e(bma2p)=1τ(χ¯)∑b=1p-1χ¯(b)e(bp)∑a=1p-1χ12(a)e(bma2p)=1τ(χ¯)∑b=1p-1χ¯(b)e(bp) ×∑a=1p-1χ1(a)(1+(ap))e(bmap)=1τ(χ¯)∑b=1p-1χ¯(b)e(bp)iiiiiiiiiiiiiiii×(χ1¯(bm)τ(χ1)iiiiiiiiiiiiiiiiiiiiii+χ1¯(bm)χ2(bm)τ(χ1χ2))=χ1¯(m)τ(χ¯)((mp)τ(χ1χ2)τ(χ1¯3χ2)τ(χ1)τ(χ1¯3)iiiiiiiiiiiiiiii+(mp)τ(χ1χ2)τ(χ1¯3χ2))=τ(χ1)τ(χ1¯3)χ1¯(m)τ(χ¯) ×(1+(mp)τ(χ1χ2)τ(χ1¯3χ2)τ(χ1)τ(χ1¯3)).
This proves Lemma 1.
Lemma 2.
Let p be an odd prime, let χ be any nonprincipal even character modp, χ=χ12, and χ3≠χ0, the principal character modp. Then for any integer m and any quadratic nonresidue rmodp with (m,p)=1, we have the identity
(10)|∑a=1p-1χ(ma3+a)|2=2p+(mp)τ2(χ2)2p∑a=1p-1(χ(a)+χ¯(a)) ×∑b=1p-1(1-a2b3p)(1-bp) +(mp)τ2(χ2)2p ×∑a=1p-1(χ1(r)χ(a)+χ1¯(r)χ¯(a)) ×∑b=1p-1(1-ra2b3p)(1-bp).
Proof.
From the properties of Gauss sums we have
(11)τ(χ1)¯τ(χ1χ2)=∑a=1p-1χ¯1(a)∑b=1p-1χ1(b)χ2(b)e(b-ap)=∑a=1p-1χ¯1(a)∑b=1p-1χ2(b)e(b(1-a)p)=τ(χ2)∑a=1p-1χ¯1(a)(1-ap).
So from (11) we have
(12)τ(χ1χ2)τ(χ1¯3χ2)τ(χ1)τ(χ1¯3) =1p2τ(χ1)τ(χ1¯3)¯τ(χ1χ2)τ(χ1¯3χ2) =τ2(χ2)p2∑a=1p-1χ¯1(a)(1-ap)∑b=1p-1χ13(b)(1-bp) =τ2(χ2)p2∑a=1p-1χ¯1(a)∑b=1p-1(1-ab3p)(1-bp) =τ2(χ2)2p2∑a=1p-1χ¯(a)∑b=1p-1(1-a2b3p)(1-bp) +χ¯1(r)τ2(χ2)2p2∑a=1p-1χ¯(a)∑b=1p-1(1-ra2b3p)(1-bp).
Note that |τ(χ)|=|τ(χ1)|=|τ(χ13)|=p and τ2(χ2)=±p; from (12) and Lemma 1 we may immediately deduce the identity
(13)|∑a=1p-1χ(ma3+a)|2 =p·|1+(mp)τ(χ1χ2)τ(χ1¯3χ2)τ(χ1)τ(χ1¯3)|2 =2p+(mp)τ2(χ2)2p∑a=1p-1(χ(a)+χ¯(a)) ×∑b=1p-1(1-a2b3p)(1-bp)+(mp)τ2(χ2)2p ×∑a=1p-1(χ1(r)χ(a)+χ1¯(r)χ¯(a)) ×∑b=1p-1(1-ra2b3p)(1-bp).
This proves Lemma 2.
Lemma 3.
Let p be an odd prime, let χ be any nonprincipal even character modp, χ=χ12, and χ3≠χ0, the principal character modp. Then for any integer m and any quadratic nonresidue rmodp with (m,p)=1, one has the estimate
(14)|∑a=1p-1(χ1(r)χ(a)+χ1¯(r)χ¯(a))∑b=1p-1(1-ra2b3p)(1-bp)iiiiiiii+∑a=1p-1(χ(a)+χ¯(a))∑b=1p-1(1-a2b3p)(1-bp)|≤4p.
Proof.
Let n be any integer such that (mn/p)=-1 or (m/p)+(n/p)=0. Then from Lemma 2 we have
(15)|∑a=1p-1χ(ma3+a)|2+|∑a=1p-1χ(na3+a)|2=4p.
Note that |(m/p)(τ2(χ2)/p)|=1; applying (15) and Lemma 2 we have the estimate
(16)|∑a=1p-1(χ1(r)χ(a)+χ1¯(r)χ¯(a))∑b=1p-1(1-ra2b3p)(1-bp)iiiiiiiiii+∑a=1p-1(χ(a)+χ¯(a))∑b=1p-1(1-a2b3p)(1-bp)| =||∑a=1p-1χ(ma3+a)|2-|∑a=1p-1χ(na3+a)|2| ≤|∑a=1p-1χ(ma3+a)|2+|∑a=1p-1χ(na3+a)|2≤4p.
This proves Lemma 3.
Lemma 4.
Let p>3 be a prime. Then we have the identity
(17)∑m=1p-1(mp)|∑a=1p-1e(ma3+ap)|2=-τ2(χ2)(2+(3p)),
where (*/p)=χ2 denotes the Legendre symbol.
Proof.
For any odd prime p and integer n with (n,p)=1, from Hua's book [11] (Section 7.8, Theorem 8.2) we know that
(18)∑a=1p(a2+np)=-1.
From this identity and the definition and properties of Gauss sums we have
(19)∑m=1p-1(mp)|∑a=1p-1e(ma3+ap)|2 =∑a=1p-1∑b=1p-1∑m=1p-1(mp)e(m(a3-b3)+a-bp) =∑a=1p-1∑b=1p-1∑m=1p-1(mp)e(mb3(a3-1)+b(a-1)p) =τ(χ2)∑a=1p-1∑b=1p-1(b3(a3-1)p)e(b(a-1)p) =τ(χ2)∑a=1p-1(a3-1p)∑b=1p-1(bp)e(b(a-1)p) =τ2(χ2)∑a=1p-1((a3-1)(a-1)p) =τ2(χ2)(∑a=1p(4a2+4a+4p)-1-(3p)) =τ2(χ2)(∑a=1p((2a+1)2+3p)-1-(3p)) =τ2(χ2)(∑a=1p(a2+3p)-1-(3p)) =-τ2(χ2)(2+(3p)).
This proves Lemma 4.
3. Proof of the Theorem
In this section, we will complete the proof of our theorem. Note that the identities |τ(χ2)|2=p and
(20)∑m=1p-1|∑a=1p-1e(ma3+ap)|2 =∑m=1p|∑a=1p-1e(ma3+ap)|2-1 =∑a=1p-1∑b=1p-1∑m=1pe(m(a3-b3)+a-bp)-1 =∑a=1p-1∑b=1p-1∑m=1pe(m(a3-1)+b(a-1)p)-1 ={p2-p-1,if 3†p-1,p2-3p-1,if 3∣p-1.
So from (20), Lemmas 2, 3, and 4, and noting that |τ(χ2)|2=p we have
(21)∑m=1p-1|∑a=1p-1χ(ma3+a)|2·|∑b=1p-1e(mb3+bp)|2 =2p·∑m=1p-1|∑c=1p-1e(mc3+cp)|2 +τ2(χ2)2p∑a=1p-1(χ(a)+χ¯(a)) ×∑b=1p-1(1-a2b3p)(1-bp) ×∑m=1p-1(mp)|∑c=1p-1e(mc3+cp)|2+τ2(χ2)2p ×∑a=1p-1(χ1(r)χ(a)+χ¯1(r)χ¯(a)) ×∑b=1p-1(1-a2b3p)(1-bp) ×∑m=1p-1(mp)|∑c=1p-1e(mc3+cp)|2 =2p3+E(p),
where E(p) satisfies the inequalities -12p2-2p≤E(p)≤4p2-2p.
This completes the proof of our theorem.