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≡1 mod 4, there exists two non-zero integers αp and βp such that (see [15]: Theorems 4–11) (1)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≡1 mod 6, and prove the following result:

Theorem. Let p be a prime with p≡1 mod 6, 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≡1 mod 3 and b>0.

From this Theorem, we can immediately deduce the following four Corollaries:

Corollary 1. Let p be a prime with p≡1 mod 6, 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≡1 mod 6, 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≡1 mod 6, 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≡1 mod 6, 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≡1 mod 6. If p≡5 mod 6, 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≡5 mod 6 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≡1 mod 6 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≡1 mod 6 and d≡1 mod 3. The exact results of calculation are summarised in Table 1.

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≡1 mod 3, 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 (2)∑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 (3)∑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≡1 mod 6, 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 (4)13⋅(1+λ(a)+λ¯(a))={1ifaisacubicresiduemodulop;0otherwise.

Applying Eq. (4) we have (5)∑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 (6)∑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 (7)τ(λ¯)=λ¯(2)τ(χ2)⋅τ(λ)⋅τ(λχ2).

Note that τ(λ)⋅τ(λ¯)=p, from Eqs. (6) and (7) we have (8)∑a=1p−1λ(a)χ2(a)χ2(a+1)=λ(2)p⋅τ3(λ¯).

Similarly, we also have (9)∑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≡1 mod 6, 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 (10)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≡1 mod 6, then we have S0(p)=1,S1(p)=0,S2(p)=2p,S3(p)=d⋅(d2−3p)and Sk(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 (11)S0(p)=1,S1(p)=1p(p−1)∑r=1p−1(λ¯(2r)⋅τ3(λ)+λ(2r)⋅τ3(λ¯))=0, (12)S2(p)=1p2(p−1)∑r=1p−1(λ¯(2r)⋅τ3(λ)+λ(2r)⋅τ3(λ¯))2=2p.

From Eq. (10) we also have (13)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 (14)Sk(p)=3p⋅Sk−2(p)+(d3−3dp)⋅Sk−3(p).

Now Lemma 5 follows from Eqs. (11)–(14).

In this section, we complete the proof of our Theorem. It is clear that the characteristic equation of the third order linear recursive formula (15)Sk(p)=3p⋅Sk−2(p)+(d3−3dp)⋅Sk−3(p)is (16)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 (17)Sk(p)=C1⋅dk+C2⋅(−d+9b2)k+C3⋅(−d−9b2)k,k≥0.

From Lemma 5 we have (18){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.

In this paper, we give an exact computational formula for Sk(p) with p≡1 mod 6, 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≡1 mod 6 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≡1 mod 6 and d≡1 mod 3, as shown in Table 1. Its application can also extend to Sk(p) that satisfies conditions p≡1 mod 6 and d≡1 mod 3 (where 4p=d2+27b2) for any k. See the Appendix A for this specific computer program.