Open Math. 2018; 16: 955–966
Open Mathematics Research Article Shimeng Shen*
On the recursive properties of one kind hybrid power mean involving two-term exponential sums and Gauss sums https://doi.org/10.1515/math-2018-0081 Received May 24, 2018; accepted June 29, 2018.
Abstract: The main purpose of this paper is to study the computational problem of one kind hybrid power mean involving two-term exponential sums and quartic Gauss sums using the analytic method and the properties of the classical Gauss sums, and to prove some interesting fourth-order linear recurrence formulae for this problem. As an application of our result, we can also obtain an exact computational formula for one kind congruence equation mod p, an odd prime. Keywords: The quartic Gauss sums, Two-term exponential sums, Hybrid power mean, The fourth-order linear recurrence formula MSC: 11L05, 11L07
1 Introduction Let p ≥ 3 be an odd prime. For any integer m with (m, p) = 1, the quartic Gauss sums B(m) = B(m, p) is defined as p−1 ma4 B(m) = ∑ e ( ), p a=0 where as usual, e(y) = e2π iy . Recently, some scholars have studied the hybrid power mean problems of various trigonometric sums, and obtained many interesting results. For example, Chen Li and Hu Jiayuan [1] studied the computational problem of the hybrid power mean k
2
p−1 ma3 mc + c S k (p) = ∑ ( ∑ e ( )) ⋅ ∣ ∑ e ( )∣ , p p m=1 a=0 c=1 p−1
p−1
where c denotes the multiplicative inverse of c mod p. That is, c ⋅ c ≡ 1 mod p. For p ≡ 1 mod 3, they used the elementary method to obtain an interesting third-order linear recurrence formula for S k (p). Li Xiaoxue and Hu Jiayuan [2] studied the computational problem of the hybrid power mean p−1 p−1
∑ ∣∑ e ( b=1 a=0
2
2
p−1 ba4 bc + c )∣ ⋅ ∣ ∑ e ( )∣ , p p c=1
(1)
and proved an exact computational formula for (1).
*Corresponding Author: Shimeng Shen: School of Mathematics, Northwest University, Xi’an, Shaanxi, China, E-mail:
[email protected] Open Access. © 2018 Shen, published by De Gruyter. NonCommercial-NoDerivs 4.0 License.
This work is licensed under the Creative Commons Attribution-
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956 | S. Shen Zhang Han and Zhang Wenpeng [3] proved the identity 4
2p3 − p2 if 3 ∤ p − 1, ma3 + na )∣ = { 3 ∑ ∣∑ e ( p 2p − 7p2 if 3 ∣ p − 1. m=1 a=0 p−1 p−1
Other related results can also be found in references [4-13]. In this paper, we will consider the calculating problem of the following hybrid power mean: k
3
p−1 ma4 mb4 + b V k (p) = ∑ ( ∑ e ( )) ⋅ ( ∑ e ( )) , p p m=1 a=0 b=0 p−1
p−1
(2)
where k ≥ 0 is an integer. If p = 4h + 3, then from the properties of the Legendre’s symbol mod p we have (see [14], formula (30) in Chapter 9) p−1 p−1 p−1 √ ma2 ma2 ma4 ) = 1 + ∑ (1 + χ2 (a)) e ( ) = ∑ e( ) = iχ2 (m) p, ∑ e( p p p a=1 a=0 a=0 where χ2 = ( ∗p ) denotes the Legendre’s symbol mod p. So in this case, the problem we considered in (2) is trivial. If p = 4h + 1, then the situation is more complicated. We will use the analytic method and the properties of classical Gauss sums to study this problem, and prove some new interesting fourth-order linear recurrence formulae for (2) with p = 4h + 1. That is, we will give the following four results. Theorem 1.1. Let p be a prime with p = 24h + 1. Then for any integer k ≥ 4, we have the fourth-order linear recurrence formula V k (p) = 6pV k−2 (p) + 8pαV k−3 (p) − p (p − 4α2 ) V k−4 (p), where the first four values are V0 (p) = p2 − 6pα, V1 (p) = p (p2 − 16p − 4α2 ), V2 (p) = p2 (2pα + 3p − 58α) 2
2
p−1 2
2
and V3 (p) = p (7p + 4pα − 92p − 72α ), α = α(p) = ∑ ( a=1
a+a ) is an integer, which satisfies the identity p
(see Theorem 4-11 in [15]) p−1
2
2
p−1
⎛ 2 a + a ⎞ ⎛ 2 a + ra ⎞ p = α + β ≡ ⎜∑ ( )⎟ + ⎜ ∑ ( )⎟ , p p ⎝a=1 ⎠ ⎝a=1 ⎠ 2
2
which r is any quadratic non-residue mod p. Theorem 1.2. Let p be a prime with p = 24h + 17. Then for any integer k ≥ 4, we have the fourth-order linear recurrence formula V k (p) = 6pV k−2 (p) + 8pαV k−3 (p) − p (p − 4α2 ) V k−4 (p), where the first four values are V0 (p) = −p2 − 6pα, V1 (p) = p (p2 − 18p − 4α2 ), V2 (p) = p2 (2pα − 3p − 62α) and V3 (p) = p2 (7p2 − 4pα − 106p − 72α2 ). Theorem 1.3. Let p be a prime with p = 24h + 5. Then for any integer k ≥ 4, we have the fourth-order linear recurrence formula V k (p) = −2pV k−2 (p) + 8pαV k−3 (p) − p (9p − 4α2 ) V k−4 (p), where the first four terms are V0 (p) = − (p2 + 6pα), V1 (p) −p2 (2pα − p − 22α) and V3 (p) = p2 (5p2 − 6pα − 28p − 36α2 ).
=
−p (p2 − 8p + 4α2 ), V2 (p)
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=
On the recursive properties of one kind hybrid power mean |
957
Theorem 1.4. Let p be a prime with p = 24h + 13. Then for any integer k ≥ 4, we have the fourth-order linear recurrence formula V k (p) = −2pV k−2 (p) + 8pαV k−3 (p) − p (9p − 4α2 ) V k−4 (p), where the first four terms are V0 (p) = p2 − 6pα, V1 (p) = −p (p2 − 6p + 4α2 ), V2 (p) = −p2 (2pα + p − 18α) and V3 (p) = p2 (5p2 + 6pα − 18p − 36α2 ). From our theorems we may immediately deduce the following: Corollary 1.5. Let p be a prime with p ≡ 1 mod 4, then we have the identity p−1
p−1
∑ (∑ e ( m=1
a=0
3
3
p−1 ma4 mb4 + b )) ⋅ ( ∑ e ( )) p p b=0
⎧ ⎪ p2 (7p2 + 4pα − 92p − 72α2 ) if p = 24h + 1, ⎪ ⎪ ⎪ ⎪ ⎪ 2 2 2 ⎪ ⎪ ⎪ p (7p − 4pα − 106p − 72α ) if p = 24h + 17, =⎨ 2 ⎪ p (5p2 − 6pα − 28p − 36α2 ) if p = 24h + 5, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p2 (5p2 + 6pα − 18p − 36α2 ) if p = 24h + 13. ⎪ ⎩ √ Note that the estimate ∣α∣ ≤ p, from Corollary 1.5 we also have the following: Corollary 1.6. Let p be a prime with p ≡ 1 mod 8, then we have the asymptotic formula 3
3
p−1 7 ma4 mb4 + b )) ⋅ ( ∑ e ( )) = 7p4 + O (p 2 ) . ∑ (∑ e ( p p m=1 a=0 b=0 p−1
p−1
Corollary 1.7. Let p be a prime with p ≡ 5 mod 8, then we have the asymptotic formula p−1
3
p−1
∑ (∑ e ( m=1
a=0
3
p−1 7 mb4 + b ma4 )) ⋅ ( ∑ e ( )) = 5p4 + O (p 2 ) . p p b=0
For any prime p with p ≡ 1 mod 4 and any positive integer k, let M k (p) denote the number of the solutions of the congruence equation x41 + x42 + ⋯ + x4k + y41 + y42 + y43 ≡ 0 mod p, y1 + y2 + y3 ≡ 0 mod p, where 0 ≤ x i , y j ≤ p − 1, i = 1, 2, ⋯, k, j = 1, 2, 3. Then from our theorems we can give an exact computational formula for M k (p). For example, let H s (p) denote the number of the congruence equation x41 + x42 + ⋯ + x4s ≡ 0 mod p, 0 ≤ x i ≤ p − 1, i = 1, 2, ⋯s. Then we have the identity V k (p) =
p p2 ⋅ M k (p) − ⋅ H k (p). p−1 p−1
Since H k (p) has a fourth-order linear recurrence formula (see [8]), so from the above formula and our theorems we can deduce the exact value of M k (p).
2 Several lemmas To complete the proofs of our theorems, we need to prove four simple lemmas. Hereafter, we will use many properties of the classical Gauss sums and the fourth-order character mod p, all of which can be found in
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958 | S. Shen books concerning Elementary Number Theory or Analytic Number Theory, such as references [7], [14] or [15]. Some important results related to Gauss sums can also be found in [16] and [17]. These contents will not be repeated here. First we have the following: Lemma 2.1. Let p be a prime with p ≡ 1 mod 4, λ be any fourth-order character mod p, then we have 2
2
τ (λ) + τ (λ) =
√
p−1
p⋅ ∑( a=1
√ a+a ) = 2 p ⋅ α, p
p−1 a where τ (λ) = ∑ λ(a)e ( ) denotes the classical Gauss sums, and ( ∗p ) is the Legendre’s symbol mod p. p a=1
Proof. In fact this is Lemma 2 of [18], so its proof is omitted. Lemma 2.2. Let p be a prime with p ≡ 1 mod 4, then for any fourth-order character λ mod p, we have the identity 3 √ p−1 p−1 −5pτ (λ) − 2 pατ (λ) if p ≡ 1 mod 8, ma4 + a )) = { √ ∑ λ(m) ( ∑ e ( p −pτ (λ) − 2 pατ (λ) if p ≡ 5 mod 8, m=1 a=0 where α is the same as in Lemma 2.1. Proof. First applying trigonometric identity q
∑ e( m=1
q if q ∣ n, nm )={ q 0 if q ∤ n
(3)
and note that λ4 = χ0 , the principal character mod p, we have 3
2
p−1 p−1 ma4 + a ma4 + a )) = ∑ λ(m) ( ∑ e ( )) ∑ λ(m) ( ∑ e ( p p m=1 a=0 m=1 a=0 p−1
p−1
2
p−1 ma4 + a ma4 + a + ∑ λ(m) ( ∑ e ( )) ( ∑ e ( )) p p m=1 a=0 a=1 p−1
p−1
p−1 p−1
p−1
a=0 b=0
c=1
= τ (λ) ∑ ∑ λ (a4 + b4 + 1) ∑ e ( p−1 p−1
+τ (λ) ∑ ∑ λ (a4 + b4 ) e ( a=0 b=0 p−1 p−1
= τ (λ)p
∑∑
4
c(a + b + 1) ) p
a+b ) p 4
p−1 p−1
4
4
λ (a + b + 1) − τ (λ) ∑ ∑ λ (a + b + 1)
a=0 b=0 a+b+1≡0 mod p
a=0 b=0
p−1
p−1
a=0
b=1
−τ (λ) + τ (λ) ∑ λ (a4 + 1) ∑ e (
b(a + 1) ). p
(4)
From (3) we have p−1
p−1
4
τ (λ) ∑ λ (a + 1) ∑ e ( a=0
b=1
p−2 b(a + 1) ) = λ(2)τ (λ)(p − 1) − τ (λ) ∑ λ (a4 + 1) p a=0
p−1
= λ(2)τ (λ)p − τ (λ) ∑ λ (a4 + 1) .
(5)
a=0
Note that the identity λχ2 = λ and p−1
B(m) = ∑ e ( a=0
√ ma4 ) = χ2 (m) p + λ(m)τ (λ) + λ(m)τ (λ) . p
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(6)
On the recursive properties of one kind hybrid power mean |
959
From (6) we have p−1
p−1
4
p−1
τ (λ) ∑ λ (a + 1) = ∑ λ(b) ∑ e a=0
a=0
b=1
⎛ b (a4 + 1) ⎞ p ⎝ ⎠
p−1
√ b = ∑ λ(b) (χ2 (b) p + λ(b)τ (λ) + λ(b)τ (λ)) e ( ) p b=1 √ √ √ = pτ (λ) − τ (λ) + pτ (λ) = 2 pτ (λ) − τ (λ).
(7)
If p ≡ 5 mod 8, then note that λ(−1) = −1 and τ (λ)τ (λ) = −p, applying (6) and Lemma 2.1 we also have p−1 p−1 ca4 + cb4 + c 1 p−1 ) ∑ λ(c) ∑ ∑ e ( τ (λ) c=1 p a=0 b=0
p−1 p−1
4 4 ∑ ∑ λ (a + b + 1) =
a=0 b=0
2
=
c p−1 1 p−1 ca4 )) ∑ λ(c)e ( ) ( ∑ e ( τ (λ) c=1 p a=0 p
=
√ c 1 p−1 2 ∑ λ(c)e ( ) (χ2 (b) p + λ(b)τ (λ) + λ(b)τ (λ)) τ (λ) c=1 p
√ √ √ 1 p−1 c ∑ λ(c) (2χ2 (c) pα − p + 2λ(c) pτ (λ) + 2λ(c) pτ (λ)) e ( ) τ (λ) c=1 p √ 2 p (α − 1) τ (λ) = p+ . τ (λ)
=
(8)
2
Note that λ2 = χ2 = λ and the congruence a + b + 1 ≡ 0 mod p implies the congruence a4 + b4 + 1 ≡ 2 2 (a2 + a + 1) mod p. So we have p−1 p−1
4
p−1
4
2
2
λ (a + b + 1) = ∑ λ (2 (a + a + 1) )
∑∑ a=0 b=0 a+b+1≡0 mod p
a=0
p−1
p−1
a=0
a=0
= λ(2) ∑ χ2 (a2 + a + 1) = λ(2) ∑ χ2 (4a2 + 4a + 4) p−1
p−1
a=0
a=0
= λ(2) ∑ χ2 ((2a + 1)2 + 3) = λ(2) ∑ χ2 (a2 + 3) = −λ(2).
(9)
Combining (4), (5), (7), (8) and (9) we have the identity 3
√ ma4 + a )) = −pτ (λ) − 2 pατ (λ) . ∑ λ(m) ( ∑ e ( p m=1 a=0 p−1
p−1
(10)
If p ≡ 1 mod 8, then λ(−1) = 1 and τ (λ)τ (λ) = p, from the method of proving (8) we have p−1 p−1
4 4 ∑ ∑ λ (a + b + 1) =
a=0 b=0
=
p−1 p−1 1 p−1 ca4 + cb4 + c ) ∑ λ(c) ∑ ∑ e ( τ (λ) c=1 p a=0 b=0
√ c 1 p−1 2 ∑ λ(c)e ( ) (χ2 (b) p + λ(b)τ (λ) + λ(b)τ (λ)) τ (λ) c=1 p
√ √ √ 1 p−1 c ∑ λ(c) (3p + 2χ2 (c) pα + 2λ(c) pτ (λ) + 2λ(c) pτ (λ)) e ( ) τ (λ) c=1 p √ 2 p (α − 1) τ (λ) = 5p + . τ (λ)
=
(11)
Combining (4), (5), (7), (8) and (11) we have the identity 3
√ ma4 + a )) = −5pτ (λ) − 2 pατ (λ) . ∑ λ(m) ( ∑ e ( p m=1 a=0 p−1
p−1
(12)
Now Lemma 2.2 follows from (10) and (12).
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960 | S. Shen Lemma 2.3. Let p be a prime with p ≡ 1 mod 4, then we have the identity p−1
3
p−1
∑ (∑ e ( m=1 a=0 2
={
ma4 + a )) p
p − 6pα if p = 24h + 1 or p = 24h + 13, −p2 − 6pα if p = 24h + 5 or p = 24h + 17.
Proof. From (3) we have p−1
3
p−1
∑ (∑ e ( m=1
a=0
ma4 + a )) = p p
p−1 p−1
=p
e(
∑∑ a=0 b=0 a4 +b4 ≡0 mod p p−1
=p
∑ a=0 a4 ≡0 mod p
a e( )+p p
a+b+c ) p
p−1 p−1
p−1
∑∑
∑ e(
a=0 b=0 c=1 a4 +b4 +1≡0 mod p
p−1
p−1
∑
∑ e(
a=0 b=1 a4 +1≡0 mod p
c(a + b + +1) ) p
b(a + 1) ) p
p−1 p−1
1−p
∑∑ a=0 b=0 a4 +b4 +1≡0 mod p a+b+1≡0 mod p p−1
= p−p
e(
∑∑∑ a=0 b=0 c=0 a4 +b4 +c4 ≡0 mod p
a+b )+p p
p−1 p−1
+p2
p−1 p−1 p−1
p−1 p−1
p−1 p−1
1 + p2
∑
1
∑∑ a=0 b=0 a4 +b4 +1≡0 mod p
1−p
∑∑ a=0 b=0 a4 +b4 +1≡0 mod p a+b+1≡0 mod p
a=0 a4 +1≡0 mod p
∑∑
1.
(13)
a=0 b=0 a4 +b4 +1≡0 mod p
Now we calculate each term in (13). If p ≡ 5 mod 8, then note that λ(−1) = −1 we have p−1
p
1 = 0.
∑
(14)
a=0 mod p
a4 +1≡0
Applying (6) and Lemma 2.1 we have p−1 p−1 p−1
p−1 p−1
p
1= ∑ ∑ ∑ e
∑∑ a=0 b=0 mod p
a=0 b=0 m=0
⎛ m (a4 + b4 + 1) ⎞ p ⎝ ⎠
a4 +b4 +1≡0
2
ma4 m = p + ∑ (∑ e ( )) e ( ) p p m=1 a=0 2
p−1
p−1
p−1 √ √ √ m = p2 + ∑ (2χ2 (c) pα − p + 2λ(c) pτ (λ) + 2λ(c) pτ (λ)) e ( ) p m=1 √ 2 √ 2 2 2 = p + 2pα + p + 2 pτ (λ) + 2 pτ (λ) = p + p + 6pα.
(15)
It is clear that the congruences a4 + b4 + 1 ≡ 0 mod p and a + b + 1 ≡ 0 mod p implies that ab ≡ 1 mod p and a3 ≡ b3 ≡ 1 mod p with a ≠ b. So we have p−1 p−1
p2
1 = p2
∑∑ a=0 b=0 a4 +b4 +1≡0 mod p a+b+1≡0 mod p
p−1 p−1
∑∑ a=2 b=2 a3 ≡b3 ≡1 mod p ab≡1 mod p
1={
2p2 if p ≡ 1 mod 3, 0 if p ≡ 2 mod 3.
(16)
Applying (13), (14), (15) and (16) we have the identity p−1
p−1
∑ (∑ e ( m=1
a=0
3
p2 − 6pα if p = 24h + 13, ma4 + a )) = { 2 p −p − 6pα if p = 24h + 5.
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(17)
On the recursive properties of one kind hybrid power mean
| 961
If p ≡ 1 mod 8, then we also have p−1
p
1 = 4p.
∑
(18)
a=0 a4 +1≡0 mod p
p−1 p−1
p
p−1 p−1 p−1
1= ∑ ∑ ∑ e
∑∑ a=0 b=0 a4 +b4 +1≡0 mod p
a=0 b=0 m=0
⎛ m (a4 + b4 + 1) ⎞ p ⎝ ⎠
p−1 √ √ √ m = p2 + ∑ (3p + 2χ2 (c) pα + 2λ(c) pτ (λ) + 2λ(c) pτ (λ)) e ( ) p m=1 √ 2 √ 2 2 2 = p + 2pα − 3p + 2 pτ (λ) + 2 pτ (λ) = p − 3p + 6pα.
p−1 p−1
p2
p−1 p−1
1 = p2
∑∑ a=0 b=0 a4 +b4 +1≡0 mod p a+b+1≡0 mod p
∑∑ a=2 b=2 a3 ≡b3 ≡1 mod p ab≡1 mod p
1={
2p2 if p = 24h + 1, 0 if p = 24h + 17.
(19)
(20)
Applying (13), (18), (19) and (20) we have 3
p2 − 6pα if p = 24h + 1, ma4 + a )) = { 2 ∑ (∑ e ( p −p − 6pα if p = 24h + 17. m=1 a=0 p−1
p−1
(21)
It is clear that Lemma 2.3 follows from (17) and (21). Lemma 2.4. Let p be a prime with p ≡ 1 mod 4, then we have the identity 3 ⎧ p 2 (p − 6) ⎪ ⎪ ⎪ 3 3 ⎪ ⎪ p−1 p−1 ⎪ ma4 + a ⎪ p 2 (p − 8) )) = ⎨ 3 ∑ χ2 (m) ( ∑ e ( ⎪ p −p 2 (p − 4) ⎪ m=1 a=0 ⎪ ⎪ 3 ⎪ ⎪ ⎪ ⎩ −p 2 (p − 6)
if p = 24h + 1, if p = 24h + 17, if p = 24h + 13, if p = 24h + 5.
Proof. From the properties of the Legendre’s symbol mod p we have p−1
3
p−1
∑ χ2 (m) ( ∑ e ( m=1
a=0
p−1
2
p−1
+ ∑ χ2 (m) ( ∑ e ( m=1
√ =
a=0
2
p−1 p−1 ma4 + a ma4 + a )) = ∑ χ2 (m) ( ∑ e ( )) p p m=1 a=0 p−1 ma4 + a mc4 + c )) ( ∑ e ( )) p p c=1
p−1
p−1 √ p−1 b(a + 1) a p ∑ e ( ) + p ∑ χ2 (a4 + 1) ∑ e ( ) p p a=1 a=0 b=1
p−1 √ p−1 p−1 c(a + b + 1) + p ∑ ∑ χ2 (a4 + b4 + 1) ∑ e ( ) p a=0 b=0 c=1 3 √ √ p−1 √ p−1 p−1 = − p + χ2 (2)p 2 − p ∑ χ2 (a4 + 1) − p ∑ ∑ χ2 (a4 + b4 + 1)
a=0 p−1 p−1
3
+p 2
∑∑
4
a=0 b=0 4
χ2 (a + b + 1) .
(22)
a=0 b=0 a+b+1≡0 mod p
From the properties of fourth-order mod p and Lemma 2.1 we have p−1
p−1
a=0
a=1
4 ∑ χ2 (a + 1) = 1 + ∑ χ2 (a + 1) (1 + λ(a) + χ2 (a) + λ(a))
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962 | S. Shen 1 = √ ⋅ (τ 2 (λ) + τ 2 (λ) − 1 = 2α − 1. p p−1 p−1
∑∑
4
p−1
4
(23)
4
4
χ2 (a + b + 1) = ∑ χ2 (a + (a + 1) + 1)
a=0 b=0 a+b+1≡0 mod p
a=0
p−1
p−1
a=0
a=0
2
= χ2 (2) ∑ χ2 (a4 + 2a3 + 3a2 + 2a + 1) = χ2 (2) ∑ χ2 ((a2 + a + 1) ) ={
χ2 (2)p if p = 12h + 1, χ2 (2)(p − 2) if p = 12h + 5.
(24)
Note that τ (λ)τ (λ) = −p, if p = 8h + 5. τ (λ)τ (λ) = p, if p = 8h + 1. From the method of proving (15) and (19) we have √
p−1 p−1
p−1 p−1 p−1
a=0 b=0
a=0 b=0 m=1
p ∑ ∑ χ2 (a4 + b4 + 1) = ∑ ∑ ∑ χ2 (m)e (
ma4 + mb4 + m ) p
p−1
√ m 2 = ∑ χ2 (m) (χ2 p + λ(m)τ (λ) + λ(m)τ (λ)) e ( ) p m=1 3 √ ⎧ ⎪ ⎪ 7p 2 − 2 pα if p = 8h + 1, =⎨ 3 √ ⎪ 2 ⎪ ⎩ −5p − 2 pα if p = 8h + 5.
(25)
Combining (22), (23), (24) and (25) we have 3 ⎧ p 2 (p − 6) ⎪ ⎪ ⎪ 3 ⎪ ⎪ p−1 p−1 ⎪ ma4 + a ⎪ p 2 (p − 8) )) = ⎨ 3 ∑ χ2 (m) ( ∑ e ( ⎪ p −p 2 (p − 4) ⎪ m=1 a=0 ⎪ ⎪ 3 ⎪ ⎪ ⎪ ⎩ −p 2 (p − 6)
3
if p = 24h + 1, if p = 24h + 17, if p = 24h + 13, if p = 24h + 5.
This proves Lemma 2.4.
3 Proofs of the theorems Now we prove our main results. First we prove Theorem 1.1. If p = 24h + 1, then from Lemmas 2.1, 2.2 and 2.4 we have 3
ma4 + a V1 (p) = ∑ B(m) ( ∑ e ( )) p m=1 a=0 p−1
p−1
3
p−1 √ ma4 + a = ∑ (χ2 (m) p + λ(m)τ (λ) + λ(m)τ (λ)) ( ∑ e ( )) p m=1 a=0 √ √ = p2 (p − 6) − 5p2 − 2 pατ 2 (λ) − 5p2 − 2 pατ 2 (λ) p−1
= p (p2 − 16p − 4α2 ) .
(26)
Applying Lemmas 2.1–2.4 we also have p−1
p−1
m=1
a=0
V2 (p) = ∑ B2 (m) ( ∑ e (
3
ma4 + a )) p
3
p−1 p−1 √ ma4 + a 2 = ∑ (χ2 (m) p + λ(m)τ (λ) + λ(m)τ (λ)) ( ∑ e ( )) p a=0 m=1
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On the recursive properties of one kind hybrid power mean
| 963
3
p−1 p−1 √ √ √ ma4 + a = ∑ (3p + 2χ2 (m) pα + 2λ(m) pτ (λ) + 2λ(m) pτ (λ)) ( ∑ e ( )) p m=1 a=0
= p2 (2pα + 3p − 58α) .
(27)
If p = 8h + 1, then from (6) we have √ 3 B3 (m) = (χ2 (m) p + λ(m)τ (λ) + λ(m)τ (λ)) 3
= 7χ2 (m)p 2 + 4pα + 5p (λ(m)τ (λ) + λ(m)τ (λ)) √ +2 (λ(m)τ (λ) + λ(m)τ (λ)) pα.
(28)
So if p = 24h + 1, then from (28), Lemmas 2.1–2.4 we have 3
ma4 + a V3 (p) = ∑ B (m) ( ∑ e ( )) p m=1 a=0 p−1
p−1
3
√ = 7p3 (p − 6) + 4pα (p2 − 6pα) − 5pτ (λ) (5pτ (λ) + 2 pατ (λ)) √ √ √ −5pτ (λ) (5pτ (λ) + 2 pατ (λ)) − 2 pατ (λ) (5pτ (λ) + 2 pατ (λ)) √ √ −2 pατ (λ) (5pτ (λ) + 2 pατ (λ)) = p2 (7p2 + 4pα − 92p − 72α2 ) .
(29)
If p = 24h + 17, then from Lemmas 2.1–2.4 we have 3
ma4 + a V1 (p) = ∑ B(m) ( ∑ e ( )) p m=1 a=0 √ √ = p2 (p − 8) − 5p2 − 2 pατ 2 (λ) − 5p2 − 2 pατ 2 (λ) p−1
p−1
= p (p2 − 18p − 4α2 ) .
p−1
p−1
m=1
a=0
(30) 3
V2 (p) = ∑ B2 (m) ( ∑ e (
ma4 + a )) p
3
p−1 √ √ √ ma4 + a = ∑ (3p + 2χ2 (c) pα + 2λ(c) pτ (λ) + 2λ(c) pτ (λ)) ( ∑ e ( )) p a=0 m=1 p−1
= p2 (2pα − 3p − 62α) .
(31)
Applying (28) and the method of proving (29) we also have p−1
p−1
m=1
a=0
V3 (p) = ∑ B3 (m) ( ∑ e (
3
ma4 + a )) p
√ = 7p3 (p − 8) − 4pα (p2 + 6pα) − 5pτ (λ) (5pτ (λ) + 2 pατ (λ)) √ √ √ −5pτ (λ) (5pτ (λ) + 2 pατ (λ)) − 2 pατ (λ) (5pτ (λ) + 2 pατ (λ)) √ √ −2 pατ (λ) (5pτ (λ) + 2 pατ (λ)) = p2 (7p2 − 4pα − 106p − 72α2 ) .
(32)
Similarly, if p = 24h + 5, then we have p−1
p−1
V1 (p) = ∑ B(m) ( ∑ e ( m=1
a=0
3
ma4 + a )) p
√ √ = −p2 (p − 6) + p2 − 2 pατ 2 (λ) + p2 − 2 pατ 2 (λ) = −p (p2 − 8p + 4α2 ) .
(33)
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964 | S. Shen p−1
p−1
m=1
a=0
3
V2 (p) = ∑ B2 (m) ( ∑ e (
ma4 + a )) p
3
p−1 √ ma4 + a 2 = ∑ (χ2 (m) p + λ(m)τ (λ) + λ(m)τ (λ)) ( ∑ e ( )) p m=1 a=0 p−1
3
p−1 p−1 √ √ √ ma4 + a = ∑ (2χ2 (c) pα − p + 2λ(c) pτ (λ) + 2λ(c) pτ (λ)) ( ∑ e ( )) p a=0 m=1
= −p2 (2pα − p − 22α) .
(34)
If p = 24h + 5, then from (6) we have √ 3 B3 (m) = (χ2 (m) p + λ(m)τ (λ) + λ(m)τ (λ)) 3
= −5χ2 (m)p 2 + 6pα + p (λ(m)τ (λ) + λ(m)τ (λ)) √ +2 (λ(m)τ (λ) + λ(m)τ (λ)) pα.
(35)
So from (35) and the method of proving (29) we have p−1
p−1
V3 (p) = ∑ B3 (m) ( ∑ e ( m=1
a=0
3
ma4 + a )) p
√ = 5p (p − 6) − 6pα (p + 6pα) − pτ (λ) (pτ (λ) + 2 pατ (λ)) √ √ √ −pτ (λ) (pτ (λ) + 2 pατ (λ)) − 2 pατ (λ) (pτ (λ) + 2 pατ (λ)) √ √ −2 pατ (λ) (pτ (λ) + 2 pατ (λ)) 3
2
= p2 (5p2 − 6pα − 28p − 36α2 ) .
(36)
If p = 24h + 13, then (35), Lemmas 2.1–2.4 we have p−1
3
p−1
V1 (p) = ∑ B(m) ( ∑ e ( m=1
a=0
ma4 + a )) p
√ √ = −p (p − 4) + p − 2 pατ 2 (λ) + p2 − 2 pατ 2 (λ) 2
2
= −p (p2 − 6p + 4α2 ) .
p−1
p−1
m=1
a=0
(37) 3
V2 (p) = ∑ B2 (m) ( ∑ e (
ma4 + a )) p
3
p−1 √ ma4 + a 2 )) = ∑ (χ2 (m) p + λ(m)τ (λ) + λ(m)τ (λ)) ( ∑ e ( p m=1 a=0 p−1
3
p−1 p−1 √ √ √ ma4 + a = ∑ (2χ2 (c) pα − p + 2λ(c) pτ (λ) + 2λ(c) pτ (λ)) ( ∑ e ( )) p a=0 m=1
= −p2 (2pα + p − 18α) . p−1
(38) p−1
V3 (p) = ∑ B3 (m) ( ∑ e ( m=1
a=0
3
ma4 + a )) p
√ = 5p (p − 4) + 6pα (p − 6pα) − pτ (λ) (pτ (λ) + 2 pατ (λ)) √ √ √ −pτ (λ) (pτ (λ) + 2 pατ (λ)) − 2 pατ (λ) (pτ (λ) + 2 pατ (λ)) √ √ −2 pατ (λ) (pτ (λ) + 2 pατ (λ)) 3
2
= p2 (5p2 + 6pα − 18p − 36α2 ) .
(39)
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On the recursive properties of one kind hybrid power mean
| 965
Finally, note that if p = 8h +1, then from (6) and direct calculation (or see Lemma 3 in [7]) we have the identity B4 (m) = 6pB2 (m) + 8pαB(m) − p (p − 4α2 ) .
(40)
For any prime p = 24h + 1 and integer k ≥ 4, from (26), (27), (29) and (40) we may immediately deduce the fourth-order linear recurrence formula p−1
p−1
m=1
a=0
V k (p) = ∑ B k (m) ( ∑ e (
3
ma4 + a )) p
p−1
3
p−1
= ∑ B k−4 (m) (6pB2 (m) + 8pαB(m) − p (p − 4α2 )) ( ∑ e ( m=1
a=0
ma4 + a )) p
2
= 6pV k−2 (p) + 8pαV k−3 (p) − p (p − 4α ) V k−4 (p), where the first four values V0 (p) = p2 − 6pα, V1 (p) = p (p2 − 16p − 4α2 ), V2 (p) = p2 (2pα + 3p − 58α) and V3 (p) = p2 (7p2 + 4pα − 92p − 72α2 ). This proves Theorem 1.1. If p = 24h + 17, then from (30), (31), (32) and (40) we have V k (p) = 6pV k−2 (p) + 8pαV k−3 (p) − p (p − 4α2 ) V k−4 (p), where the first four values V0 (p) = −p2 − 6pα, V1 (p) = p (p2 − 18p − 4α2 ), V2 (p) = p2 (2pα − 3p − 62α) and V3 (p) = p2 (7p2 − 4pα − 106p − 72α2 ). This proves Theorem 1.2. If p = 8h + 5, then from (6) and direct calculation (or see Lemma 3 in [7]) we also have B4 (m) = −2pB2 (m) + 8pαB(m) − p (9p − 4α2 ) .
(41)
For any prime p = 24h + 5 and integer k ≥ 4, from (33), (34), (35) and (41) we can deduce the fourth-order linear recurrence formula V k (p) = −2pV k−2 (p) + 8pαV k−3 (p) − p (9p − 4α2 ) V k−4 (p), where the first four terms are V0 (p) = − (p2 + 6pα), V1 (p) −p2 (2pα − p − 22α) and V3 (p) = p2 (5p2 − 6pα − 28p − 36α2 ). This proves Theorem 1.3. If p = 24h + 13, then from (37), (38), (39) and (41) we also have
=
−p (p2 − 8p + 4α2 ), V2 (p)
=
V k (p) = −2pV k−2 (p) + 8pαV k−3 (p) − p (9p − 4α2 ) V k−4 (p), where the first four terms are V0 (p) = p2 − 6pα, V1 (p) = −p (p2 − 6p + 4α2 ), V2 (p) = −p2 (2pα + p − 18α) and V3 (p) = p2 (5p2 + 6pα − 18p − 36α2 ). This completes the proofs of our all results. Acknowledgement: The author would like to thank the referees for their very helpful and detailed comments, which have significantly contributed to improving the presentation of this paper. This work is supported by the N. S. F. (11771351) of P. R. China and Northwest University Graduate Innovation and Creativity Founds (YZZ17086).
References [1]
Chen L., Hu J. Y., A linear recurrence formula involving cubic Gauss sums and Kloosterman sums, Acta Mathematica Sinica (Chinese Series), 2018, 61, 67–72.
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966 | S. Shen [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]
Li X. X., Hu J. Y., The hybrid power mean quartic Gauss sums and Kloosterman sums, Open Mathematics, 2017, 15, 151– 156. Zhang H., Zhang W. P., The fourth power mean of two-term exponential sums and its application, Mathematical Reports, 2017, 19, 75–83. Chowla S., Cowles J. and Cowles M., On the number of zeros of diagonal cubic forms, Journal of Number Theory, 1977, 9, 502–506. Cochrane T. Pinner C., Using Stepanov’s method for exponential sums involving rational functions, Journal of Number Theory, 2006, 116, 270–292. Han D., A Hybrid mean value involving two-term exponential sums and polynomial character sums, Czechoslovak Mathematical Journal, 2014, 64, 53–62. Hua L. K., Introduction to Number Theory, Science Press, Beijing, 1979. Shen S. M., Zhang W. P., On the quartic Gauss sums and their recurrence property, Advances in Difference Equations, 2017, 2017: 43 Weil A., On some exponential sums, Proc. Nat. Acad. Sci. U.S.A., 1948, 34, 204–207. Zhang W. P., On the fourth power mean of the general Kloosterman sums, Journal of Number Theory, 2016, 169, 315–326. Zhang W. P., Han D., On the sixth power mean of the two-term exponential sums, Journal of Number Theory, 2014, 136, 403–413. Zhang W. P., Hu J. Y., The number of solutions of the diagonal cubic congruence equation mod p, Mathematical Reports, 2018, 20, 73–80. Zhang W. P., Liu H. N., On the general Gauss sums and their fourth power mean, Osaka Journal of Mathematics, 2005, 42, 189–199. Apostol T. M., Introduction to Analytic Number Theory, Springer-Verlag, New York, 1976. Zhang W. P., Li H. L., Elementary Number Theory, Shaanxi Normal University Press, Xi’an, Shaanxi, 2008. Berndt B. C., Evans R. J., Sums of Gauss, Jacobi, and Jacobsthal, Journal of Number Theory, 1979, 11, 349–389. Berndt B. C., Evans R. J., The determination of Gauss sums, Bulletin of the American Mathematical Society, 1981, 5, 107– 128. Chen Z. Y., Zhang W. P., On the fourth-order linear recurrence formula related to classical Gauss sums, Open Mathematics, 2017, 15, 1251–1255.
Unauthenticated Download Date | 8/25/18 2:30 PM
