Gauss sums and jacobi sums

MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY N2 DEPARTMENT OF MATHEMATICS NGUYEN THI TRANG GAUSS SUMS AND JACOBI SUMS GRADUATION THESIS Major: Algebra HA NOI - 2019 MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY N2 DEPARTMENT OF MATHEMATICS NGUYEN THI TRANG GAUSS SUMS AND JACOBI SUMS GRADUATION THESIS Major: Algebra Supervisor: Dr NGUYEN DUY TAN HA NOI – 2019 Thesis Acknowledgement I would like to express my gratitudes to the teachers of the Department of Mathematics, Hanoi Pedagogical University 2, the teachers in the algebra group as well as the teachers involved The lecturers have imparted valuable knowledge and facilitated for me to complete the course and the thesis In particular, I would like to express my deep respect and gratitude to Dr Nguyen Duy Tan, who has direct guidance, help me complete this thesis Due to time, capacity and conditions are limited, so the thesis can not avoid errors Then, I look forward to receiving valuable comments from teachers and friends Ha Noi, May 5, 2019 Student Nguyen Thi Trang Graduation thesis NGUYEN THI TRANG Thesis Assurance I assure that the data and the results of this thesis are true and not identical to other topics I also assure that all the help for this thesis has been acknowledged and that the results presented in the thesis has been identified clearly Ha Noi, May 5, 2019 Student Nguyen Thi Trang i Contents Preface 1 Preliminaries 1.1 Quadratic residues 1.2 The matrix and its characteristic function 1.3 Multiplicative characters Quadratic Gauss sums 13 2.1 The absolute value of quadratic Gauss sums 13 2.2 The sign of the quadratic Gauss sums 17 2.3 Extensions to composite moduli 30 2.4 Application to the quadratic Gauss sums 33 2.4.1 The law of quadratic reciprocity 33 2.4.2 Some related trigonometric problems 34 Gauss sums and Jacobi sums 36 3.1 Gauss sums 36 3.2 Jacobi sums 38 3.3 The equation xn + y n = in Fp 46 3.4 Exercises 47 Conclusion 58 References 58 ii Graduation thesis NGUYEN THI TRANG Preface Devised in the 19th century, Gauss and Jacobi Sums are classical formulas that form the basis for contemporary research in many of today’s sciences In algebraic number theory, a Gauss sum is a particular kind of finite sum of roots of unity Gauss utilized the properties of the sum to solve certain problem in number theory; a particular case is one of the proofs of the quadratic reciprocity law A Jacobi sum is a type of character sum formed with Dirichlet characters Simple examples would be Jacobi sums J(χ, ψ) for Dirichlet chracters χ, ψ modulo a prime number p, defined by J(χ, ψ) = χ(a)ψ(1 − a), where the summation runs over all residues a = 2; 3; · · · ; p − (mod p) (for which neither a nor − a is 0) The notions of Gauss and Jacobi sums are defined and applied to investigate the number of solutions of polynomial equations over finite field Base on the basic knowledge about algebraic and desiring comprehensive improvement of Mathematics, we has selected a topic “Gauss sums and Jacobi sums” for Graduation Thesis This thesis is the general result of books [2],[3],[4] The main aim in this thesis is to know how to be Gauss sums and Jacobi sums, understand some proofs about The sign of the Quadratic Gauss Sum, prove the laws of quadratic Reciprocity and find the number of solutions to congruence equations of the form xn + y n = (mod p) Let us describe the content of this thesis In Chapter 1, we will review the definition and some basic properties of quadratic residues, Legendre symbol, prove Wilson’s theorem and recall some knowledge about the matrix and its characteristic function Then we introduce the notion of multiplicative characters needed for the results of this thesis Graduation thesis NGUYEN THI TRANG In Chapter 2, we introduce quadratic Gauss sums We will present the notion of Gauss sums, hence quadratic Gauss sums We will give two methods to prove the formula to calculate the absolute value of quadratic Gauss sums In particular, we also present three methods to prove about the sign of the quadratic Gauss sums Moreover, we also prove for two remaining cases n ≡ (mod 4) and n ≡ (mod 4) Next, application to the quadratic Gauss sums is given such as the law of quadratic reciprocity, some related trigonometric problems In Chapter 3, a more general notion of Gauss sum and Jacobi sums will be introduced Here we shall consider a problem of counting the number of solutions of equations with coefficients in a finite field Finally, by using Jacobi sums, we give some exercises to practice Chapter Preliminaries In this Chapter, we will review some results in number theory which will be used in the other chapters We begin with the notion of quadratic residues, matrices and their characteristic functions Multiplicative characters are also introduced 1.1 Quadratic residues Theorem 1.1.1 (Wilson’s Theorem) A positive integer n > is a prime if and only if (n − 1)! ≡ −1 (mod n) Proof Firstly, we prove that: If a positive integer n > is a prime, then (n − 1)! ≡ −1 (mod n) Consider the field Z/nZ - the set of integers modulo n By Fermat’s little theorem for n is a prime, we have xn−1 − ≡ (mod n) for all ≤ x ≤ n − So in Z/nZ, there must be the n − roots of f Hence, we can write n−1 x n−1 −1= n−1 (x − k) = (−1) k=1 n−1 · (k − x) k=1 Graduation thesis NGUYEN THI TRANG For odd prime n, n − is even, and (−1)n−1 = For even prime n = 2, we get (−1)n−1 = −1 ≡ (mod 2) Thus, xn−1 − = Let x = we get n−1 k=1 k n−1 k=1 (k − x) = −1 = (n − 1)! in Z/nZ That means (n − 1)! ≡ −1 (mod n) Secondly, we must prove that: For a positive integer n > 1, if (n − 1)! ≡ −1 (mod n), then n is a prime Indeed, assume that n is a composite number, then it has at least one divisor d less than n, that is ≤ d ≤ n − But since (n − 1)! is the product of all positive integers from to n − 1, the product must contain d and thus (n − 1)! is divisible by d So we have (n − 1)! ≡ (mod d) Since d | n, (n − 1)! ≡ ≡ −1 (mod d), contradicting the hypothesis Then n can not be composite, hence prime Let p be a prime number We consider the equation x2 ≡ k (mod p) Definition 1.1.2 We say that an integer k which is not divisible by p is a quadratic residue modulo p if there exists an integer x such that x2 ≡ k (mod p); we say that k is a quadratic non-residue if no such x exists The case p = is trivial, so we assume p > 2, means that p is an odd prime number From Fermat’s little theorem, we have xp−1 ≡ (mod p), with x ≡ (mod p) Then (xp−1 − 1)(xp−1 + 1) ≡ Theorem 1.1.3 The congruent equation x (mod p) p−1 p−1 roots, are congruent classes , , · · · , 2 − ≡ (mod p) has p−1 (mod p) Other congru- Graduation thesis NGUYEN THI TRANG ent classes (not class 0) are roots of the equation x p−1 +1≡0 (mod p) p−1 Proof Indeed, we know that the equation x − ≡ (mod p) has at p−1 most roots in a residue system modulo p 2 p−1 2 We see that the congruent classes , , · · · , (mod p) are dis2 tinguish and from Fermat’s little theorem, they are also roots of the equation x p−1 roots of x − ≡ (mod p) From that, other congruent classes are not p−1 −1≡0 (mod p) Moreover, (xp−1 − 1) = (x are roots of the equation x p−1 p−1 + 1)(x p−1 − 1) So, other congruent classes + ≡ (mod p) Corollary 1.1.4 The number a is a quadratic residue modulo p if and only if a p−1 ≡1 (mod p) Proof Suppose that a is a quadratic residue modulo p, then a ≡ x2 (mod p) for some x ∈ Z Thus, a p−1 ≡ xp−1 ≡ (mod p) Now, if a is not a quadratic residue modulo p, then from Theorem 1.1.3, a is a root of the equation x p−1 +1 ≡ (mod p), this means that a (mod p) Definition 1.1.5 (Legendre symbol) We write    if a is a quadratic residue modulo p    a = −1 if a is a quadratic non-residue modulo p  p    0 if a ≡ (mod p) a p Corollary 1.1.6 (Euler criterion) ≡a p−1 (mod p) p−1 ≡ −1 Graduation thesis NGUYEN THI TRANG Proof We have g(χ)3 = p · J(χ, χ) ≡ a + bω (mod 3) (because p ≡ (mod 3)) On the other hand, g(χ)3 = χ(t) · ξ t t χ(t)3 · ξ 3t ≡ (mod 3) t ξ 3t = −1 χ(t)3 · ξ 3t = = χ(0) · ξ + t=0 (mod 3) t=0 (Since χ(t)3 = 1) Hence, g(χ)3 ≡ a + bω ≡ −1 (mod 3) Working with χ instead of χ and noting that g(χ) = χ(−1)g(χ), we get g(χ)3 = p · J(χ, χ) = g(χ) ≡ a + bω ≡ −1 Then b(ω − ω) ≡ (mod 3) So, b · √ (mod 3) −3 ≡ (mod 3) It implies that −3b2 ≡ (mod 9) and thus | b2 and hence | b Since | b and a + bω ≡ −1 (mod 3), we have a ≡ −1 (mod 3) b Corollary 3.2.11 Let A = 2a − b and B = Then A ≡ (mod 3) and 4p = A2 + 27B Proof Since p ≡ (mod 3), there are integers a and b such that p = a2 − ab + b2 Hence 4p = 4(a2 − ab + b2 ) = (2a − b)2 + 3b2 = A2 + 27B Since | b, a ≡ −1 (mod 3), we get A = 2a − b ≡ (mod 3) Theorem 3.2.12 Suppose that p ≡ (mod 3) Then there are integers A and B such that 4p = A2 + 27B If we have A ≡ (mod 3), A is uniquely determined and N (x3 + y = 1) = p − + A Proof We already proved that there are A, B such that 4p = A2 + 27B By replacing A by −A if necessary, we may assume that A ≡ −1 (mod 3) Now suppose that 4p = C + 27D2 , for some C, D ∈ Z with C ≡ −1 (mod 3) By Proposition 3.2.7, A = ±C Because A ≡ C ≡ −1, we get A = C and A is uniquely determined 45 Graduation thesis NGUYEN THI TRANG Let χ be a cubic character We get χi (a) · χj (b) N (x3 + y = 1) = i a a χ2 (b) + a+b=1 χ2 (a) · χ2 (b) χ(a) · χ(b) + a+b=1 a+b=1 b χ(a) · χ2 (b) χ2 (a) · χ(b) + a+b=1 b χ2 (a) + + a+b=1 χ(b) + χ(a) + =p+ j Since χ2 = χ−1 , we have χ(a) · χ2 (b) = J(χ, χ−1 ) = −χ(−1) χ2 (a) · χ(b) = a+b=1 a+b=1 Therefore, N (x3 + y = 1) = p − · χ(−1) + J(χ, χ) + J(χ2 , χ2 ) Moreover, χ(−1) = and χ2 = χ−1 = χ Thus N (x3 + y = 1) = p − + · ReJ(χ, χ) Since J(χ, χ) = a + bω, 2· Re J(χ, χ) = 2a − b = A Then, N (x3 + y = 1) = p − + A 3.3 The equation xn + y n = in Fp We will provide the formula to find the number of solutions to the equation xn + y n = over Fp with p ≡ (mod n) One has N (xn + y n = 1) = N (xn = a) · N (y n = b) a+b=1 n−1 n−1 i = χj (b) χ (a) i=0 a+b=1 n−1 n−1 J(χi , χj ) = i=0 j=0 46 j=0 Graduation thesis NGUYEN THI TRANG If i = j = 0, then J(χ0 , χ0 ) = J(ε, ε) = p If i + j = n, then i = n − j and χi = χn−j = (χj )−1 Then, J(χi , χj ) = J((χj )−1 , χj ) = −χj (−1) In this case, N (xn + y n = 1) = If −1 is an nth power, then χ(−1) = 1, then otherwise, then n−1 j j=0 χ (−1) = Therefore, n−1 j j=1 χ (−1) n−1 j j=0 χ (−1) n−1 j j=1 χ (−1) = n If = − δn (−1)n,  1 if − is an nth power where δn (−1) = 0 if otherwise If i = 0, j = or i = 0, j = then J(χi , χj ) = Therefore N (xn + y n = 1) = p + − δn (−1)n + J(χi , χj ) i,j The sum is over indices i and j satisfying ≤ i, j ≤ n − and i + j = n 3.4 Exercises Exercise 1: Use Gauss’s theorem to find the number of solutions to x3 + y = in Fp for p = 13, 19, 37, and 97 Solution (a) p = 13 ≡ (mod 3) We have · 13 = 52 + 27 · 12 Since ≡ (mod 3), we must choose A = −5 Thus, N (x3 + y = 1) = p − + A = 13 − − = (b) p = 19 ≡ (mod 3) We see that · 19 = 72 + 27 · 12 , so A = Thus, N (x3 + y = 1) = 19 − + = 24 (c) p = 37 ≡ (mod 3) We have 4p = · 37 = 148 = 112 + 27 · 12 , so A = −11 ≡ (mod 3) Thus, N (x3 + y = 1) = 37 − − 11 = 24 (d) p = 97 ≡ (mod 3) 47 Graduation thesis NGUYEN THI TRANG One has 4p = · 97 = 388 = 192 + 27 · 12 , so A = 19 Thus, N (x3 + y = 1) = 97 − + 19 = 114 Exercise 2: Let χ be a nontrivial multiplicative character of Fp and ρ be the character of order Show that t χ(1 − t2 ) = J(χ, ρ) Solution One has χ(1 − t2 ) = t χ(1 − a) · N (t2 = a) = a a χ(1 − a)ρ(a) χ(1 − a) + = a a Since a χ(1 χ(1 − a)(1 + ρ(a)) − a) = Thus, χ(1 − t2 ) = χ(1 − a)ρ(a) = J(χ, ρ) t a Exercise 3: Show, if k ∈ Fp , k = 0, that t χ(t(k − t)) = χ(k /22 )J(χ, ρ) Solution We have (t(k − t)) = χ t Put t t k 1− t k · χ(k ) t = t , then k χ(2t · (2 − 2t )) · χ(k /22 ) χ(t(k − t)) = t t χ(1 − (2t − 1)2 ) · χ(k /22 ) = t Put 2t − = u, by Exercise we get u χ(1 − u2 ) = J(χ, ρ) Thus χ(t(k − t)) = χ(k /22 )J(χ, ρ) t 48 Graduation thesis NGUYEN THI TRANG Exercise 4: If χ2 = ε, show that g(χ)2 = χ(2)−2 J(χ, ρ)g(χ2 ) Solution We apply Exercise with k = 1, we get χ(t(1 − t)) = χ(2)−2 · J(χ, ρ) J(χ, χ) = t g(χ)2 Moreover, J(χ, χ) = Thus,g(χ)2 = χ(2)−2 J(χ, ρ)g(χ2 ) g(χ ) Exercise 5: Suppose that p ≡ (mod 4) and that χ is a character of order Then χ2 = ρ and J(χ, χ) = χ(−1)J(χ, ρ) × Solution Firstly, we prove that χ2 = ρ Indeed, we have ρ : F× p → C 2 Since F× p =< g >, [ρ(g)] = ρ (g) = Moreover, ρ is a character of order 2, so ρ(g) = −1 Hence, ρ is the unique character of order As χ2 is a character of order 2, we get χ2 = ρ Secondly, we must prove J(χ, χ) = χ(−1)J(χ, ρ) From the result of Exercise 4, we get g(χ)2 = χ(2)−2 J(χ, ρ)g(χ2 ) Thus g(χ)4 = χ(2)−4 J(χ, ρ) · J(χ, ρ) · g(χ2 )2 = J(χ, ρ) · J(χ, ρ) · g(ρ)2 (Note that χ2 = ρ and χ4 = ε) We have g(ρ)2 = (−1)(p−1)/2 p = p Since J(χ, χ) = χ(2)−2 J(χ, ρ), we get J(χ, ρ) = χ(2)2 · J(χ, χ) Thus g(χ)4 = χ(2)4 · J(χ, χ) · J(χ, χ) · p = J(χ, χ) · J(χ, χ) · p Moreover, we get g(χ)4 = χ(−1) · p · J(χ, χ) · J(χ, χ2 ) = χ(−1) · p · J(χ, χ) · J(χ, ρ) Hence J(χ, χ) = χ(−1) · J(χ, ρ) 49 Graduation thesis NGUYEN THI TRANG Exercise 6: Suppose that p ≡ (mod 3) and that χ is a character of order Prove that g(χ)3 = pπ, where π = χ(2)J(χ, ρ) Solution We have g(χ)3 = p · J(χ, χ) = pχ(2)−2 J(χ, ρ) = pχ−2 (2)J(χ, ρ) = pχ(2)J(χ, ρ) Exercise 7: (continuation) Show that χρ is a character of order and that g(χρ)6 = (−1)(p−1)/2 pπ Solution Suppose that ord(χρ) = m, m ∈ N∗ It implies that (χρ)m = ε Since (χρ)m = χm ρm = ε, hence χm = ρ−m Then ε = χ3m = ρ−3m Thus | 3m, thus | m Similarly, one has χ2m = ρ−2m = ε, hence | 2m, and | m As (3, 2) = 1, it implies that | m Thus, m = and χρ is a character of order We have g(χ)6 · g(ρ)6 g(χρ) = J(χ, ρ)6 We know that g(χ)3 = pπ, g(ρ)2 = (−1)(p−1)/2 · p and J(χ, ρ) = π χ(2) Therefore p2 π · (−1)(p−1)/2 · p3 g(χρ) = π /χ6 (2) p = p · (−1)(p−1)/2 · π = p · (−1)(p−1)/2 · π (Note that p = π · π) Exercise 8: Suppose that p ≡ (mod 4) and that χ is a character of order Let N be the number of solutions to x4 + y = in Fp Show that N = p + − δ4 (−1)4 + 2ReJ(χ, χ) + 4ReJ(χ, ρ) 50 Graduation thesis NGUYEN THI TRANG Solution We have N (x4 + y = 1) = p + − δ4 (−1)4 + J(χi , χj ), i,j where i,j J(χ i , χj ) = J(χ, χ) + 2J(χ, χ2 ) + 2J(χ2 , χ3 ) + J(χ3 , χ3 ) Since χ = ρ, and χ3 = χ−1 = χ, J(χi , χj ) = J(χ, χ) + 2J(χ, ρ) + 2J(χ, ρ) + J(χ, χ) i,j = 2ReJ(χ, χ) + 4ReJ(χ, ρ) Thus N = p + − δ4 (−1)4 + 2ReJ(χ, χ) + 4ReJ(χ, ρ) Exercise 9: (continuation) By Exercise 8, J(χ, χ) = χ(−1)J(χ, ρ) Let π = −J(χ, ρ) Show that (a) N = p − − 6Reπ if p ≡ (mod 8), (b) N = p + − 2Reπ if p ≡ (mod 8) Solution (a) Since p ≡ (mod 8), one has p − = 8k, for some k ∈ N× × × Since F× p is a cyclic group, we may write Fp = g , for some g ∈ Fp 4k Let x = g k Then x4 = g 4k has order in F× = −1 = x4 p Thus g 4 in F× p Therefore, χ(−1) = χ(x ) = χ(x) = and δ4 (−1) = So J(χ, χ) = J(χ, ρ) It implies that N = p + − + 6J(χ, ρ) − p − − 6Reπ (b) Because p ≡ (mod 8), one has p − = 8k, for some k ∈ N The order of g is p−1 = 8k +4 Hence the order of g 4k+2 is Thus g 4k+2 = −1 Therefore χ(−1) = χ(g 4k+2 ) = χ(g)4k+2 = χ(g)2 m Suppose that χ(g)2 = Then for every a ∈ F× p , one has a = g , for some 51 Graduation thesis NGUYEN THI TRANG m ∈ Z and hence χ(a)2 = χ(g)2m = In this case χ2 (a) = and χ is not a character of order 4, a contradiction Thus χ(−1) = −1, and J(χ, χ) = −J(χ, ρ) and δ4 (−1) = Therefore N = p + + 2ReJ(χ, ρ) = p + − 2Reπ Exercise 10 (continuation) Letπ = a + bi One can show that a is odd, b is even, and a ≡ (mod 4) if | b and a ≡ −1 (mod 4) if b Let p = A2 + B and fix A by requiring that A ≡ (mod 4) Then show that (a) N = p − − 6A if p ≡ (mod 8) (b) N = p + + 2A if p ≡ (mod 8) Solution Notice that ρ(a)−1 ≡ (mod 2) and that χ(a)−1 ≡ (mod 1+ i) for all a = in Fp Thus if a = and b = 0, (ρ(a) − 1)(χ(b) − 1) ≡ (mod + 2i) This congruence is trivially true for the pairs a = 0, b = and a = 1, b = Therefore (ρ(a) − 1)(χ(b) − 1) ≡ (mod + 2i) a+b=1 Expanding we see that −π − χ(b) − ρ(a) + p ≡ (mod + 2i) a b The second and third terms are zero Thus π≡p≡1 (mod + 2i) The last step follows because p ≡ (mod 4) by hypothesis, and + 2i divides 4; indeed = (1 − i)(2 + 2i) It implies that a + bi ≡ (mod 2) and so a is odd and b is even Since = −2(i − 1)(i + 1) it follows that if | b, then a + bi ≡ a ≡ (mod 2+2i) Taking conjugates a ≡ (mod 2−2i) Thus (2−2i)(2+2i) = 52 Graduation thesis NGUYEN THI TRANG | (a − 1)2 and a ≡ (mod 4) If b, then b = 4k+2 for some k Thus a+bi ≡ a+2i ≡ (mod 2+2i) Since 2i ≡ −2 (mod + 2i), we have a ≡ ≡ −1 (mod + 2i) As before | (a + 1)2 and so a ≡ −1 (mod 4) (a) We have π · π = p = a2 + b2 = A2 + B Since b is even, b = 2k, k ∈ N Then, p = a2 + 4k From p ≡ (mod 8) and a is odd, so a2 ≡ (mod 8) It implies that 4k ≡ (mod 8), then k ≡ (mod 2), k ≡ (mod 2) Therefore, | b and a ≡ (mod 4) Hence a = A By Exercise we get N = p − − 6Reπ = p − − 6a = p − − 6A (b) Similarly, we put b = 2k, so p = a2 + 4k ≡ (mod 8) And from a2 ≡ (mod 8), it follows that 4k ≡ (mod 8) Thus, k ≡ (mod 8), then k is odd Therefore, b and a ≡ −1 (mod 4) Thus A = −a, and we have N = p + − 2a = p + + 2A Exercise 11 Let p be a prime, p ≡ (mod 4), χ a multiplicative character of order on Fp , and ρ the Legendre symbol Put J(χ, ρ) = a + bi Show (a) N (y + x4 = 1) = p − + 2a (b) N (y = − x4 ) = p + (c) 2a ≡ −(−1)(p−1)/4 · ρ(1 − x4 ) 2m m (mod p) where m = (p − 1)/4 53 Graduation thesis NGUYEN THI TRANG Solution (a) One has N (y = a) · N (x4 = b) N (y + x4 = 1) = a+b=1 (1 + ρ(a))(1 + χ(b) + χ2 (b) + χ3 (b)) = a+b=1 (1 + χ(b) + χ2 (b) + χ3 (b) + ρ(a) + ρ(a)χ(b)+ = a+b=1 + ρ(a)χ2 (b) + ρ(a)χ3 (b)) We have χ2 = ρ, χ3 = χ−1 = χ and χ(b) = χ3 (b) = ρ(b) = b ρ(a) = b Therefore N (y + x4 = 1) = p + J(χ, ρ) + ρ(ab) + J(χ, ρ) a+b=1 = p − + · ReJ(χ, ρ) = p − + 2a (b) We have N (y = − x4 ) = (1 + ρ(1 − x4 )) = p + x ρ(1 − x4 ) x (c) One gets ρ(1 − x4 ) ≡ (1 − x4 )(p−1)/2 (mod p) Put m = p−1 By Newton’s binomial Theorem, we have 2m 2m (1 − x ) = 2m k k=0 Thus, x ρ(1 − x4 ) ≡ 2m k=0 2m k · (−1)k · x4k · (−1)k · 54 4k xx p−1 , 2m = Graduation thesis NGUYEN THI TRANG We know that x4k x  p − if p − | 4k = 0 otherwise Therrefore 2m ρ(1 − x4 ) ≡ (−1)m · m x ≡ −(−1)m · 2m m · (p − 1) (mod p) (mod p) Remark 3.4.1 From Exercise 10, if we choose a, b such that p = a2 + b2 , then a ≡ −1 (mod 4) if | b and a ≡ (mod 4) if b, b is even.This is 2m equivalent with 2a ≡ −(−1)(p−1)/4 · (mod p), where m = (p − 1)/4 m Indeed, for example, for p = We see that p = 12 + 22 , so a = 1, b = 2 Moreover, 2a = −(−1)1 · = · 1 For p = 17, m = We have p = 12 + 42 , then a = −1 We see that 2a ≡ −(−1)4 = −70 ≡ · (−1) (mod 17) Corollary 3.4.2 Let p be a prime, p ≡ (mod 4) and p = a2 + b2 Let a ≡ (mod 4) Then we have 2a ≡ (p − 1)/2 (p − 1)/4 (mod p) Exercise 12 Let p ≡ (mod 3), χ a character of order 3, ρ the Legendre symbol Show 55 Graduation thesis NGUYEN THI TRANG (a) N (y = − x3 ) = p + ρ(1 − x3 ) (b) N (y + x3 = 1) = p + 2ReJ(χ, ρ) (c) 2a − b ≡ − (p − 1)/2 (mod p), where J(χ, ρ) = a + bω (p − 1)/3 Solution (a) We have ρ(1 − x3 ) + ρ(1 − x3 ) = p + N (y = − x3 ) = x x (b )One has N (y + x3 = 1) = N (y = a) · N (x3 = b) a+b=1 (1 + ρ(a)) + χ(b) + χ2 (b) = a+b=1 ρ(a) · χ(b) + =p+ a+b=1 ρ(a) · χ(b) a+b=1 = p + 2ReJ(χ, ρ) b (c) We have ReJ(χ, ρ) = a − , and 2ReJ(χ, ρ) = 2a − b = x ρ(1 − x3 ) (p − 1)/2 Thus we must prove that x ρ(1 − x3 ) ≡ − (mod p) (p − 1)/3 Indeed, we have ρ(1 − x3 ) ≡ (1 − x3 )(p−1)/2 (mod p) Putting p − = p−1 p−1 6m(m ∈ N), then = 3m, = 2m From Newton’s binomial formula, we have 3m 3m (1 − x ) 3m = k k=0 56 (−1)k · x3k Graduation thesis NGUYEN THI TRANG Then 3m ρ(1 − x ) ≡ x k=0 3m k x ≡ (−1)2m · (p − 1) · ≡− x3k (−1)k · (p − 1)/2 (p − 1)/3 57 3m 2m (mod p) (mod p) Conclusion In this thesis, we have presented systematically the following results (1) Recall the definition and basic properties of quadratic residues, some knowledge related to matrices, its characteristic function and multiplicative characters (2) Study quadratic Gauss sums: the absolute value of quadratic Gauss sums, the sign of the quadratic Gauss sums, extensions to composite moduli and some applications (3) Study general Gauss sums and Jacobi sums, and use these sums to obtain a formula to calculate the number solutions of the equation xn + y n = overFp (4) We have done some exercises to illustrate the applications of Jacobi sums 58 Bibliography [1] E Landau, Elementary number theory, Chelsea Publishing Co., New York, 1958 [2] T Nagell, Introduction to Number Theory, John Wiley & Sons, Inc., New York; Almqvist & Wiksell, Stockholm, 1951 [3] K Ireland and M Rosen, A classical introduction to Modern number theory, Second edition Graduate Texts in Mathematics, 84 Springer-Verlag, New York, 1990 [4] B.C Berndt and R.J Evans, The Determination of Gauss sums, Bull Amer Math Soc (1981), 107-129 [5] M R Murty and S Pathak, Evaluation of the quadratic Gauss sum, Math Student 86 (2017), no 1-2, 139-150 59 ... ? ?Gauss sums and Jacobi sums? ?? for Graduation Thesis This thesis is the general result of books [2],[3],[4] The main aim in this thesis is to know how to be Gauss sums and Jacobi sums, understand... 2.4.2 Some related trigonometric problems 34 Gauss sums and Jacobi sums 36 3.1 Gauss sums 36 3.2 Jacobi sums 38 3.3 The equation xn... sin + sin = 7 35 Chapter Gauss sums and Jacobi sums In Chapter we introduced the notion of a quadratic Gauss sum In this chapter a more general notion of Gauss sum and Jacobi sum will be introduced