Mathematical Reasoning and the Inductive Process: An Examination of The Law of Quadratic Reciprocity

Transcription

1 California State University, San Bernardino CSUSB ScholarWorks Electronic Theses, Projects, and Dissertations Office of Graduate Studies Mathematical Reasoning and the Inductive Process: An Examination of The Law of Quadratic Recirocity Nitish Mittal California State University - San Bernardino, niimits1@gmail.com Follow this and additional works at: htt://scholarworks.lib.csusb.edu/etd Part of the Number Theory Commons, and the Set Theory Commons Recommended Citation Mittal, Nitish, "Mathematical Reasoning and the Inductive Process: An Examination of The Law of Quadratic Recirocity" 016. Electronic Theses, Projects, and Dissertations. Paer 8. This Thesis is brought to you for free and oen access by the Office of Graduate Studies at CSUSB ScholarWorks. It has been acceted for inclusion in Electronic Theses, Projects, and Dissertations by an authorized administrator of CSUSB ScholarWorks. For more information, lease contact scholarworks@csusb.edu.

2 Mathematical Reasoning and the Inductive Process: An Examination of the law of Quadratic Recirocity A Thesis Presented to the Faculty of California State University, San Bernardino In Partial Fulfillment of the Reuirements for the Degree Master of Arts in Mathematics by Nitish Mittal June 016

3 Mathematical Reasoning and the Inductive Process: An Examination of the law of Quadratic Recirocity A Thesis Presented to the Faculty of California State University, San Bernardino by Nitish Mittal June 016 Aroved by: Dr. James Paul Vicknair, Committee Chair Date Dr. Zahid Hassan, Committee Member Dr. Rolland Tra, Committee Member Dr. Charles Stanton, Chair, Deartment of Mathematics Dr. Corey Dunn Graduate Coordinator, Deartment of Mathematics

4 iii Abstract This roject investigates the develoment of four different roofs of the law of uadratic recirocity, in order to study the critical reasoning rocess that drives discovery in mathematics. We begin with an examination of the first roof of this law given by Gauss. We then describe Gauss fourth roof of this law based on Gauss sums, followed by a look at Eisenstein s geometric simlification of Gauss third roof. Finally, we finish with an examination of one of the modern roofs of this theorem ublished in 1991 by Rousseau. Through this investigation we aim to analyze the different strategies used in the develoment of each of these roofs, and in the rocess gain a better understanding of this theorem.

5 iv Acknowledgements I wish to thank Dr. James Paul Vicknair for all his hel and suort during the rearation of this aer. I would also like to thank Dr. Hassan and Dr. Tra for their comments and contribution to this thesis. I would esecially like to thank Dr. Stanton for heling me fully assimilate into the graduate rogram, and Dr. Dunn in heling me re-enter the rogram after an extended leave of absence. I would further like to acknowledge my classmates and friends Leonard Lam, Kevin Baccari, and Avi Misra whose hel and suort layed a ivotal role in my learning and develoment throughout my tenure in the masters rogram.

6 v Table of Contents Abstract Acknowledgements List of Figures iii iv vi 1 Introduction Observation and Induction in Mathematics The Law of Quadratic Recirocity What is Covered in this Project Background Definitions and Theorems 4.1 Theory of Congruences Wilson s and Fermat s Theorems Euler s Criterion and Legendre Symbol Gauss Lemma Gauss Sums Normal Subgrous and Quotient Grous The Chinese Remainder Theorem Proofs of The Law of Quadratic Recirocity The Law of Quadratic Recirocity Gauss s First Proof of Quadratic Recirocity Gauss s Fouth Proof of Quadratic Recirocity Eisenstein s Geometric Proof of Quadratic Recirocity Rousseau s Proof of Quadratic Recirocity Conclusion 50 Bibliograhy 51

7 vi List of Figures 3.1 Eisenstein s Lattice Points

8 1 Chater 1 Introduction 1.1 Observation and Induction in Mathematics There are even many roerties of numbers with which we are well acuainted, but which we are not yet able to rove; only observations have led us to their knowledge. Hence we see that in the theory of numbers, which is still very imerfect, we can lace our highest hoes in observations; they will lead us continually to new roerties which we shall endeavor to rove afterwards. The kind of knowledge which is suorted only by observations and is not yet roved must be carefully distinguished from the truth; it is gained by induction as we usually say. Yet we have seen cases in which mere induction led to error. Therefore we should take care not to accet as true such roerties of the numbers which we have discovered by observation and which are suorted by induction alone. Indeed, we should use such a discovery as an oortunity to investigate more than exactly the roerties discovered and to rove or disrove them; in both cases we may learn something useful. Leonhard Euler in [Pól54] Euler s uote about the use of observation in mathematics is a great commentary on mathematical reasoning and the inductive rocess itself. As Euler mentions above even in fields as abstract as ure mathematics and the theory of numbers, observation and induction are imortant tools in heling identify the various intriguing behaviors and atterns of numbers. However, observation alone is not sufficient and the true strength of mathematical discovery lies in the use of mathematical tools to rove without excetion what is being observed. As students of mathematics we are no strangers to this rocess. Identifying a attern in a given examle and alying inductive reasoning to generalize

9 said attern is the basis of modern mathematics; however, as Euler notes, we should be vary of roofs by induction alone. The strongest conjectures are those which can not only be suorted by inductive reasoning, but roved and reroved using other methods as well. The Law of Quadratic Recirocity is one such examle in the theory of numbers. [Pól54] 1. The Law of Quadratic Recirocity Dubbed the golden theorem of number theory by the rince of mathematics, Carl Friedrich Gauss, the law of uadratic recirocity is one of the most ursued theorems of 18th and 19th century mathematics. The theorem was first formulated by Euler in 1783 and later tackled by Legendre in 1785, and again in Essai sur la Theorie des Nombres in Though both his roofs were later shown to be invalid, the elegant notation emloyed by Legendre eventually became the modern Law of Quadratic Recirocity. The first comlete roof of the theorem was written by Gauss when he was 18, and ublished in his book Disuisitiones Arithmaticae in In his first attemt Gauss looked at individual cases and used elementary techniues to rove the law and subseuently generalized it using mathematical induction. Gauss later ublished 6 more roofs of the same theorem, each time emloying a different method, refining the roof and making it more elegant. [Büh81] We will begin with examining thoroughly Gauss s first attemt, which, though rather long, uses only basic techniues. We will then examine two of his latter roofs to see the develoments he made over time to further refine his roof. We will look at iterations or simlifications of his third and fourth roofs of the law of uadratic recirocity. In his fourth roof Gauss used Gaussian sums to rove the law. We will describe his roof in comlete detail and examine the differences between this roof and his first attemt, the most aarent of which is the sheer difference in length between the two. We will then analyze his third roof of this theorem. In this roof Gauss emloyed the use of Gauss s lemma to rove the fundamental theorem in a very concise and elegant fashion. Eisenstein s simlification of this roof is erhas one of the most commonly used in elementary number theory courses. Finally, we will conclude with a modern elementary roof of the law of uadratic recirocity.

10 3 1.3 What is Covered in this Project This roject will take us through the first known roof of one of the fundamental laws of the theory of numbers and one of the most scrutinized theorems in all of mathematics, and illustrate the ways in which it was reformulated and refined over time. The roject is an interesting examination of the essence of the rocess of inductive reasoning described by Polya and rovides us a firsthand exerience whilst also examining one of Gauss celebrated contributions to the field of mathematics. While the significance of the latter cannot be denied we stand to gain much more from the former. The inductive attitude has been a major driving force for continuous investigation and invention in mathematics and other natural sciences. The rocess is a telling tale of how the human mind works and how new knowledge is discovered. A journey that is not very different from that of a diamond, starting off as a ebble in a mine and constantly cleaned and refined along the way until it is finally cut and olished to reveal the elegant jewel that it is. This roject outlines a aradigm that has been emloyed in the discovery of countless other theorems in the ast and will continue to hel discover countless more in the future. It highlights the rocess from the recognition of atterns observed by the investigation of secial cases, to forming a conjecture and eventually using known techniues to simlify, generalize and rove our conjecture. It further illustrates the imortance of continued investigation in heling uncover new imlications and interretations of a theorem. This roject ultimately highlights the oint that while the saying necessity is the mother of all invention may hold true in other natural sciences, the inductive attitude and reasoning are certainly the root of all discovery in the abstract field of ure mathematics.

11 4 Chater Background Definitions and Theorems We will begin by giving some elementary definitions and theorems in number theory, which can be found in the following sources: [Bur07] [Nag51].1 Theory of Congruences Definition.1. Let n be a fixed ositive integer. Two integers a and b are said to be congruent modulo n, symbolized by a bmod n if n divides the difference a b; that is, given that a b = kn for some integer k. Lemma.. Let a, b and n > 0 be integers, then a bmod n n a b. Proof. If a bmod n, then a and b have the same remainder when divided by n. Thus, by the division algorithm a = n+r and b = n+r, for integers, and r with 0 r n. Now we have, a n = b n a b = n n a b = n

12 5 Thus, n a b. Conversely, If n a b, then there exist integers x and y such that, a = nx + r 1 and b = ny = r, where 0 r 1 n and 0 r n. Then, a b = nx + r 1 ny + r a b = nx + y r 1 r Thus, since n a b, n must also divide r 1 r. However r 1 r < n, thus n < r 1 r < n, therefore r 1 r = 0 and r 1 = r, and so a bmod n. Examle.3. Let n = 7, then 10 4mod 7 since, 4 10 = 7. On the other hand, 5 1mod 7, since 5 1 k7 for any integer k. Theorem.4. Let n > 1 and a, b, c, d be integers, then the following roerties hold: a a amod n. b If a bmod n, then b amod n. c If a bmod n and b cmod n, then a cmod n. d If a bmod n and c dmod n, then a + c b + dmod n and ac bdmod n. e If a bmod n, then a + c b + cmod n and ac bcmod n. f If a bmod n, then a k b k mod n for any ositive integer k. Proof. For any integer a, a a = 0 n, so that a amod n. Furthermore, if a bmod n, then a b = kn for some integer k. And, b a = kn = kn and since k is an integer roerty b holds. Now suose a bmod n and also b cmod n, then there exist integers, such that a b = n and b c = n. Then, a c = a b + b c = n + n = + n a cmod n Similarly, if a bmod n and c dmod n, then a b = n and c d = n for some, and a + c b + d = a b + c d = n + n = + n a + c b + dmod n

13 6 Also, ac = b + nd + n = bd + b + d + nn n ac bd and ac bdmod n The result of roerty e follows from roerty d and roerty a. For roerty f we can use mathematical induction. Since the argument holds for k = 1, we can assume a k b k mod n. Given that a bmod n, d imlies: aa k bb k mod n a k+1 b k+1 mod n Lemma.5. If gcda, n = 1 and n > 0, then there exists a uniue integer x modulo n, such that ax 1mod n. Proof. Given gcda, n = 1, we can write ax + ny = 1 for some integers x and y. This can be rewritted as ax 1 = n y, thus n ax 1 and ax 1mod n. Suose ax 1mod n and ax 1mod n, then ax ax mod n n ax ax n ax x Since gcda, n = 1, n must divide x x and x x mod n. Therefore the solution x is uniue modulo n. Examle.6. Let a = 3 and n = 5 then, we know that gcda, n = 1 and 3 = 6 1mod 5. Theorem.7. If ca cbmod n, then a bmod n d, where d = gcdc, n. Proof. Let, ca b = ca cb = kn for som integer k. Given that gcdc, n = d, there exist relatively rime integers and such that c = d and n = d. Then, da b = kd a b = k Thus a b, and since gcd, = 1, then by Lemma. a b a bmod or a bmod n d Examle.8. Let a = 7, b = and n = 0, then for c = 4, 8 8mod 0 and gcdc, n = d = 4. Therefore, 7 mod mod 5.

14 7. Wilson s and Fermat s Theorems Theorem.9. Wilson s Theorem. If is a rime, then 1! 1mod. Proof. It is aarent that this theorem holds for = or 3, therefore let > 3. Suose a is an integer from 1,, 3,..., 1 Since is a rime number, gcda, = 1 and by Lemma.5 there is a uniue integer a such that, 1 a 1, that satisfies aa 1mod. Since is rime, a = a if and only if a = 1 or 1. As we can see a 1mod is euivalent to a 1a + 1 0mod. Thus, either a 1 0mod and a = 1 or a + 1 0mod and a = 1. If we remove 1 and 1 the remaining integers, 3,..., can be groued in to airs a, a such that a a and aa 1mod. Thus,! 1mod multilying both sides by 1, we get 1! 1mod Examle.10. Let = 7, then 1! = 6! = 70 and = mod 7. Similarly, for = 19, then 1! = 18! = and, = mod 19. Theorem.11. Fermat s little Theorem. Let be a rime and suose that a. Then a 1 1mod. Proof. Let us consider the first 1 ostitive multile of a: a, a, 3a,..., 1a

15 8 These numbers are not congruent modulo to each other or to zero. Otherwise, ma namod where, 1 m < n 1 m nmod which is imossible. Thus it is aarent that these integer multiles of a are distinct and must be congruent modulo to 1,, 3,..., 1. Multilying these together we get, a a 3a 1a 1 3 1mod a 1 1! 1!mod By cancelling 1! from both sides we get the desired result. Examle.1. Let = 7 and a = 5, then 7 5 and, = 5 6 = 1565 and = mod 7 Similarly, for = 13 and a = 8, then 13 8 and, = 8 1 = = mod 13.3 Euler s Criterion and Legendre Symbol Definition.13. Let be an odd rime and gcda, = 1. If the uadratic congruence x amod has a solution, then a is said to be a uadratic residue of. Otherwise, a is called a uadratic nonresidue of. Examle.14. Let = 11 then, , 9 4, , 5 6 3mod 11 Thus 1, 3, 4, 5, and 9 are uadratic residues of 11 and, 6, 7, 8, and 10 and nonresidues of 11.

16 9 Definition.15. If gcda, n = 1 and a is of order φn modulo n, then a is a rimitive root of the integer n. Examle.16. The definition can be restated as, a φn 1mod n and a k 1mod n for all k < φn if a is a rimitive root of n. Furthermore, if n is a rime number, φn = n 1, for all n. Then for a = and n = 13 we have, 1, 4, 3 8, 4 3, 5 6, 6 1, , 9 5, 10 10, 11 7, 1 1mod 13 Theorem.17. Euler s Criterion. Let be an odd rime and gcda, = 1. Then a is a uadratic residue of if and only if a 1 1mod. Proof. Suose a is a uadratic residue of, such that x amod has a solution. Lets call it x 1. Then, gcdx 1, = 1 since gcda, = 1. Thus by Fermat s little theorem a 1 x 1 1 1mod If we assume that the congruence a 1 1mod holds, and let r be a rimitive root of. Then a r k mod for some integer k, with 1 k 1. It follows that r k 1 a 1 1mod The order of r = 1 must divide the exonent k 1 imlying k is even. Lets say k = j, then r j = r j = r k amod making r j a solution of the congruence x amod. Corollary.18. Let be an odd rime and gcda, = 1. Then a is a uadratic residue or non residue of according to whether a 1 1mod or a 1 1mod Proof. Since is always an odd rime and gcda, = 1, then a 1 1a = a 1 1mod

17 10 Hence, either a 1 1mod or a 1 1mod, but not both, otherwise 1 1mod and. therefore it must satisfy a 1 1mod. A uadratic nonresidue of does not satisfy a 1 1mod, Definition.19. Let be an odd rime and let gcda, = 1. The Legendre symbol a is defined by a = 1 if a is a uadratic residue of 1 if a is a uadratic nonresidue of 0 if a 0mod Legendre s symbol is only defined for rimes. Jacobi later introduced a more general symbol known as the Jacobi Symbol a P, for all natural odd numbers P, when: P = e 1 1 e em m is a roduct of rimes e 1 1, e,..., em m, and when a is relatively rime to P. Then: a a e1 a = P 1 e a em m where the factors on the right hand side are Legendre symbols. Thus when P is a uadratic residue of a, a P = 1 since all the factors on the right hand side eual 1. However, when P is not a uadratic residue it is not necessarily true that a P = 1. This is because when an even number of factors on the right hand side have the value 1, the resulting roduct will be +1. We will use Jacobi symbol in Gauss first roof of uadratic recirocity and we note that the Jacobi symbol in not defined for the integers P < 0 or for even P. Examle.0. Using legendre symbol, we can rewrite Examle.14 as: = = = = = and = = = = = 1 11 Theorem.1. Let be and odd rime and let a and b be integers that are relatively rime to. Then the Legendre symbol has the following roerties: a If a bmod, then a = b.

18 11 b a = 1 c a a 1 mod. d ab = a b. e 1 = 1, 1 = 1 1, and 0 = 0. f ab = a b = a. Proof. If a bmod, then the two congruences x amod and x bmod have the same exact solutions, and thus either both are solvable or both unsolvable, hence a = b. For roerty b integer a trivially satisfies the congruence x amod, hence = 1. Proerty c is a direct result of Euler s criterion. Using c we get a ab ab 1 a 1 b 1 a b Since Legendre symbol assumes only values 1 or 1, if 1 1mod or 0mod, but >. Thus, ab a b = ab mod a b we would get The first art of roerty e is a secial case of roerty b, when a = 1, and the second art is derived from roerty c when a = 1. The result for roerty f follows directly from roerties b and d. Theorem.. If is an odd rime, then 1 a=1 a = 0 Hence, there are recisely 1/ uadratic residues and 1/ uadratic nonresidues of. Proof. Let r be a rimitive root of. Then modulo, the owers r, r,..., r 1 are just a ermutation of the integers 1,,..., 1. Thus for any a between 1 and 1, there exists a uniue ositive integer k1 k 1, such that a r k mod. By Euler s criterion we have a r k = r k 1 = r 1 k 1 k mod

19 1 where, r 1 1mod because r is a rimitive root of. We can then add u the Legendre symbols to obtain 1 a 1 = 1 k = 0 a=1 k=1 Corollary.3. The uadratic residues of an odd rime are congruent modulo to the even owers of a rimitive root r of ; the uadratic nonresidues are congruent to the odd owers of r..4 Gauss Lemma Theorem.4. Gauss Lemma. Let be an odd rime and let gcda, = 1. If n denotes the number of integers in the set { } 1 S = a, a, 3a,..., a whose remainders uon division by exceed /, then a = 1 n Proof. Because gcda, = 1, none of the 1 integers in S are congruent to zero or to one another modulo. Let r 1,..., r m be those remainders uon division by such that 0 < r i <, and let s 1,..., s n be those remainders such that > s i >. m + n = 1, and the integers Then r 1,..., r m s 1,..., s n are all ositive and less than. To rove that these integers are all distinct, it suffices to show that no s i is eual to any r j. Let us assume that s i = r j, for some i and j, then there exist u and v 1 u, v 1 such that s i uamod and r j vamod. Thus, u + va s i + r j = 0mod

20 13 However, u + v 0mod since 1 < u + v 1. The central oint is that the 1 numbers r 1,..., r m s 1,... s n are the integers 1,,..., 1 in some order. Thus, their roduct is 1!: 1! = r 1 r m s 1 s n r 1 r m s 1 s n mod 1 n r 1 r m s 1 s n mod We know that r 1,..., r m, s 1,..., s n are congruent modulo to a, a,..., 1 a in some order, thus 1! 1 n a a 1 Since 1 n a 1 1!mod amod 1! is relatively rime to, we can cancel it from both sides: multilying both sides by 1 n, we get Using Euler s criterion we get: a 1 1 n a 1 mod a 1 1 n mod a 1 1 n mod a = 1 n Examle.5. Now we can look at Gauss s lemma with an examle, where a = 7 and = 17. Then 1/ = 8 and: S = {7, 14, 1, 8, 35, 4, 49, 56}

21 14 Modulo 17, we can rewrite S as the following: S = {7, 14, 4, 11, 1, 8, 15, 5} Three of these are greater that 17/; therefore, n = 3, and according to Theorem.4: 7 = 1 3 = 1 17 We can also confirm this using the Corollary.18 to Euler s criterion: mod 17 Theorem.6. Using Gauss s lemma, we can show that if is an odd rime, then 1 if ±1mod 8 = 1 if ±3mod 8 Proof. According to Gauss s lemma, = 1 n where n is the number of integers in the set { } 1 S =, 4, 6,..., whose remainders uon division by are greater than /. Since all members of S are less than modulo it suffices to count the number of even integers 1 < k < 1/ that exceed /. We see that k < / when k < /4; therefore, if we let [/4] be the largest even integer less than /, then n = 1 [ 4] Now we can look at the individual cases for the four different forms of : [ If = 8M + 1, then n = 4M M + 1 ] = 4M M = M 4 [ If = 8M + 3, then n = 4M + 1 M + 3 ] = 4M + 1 M = M [ If = 8M + 5, then n = 4M + M ] = 4M + M 1 = M [ If = 8M + 7, then n = 4M + 3 M ] = 4M + 3 M 1 = M + 4 Thus, when ±1 mod 8 = 8M + 1 or 8M + 7, n is even and 1 n is 1. Conversely, if ±3 mod 8 = 8M + 3 or 8M + 5, n is odd and 1 n is 1.

22 15.5 Gauss Sums More information on the Gauss Sum can be found in the following text: [Lem00] Definition.7. An n th root of unity is a comlex number ζ n such that, ζ n n = 1. Thus, Examle.8. Let ζ 3 = e i π 3. Now, ζ n = cos π n + i sin π n = ei π n ζ 3 3 = e i π 3 3 = e iπ = cos π + i sin π = 1 + i 0 = 1 Definition.9. If we fix an odd rime, then the Gauss Sum associated with an integer a is: 1 n G a = ζ an n=0 Examle.30. Let = 5 and a = 3, now the Gauss sum G 3 is: 4 n=0 n ζ5 3n = Theorem.31. For any integer a 1 ζ n=0 ζ = ζ 3 5 ζ 5 ζ ζ 5 ζ an = if a 0 if a Proof. 1. When a we can write a = x, and we get: Now, since ζ = 1, we have: n=0 1 n=0 1 ζ an = n=0 ζ xn 1 1 ζ xn = 1 xn = n=0. When a. First we look at the following identity: 3 ζ x 1 = x 1 x ζ5 5

23 16 Now, if we substitute ζ a for x, we get: 1 n=0 Now, since ζ = 1, we get: x = x 1 x 1 ζ a n = ζ a = ζa 1 ζ a 1 ζ a 1 ζ a 1 = 1a 1 ζ a 1 = 0 Corollary.3. For any integer x, y 1 n=0 ζ x yn = if x ymod 0 if x ymod The roof for Corollary.31 is exactly the same as that of Theorem.30, and can be acheived by relacing x y with a. Examle.33. Let = 3 and a = 6, then a and: n=0 ζ 6n 3 = ζ 3 n 3 = 1 n = = 3 = n=0 n=0 Examle.34. Let = 3 and a =, then a and: n=0 ζ n 3 = 1 + ζ 3 + ζ 3.1 We know that ζ 3 is a solution for the euation x 3 1 = 0, which can be exanded to x 1x + x + 1. Here we take the right hand side of the roduct to be 0 since ζ 3 1, then x + x + 1 = 0 and x = x 1. Substituting ζ 3 back in for x, we get ζ 3 = ζ 3 1. Alying this to Euation.1 above, we get: n=0 Theorem.35. The Gauss Sum G 0 = 0. ζ n 3 = 1 ζ ζ 3 = 0

24 17 Proof. 1 n G 0 = ζ 0n = n=0 1 n=0 n Now by alying Theorem. to the right hand side, we get: 1 n=0 n = 0 since there are exactly 1/ uadratic residues and 1/ uadratic nonresidues of. Examle.36. Let = 3 and a = 0, then: G 0 = n=0 n 3 ζ 0n 3 = = Normal Subgrous and Quotient Grous More information on the theorems and definitions of grou theory delineated below can be found in the following text: [Rom05] Definition.37. An abelian grou G is a set G with a binary oeration such that the following roerties hold: 1. Closure If x, y G, then x y G.. Associativity x y z = x y z, for all x, y, z G. 3. Identity There exists an element e G, such that e x = x for all x G. 4. Inverse For all x G, there exists y G, such that x y = e. 5. Commutativity x y = y x for all x, y G. Examle.38. The grou of all integers Z is an abelian grou under addition. 1. The sum of any two integers is an integer.. The associativity law alies to all integers. Eg: = The additive identity for all integers is 0.

25 18 4. The additive inverse for any integer x is x, which is also an integer. 5. Integer addition is commutative. Eg: + 3 = 3 + Examle.39. The grou of rational numbers Q, without 0, is an abelian grou under multilication. We need to remove 0, because 0 does not have a multilicative inverse. 1. The roducts of two rational numbers is a rational number.. Integers are a subgrou of rationals, thus associativity alies as seen in the examle above. 3. The multilicative identitiy for rationals is The multilicative inverse for any rational x y, where x, y Z is y x. 5. Rational multilication is commutative. Eg. 3 = 3. Definition.40. A subset S of a grou G, is said to be a subgrou of G, if it is a grou itself. Examle.41. Consider the set of real numbers R, which is a grou under addition. Then the integers Z, which are a subset of real numbers, and also form a grou under addition see Examle.38, are said to a be a subgrou of R. Definition.4. Let H be a subgrou of G, and let x G be an element of G, then we define x H, as the subset {x h h H}, to be a left coset of H, and H x, as the subset {h x h H} to be the right coset of H. Where is a binary oeration, such as addition or multilication, deending on the definition of the grou G. Examle.43. Consider the set of all even integers Z, it is clear that this is a grou under addition. We also know that the set of all integers Z is a grou, thus Z is clearly a subgrou of Z. Then we can say that, 1 + Z is a left coset of Z, and Z + 1 is a right coset of Z. Definition.44. A subgrou H is said to be a normal subgrou of G, if the left cosets of H are eual to the right cosets of H. That is x H = H x for all x G. Examle.45. Let s look at the subgrou Z from Examle.43 above. We can see that the left coset of Z, 1 + Z, is eual to the set of odd integers {..., 3, 1, 1, 3,... }. Furthermore, the right coset of Z, Z + 1, is also eual to the set of odd integers {..., 3, 1, 1, 3,... }. Thus, Z is a normal subgrou of Z.

26 19 Definition.46. If H is a normal subgrou of G, then we can construct a grou G/H by multilying the left cosets of H, such that for all α, β G, αhβh = αβh. The grou G/H is called the uotient grou of G by H. Note: Here we say left cosets for the sake of illustration. However, since H is a normal subgrou of G, the left and the right cosets of H are eual. Examle.47. Let G = Z 6 = {0, 1,, 3, 4, 5} and let H be the normal subgrou {0, 3}. The cosets of H are: {0, 3}, 1 + {0, 3} = {1, 4}, + {0, 3} = {, 5} Then the uotient grou G/H is a grou of order 3 containing the following elements: {0, 3}, {1, 4}, {, 5}.

27 0.7 The Chinese Remainder Theorem Fore more information on the Chinese Remainder Theorem, lease refer to the following text: [Sti94] Definition.48. A non-emty set R, along with the binary oerations of multilication and addition, is called a ring, if it satisfies the following roerties: 1. R is an abelian grou under the order addition.. Multilication in R is associative. That is αβγ = αβγ for all α, β, γ R. 3. Multilication in R is distributive. That is αβ + γ = αβ + αγ for all α, β, γ R. Examle.49. Consider the set Z/nZ of integers modulo n. Let n = 7 and we can see that Z/7Z is a ring. 1. We can check that Z/7Z is abelian under addition using Definition.37. i = 8 1mod 7 Z/7Z. ii = mod 7 Z/7Z. iii. The additive identity for all elements in Z/7Z is 0. iv. The additive inverse for any element x Z/7Z is x Z/7Z. v. + 3 = 3 + = 5mod 7 Z/7Z = 1 3 6mod 7 Z/7Z = mod 7 Z/7Z. Definition.50. Let G and H be two grous. isomorhism between G and H, if A function f : G H is called an 1. f is a homomorhism, that is for any a, b G, fab = fafb.. f is a one-to-one and onto maing from G to H. Examle.51. Let G be the ositive real numbers under addition, and H be the real numbers under multilication. Then f = log: G H is an isomorhism. 1. log xy = log x + log y.. Let log x = log y, then e log x = e log y x = y. Thus f is one-to-one. 3. Since, the log function sans all real numbers it is clear that f is onto.

28 1 Theorem.5. The Chinese Remainder Theorem. If gcdm, n = 1, then the ma fx = x mod m, x mod n is an isomorhism of Z/mnZ onto Z/mZ Z/nZ. Proof. Let m be a non-zero integer, then there is a ring homomorhism g : Z Z/mZ such that gx = x mod m. We have already seen in an examle above that Z/mZ is indeed a ring. We can also see that it is a homomorhism: gx + y = x + ymod m = xmod m + ymod m = gx + gy and gxy = xymod m = xmod mymod m = gxgy Similarly, if n is another non-zero integer, then hx = x mod n is another ring homomorhism that takes Z Z/nZ. Let f be a maing that combines these two homomorhisms, such that f : Z Z/mZ Z/nZ, by defining: fx = gx, hx = xmod m, xmod n Now, the ring oerations on Z/mZ Z/nZ are comonent-wise addition and multilication. x, u + y, v = x + y, u + v x, uy, v = xy, uv where x, y Z/mZ and u, v Z/nZ. Now, we can see that f is a homomorhism, since:

29 fx + y = gx + y, hx + y = gx + gy, hx + hy = gx, hx + gy, hy = fx + fy and fxy = gxy, hxy = gxgy, hxhy = gx, hxgy, hy = fxfy Moreover, we also note that: fx + mn = x + mnmod m, x + mnmod n = xmod m, xmod n = fx Thus it is clear that fx only deends on x mod mn, and we can say that f is a homomorhism from Z/mnZ to Z/mZ Z/nZ. Now we only need to show that f is one-to-one and onto from Z/mnZ to Z/mZ Z/nZ. We can show that f is one-to-one if fx = 0 x = 0. Let us assume that fx = 0, then: fx = 0 = 0 mod m, 0 mod n = x mod m, x mod n x = 0 mod m, and 0 mod n Since, gcdm, n = 1, x must be congruent to 0 mod mn. Thus, x = 0 in Z/mnZ and f is one-to-one. In order to show that f is onto, for any two integers a, b, there exists an integer x, such that:

30 3 fx = a mod m, b mod n Thus, we have: x = a mod m and x = b mod n. Now, since m and n are relatively rime, we know that there exist uniue integers u and v, such that: We claim that um + nv = 1.3 x = bum + anv.4 is a solution to Euation.. We can test this by first multilying a to both sides of Euation.3: Now, combining Euation.4 and.5, it is clear that: aum + anv = a.5 x mod m anv mod m a mod m A similar calculation can be done by multilying Euation.3 with b: Now, combining Euation.5 and.6, we get: bum + bnv = b.6 x mod n bum mod n b mod n This shows that x = a mod m and x = b mod n, therefore f is also an onto maing. This roves that f is indeed an isomorhism and comletes our roof.

31 4 Chater 3 Proofs of The Law of Quadratic Recirocity 3.1 The Law of Quadratic Recirocity The Law of Quadratic Recirocity is one of the fundamental theorems of number theory. Legendre attemted two incomlete roofs of the law in 1785 and 1798 resectively, and it was eventually roved by Gauss in Desite Legendre s incomlete attemts to rove the law, his elegant notation, most imortantly the Legendre symbol, eventually became the modern Law of Quadratic Recirocity. The iteration of Gauss s first roof of uadratic recirocity described here was comosed with information from the following sources: [Bau15] [Dir91] [GC86] [Lem00] Theorem 3.1. Quadratic Recirocity Law. If and are distinct odd rimes, then = Gauss s First Proof of Quadratic Recirocity In his first roof Gauss utilized the method of induction to rove the Generalized Quadratic Recirocity Law. The startegy he used was to show that if we assume the Quadratic Recirocity Law to be true for every distinct air of odd rimes less than a rime then it must also hold true for every combination of those rimes with. Since,

32 5 the theorem holds true for the two smallest odd rimes 3 and 5, that is: 3 5 = 5 3 = 1, then it must also hold for every combination of 3 and 5 with the next largest rime 7. Conseuently, if it holds true for every combination of the rimes 3, 5 and 7, then it must also hold true for every combination of 3, 5 and 7 with the rime 11 and so on. Thus by mathematical induction it will hold true for every air of distinct odd rimes. In the first roof, Gauss looked at each of the following eight cases for the rimes and, where is an odd rime less than and we assume that Theorem 3.1 holds for each air of distinct odd rimes less than. The eight cases are as follows: 1. If = 4n + 1, = 4n + 1 and = 1, then we have to rove that = 1;. If = 4n + 1, = 4n + 3 and = 1, then we have to rove that = 1; 3. If = 4n + 1, = 4n + 1 and = 1, then we have to rove that = 1; 4. If = 4n + 1, = 4n + 3 and = 1, then we have to rove that = 1; 5. If = 4n + 3, = 4n + 3 and = 1, then we have to rove that = 1; 6. If = 4n + 3, = 4n + 1 and = 1, then we have to rove that = 1; 7. If = 4n + 3, = 4n + 3 and = 1, then we have to rove that = 1; 8. If = 4n + 3, = 4n + 1 and = 1, then we have to rove that = 1; Later demonstrations by Dedekind and Bachmann showed that these 8 ossibilites can be collased into the following two mutually exclusive cases, which encomass all of the ossibilities listed above: i. at least one of or is 1, then = ; ii. is of the form 4M + 1 and = 1, then = 1 We know from the Corollary.18 to Euler s criterion that case i covers the following ossibilities: 1 = 1 1, therefore is of the form 4M + 3 and = 1; is of the form 4M + 3 and = 1;

33 6 When = 4M + 3, Thus, either = 1 or is of the form 4M + 1 and = 1; = 1 1 = = 1. Conversely, when = 4M + 1, Therefore, either both = 1 1 = = = 1 or = = 1 which constitutes case ii. Proof. We desire to show that if and are distinct odd rimes and the Quadratic Recirocity Law holds for all rimes less than, then when < : = Proving cases i and ii listed above will thus rove the Generalized Quadratic Recirocity law, which simly states that the theorme is also true for two relatively rime odd integers P and Q, given that all rime factors of P and Q are less than ; that is, P Q Q P = 1 P 1 Q In order to rove case i, we assume that at least one or both of and and we need to show that = is 1, 3. Let w = ±, for which w = 1. Now, there are two distinct solutions x for the euation x wmod. Let these solutions be ositive and <. If x 0 is one such integer, then the other is x 0. Since, x 0 = x and x 0 = x 0 = x. Let e be the solution which is even. Since, is odd, one of x 0 or x 0 must be even. Then, for 0 < e <, e w = f 3.3

34 7 Now we can see that f cannot be negative. In order for f to be negative, would have to be ositive and greater than e. Conseuently, e = f, where e is divisible by ; however, since e < and < this is imossible because is a rime and cannot divide any number smaller than itself. Furthermore, f < because both e and w are less than and therefore: f = e w 1 w = w 1 < 3 = 3. Therefore, f <. Moreover, f is odd, since f = e w is odd. Now there are two ossibilities for Euation e and f are corime to w. Since, e f mod w, f w = 1, and f w = w. Also, e w mod f, therefore, w f = 1. Now, both f and w are relatively rime odd integers less than, therefore we can aly Euation 3.1 and we get: f w = = w w f 1 w 1 f 1 = 1 w 1 Since e is even, e is divisible by 4. Therefore, w f mod 4. Also: w 1 = f 1mod 4 and Setting f = f 1/ and = 1/, we get w 1 f = f 1 mod f 1 = f = 4f + f + Now, f + 1 = f + f + w + 1 = f + 1 = f + mod w + 1 = f mod Multilying both sides by w 1/, we get w + 1 w 1 = f 1 w w 1 mod Since, w + 1/ and w 1/ are consecutive integers, their roduct is even and we have: f 1 w 1 = 1 w 1 mod

35 8 Since, 1 mod = 1 mod, we can remove the negative sign from the left side of the euation above. Thus, f 1 w 1 = 1 w 1 mod Alying this to Euation 3.4, we get: w Now, when w = and w = = 1, we have: When w = and w = = 1 w = 1 w 1 1 = 1, we have: = w w+1 1 = 1 w 1 1 w = 1 w 1 = 1 w Since, the only two solutions for this euation are 1 and 1 and raising either of them to the 1 ower does not change the result, we can remove the negative sign from the exonent without affecting the solution. Therefore, when w = we also have: = 1 w 1 1 This roves Case i, when e and f are corime to w. ossiblity: Next we look at the second

36 9. f and e are divisible by w Let f = wf 1, for some odd number f 1 < f and let e = we 1 for some even number e 1 < e. Now we can rewrite Euation 3.3 as: e 1w w = wf 1 we 1 w 1 w = f 1 Then, e 1 w 1 mod f 1 and therefore, = f1 w w e 1w 1 = f 1 or e 1w = 1 + f e 1 f and get: f1 w Now, we can see that: Therefore, now we have: w f 1 = = 1. Moreover, 1 f 1 mod w;. Since f 1 and w are both less than, we can aly Euation = 1 w 1 f 1 1 f1 w w f 1 = 1 w 1 f 1 1 f1 1 = = w w w 1 = w w = 1 w 1 w 1 w 1 f 1 1 w f1 w 1 w 1 f 1 1 = 1 w 1 w 1 f = 1 f 1 +1 w We know that e 1 is even, therefore e 1 0 mod 4. Now from Euation 3.6 we have e 1 w 0 mod 4 f mod 4. Since, both f 1 and are odd f mod 4, if and only if one of f 1 or is of the form 4M + 1 and the other is of the form 4M + 3. In either case we have: f mod

37 30 Alying this to Euation 3.7, we get: w = 1 1 w Now we can see that Euation 3.8 is analogous to Euation 3.5 above and holds true for both w = and w =. This concludes the roof for Case i. Now we look at Case ii. In Case ii we need to show that when is of the form 4M + 1 and = 1 and is not a uadratic residue mod, then is also a non residue mod and = 1. In order to rove this, we will first rove the following lemma give by Gauss. Lemma 3.. If is a rime of the form 4N + 1, there exists an odd rime < for which = 1. Proof. 1. This is aarent when is of the form 8N + 5. In this case, +1 is of the form 4N + 3. Now since, not all of its rime factors can be of the form 4N + 1. There must be a rime factor of the form 4N + 3, which divides + 1 such that mod and 1 mod. Thus: = 1 = 1 1 = 1. When is of the form 8N + 1, if we assume that is a uadratic residue of every odd rime less than m + 1 <. Then since we know from Theorem.6 that = 1, we can see that is also a uadratic residue of every ositive integer which is a roduct of numbers m + 1. Now if M = 1 3 mm + 1, then, the congruence x mod M is solvable. Let k = x be one of its solutions, then k and are relatively rime to M, and: Moreover, k 1 k m 1 m mod M k k 1 k m k+mk+m 1 k+1kk 1 k m 1k mmod M

38 31 The right side of the euation is a roduct of M + 1 consecutive integers, it must be divisible by M, thus the roduct: 1 m 1 m + 1 is an integer. Furthermore, since m + 1! can be rewritten as: [m + 1 m] [m + 1 m 1] [m + 1 1] [m + 1 0] [m m] [m m 1] [m ] Therefore, the following is an integer: m + 1[m + 1 m ] [m ] 1 1 m + 1 m m + 1 m m + 1 m 3.9 Now, if we choose m = [ ] to be the largest integer less than, such that m < < m + 1, then m < < m + 1, and every factor of the roduct in Euation 3.9 is a roer fraction, which is a contradiction. Furthermore, since is of the form 8N + 1, and the smallest ossible rime of that form is 17, 17, and it is aarent that 8 < 3. Then, 4 < + 1 = 1, and < 1 or + 1 <. Since we chose m = [ ] to be the largest integer less than, it follows that m + 1 <. Thus, there must exist some rime < m + 1 <, which is a uadratic non residue of if is of the form 8M + 1. This comletes the roof for Lemma 3.. Now, if = 4N + 1, there is an odd rime < with = 1, then be 1. If = 1, then we can aly Case i, and we get: = = 1 must also This contradicts the assumtion that = 1. In order to comlete the roof of Case ii, we only need to show that if there exists another rime < searate from the existing rime, such that = 1, then also = 1 or in other words: =

39 Now we know that = 1, and by assumtion we have = 1, thus = +1, and the congruence x mod is solvable. Then the solutions for x are x 0 and x o, let e be the solution which is even, then: 3 e = + f 3.11 where f is an odd integer less than. Now we can look at the following cases: 1. e and f are not divisible by or. Then we can see that e mod f, and f = 1. Moreover, e f mod, therefore, f = 1, and f =. Now when f > 0 we get: f = = 1 1 f Furthermore, when f is negative, we get: f 1 = = 1 1 f+1 = 1 1 f = 1 1 Since e is even, we have f mod 4. Moreover, since 1 mod 4, f mod 4: f mod Since +1 and 1 are consecutive integers, their roduct is even. Thus relacing f 1 with +1 in Euation 3.1 gives us an even exonent and yields: = 1 which is what we wanted to show.. e and f are divisible by, but not. f 1 Then we can set f = f 1, and e = e 1. Then Euation 3.11 takes the form: e 1 = f From Euation 3.13 we see that e 1 mod f 1. Therefore, e 1 e = 1 = = f 1 f 1 f 1 f 1 f 1

40 33 Now, f 1 = 1. Moreover, f1 = = Also, = f1 = f1 f1 Since f 1 and are less than, by our assumtion the generalized uadratic recirocity law holds for f 1 and, and we get: f1 = 1 f f 1 Now if f 1 > 0, we have: = 1 f = 1 f = 1 f = 1 f f1 f1 Since the uadratic recirocity law also holds for and, we get: = 1 f If f 1 < 0, we have: = 1 f = 1 f = 1 f = 1 f f1 f Since e 1 in Euation 3.13 is even and 1 mod4, we get f 1 mod 4 mod 4. Thus: f mod +1 1

41 34 Now, Moreover, f Therefore, f mod f Since +1 and 1 are consecutive integers, their roduct is even. Thus in both cases where either f1 > 0 or f 1 < 0, we have: = 1 which is what we wanted to show. 3. Since in the roof of. above, we did not utilize the fact that = 1, we can see that simly interchanging with in the roof above will similarly rove the case where e and f are divisible by, but not. 4. The final case is where e and f are divisible by both and. Let f = f 1, and e = e 1. Then Euation 3.11 takes the form: e 1 1 = f From Euation 3.15, we can see that: e 1 1 = = f 1 f 1 and f1 = 1 Thus, we can see that: = f1 = 1 f1

42 35 Since, f 1 and are less than, we can aly the generalized uadratic recirocity law and we get: f1 = f 1 Thus when f 1 > 0, And when f 1 < 0, = 1 1 = f f f = 1 f 1 +1 = 1 f Now, since e 1 is even and 1 mod 4, we can see from Euation 3.15 that f 1 1 mod 4, thus f 1+1 is even and conseuently the exonent is even in both cases for f 1. Therefore, which is what we wanted to show. = 1 This concludes the roof for Case ii, and hence comletes Gauss s first roof of the Law of Quadratic Recirocity by induction. 1 Although Gauss looked at each of the eight cases outlined at the beginning of this roof in his seminal demonstration, here we chose a slightly smaller version of the roof, which allowed us to collase several of these cases. This by no means takes aways from the urose of this demonstration, which was to show that such a fundamental theorem of number theory could be roven via observation and the very basic techniue of induction. We also add that while Gauss listed each of the eight cases searately in his own roof in Disuisitiones Arithmeticae, he also chose to forgo reetition of the roof in cases where the strategy was the same as one of the earlier cases. This brings u an imortant insight that can be gleaned from this roof. Desite the fact that we were able to collase our version of the roof into two mutually exclusive cases, these two cases still resented us with several distinctions and subdivisions. Yet many of these subdivisions utilized very similar strategies, secifically, they entailed euating the uadratic character of to another odd rime less than and then determining whether the resulting combination

43 36 of exonents would be even or odd. This should suggest that it would be ossible to further collase this roof and identify aroaches which are even shorter. Indeed, in his second roof Gauss used the genus theory of uadratic binary forms, which consisted of establishing a bound on the number of existing genera of the uadratic forms of a given determinant and subseuently investigating the only two cases for the rimes and. The resulting roof is far shorter than his first roof. We will not look at his second roof here, but the original roof can be found in Disuisitiones Arithmeticae, and another version of it can be seen in The Quadratic Recirocity Law: A Collection of Classical Proofs. Instead we will investigate Gauss s fourth roof, which made use of uadratic Gauss sums and eventually heled advance the field of algebraic number theory. The first ste of our journey showed us how observation and induction heled establish one of the fundamental theorems of number theory. Next, we will look at how further attemts to refine and strengthen this theorem resulted in the discovery of new territories and gave rise to new features of the mathematical landscae. 3.3 Gauss s Fouth Proof of Quadratic Recirocity In his fourth roof Gauss used uadratic Gauss sums to investigate the Law of Quadratic Recirocity. This roof extended the law of uadratic recirocity to cyclotomic fields and in so doing contributed greatly to the develoment of this field. We have already defined Gauss Sums and given some of their fundamental characteristics in the revious chater. In order to give the roof of uadratic recirocity, we first need to rove two additional roositions, which will then allow us to roceed with the comlete roof using Gauss Sums. For more information on this roof lease refer to the following texts: [Bau15] [Lan94] [Lem00] Proosition 3.3. For any integer a, G a = a G 1 Proof. 1. When a, a 0 mod and: 0 G 0 = G 1 = 0

44 37. In the more difficult case, where a, we need to show that a a G a = G 1 G a = G 1 Now, a G a = a 1 n=0 n ζ an = 1 n=0 an ζ an Since a, we can see that the roduct an will ermute all the numbers 0 < n < modulo. Therefore, the sum: 1 n=0 an ζ an = 1 n=0 n ζ n = G 1 Examle 3.4. Let a = 3 and = 5, then: G 3 = G 1 = 4 n=0 4 n=0 n 5 ζ 3 5 = 1 ζ ζ ζ ζ5 5 G 3 = ζ5 3 ζ 5 ζ5 4 + ζ n ζ5 n = ζ 5 + ζ5 + ζ5 3 + ζ G 1 = ζ 5 ζ 5 ζ ζ We know that 3 5 = 1. Therefore, from Euations 3.17 and 3.18, we can see that: G 3 = 1G 1 Proosition 3.5. For any integer a, such that a: G a = 1 1

45 38 Proof. Since a, from the definition of Gauss Sum, we have: G a = 1 a,b=1 ab ζ a+b Moreover, because a, b and both a and b range over 1,..., 1. We can rewrite b ac mod where c also ranges over 1,..., 1. Then: Since 1 1 a G c a = a=1 c=1 a = 1, we can further simlify to: G a = 1 1 c=1 a=1 ζ a1+c ζ a+ac c Now if 1 + c 0 mod, then the sum of the series 1, ζ 1+c ζ 1 ζ 1 = 0. Thus, 1 a=1 ζ a1+c = 1 Now if 1 + c 0 mod, then we are summing 1 ones, and: 1 a=1 ζ a1+c = 1 Here we note that c = c 0 mod. Therefore, G a = 1 1 c=1 a=1 ζ a1+c We can sum from 1 to 1, if we add 1 c = 1 G a = c=1 1 c=1, ζ 1+c c + 1. Therefore, we get: c + 1,..., ζ 11+c = 1 Here we can see that the summation on the left is eual to 0, by Theorem.. Moreover, = 1 1, by Theorem.1. Thus we have: G a = 1 1

46 39 Examle 3.6. Let = 3 and a = 1, then: Using the definition of Gauss Sum, we get: G 1 = G 1 = = 3 1 ζ ζ3 = ζ 3 ζ3 3 We know that by definition of the n th root of unity ζ 3 3 = 1 or ζ3 3 1 = 0. Moreover, x 3 1 = x 1x + x + 1 = 0, therefore x = x 1. Substituting x for ζ 3 we get ζ 3 = ζ 3 1. Now, and: G 1 = ζ 3 ζ 3 = ζ 3 + ζ = ζ Substituting for ζ 3 = ζ 3 1 again, we get: G 1 = ζ = 4ζ 3 + 4ζ G 1 = 4ζ ζ = 3 Using Proositions 3.3 and 3.5, we can now rove the law of Quadratic Recirocity via Gauss Sums. In order to rove the law we will calculate G 1 in two different ways and show them to be eual. Let us restate the theorem here before we begin the roof: Theorem 3.7. If and are distinct odd rimes, then = Proof. i Let = 1 1. Let G = G 1 = 1 n n=0 ζ n C. Then from Proosition 3.5 we know that: G = Moreover, from Corollary.18 we know that: 1 mod

47 40 Now, we can see that: G = G 1 G = G 1 G Thus, G 1 Gmod ii G = G 1 = 1 n n=0 ζ n C. Therefore, G = 1 n n=0 Since, is an odd rime, we know that ζ n = n n=0 1 n=0 Gmod 3.19 n ζ n n. Therefore: 1 n G ζ n G Now we can aly Proosition 3.3 to the euation above and we get: G G By combining Euations 3.19 and 3.0, we can see that: G Gmod By cancelling G from both sides, we have: Gmod 3.0 Gmod Since both residue symbols are ±1 and is odd, we can say that = 1 Multilying both sides by we get: = 1 1 = = and: =

48 41 which is what we wanted to show. The roof using Gauss sums can also be restated using elementary techniues from basic algebraic number theory. In articular, one can use Galois theory and algebraic number theory to first define the following concets: uadratic subfields of cyclotomic fields, the silitting of rime ideals and the Frobenius element. Then, using these roerties one can then roduce a relatively simle roof of uadratic recirocity, similar in rincile to the one above. Interestingly, these theories did not fully develo until the late 19th century after Gauss time. Yet Gauss was able to utilize Gauss sums to construct a uniue uadratic subfield and indetify the slitting of the rime, without using any of the definitions or language from either of these theories. Gauss had done some early work with cyclotomic fields in connection to the construction of a 17-gon. His later work, which generalized the law of uadratic recirocity to cyclotomic fields, heled demonstrate roerties which eventually became an imortant art of the theory of cyclotomy. This may shed some light on Gauss motivation to exlore his fundamental theorem using different techniues. He was not simly looking for more arguments in suort of his theorem, instead his ultimate goal was to exlore new branches of mathematics as he worked through his various roofs, in order to identify techniues and strategies which would eventually form the basis of modern number theory. Next, we will look at a variation of Gauss third roof of the law of uadratic recirocity. Gauss considered his third roof to be the most direct and natural of the eight roofs of this theorem rovided by him. This third roof greatly simlifies and reduces the number of stes reuired to acheive the desired conclusion in his first roof. However, rather than look at the roof rovided by Gauss himself, here we will instead focus on a further version on Gauss third roof rovided by Gotthold Eisenstein. Although Gauss third roof was simler and more direct than his first roof, it utilized Gauss lemma and reuired several technical maniulations before his lemma could be successfully alied. In the version resented here, Eisenstein follows a very similar outline to the one used by Gauss; however, is able to use a geometric transformation to greatly simlify some of the stes, which otherwise reuired Gauss to consider several different cases to arrive at the desired conclusion. For more information on this roof, lease refer to the following sources: [Bau15] [Bur07] [LP94]