Degree in Mathematics

Transcription

1 Degree in Mathematics Title: Gauss s roofs of the uadratic recirocity law Author: Anna Febrer Galvany Advisor: Jordi Quer Bosor Deartment: Matemàtica Alicada II Academic year: 01/013

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3 Universitat Politècnica de Catalunya Facultat de Matemàtiues i Estadística Bachelor s degree thesis Gauss s roofs of the uadratic recirocity law Anna Febrer Galvany Advisor: Jordi Quer Bosor Matemàtica Alicada II

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5 To all those eole who had faith in me from the beginning and now can see how I become a mathematician

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7 Contents Introduction 1 Chater 1 Preliminaries 3 11 Historic introduction 3 1 Quadratic residues 7 13 Legendre symbol 9 1 Quadratic recirocity law Jacobi symbol 1 Chater Gauss s first roof 15 1 First and second sulementary laws 16 Technical lemmas 18 3 Quadratic recirocity law for odd integers Quadratic recirocity law for odd rimes 8 Chater 3 Proofs based on Gauss s Lemma 3 31 Gauss s Lemma 3 3 Gauss s third roof 33 Gauss s fifth roof 8 Chater Proofs based on trigonometric sums 53 1 Gauss s fourth roof 53 Gauss s sixth roof 6 References 71 9

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9 Introduction I find very interesting that a single statement can be roven in many different ways For this reason I decided to dedicate my bachelor s degree thesis to the study of the uadratic recirocity law, robably the statement that admitted the greatest number of different roofs: u to now, several hundred roofs of the uadratic recirocity law have been found Franz Lemmermeyer, one of the leading exerts on the subject maintains a web age [Lem1] with a list of ublished roofs that today contains 0 entries At the age of 19 Gauss roved for the first time the uadratic recirocity law, that had stayed unroven for nearly 150 years, resisting the efforts of some of the best mathematicians including Fermat, Euler and Legendre Gauss believed that the uadratic recirocity law was a very imortant mathematical result, so imortant that he dared to call it fundamental theorem Gauss was very roud of his roof of the fundamental theorem, so he tried to develo a bit more this field by looking for similar theorems for recirocity of higher owers His attemts to generalize the fundamental theorem to higher owers led him to found seven more roofs of the uadratic recirocity law The goal of this thesis is to understand the uadratic recirocity law, study the roofs that Gauss gave of it in its original sources most of them written in Latin and not translated into other languages and exlain them in a more modern language Due to time limitation, this roject only resents five of the eight roofs of Gauss, which are the first, the third, the fourth, the fifth and the sixth We structured the thesis in four chaters Chater 1 introduces the toic, giving a bit of history of the develoment of the uadratic recirocity law and basic notation, concets and statement of uadratic recirocity Chater, the largest chater, resents the first roof which was made by induction and is by far the largest, tricky and more intricate roof found by Gauss In Chater 3 we show the third and the fifth roofs, the two ones based on Gauss s Lemma And finally in Chater we exlain the roofs based on trigonometric sums, which are the fourth and the sixth All this roject is sourced from Gauss s original works, basically from the book Disuisitiones arithmeticae [Gau1] and the original aers of Gauss, collected in Werke Vol [Gau3],[Gau],[Gau5] and [Gau6] We tried to exose the five roofs using more modern notation in order to make them more understandable 1

10 INTRODUCTION and easy to follow However, some times we adoted Gauss notation which is not used any longer because it was useful in order to follow Gauss stes

11 Chater 1 Preliminaries 11 Historic introduction The uadratic recirocity law was born when Fermat became interested in the following roblem: Determine the rimes that can be written in the form: where n is a fixed integer x + ny, Since the first results were stated until the uadratic recirocity law was roved, two hundred years later, there have been four mathematicians who layed a key role in the develoment of the law Fermat Pierre de Fermat was the mathematician who started studying recirocity uestions All we know about Fermat s results is contained in his letters to other mathematicians or in the margins of some books he read since in Fermat s time mathematical journals did not exist The first result related with uadratic recirocity aears in a letter to Mersenne, and it stated: Tout nombre remier, ui surasse de l unité un multile du uaternaire, est une seule fois la somme de deux carrés 1 This statement is euivalent to say that for every rime number of the form n + 1, say, there exist two ositive integers x and y such that x +y Using Bézouts s identity, Fermat s little theorem and some basic roerties of congruences, it can be seen that a rime is the sum of two suares x + y if and only if the euation X 1 mod is solvable Since Fermat knew that every rime number of the form n + 1 and no other odd rimes can be exressed as a sum of two suares, it follows that the congruence X 1 mod is solvable if an 1 Every rime which is one more than a multile of is a sum of two suares in one and only one way 3

12 1 PRELIMINARIES only if is congruent to 1 modulo What Fermat stated and roved was what we nowadays know as the first sulementary law of uadratic recirocity For other integers n, the study of the rimes of the form x + ny leads to the congruence X n mod, whose solutions for all reuires the uadratic recirocity law Fermat was able to deduce and conjecture the uadratic character of ± and ±3 but he couldn t rove them Euler Euler first stes in number theory were guided by Fermat s work He roved some of Fermat conjectures and he also refuted some of them Euler used to corresond with Goldbach and to discuss with him themes related to number theory In one of his letters to Euler, Goldbach asked: PS Notane Tibi est Fermatii observatio omnes numeros hujus formulae x 1 + 1, neme 3,5,17, etc esse rimos, uam tamen ise fatebatur se demonstrare non osse et ost eum nemo, uod sciam, demonstravit After that letter, Euler started to study Fermat numbers n + 1 and Marsenne numbers 1; and his work on divisors of these numbers and, more generally, on divisors of number reresented by binary uadratic forms nx + my led Euler to realize the uadratic recirocity law His first result directly connected with uadratic recirocity law was: Euler s criterion For integers a and odd rimes such that a we have { a 1 +1 mod, if a is a uadratic residue modulo, 1 mod, if a is a uadratic nonresidue modulo After some time working in other roblems, the research of Lagrange in the years 1773/75 about uadratic recirocity made Euler to take u the subject again, and this time he discovered the comlete uadratic recirocity law, that, translated into modern notation, he stated in the following form: 1 If 1 mod is rime and x mod for some rime, then ± y mod If 3 mod is rime and x mod for some rime, then y mod but y mod 3 If 3 mod is rime and x mod for some rime, then y mod but y mod If 1 mod is rime and x mod for some rime, then ± y mod PS Is Fermat s observation known to you that all numbers of the form x 1 + 1, namely 3,5,17, etc are rimes, which he himself could not rove and which no one after him, to the best of my knowledge, has ever roved

13 11 HISTORIC INTRODUCTION 5 This statement was finally ublished in 1783, after Euler s death, without him had been being able to roduce a roof Euler also conjectured the following theorem: Theorem Let f, g Z be corime suarefree integers that are not both negative If h and h are ositive rimes such that h h mod fg, then the euation fx + gy hz is solvable if and only if the euation fx + gy h z is also solvable The connection between this theorem and the uadratic recirocity law comes by utting g a and f 1 Then Euler s conjecture says that if mod a, then the euation x ay z is solvable if and only if x ay z Legendre The uadratic recirocity law was ublished in a form that is more familiar to us in 1785 by Legendre In his aer see [Leg1] he considers rimes a, A 1 mod and b, B 3 mod and states: Théorème I Théorème II Théorème III Théorème IV Théorème V Théorème VI Théorème VII Théorème VIII Si b a 1 +1, il s ensuit a b 1 +1 Si a b 1 1, il s ensuit b a 1 1 Si a A 1 +1, il s ensuit A a 1 +1 Si a A 1 1, il s ensuit A a 1 1 Si a b 1 +1, il s ensuit b a 1 +1 Si b a 1 1, il s ensuit a b 1 1 Si b B 1 +1, il s ensuit B b 1 1 Si b B 1 1, il s ensuit B b 1 +1 Actually, Legendre wrote instead of in order to denote euality u to certain multiles; the notion of congruence and the symbol was later introduced by Gauss It is worth to note that both the notation, using letters of the same tye to denote rimes 1 or 3 mod and also the way to state the result are the ones used by Gauss in the first roof In later roofs Gauss moved to more comact and elegant ways to state the uadratic recirocity law In 1798, Legendre introduces the Legendre symbol see 186 of [Leg]: Comme les uantités analogues N c 1 se rencontreront fréuemment dans N le cours de nos recherches, nous emloierons le caractère abrégé c our exrimer le reste ue donne N c 1 divisé ar c, reste ui suivant ce u on vient de voir ne eut être ue +1 ou 1 3 c 1 3 Since the analogous uantities N will occur often in our researches, we shall emloy the N abbreviation for exressing the residue that N c 1 gives uon division by c, and which, c according to what we just have seen, only assumes the values +1 or 1

14 6 1 PRELIMINARIES And on age 1, he continues Quels ue soient les nombres remiers m et n, s ils ne sont as tous deux n m de la forme x + 1, on aura toujours ; et s ils sont les deus m n n m de la forme x 1, on aura Ces deux cas géneraux sont m n comris dans la formule n m 1 n 1 m 1 m n Trying to rove the uadratic recirocity law, Legendre found that the following statement must be true: For each rime a 1 mod there exists a rime β 3 mod such α that 1 β However he could not rigorously rove the existence of such a rime β, and this revented him from roving the result The first roof by Gauss in fact uses a technical nontrivial lemma similar to this result Gauss On 8 Aril 1796, Gauss finally could rove the uadratic recirocity law at the age of 19 But his desire to find similar theorems for recirocity of higher owers made him look for roofs which would generalize, and by 1818 he had ublished six different roofs, and two more were found in his unublished aers and were ublished after his death But not only did Gauss give the first comlete roofs of the uadratic recirocity law, he also extended it to comosite values of and This generalization can be stated using a generalization of the Legendre symbol for odd ositive integers This symbol was introduced by Jacobi in 1837 and is best known as Jacobi symbol Using Jacobi symbols, the Gauss generalization for the uadratic recirocity law and its sulementary laws can be stated as follows: Generalization of the uadratic recirocity for odd ositive integers: Let n, m N be relatively rime and odd, then m 1 m 1 n 1 n 1, 1 n m n 1, 1 n 1 8 n n This was a great imrovement on Euler s and Legendre s version of uadratic recirocity, as far as the comutation of residue symbols was concerned: instead m n Whatever the rime numbers m and n are, if they are both of the form x + 1, one always has n m n m ; and if both are of the form x 1, one has These two general m n m n cases are contained in the formula n m 1 n 1 m 1 m n

15 1 QUADRATIC RESIDUES 7 of having to factor the residue of m mod n before inverting the occurring Legendre symbols one could simly invert the Jacobi symbols and aly a comutationally chea Euclidean algorithm 1 Quadratic residues In the article 95 of Disuisitiones Arithmeticae see [?] Gauss introduced the terminology uadratic residue and nonuadratic residue as follows: Definition 11 An integer is called a uadratic residue modulo n if it is congruent to a erfect suare modulo n; ie, if there is an integer x such that: x mod n Otherwise, is called a uadratic nonresidue modulo n When this can not cause any ambiguity, we will simly call them residues and nonresidues The modulo n can be any integer, but for simlicity here we will only consider odd modulus Proosition 1 Let be an odd rime number and a a residue of Considering only the solutions smaller than, the euation: x a mod has exactly two different solutions, one odd and one even Proof Since a is a residue of, the euation x a mod has at least a solution Suose that m < is a solution of the euation, ie m a mod Then m is also a solution of the euation, because m m a mod But m m mod, so m is another solution of the euation and it is also smaller than Note that m + m, so m and m must have different arity since their sum is an odd number Now we shall show that, aart from m and m, the euation has no other solutions Let us consider the set { A i mod i 1,, 1 } Note that A #{residues of less than }, and by Proosition 1 we know that #{residues of less than } 1 1 So A Hence all elements of A must be different, ie there isn t any number n < 1 different from m and m such that m n mod Conseuently, there is not any number less than 1, aart from m and m, satisfying the euation x a mod Therefore we

16 8 1 PRELIMINARIES roved that the only solutions of x a mod are m and m, one odd and the other even Given a ower of an odd rime n, it is easy to count how many numbers less than n are a residue of it The following Proositions 13 and 1 state this Proosition 13 Let be a odd rime number Then half of the numbers less than n and not divisible by are residue of n The other half are nonresidue of n Ie there are 1 1n 1 residues of n less than n and not divisible by and 1 1n 1 nonresidues of n less than n and not divisible by Proof A residue r modulo n needs to be congruent to a suare whose suare root is less than n / We have 1 1n 1 numbers less than n / and not divisible by and their suares are all different modulo n Indeed, if we have a b a ba + b 0 mod then n m a b and m a + b for some m {0, 1,, n} If m 0 or m n we have that a b or a + b is divisible by n but this cannot be ossible since a, b < n / If m 0, n both a + b and a b are divisible by, so also the sum, a, and the difference, b, are divisible by, so a and b need to be divisible by and that contradicts our hyothesis So we can conclude that there are exactly 1 1n 1 residues of n less than n and not divisible by Corollary 1 Let be an odd rime number Then in the set {1,, 3,, 1} there are 1 1 residues of and 1 1 nonresidues of For odd comosite numbers, P, there exists a relation between the integers that are residues of P and the integers that are residues of all rime factors of P This relation is stated below, in Proosition 15 and Proosition 16 Proosition 15 Let be an odd rime and M any number corime to Then M is a residue of if and only if M is a residue of n for all n N Proof Clearly if M is a residue of n then M is a residue of, since x M mod n x M mod To rove the other imlication let us suose that M a residue of n for some n N and we shall show that M is also a residue of n+1 Since M is a residue of n there is a number x such that x M mod n Let us take y x + n t and imose M y x + n t x + n t + x n t mod n+1 Therefore: M x + x n t mod n+1 M x x n t mod n+1 M x n xt mod M x n x 1 t mod Then, choosing t satisfying t M x n x 1 mod we obtain an integer y such that y M mod n+1, and this roves that M is residue of

17 13 LEGENDRE SYMBOL 9 Hence, if M is a residue of then M will be a residue of But M will also be a residue of 3 and so on In conclusion, if M is a residue of then M will be a residue of n for all n N Proosition 16 Let P be a comosite odd number, and P α1 1 α αn n its rime factorization Let M be any odd number Then M is a residue of P if and only if M is a residue of all rime factors of P, ie iff M is a residue of 1,,, n Proof If M is a residue of P then it is also a residue of every rime i Assume that M is a residue of each i Then, by the Proosition 15 we know that if M is a residue of i then M is also a residue of αi i for i 1,, n Hence: M residue of 1 M residue of α1 1 x 1 such that x 1 M mod α1 1, M residue of M residue of α x such that x M mod α, M residue of n M residue of αn n x n such that x n M mod αn n Let us consider the following euations on X: x 1 X x X x n X mod α1 1, mod α, mod αn n By the Chinese remainder theorem we know that there is an integer N such that N X mod α1 1 α αn n, ie N X mod P Hence: N X x 1 mod α1 1 N x 1 M mod α1 1 N X x mod α N x M mod α N X x n mod αn n N x n M mod αn n Therefore N M mod P {}}{ α1 1 α αn n, so M is a residue of P 13 Legendre symbol The Legendre symbol was introduced by Legendre in 1798 in the book Essai sur la théorie des nombres see [Leg], in the course of his attemts at roving the uadratic recirocity law Definition 131 Legendre symbol Let be an odd rime number and a an integer The Legendre symbol is a function of a and defined as follows:

18 10 1 PRELIMINARIES a 0 if a 1 if a is a uadratic residue modulo 1 if a is a uadratic nonresidue modulo Legendre s original definition was the exlicit formula: a a 1/ mod Nowadays this symbol is widely known and, in fact, is the one used when talking about the uadratic recirocity law But Legendre introduced the symbol two years later than the year Gauss roved the law for first time 1796 So, in his aers, Gauss introduced another notation for defining a residue and a nonresidue Gauss s notation was the following: ar if a is a residue of, an if a is a nonresidue of After being defined the Legendre symbol we will show some results related with it Proosition 13 The Legendre symbol is multilicative, ie a b ab ab Proof If ab is divisible by, 0 Moreover, as is rime, has to divide a b a, b or both So we always have 0 Then, if divides ab, the identity a b ab is true Now we will study the case that ab If a and b are residues modulo then } α a mod β αβ ab mod b mod So the roduct of two residues is a residue Note that if a is a residue modulo then 1 a α a mod α 1 a 1 will also be a residue modulo mod Now suose that a is a residue and b is a nonresidue The roosition says that ab 1, ie ab has to be a nonresidue and we rove it by contradiction Let us suose that ab is a residue so α ab mod Multilying both sides of the euation by 1 a we get α /a b mod And since 1 a is a residue and the roduct of two residues is a residue, we have that b must be a residue and this is a contradiction

19 1 QUADRATIC RECIPROCITY LAW 11 And finally, if a and b are nonresidues, then the roduct ab has to be a residue Let us rove it by contradiction Let us suose ab is a nonresidue Now multily a by all the residues r modulo and we will get all the residues of the form ar So it must exist r such that ab ar mod Multilying both sides of the euation by 1 a we have b r mod And this is a contradiction because r is a residue and b a nonresidue Proosition 133 Euler s criterion Let be a rime number and a an integer, then a a 1/ mod Proof If a is divisible by then 0 a a 1/ 0 mod Now consider the case in which a is not divisible by We will work with the elements a Z/Z Let g be a generator of Z/Z, then g is a nonresidue because if not all the elements of Z/Z would be residues and this cannot haen because we saw in Corollary 1 that half of the elements of Z/Z are residues and half are nonresidues g has order 1 in Z/Z, so g 1/ has order in Z/Z Then g 1/ 1 because this is the only element of order in this grou Therefore, every element a Z/Z is a ower g e of the generator, and: a 1/ g e 1/ g 1/ e 1 e g e g e a 1 Quadratic recirocity law The law of uadratic recirocity is a law that gives conditions for the solvability of uadratic euations of the form x mod where and are odd rime numbers This law also includes two sulementary laws that give conditions for the solvability of uadratic euations of the form x 1 and x both modulo an odd rime Theorem 11 Quadratic recirocity law Let be an odd rime number Then 1st sulementary law x 1 mod is solvable 1 mod nd sulementary law x mod is solvable ±1 mod 8 Quadratic recirocity law Let > be another odd rime number different from Let ± where the sign is lus if 1 mod and minus if 3 mod Then x mod is solvable if and only if x mod is solvable

20 1 1 PRELIMINARIES Using Legendre symbols, this theorem can be written as follows Theorem 1 Quadratic recirocity law [Written using the Legendre symbols] Let and be odd rime numbers Then 1st sulementary law 1 1 1/ nd sulementary law 1 1/8 Quadratic recirocity law 1 1 1/ Indeed, Theorem 11 and Theorem 1 are euivalent In the first sulementary law, x 1 mod if and only if and this is satisfied when 1/ is even 1 1, 1/ is even 1 0 mod 1 mod In the second sulementary law, x mod if and only if 1, and this is satisfied when 1/8 is even 1/8 is even 1/8 Z 1 mod 8 ±1 mod 8 And conversely ±1 mod 8 n N such that 8n ± 1 1/8 6n ± 16n /8 n ± n is even The uadratic recirocity law states Hence, when 1 mod we have that and, indeed, x mod is solvable if and only if x mod is solvable But when 3 mod we have: Therefore x mod is solvable if and only if x mod is solvable 15 Jacobi symbol The Jacobi symbol is a generalization of the Legendre symbol It was introduced by Jacobi in 1837 in the book Über die Kreistheilung und ihre Anwendung auf die Zahlentheorie see [Jac]

21 15 JACOBI SYMBOL 13 Definition 151 Let n be a ositive odd integer with rime factorization n ri i For every a Z let us define the Jacobi symbol a n a i Ie the roduct of the Legendre symbols for the rimes that divide n, counted by multilicity The main difference between the Legendre and the Jacobi symbol is that in this last one, the bottom number must be ositive and odd, but does not need to be rime as in the Legendre symbol If it is rime, the two symbols agree Using the Jacobi symbol, the uadratic recirocity law can be generalized to ositive odd integers Theorem 15 Quadratic recirocity law for the Jacobi symbol Let n and m be two ositive odd integers Then 1st sulementary law 1 1 n 1/ n nd sulementary law 1 n 1/8 n Quadratic recirocity law m 1 n 1m 1/ n n m m Remark 153 When n is a rime number then the Jacobi symbol agrees n with the Legendre symbol m Remark 15 The symbol is 0 when gcdm, n 1 n m Remark 155 When m is a residue of n then 1 but the converse is not n m true In this case 1 means that there are an even number of rime factors n ri of n, counted with multilicity, such that m is nonresidue of them Proosition 156 If either the to or the bottom argument is fixed, the Jacobi symbol is a comletely multilicative function in the other argument, ie ab a b Fixed n, n n n a a a Fixed a, nm n m Proof The Jacobi symbol is multilicative in the bottom argument by definition and in the to argument by the multilicativity of the Legendre symbol

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23 Chater Gauss s first roof In this chater we will exlain the first roof of the uadratic recirocity law This roof can be found in Gauss s book Disuisitiones arithmeticae see [Gau1] Originally, Gauss called the uadratic recirocity law fundamental theorem and he stated it, in the article 15 of [Gau1], as follows: If is a rime number of the form n + 1, then + will be a residue or a nonresidue of any rime number which, taken ositively, is a residue or a nonresidue of But if is of the form n + 3, then it will be the one that will have the roerties described above Previously, in the articles 108 to 1 he had stated some articular cases, including the two sulementary laws These had been stated as follows: 1 is a uadratic residue of any rime number of the form n + 1, but is a nonresidue of any rime number of the form n + 3 is a residue of any rime number of the form 8n + 1 or 8n + 7, but is a nonresidue of any rime number of the form 8n + 3 or 8n + 5 Note that what Gauss stated is euivalent to the theorem in the form that we know it nowadays 1 Indeed, by the first sulementary law we know that 1 1/ Therefore, if is of the form n + 1 then 1 n 1, so 1 is a residue of 1 1 But if is of the form n + 3 then 1 n+1 1, so 1 is a nonresidue of The second sulementary law states that 1 1/8 Therefore, if is of the form 8n+1 or 8n+7, ie is of the form 8n±1, then n ±n 1,

24 16 GAUSS S FIRST PROOF so is a residue of While if is of the form 8n + 3 or 8n + 5, ie is of the form 8n ± 3, then 1 8n ±6n+1 1 The uadratic recirocity law says that 1 1 1/ where and are two different odd rime numbers Then, if is of the form n + 1, But if is of the form n Gauss s aim was to rove the fundamental theorem for rime numbers, but in fact he ended roving the generalisation of it for any odd number in which the Legendre symbol is relaced by the Jacobi symbol 1 First and second sulementary laws In this section we shall rove the two sulementary laws of the uadratic recirocity law as Gauss did in his book Disuisitiones arithmeticae see [Gau1] Theorem 11 First sulementary law 1 is a uadratic residue of any rime number of the form n + 1, but is a nonresidue of any rime number of the form n + 3 Proof This theorem can be easily roved using Euler s criterion [Theorem 133] If we have a rime number of the form n + 1, then: 1 1 1/ 1 n+1 1/ 1 n 1 mod But, if we have a rime number of the form n + 3, then: 1 1 1/ 1 n+3 1/ 1 n+1 1 mod Theorem 1 Second sulementary law is a residue of any rime number of the form 8n + 1 or 8n + 7, but is a nonresidue of any rime number of the form 8n + 3 or 8n + 5 Proof Let us start roving that is a nonresidue of any rime number of the form 8n ± 3 We shall rove it by contradiction Suose that there exists a rime number of the form 8n ± 3 such that is a residue of it Let t be the smallest rime number of the form 8n ± 3 satisfying 1, ie is a nonresidue of all rimes t of the form 8n ± 3 less than t Moreover is a nonresidue of all t < t of the form 8n ± 3 Indeed, by the Proosition 16, is a residue of t if and only if is a

25 1 FIRST AND SECOND SUPPLEMENTARY LAWS 17 residue of all rime factors of t, but for being of the form 8n ± 3, t has at least a rime factor of the form 8n ± 3, so is a nonresidue of at least a rime factor of t, therefore is a nonresidue of t Then t is not only the smallest rime number of the form 8n ± 3 such that is nonresidue of t but also the smallest number of the form 8n ± 3 such that is nonresidue of it Hence there is an odd number a less than t by Proosition 1 such that a mod t It imlies that a + tu with u < t Since a is odd, a is of the form 8n + 1 because, utting a m + 1, we have: a m mm + 1/ + 1 Then a tu, and this euality modulo 8 becomes 1 ±3u mod 8 which imlies that u 3 mod 8 Note now that is also a residue of u, since a mod u So we found a number u < t of the form 8n ± 3 such that is a residue of it and this contradicts the fact that t is the minimum number of the form 8n±3 satisfying this Hence is nonresidue of any rime number of the form 8n ± 3 Now let us rove that is a residue of any rime number of the form 8n + 7 In order to do that, we first rove the auxiliary case that states that is a nonresidue of any rime number of the form 8n + 5 or 8n + 7 As the revious case, we will rove it by contradiction Suose that there exists a rime number of the form 8n + 5 or 8n + 7 such that is a residue of it Let t be the minimum rime number of the form 8n + 5 or 8n + 7 satisfying 1, ie is a nonresidue t of all rimes of the form 8n + 5 or 8n + 7 less than t Moreover is a nonresidue of all t < t of the form 8n + 5 or 8n + 7 Indeed, by the Proosition 16, is residue of t if and only if is a residue of all rime factors of t, but all number of the form 8n + 5 or 8n + 7 has at least a rime factor of the form 8n + 5 or 8n + 7 resectively, so is a nonresidue of at least a rime factor of t, therefore is a nonresidue of t Then t is not only the minimum rime number of the form 8n + 5 or 8n + 7 satisfying that is a residue of t but also the minimum number of the form 8n + 5 or 8n + 7 satisfying that is a residue of it Then there is an odd number a less than t such that a mod t This imlies that a +tu with u < t Since a is odd, a is of the form 8n+1 So a + tu, and this euality modulo 8 becomes 1 + 5u mod 8 if t is of the form 8n + 5 or 1 + 7u mod 8 if t is of the form 8n + 7 This imlies that u 7 mod 8 or u 5 mod 8 resectively Note that is also a residue of u, since a mod u So we found a number u < t of the form 8n + 5 or 8n + 7 such that is residue of it and this contradicts the fact that t is the minimum number of the form 8n + 5 or 8n + 7 satisfying this Hence is a nonresidue of any rime number of the form 8n + 5 or 8n + 7 Using the first sulementary law and the multilicativity roerty of the Legendre symbol we have 1 8n n + 7 8n n + 7 8n + 7 This roves that is a residue of any rime number of the form 8n + 7

26 18 GAUSS S FIRST PROOF And finally, let us rove that is a residue of any rime number of the form 8n + 1 This case is roved in a totally different way that the revious cases, involving rimitive roots of rime numbers Let a be any rimitive root of a rime of the form 8n + 1 So a 8n 1 mod 8 and a n 1 mod 8 Indeed the last congruence may be exressed as a n +1 a n mod 8 Then a n is a residue of 8n + 1 As a n is a suare not divisible by 8n + 1, a n is also a residue of 8n + 1 Hence, by the multilicativity roerty of the Legendre symbols we can conclude a n a n n + 1 8n + 1 8n + 1 8n + 1 8n + 1 Technical lemmas After having roved the first and the second sulementary laws, we want to rove the uadratic recirocity law The roblem that for this roof we need to use that every rime number of the form n+1 taken ositively or negatively, excet +1 is a nonresidue of some rime number, and if > 5 then it is a nonresidue of some rime number less than The aim of this section is to rove the above statement This roof is based in some technical lemmas that we will state below Lemma 1 Let a 1, a, a 3,, a n I and b 1, b, b 3,, b m II be two series of numbers not necessarily with the same number of terms If for every integer that is a rime ower, divides at least as many terms in I as the number of terms in II that can be divided by then: a 1 a a 3 a n b 1 b b 3 b m Z Proof Let A n m a i and B b i According to the hyothesis of the lemma, i1 i1 it is clear that any rime number such that B, also satisfies A Conseuently we may exress A and B as follows A α1 1 α αr r M, B β1 1 β βr r, where M is an integer such that gcdm, i 1 for i 1,, r Now we will rove that α i β i for i 1 r Let us denote c ij to the number of terms in the series I that are divisible by j i, and c ij to the number of terms in the series II that are divisible by j i We have that α i c ij and β i c ij j 1 j 1 By hyothesis we have that c ij c ij for every i, j, hence α i β i for i 1,, r And we have roved that A is divisible by B Lemma Let 1,, 3,, n I and a, a + 1, a +,, a + n 1 II be two rogressions of n terms Then the number of terms in the rogression I that are divisible by any number h needs to be greater than or eual to the number of terms in the rogression II that are divisible by h

27 TECHNICAL LEMMAS 19 Proof Let us distinguish two cases: Case 1: n hm In this case there are m terms in the rogression I divisible by h Then if a hs we have a+n 1 hs+m 1, so h s, hs+1,, hs+m 1 are in II and these are the only ones in II which are divisible by h Conseuently, if a hs the rogression II has m terms divisible by h If hs + 1 > a > h s we have hs+1+m 1 > a+n 1 > hs+m 1, so hs+1, hs+,, hs+m are in II and these are the only ones in II which are divisible by h Conseuently, if a h s, there are m terms in the rogression II divisible by h Case : n hm + r, 0 < r < h In this case there are also m numbers divisible by h in the rogression I Now if hs a hs r + 1 we have hs + m + r 1 a + n 1 h s + m, so the only terms divisible by h in the rogression II are hs, hs + 1,, hs + m, in all m + 1 terms But if hs 1 < a < hs r + 1 we have hs + m 1 + r 1 < a + n 1 < hs + m, so hs, hs + 1,, hs + m 1 are the terms in I which are divisible by h, in all, m terms The Lemma that we have just roved can be stated euivalently as follows: Lemma 3 Euivalent to Lemma In the rogression a, a+1,, a+n 1 there are at least as many terms congruent to any given number r modulo h as terms in the rogression 1,,, n divisible by h Lemma Let a be any number of the form 8n + 1 Let be any number a such that, a 1 and 1 Given any integer m we take the rogression a, 1 a 1, a, 1 a 9, a 16,, a m if m is even or the rogression a, 1 a 1, a, 1 a 9, a 16,, 1 a m if m is odd Then the number of terms of this rogression that are divisible by is greater than or eual to the number of terms of the rogression 1,, 3,, m + 1 that are divisible by Proof In order to simlify the notation we will refer to the rogression a, 1 a 1,, a m or 1 a m as I and to the rogression 1,,, m+1 as II Furthermore A will denote the number of terms in the rogression I that are divisible by and A the number of terms in the rogression II divisible by Let us distinguish three cases: Case 1: In this case A m because all the terms in I are even excet a Indeed, if i is even then a i is divisible by and if i is odd then 1 a i 1 8n + 1 n n n n n n n + 1 is also divisible by Now note that between 1 and m + 1 we have m even numbers, then A m So in this case A A a Case : is of the form i t + 1 with i 0, 1 or We know that 1, so there is some r such that a r mod Now we take the rogression m, m 1, m,, m 1, m which have the same number of terms as II Let us denote this rogression as III Then, by Lemma 3, we have that the number of terms in III that are congruent to r modulo is greater than or

28 0 GAUSS S FIRST PROOF eual to the number of terms in II divisible by Now let us see that for each term in III congruent to r modulo we have a term in I divisible by First of all note that if b is a term of III such that b r mod then b can t be congruent to r modulo Suose that b b r mod, then b 0 mod and conseuently a rr bb 0 mod ; but, as a and are corime, this only can occur when, and we discussed this case aart Continuing with what we wanted to rove, we have that if b is a term of III congruent with r modulo then a bb rr rr 0 mod, ie a bb is divisible by Therefore if b is even, the term a bb of the rogression I is divisible by, and if b is odd we have that b is of the form ss + 1 b s + 1 ss so a bb is divisible by 8 and, as i t + 1 with i 0, 1,, the uotient a bb 1 a bb is even and we can conclude that is an integer, ie the term 1 a b of the rogression I is divisible by Finally we conclude that in the rogression I there are as many terms divisible by as terms in III are congruent to r modulo, ie the number of terms divisible by in the rogression I is greater than or eual to the number of terms in the rogression II divisible by Case 3: is of the form 8n As a r mod, it is also true that a r mod Now as in the revious case we have that the number of terms in the rogression III congruent to r modulo is greater than or eual to the number of terms in the rogression II Note that the terms in III that are congruent to r modulo are all different if we take them in absolute value, as it haened in the case Now let us take a term in III, b, such that b r mod Then b also satisfies bb rr mod, because b rb + r is divisible by due to the fact that b r is divisible by and b + r is even for being the sum of two odd numbers Therefore if b is odd 1 a bb 1 rr bb 0 mod 1 ie a bb is divisible by, and if b is even a bb is also divisible by So as in the revious case we roved that the number of terms in the rogression I that can be divided by is greater than or eual to the number of terms divisible by in the rogression II Proosition 5 Every rime number of the form n + 1 taken ositively or negatively excet +1 is nonresidue of some rime number Moreover, if > 5 then it is a nonresidue of some rime less than Proof We have seen in the first sulementary law that 1 is nonresidue of all rimes of the form n is a rime of the form n + 3 so the roosition is true for 1

29 TECHNICAL LEMMAS 1 Now let us discuss the case 5 Note that 1 is the only suare modulo 3 and 5 mod 3, so 5 a is nonresidue of 3 And 5 is a nonresidue of 13, since the suares modulo 13 are 1, 3,, 9, 10, 1 but 5 8 mod 13 Now let us suose that > 5 and let us distinguish two different cases: Case 1: Let be a rime number of the form n + 1 taken negatively Let a be the even number closest to with a > For 5, 17 it s true that a < Indeed, exressing as m + k with m the largest suare less than, we have that a m + 1 or a m + deending on m is odd or even Hence If m is odd a m + 1 m + k m + m + 1 m k 1 k m 1 If m is even a m + m + k m + m + m k k m 1 If m is odd, since k 1, it s clear that a < 0 While if m is even, m is of the form n and k must be of the form n + 1 since is of the form n + 1 Then a > 0 when k <, that imlies k 1, and when m < 6, ie for m and m So when and when we have that a > 0, otherwise a < 0 We have excluded 5 of this roof and note that ±17 is nonresidue of 5 Hence we can continue the roof for the rest of the rimes using the statement a < 0 It imlies that a < with a of the form n + 3 Note that is a residue of a because a mod a Then if a is rime, by the first sulementary law and the roerty of multilicativity of the Legendre symbol, we have that a 1 And if a is comosite, since it is of the form n + 3, a has at least a rime factor of the form n + 3, because if not it would be of the form n + 1 Then is a residue of and is nonresidue of < Case : Let be a rime number taken ositively Now let us discuss two cases: Case i: is a rime number of the form 8n + 5 Let a be a ositive number 1 such that a < Then is easy to see that a > 0 If a is even, a is of the form: a 8n + 5 m 8n + 5 8m 8n m + 5 8n 3, and if a is odd, a is of the form: a 8n + 5 m + 1 8n + 5 8m 8m 8n m m + 3 8n + 3 It shows to us that a must have a rime factor,, of the form 8n ± 3 because the roduct of numbers of the form 8n ± 1 can t be a number of the form 8n ± 3 Then a 0 mod, ie a mod and it would be satisfied that a By the second sulementary law see Theorem 1 we know

30 GAUSS S FIRST PROOF that 1, therefore a a So we found a rime number less than such that is nonresidue of Case ii: is a rime number of the form 8n + 1 Suose that is a residue of all rime numbers less than + 1 Then will also be a residue of all comosite numbers less than Indeed if is a comosite number less than with i1 1 ir r its rime factorization, since is residue of 1,,, r because they are less than by the Proosition 16 we can conclude that is also residue of Let m be the integer closest to and > m Now let us take the series, 1 1,,, 1 m or m I deending on if m is odd or even and the series 1,, 3,, m + 1 II By Lemma we have that the number of terms in I which are divisible by any number less than is at least the number of terms in II divisible by that number And by Lemma 1 we know that the roduct of the terms of the rogression I is divisible by the roduct of the terms of II Note that the roduct of the terms of I is 1 m if m is even or 1 m if m is odd In both cases he have that 1 m Z 1 3 m + 1 As is rime, is corime to all the terms in II, therefore 1 m is also divisible by 1 m + 1 Now note that the roduct of the terms of II can be exressed as follow 1 m + 1 m + 1 m m + 1 m Indeed m + n + 1 m n 1 m + m + 1 n m + 1 n Therefore we have that 1 m m m + 1 m m + 1 m Z Note that the factors of this roduct are all less than 1 because is rime, what imlies that is irrational, and we have taken m as the number less than closest to it, so it is true that m + 1 > and conseuently m + 1 > Hence we found some numbers less than 1 that multilied are eual to 1, which is a contradiction So is a nonresidue of some rime number less than Quadratic recirocity law for odd integers In the original roof of the theorem Gauss introduced the following notation:

31 3 QUADRATIC RECIPROCITY LAW FOR ODD INTEGERS 3 a, a, a, denotes rimes of the form n + 1 b, b, b, denotes rimes of the form n + 3 A, A, A, denotes numbers, not necessarily rimes, of the form n + 1 B, B, B, denotes numbers, not necessarily rimes, of the form n + 3 xry denotes that x is residue of y xny denotes that x is nonresidue of y And using this notation the fundamental theorem can be rewritten as follows: If it will be 1 ara a Ra ana a Na 3 arb bra anb bna 5 bra arb 6 bna anb 7 brb b Rb 8 bnb b Nb Gauss roved the fundamental theorem by induction He suosed that the theorem is true until a certain rime and he roved that the theorem is still true for the next rime But in one of the stes of the roof, Gauss used that if the fundamental theorem is true until a certain rime then the fundamental theorem is also true for all air of odd integers less than the next rime, since their factorization only involves rimes such that the fundamental theorem is true for them Hence, we will dedicate this section to rove that the fundamental theorem is true for every air of integers suosing that the fundamental theorem is true for every air of rimes First of all we rove the following roosition: Proosition 31 Let a, b be odd rime numbers of the form n + 1 and n + 3 resectively, and let A, B be odd integers of the form n + 1 and n + 3 resectively Then: If it will be 1 ara ARa, arb BRa, 3 bra ARb, brb BRb Proof Let us factorise A and B we omitted the exonents so the factors may be reeated: A a 1 a a n b 1 b b m, B a 1a a n b 1b b m +1 By the Proosition 16 we know that if is an odd rime number and P is an odd integer such that is a residue of it, then is residue of all rime factors of P

32 GAUSS S FIRST PROOF In the first and the second case, a is a residue of all rime factors of A and B resectively Suose that the fundamental theorem is true, so we can assure that if a is a residue of A, then all the rime factors of A are also residues of a, and in the same way if a is a residue of B, then all the rime factors of B are also residues of a Finally by the multilicativity of the Legendre symbol we conclude that A a 1 and B a B a 1 In the third and the fourth case, b is a residue of all rime factors of A and B resectively Suose that the fundamental theorem is true, so we can assure that if b is a residue of A, then the rime factors of A of the form n + 1 are residues of b and the rime factors of the form n + 3 are nonresidues of b, in the same way if b is a residue of B, then the rime factors of B of the form n + 1 are residues of b and the rime factors of the form n + 3 are nonresidues of b Remember A must have an even number of rime factors of the form n + 3 while B must have an odd number of them So by the multilicativity of the Legendre symbol we conclude that A b 1 and B b B 1, that imlies 1 b Let P and Q be two odd numbers and let us define the following notation: [P, Q ] Number of rime factors of Q, taken ositively and counted with multiicity, such that P is nonresidue of them Note that [P, Q] [P, Q] [P, Q ] since the sign of Q doesn t matter [[P, Q]] The number of airs i, j, rime factors of Q and P resectively, i such that 1 j Proosition 3 Let P, Q be two integers Then { } { } P A P A Q A, Q A { } P A, Q B { } P A Q A, { } P A Q B { } P B Q B [P, Q ] [Q, P ] mod, ie [P, Q] and [Q, P ] have the same arity, and { } P A, Q B { } P B Q B, { } P A Q B { } P B Q B [P, Q ] [Q, P ] mod, ie [P, Q] and [Q, P ] have different arity Proof Let us factorise P and Q in rime factors P 1 n, Q 0 1 m, where 0 1 when Q is ositive and 0 1 when Q is negative

33 3 QUADRATIC RECIPROCITY LAW FOR ODD INTEGERS 5 Now note that [[Q, P ]] and [Q, P ] have the same arity Indeed, [[Q, P ]] can be exressed: n m [[Q, P ]] [ i, j ] j1 i0 } {{ } k j Since i and j are rime, [ i, j ] 0 when i is a residue of j and [ i, j ] 1 when i is a nonresidue of j Hence k j counts how many rime factors of Q are nonresidues of j If k j is even, Q is a residue of j, but if k j is odd, Q is a nonresidue of j Then [Q, P ] counts how many k j are odd Therefore [[Q, P ]] k 1 + k + + k n Rearranging the summands we get: [[Q, P ]] k k n [Q,P ] + k n [Q,P ]+1 }{{} k n + 1 }{{} k j s even k j s odd K + [Q, P ] Hence [[Q, P ]] and [Q, P ] have the same arity Now let us consider P any odd ositive integer and Q an integer of the form A or B Factorise P and Q in rime factors: P a 1 a a n b 1 b b m, Q a 1a a n b 1 b b m +1 By the fundamental theorem [[Q, P ]] [[P, Q]] Indeed, a i a j 1 1, a i b j b i b j a j b i a j a i b j b i b j a i a j 1, b i b j 1, a i b j 1 And we saw before that [Q, P ] [[Q, P ]] mod and [P, Q] [[P, Q]] mod Hence: [Q, P ] [[Q, P ]] [[P, Q]] [P, Q] mod, and we can conclude that in the cases {P A, Q A }, {P B, Q A}, {P A, Q B} and {P B, Q B }, [P, Q] and [Q, P ] have the same arity Let P and Q be any odd integers, and let P a 1 a a n b 1 b b m, Q a 1a a n b 1b b m +1 be their rime factorization Let us introduce the following notation: { Q } θ :# a i ai a rime factor of P of the form n + 1 such that 1 a i { Q } χ :# b i bi a rime factor of P of the form n + 3 such that 1 b i b i

34 6 GAUSS S FIRST PROOF { ψ :# a i ai a rime factor of P of the form n + 1 such that { ω :# b i bi a rime factor of P of the form n + 3 such that Q a i } 1 } Q 1 Then it is easy to see that [Q, P ] ψ + ω and [ Q, P ] χ + ψ Note that if P A then χ + ω is even, since P A must have an even number of rime factors of the form n + 3, and χ ω is also even, since Then, when P A, by the euation: χ ω χ + ω ω }{{}}{{} even even [ Q, P ] [Q, P ] + χ ω, we can conclude that [Q, P ] and [ Q, P ] have the same arity Now let us consider the cases: {P A, Q A } and {P A, Q B} We saw before that in both cases [P, Q] and [Q, P ] have the same arity Moreover [Q, P ] and [ Q, P ] have also the same arity, since P A Hence [P, Q] [P, Q] and [ Q, P ] have the same arity, ie { } P A Q A [P, Q] [Q, P ] mod, { } P A [P, Q] [Q, P ] mod Q B b i By the case {P A, Q A } which is euivalent to the case {P A, Q A } we have that [Q, P ] and [P, Q] have the same arity Moreover, since P A, we have that [Q, P ] and [ Q, P ] have the same arity Hence: { } P A Q A [P, Q] [Q, P ] mod Otherwise, when P B then χ + ω is odd, since P B must have an odd number of rime factors of the form n + 3, and χ ω is also odd, since By the euation χ ω χ + ω ω }{{}}{{} odd even [ Q, P ] [Q, P ] + χ ω, we can conclude that [Q, P ] and [ Q, P ] have different arity Now let us consider the cases: {P B, Q A} and {P B, Q B } We saw before that in both cases [P, Q] and [Q, P ] have the same arity, but since P B, we also have that [Q, P ] and [ Q, P ] have different arity Hence [P, Q] [P, Q] and [ Q, P ] have different arity, ie

35 3 QUADRATIC RECIPROCITY LAW FOR ODD INTEGERS 7 { } P B Q A { } P B Q B [P, Q] [Q, P ] mod, [P, Q] [Q, P ] mod By the cases {P B, Q A} and {P B, Q B } which are euivalent to the cases {P A, Q B} and {P B, Q B } we have that in both cases [Q, P ] and [P, Q] have the same arity Moreover, since P B, we have that [Q, P ] and [ Q, P ] have different arity Hence: { } P B Q A { } P B Q B [P, Q] [Q, P ] mod, [P, Q] [Q, P ] mod Note that, when P and Q are both ositive, 1 [P,Q] agrees with the Jacobi symbol P m Indeed, if i is the rime factorization of Q, Q i1 P P P P 1 [P,Q] Q 1 m Remember that the uadratic recirocity law for Jacobi symbols Theorem 15 states: P 1 P 1 Q 1 Q Q P P Q Then in the cases {P A, Q A }, {P A, Q B} and Q P {P B, Q A} note that these last two cases are virtually the same, so it suffices to consider only one of them; we will only consider {P A, Q B}, and P Q in the case {P B, Q B } But in Proosition 3 we roved Q P that in cases {P A, Q A } and {P A, Q B}, [Q, P ] and [P, Q] have the same arity, and in the case {P B, Q B }, [Q, P ] and [P, Q] have different arity Hence this roves the uadratic recirocity law for ositive integer numbers using the Jacobi symbol Notice that, in fact, Gauss roved something stronger than the uadratic recirocity law for the Jacobi symbol, since the Proosition 3 involves ositive and negative integers It is very imortant to remark that in the roof of the Proosition 3 we only used that the fundamental theorem is true for the rime factors of P and Q, the roof didn t involve any rime greater than P or Q

36 8 GAUSS S FIRST PROOF Quadratic recirocity law for odd rimes In this section we shall rove the uadratic recirocity law for odd rimes, the one that Gauss called fundamental theorem Using his notation, the fundamental theorem is stated as follows: If it will be 1 ara a Ra ana a Na 3 arb bra anb bna 5 bra arb 6 bna anb 7 brb b Rb 8 bnb b Nb 1 Gauss roved the fundamental theorem by induction So we will suose that the theorem is true until a certain rime and we will rove that the theorem is still true until the next rime But in order to aly this we need a base case For the base case it is enough to rove that the theorem is true for all odd rimes less than or eual to five: The case above is the only one that involves only odd rimes 5 and it satisfies the fundamental theorem Hence we can ass to the inductive ste Let 3, 5, 7, 11,, P t, P t+1, be the seuence of odd rime numbers And let us suose that the fundamental theorem is true until P t, ie the theorem is true for all airs of rimes, less than or eual to P t Note that from Proosition 3 we have that the fundamental theorem is true for all ositive odd integer less than P t+1 Now let us take a air of odd rime numbers, P t and P t+1 ; and try to rove that the theorem is also true for them Now we shall rove the fundamental theorem itself: Cases 1, 3, 5 and 7 of 1 In all cases is a residue of P t+1, ie there exists e such that e mod Since is odd, x mod has two solutions smaller than, one odd and one even, so let us take e < and even Now we distinguish two subcases: Subcase 1: e that imlies e We have e + f with f > 0, f < since e < and f since e Then f is of the form n + 1 when { } { } n + 3 n + 1 or, n + 1 n + 3

37 QUADRATIC RECIPROCITY LAW FOR ODD PRIMES 9 and f is of the form n + 3 when { } n + 1 or n + 1 It is true that e mod f, so f { n + 3 n + 3 } 1 Since and f are both smaller than, the fundamental theorem is true for them, hence: f 1 1 f f 1 f f Therefore, 1 when both f and are of the form n + 3, otherwise f f 1 But it is also true that e f mod, so 1 By the f f multilicativity of the Legendre symbol And for the four different cases we have: n n + 1 f n + 3 n n + 3 f n + 1 n n + 1 f n + 1 n n + 3 f n + 3 f 1 f 1 f 1 f , 1, 1, Subcase : e that imlies e We have e g and e g + h Hence simlifying in the euation we obtain g 1+h, where h is smaller than and rime with and g Then h is of the form n + 1 when { n + 3 n + 1 } or { n + 3 n + 3 and h is of the form n + 3 when { } { } n + 1 n + 3 or, n + 1 n + 1 g It is true that g 1 mod h, so 1 that imlies 1 because h h g 1 Since and h are both smaller than, the fundamental theorem is true h for them, hence: h 1 h h 1 1 h },