Investigating Proofs of the Quadratic Reciprocity Law Cuyler Warnock

Home , Quadratic reciprocity, Reciprocity law

Investigating Proofs of the Quadratic Reciprocity Law

Cuyler Warnock

December 2, 2016

Abstract

For over 300 years, number theorists have investigated quadratic residues and their properties. A quadratic residue modulo p is an integer a such that x2 a (mod p)forsome ⌘ x Zp. Number theorists such as Fermat, Euler, Legendre, and Gauss were interested in ﬁnding2 conditions for distinct primes p and q so that p would be a quadratic residue modulo q and q would be a quadratic residue modulo p. Through the contributions of these mathematicians, a surprisingly eloquent relationship was discovered: the quadratic nature of p modulo q is the same as the quadratic nature of q modulo p if and only if p 1(mod4) or q 1(mod4).Thisresult,ﬁrstprovedbyGaussinthelate1790’s,hasbecomeknown⌘ as the⌘ Quadratic Reciprocity Law. Since that time, there have been hundreds of di↵erent proofs published. These proofs exhibit an astounding variety of methods derived from various branches of mathematics. We will investigate three di↵erent proofs, each varying in proof technique and complexity.

1 Introduction

First stated by Euler, the Quadratic Reciprocity Law was discovered because of the need to determine whether p is a quadratic residue modulo q where p and q are distinct primes. In other words, Euler was interested in determining conditions for the solvability of the congruence x2 p (mod q), ⌘ where p and q are distinct primes. A century before Euler, Fermat had also attempted to discover a realtionship among these distinct primes p and q.In1658,Fermatwroteinaletter to Kenelm Digby that he already knew the conditions for which odd primes q and the numbers 1, 2, 3, 5arequadraticresiduesornonresiduesmoduloq. The results of Fermat, Euler,± and± others± were conjectured based on empirical evidence, but were given without any proofs. Eventually, Euler was able to prove the case for 1and 3, and Lagrange proved the case for 2and 5. ± ± ± By further examining these cases, Euler was able to formulate the Quadratic Reciprocity

1 Law in full generality. He gives the following four theorems that completely contain the Quadratic Reciprocity Law. Let p and q be distinct primes. 1. If p 1(mod4)andp x2 (mod q)forsomeintegerx,then q y2 (mod p)for some⌘ integer y. ⌘ ± ⌘

2. If p 3(mod4)and p x2 (mod q)forsomeintegerx,thenq y2 (mod p)for some⌘ integer y and q z⌘2 (mod p)forallintegersz. ⌘ 6⌘ 3. If p 3 (mod 4) and p x2 (mod q)forallintegersx,then q y2 (mod p)for some⌘ integer y and q z2 6⌘(mod p)forallintegersz. ⌘ 6⌘ 4. If p 1(mod4)andp x2 (mod q)forallintegersx,then q y2 (mod p)forall integers⌘ y. 6⌘ ± 6⌘ Gauss, unknowing that Euler had formulated a version of the Quadratic Reciprocity Law, credits Legendre with the discovery of this law. In 1785, Legendre explicitly stated the theorem using formulas and gave several proofs, each of which contained serious gaps pointed out by Gauss. Since Legendre noticed that the solvability of x2 p (mod q)where q⌘1 p and q are distinct primes and x is an integer depends on whether p 2 1(modq)or q 1 ⌘ p 2 1(modq), Legendre developed a useful shorthand notation to represent whether an integer⌘ is a quadratic residue modulo p where p is prime. In [7] on page 186, Legendre states,

c 1 “Since the analogous quantities N 2 will occur often in our researches, we shall N c 1 employ the abbreviation for expressing the residue that N 2 gives upon c division by c, and which,✓ according◆ to what we just have seen, only assumes the values +1 or 1.” Also, in [7] on page 214, Legendre gives the Quadratic Reciprocity Law in its commonly written form. He writes “Whatever the prime numbers m and n are, if they are not both of the form n m 4x 1, one always has = ;andifbothareoftheform4x 1, one m n ✓ ◆ ✓ ◆ n m has = .Thesetwogeneralcasesarecontainedintheformula m n ✓ ◆ ✓ ◆

n n 1 m 1 m =( 1) 2 2 .” m n ✓ ◆ ✓ ◆ Even though Legendre states the theorem using two distinct primes, the Legendre sym- a bol makes sense for composite values of a. Legendre never provided a complete and p accurate✓ ◆ proof of this theorem. However, in 1796 at the age of eighteen, Gauss was able to provide the ﬁrst complete proof by using induction (see [4] p. 92-97). The proof took Gauss an entire year to complete. Gauss wrote,

2 “For a whole year this theorem tormented me and absorbed my greatest e↵orts until, at last, I obtained the proof explained in the fourth section of the Disqui- sitiones Arithmeticae”.

Providing just one proof of the Quadratic Reciprocity Law was not enough for Gauss. By 1818, he had published ﬁve other proofs of the theorem and two more were found in his unpublished papers after his death.

Since Gauss’ ﬁrst complete proof of the Quadratic Reciprocity Law, more than 300 proofs have been published ([2] p. 131-138). These proofs exhibit an astounding variety of methods derived from various branches of mathematics, such as quaternion algebras, group theory, ge- ometry, and complex analysis. We will analyze three proofs, each varying in proof technique and complexity.

2 Essential Deﬁnitions and Theorems

Before analyzing the proofs, we give some deﬁnitions and theorems that we will use in the proofs of the Quadratic Reciprocity Law.

Deﬁnition 1 (quadratic residue). Let p be a prime, and let a Z such that p - a.Ifa is a square modulo p, then a is a quadratic residue modulo p.If2a is not a square modulo p, then a is a quadratic nonresidue modulo p.

In other words, a quadratic residue modulo p is a nonzero square modulo p. Determining whether an integer is a quadratic residue modulo p can be very tedious and time consuming when p becomes large. However, Euler provided a useful method for determining if a is a quadratic residue modulo p by performing modular exponentiation.

Theorem 1 (Euler’s Criterion). Let p be an odd prime, and let a be an integer not divisible by p.Ifa is a quadratic residue mod p, then

p 1 a 2 1(modp). ⌘ If a is a quadratic nonresidue mod p, then

p 1 a 2 1(modp). ⌘ A proof of Euler’s Criterion can be found in [3] on pages 171-172. Using Euler’s Criterion to determine if an integer a is a quadratic residue modulo p can still take be tedious; however, it is a better technique than squaring each positive integer less than (p/2) and checking if it is congruent to a modulo p.WewillshowtheusefulnessofEuler’sCriterionwithanexample.

Example: Is 74 is a quadratic residue modulo 167.

3 167 1 Solution. Since 167 is prime, we will use Euler’s Criterion. Note that 2 =83.Therefore, we will calculate ( 74)83 (mod 167). Note the following: ( 74)83 ( 74)64 ( 74)16 ( 74)2 ( 74) (mod 167) ⌘ · · · 547632 54768 5476 ( 74) (mod 167) ⌘ · · · 13232 1328 ( 9768) (mod 167) ⌘ · · 1742416 174244 ( 82) (mod 167) ⌘ · · 5616 564 ( 82) (mod 167) ⌘ · · 31368 31362 ( 82) (mod 167) ⌘ · · 1308 1302 ( 82) (mod 167) ⌘ · · 169004 16900 ( 82) (mod 167) ⌘ · · 334 ( 34) (mod 167) ⌘ · 10892 ( 34) (mod 167) ⌘ · 872 ( 34) (mod 167) ⌘ · 7569 ( 34) (mod 167) ⌘ · 1(mod167). ⌘ Therefore, we have ( 74)83 1(mod83).Thus,byEuler’sCriterionwehavethat 74 is a quadratic residue modulo 167.⌘ As stated in the introduction, Legendre developed a useful shorthand notation to represent whether an integer a is a quadratic residue modulo p.

Definition 2 (Legendre symbol). Let p be a prime, and let a Z. The Legendre symbol a 2 is defined by p ✓ ◆ 1ifa is a quadratic residue modulo p a = 0ifp a p 8 | ✓ ◆ 1ifa is a quadratic nonresidue modulo p. < In one of the proofs we will: analyze, we are concerned with finding a solution to a system of linear congruences. Therefore, the following theorem will be useful.

Theorem 2 (Chinese Remainder Theorem). Let k be a positive integer and let m1,m2,...,mk be pairwise relatively prime positive integers. Let b1,b2,...,bk be arbitrary integers. Then the system

x b (mod m ) ⌘ 1 1 x bk (mod mk) ⌘. . x b (mod m ) ⌘ k k has a unique solution modulo m m m . 1 2 ··· k 4 AproofoftheChineseRemainderTheoremcanbefoundin[3]onpages79-80.Inthis proof of the Chinese Remainder Theorem, a formula for the solution to such a system of linear congruences is constructed; and such a solution is of the form

x b M M 0 + b M M 0 + + b M M 0 (mod m m m ), ⌘ 1 1 1 2 2 2 ··· k k k 1 2 ··· k where Mi = m1m2 mi 1mi+1 mk for each i 1, 2, ,k and Mi0 is an integer such ··· ··· 2{ ··· } that M M 0 1(modm ). i i ⌘ i Wilson’s Theorem will also be important in one of the proofs of the Quadratic Reciprocity Law.

Theorem 3 (Wilson’s Theorem). If p is prime, then (p 1)! 1(modp). ⌘ A proof of Wilson’s Theorem can be found in [10] on page 186.

3 Importance of the Quadratic Reciprocity Law

As previously mentioned, the Quadratic Reciprocity Law provides a very useful method in determining whether an integer a is a quadratic residue modulo p,orequivalently,determin- a ing whether = 1. We ﬁrst present some useful theorems in helping to determine the p ± ✓ ◆ a value of . p ✓ ◆ Theorem 4. Let a Z and b Z, and let p be a prime. Then the following hold. 2 2 a b 1. If a b (mod p), then = . ⌘ p p ✓ ◆ ✓ ◆ a2 2. If a 0(modp), then =1. 6⌘ p ✓ ◆

a p 1 3. a 2 (mod p). p ⌘ ✓ ◆ ab a b 4. = . p p p ✓ ◆ ✓ ◆✓ ◆ Theorem 5. For p an odd prime, 1 1ifp 1(mod4) = ⌘ p 1ifp 1(mod4). ✓ ◆ ⇢ ⌘ Theorem 6. For p an odd prime,

5 2 1ifp 1(mod8) = ⌘± p 1ifp 3(mod8). ✓ ◆ ⇢ ⌘± A proof of Theorem 4 can be found in [3] on page 176. A proof of Theorems 5 and 6 can be found in [10] on pages 335 and 337-338 respectively. Using the results stated above, we are a able to determine the value of relatively quickly. However, the Quadratic Reciprocity p ✓ ◆ a Law provides even faster techniques for determining the value of . p ✓ ◆ Theorem 7 (Quadratic Reciprocity Law). If p and q are odd primes, then

p q p 1 q 1 =( 1) 2 2 . q p ✓ ◆✓ ◆ p 1 q 1 p 1 q 1 2 2 Note that if p 1(mod4)orq 1(mod4),then(2 )( 2 )iseven.Thus,( 1) =1. ⌘ ⌘ p 1 q 1 p 1 q 1 2 2 However, if p q 3(mod4),then(2 )( 2 )isodd.Therefore( 1) = 1. Thus, we have an alternate⌘ ⌘ form of the Quadratic Reciprocity Law.

Theorem 8 (Quadratic Reciprocity Law). If p and q are odd primes, then q if p 1(mod4)orq 1(mod4) p p ⌘ ⌘ = 8 ✓ ◆ q q ✓ ◆ > if p q 3(mod4). < p ⌘ ⌘ ✓ ◆ > a We will now rework the: earlier example to show the ease with which can be calcu- p lated by using the above theorems. ✓ ◆

6 Example: Is 74 is a quadratic residue modulo 167. Solution. Note that 167 is prime. Consider the following:

74 1 2 37 = By Theorem 4 (4) since 74 = ( 1)(2)(37) 167 167 167 167 ✓ ◆ ✓ ◆✓2 ◆✓37 ◆ =(1) By Theorem 5 since 167 1(mod4) 167 167 ⌘ ✓ ◆✓37 ◆ = (1) By Theorem 6 since 167 1(mod8) 167 ⌘ ✓167 ◆ = By QRL since 37 1(mod4) 37 ⌘ ✓ 19 ◆ = By Theorem 4 (1) since 167 19 (mod 37) 37 ⌘ ✓37◆ = By QRL since 37 1(mod4) 19 ⌘ ✓ 1◆ = By Theorem 4 (1) since 37 1(mod19) 19 ⌘ = ✓( 1)◆ By Theorem 5 since 19 1(mod4) =1 . ⌘ 74 Therefore, we have that = 1. Hence, 74 is a quadratic residue modulo 167. 167 ✓ ◆ 4Proofs

4.1 Classical Proof by Gauss The ﬁrst proof we will analyze is a variation of Gauss’ third proof which was simpliﬁed by Eisenstein. In this proof, Gauss uses what is now known as Gauss’ Lemma, which is used in more than 50 proofs of the Quadratic Reciprocity Law. Lemma 1 (Gauss’ Lemma). Let p be an odd prime and let a be an integer such that p - a. p 1 Let n denote the number of least positive residues of the integers a, 2a, . . . , 2 a that exceed a p/2. Then =( 1)n. p ✓ ◆ Proof. Let p be an odd prime and let a be an integer such that p - a.Letn denote the number p 1 p of least positive residues of the integers a, 2a, . . . , 2 a that exceed 2 .Sincegcd(a, p)=1, p 1 we have that each of the elements a, 2a, . . . , 2 a are unique modulo p.Sincep - a,we p 1 know that ak 0(modp)foreachk 1, 2,..., 2 .Letm and n be integers and let 6⌘ 2{ } p 1 r1,r2,...,rm denote the least positive residues of a, 2a, . . . , 2 a such that 0

Lemma 2 (Eisenstein’s Lemma). If p is an odd prime and a is an odd integer with gcd(a, p)= 1, then p 1 a 2 ka =( 1) k=1 p . p P ✓ ◆ ⌅ ⇧ Proof. Let p be an odd prime and let a be an odd integer such that gcd(a, p)=1.Let p 1 S = a, 2a, . . . , 2 a . Using the division algorithm, we obtain ka = pqk +bk where qk Z, { p 1 } ka b ka 2 k 1, 2,..., and b 0, 1,...,p 1 .Itfollowsthat = q + k . Hence, = q . 2 k p k p p k Therefore,2{ we have} that 2{ } b c ka ka = p + b . p k ⌫ 8 p p If bk < 2 ,thenitisoneoftheintegersri from the proof of Gauss’ Lemma. If bk > 2 ,thenit is one of the integers sj from the proof of Gauss’ Lemma. Therefore, we obtain the following:

p 1 p 1 2 2 ka m n ka = p + r + s . p k k Xk=1 Xk=1 ⌫ Xk=1 Xk=1 p 1 From the proof of Gauss’ Lemma, we have that the 2 numbers r ,r ,...,r ,p s ,p s ,...,p s 1 2 m 1 2 n p 1 are just a rearrangement of the integer 1, 2,..., 2 .Therefore,wehavethat

p 1 2 m n m n k = r + (p s )=pn + r s . k k k k Xk=1 Xk=1 Xk=1 Xk=1 Xk=1 Note the following:

p 1 p 1 p 1 2 2 2 (a 1) k = a k k k=1 k=1 k=1 X p X1 Xp 1 2 2 = ka k k=1 k=1 Xp 1 X 2 ka m n m n = p + r + s pn + r s p k k k k k=1 ⌫ k=1 k=1 k=1 k=1 ! X p 1 X X X X 2 ka n = p n +2 sk. p ! Xk=1 ⌫ Xk=1 p 1 p 1 2 2 ka n Thus, (a 1) k = p n +2 s . Since a and p are odd integers, we know that p k k=1 ✓k=1 ⌫ ◆ k=1 a 1 0(mod2)andX pX 1(mod2).Therefore,consideringtheaboveequationmodulo2X ⌘ p 1 ⌘ p 1 2 ka 2 ka yields 0 n (mod 2) which implies that n (mod 2). Therefore, by ⌘ p ⌘ p k=1 ⌫ k=1 ⌫ Gauss’ Lemma,X we obtain the following: X

p 1 a 2 ka =( 1)n =( 1) k=1 p . p P ✓ ◆ ⌅ ⇧

Theorem 7. (Quadratic Reciprocity Law) If p and q are distinct odd primes, then

p q p 1 q 1 =( 1) 2 2 . q p ✓ ◆✓ ◆

9 Proof. Let p and q be distinct odd primes. Consider the rectangle in the xy coordinate p q p q plane whose vertices are (0, 0), ( 2 , 0), (0, 2 ), and ( 2 , 2 ). Let R denote the region within the rectangle not including the boundary lines. We will count the number of lattice points inside R in two di↵erent ways. Since p and q are odd, the lattice points in R consists of all point p 1 q 1 (m, n)wherem, n Z with 1 m 2 and 1 n 2 . It follows that the number 2 p 1 q 1    p q of lattice points in R is 2 2 . Let D denote the diagonal from (0, 0) to ( 2 , 2 )excluding the endpoints. It follows that· D is a segment of the line with equation y =(q/p)x,or equivalently, py = qx.Suppose(a, b)isalatticepointonD.Thenpb = qa which implies p 1 that p a. However, we know that p - a since a 1, 2,..., .Therefore,nolattice | 2{ 2 } points lie on D.LetR1 denote the region in R that is below D and let R2 denote the region in R that is above D. We will now count the number of lattice points in both R1 and R2.

q p q (0, 2 ) ( 2 , 2 )

(k, kq ) b p c R1

p (0,0) (k,0) ( 2 ,0)

p 1 Let k be an integer such that 1 k . We have that the number of integers less   2 than kq is equal to kq . Therefore, we have that there are exactly kq lattice points in R p b p c b p c 1 on the line x = k.Consequently,wehavethatthetotalnumberoflatticepointsinR1 is given by p 1 2 kq . p Xk=1 ⌫ q 1 We use a similar argument for counting the points in R2.Foreachk 1, 2,..., 2 , there kp 2{ } are q lattice points on the line y = k in R2. Hence, the total number of lattice points in R2 is given by q 1 ⌅ ⇧ 2 kp . q Xk=1 j k Since this accounts for all of the lattice points in R, we have that

p 1 q 1 p 1 q 1 2 kq 2 kp = + . 2 · 2 p q Xk=1 j k Xk=1 j k 10 Using Lemma 2, we obtain the following:

q 1 p 1 p q 2 kp 2 kq =(1) k=1 q ( 1) k=1 p q p P P ✓ ◆✓ ◆ q 1 ⌅ ⇧ p 1 ⌅ ⇧ 2 kp + 2 kq =(1) k=1 q k=1 p Pp 1 q 1 P =(1) 2 2 ⌅. ⇧ ⌅ ⇧

p q p 1 q 1 Therefore, we have =( 1) 2 2 . q p ✓ ◆✓ ◆ 4.2 Kim In 2004, Sey Y. Kim published a proof of the Quadratic Reciprocity Law that uses elementary number theory results such as Wilson’s Theorem and the Chinese Remainder Theorem. This proof is similar to a proof by George Rousseau in 1991.

Let p and q be distinct odd primes and let

pq 1 = a Z :1 a and gcd(a, pq)=1 , 2   2 ⇢ and let A = a. We will ﬁrst prove the following lemma. a Y2

q 1 q p 1 p Lemma 3. A ( 1) 2 (mod p) and A ( 1) 2 (mod q). ⌘ p ⌘ q ✓ ◆ ✓ ◆ Proof. Let

pq 1 p 1 S = a Z :1 a and gcd(a, p)=1 and T = q 1,...,q . 2   2 · · 2 ⇢ ⇢ p 1 We will show T S.Lett T .Itfollowsthatt = qb for some b Z such that 1 b 2 . ✓ 2 2 p 1   Since p and q are distinct prime, we know that p - q. Also, since b 2 ,weknowthatp - b.  pq 1 Therefore, p - t.Itfollowsthatgcd(t, p)=1.Wenowshowthatt 2 .Sinceq>1, we pq q pq 1 pq q know that pq q

11 that

a = a a · a S a T a Y2 Y2 Y2 p 1 =(q 1)(q 2) (q ) A · · ··· · 2 · p 1 p 1 = q 2 ! A 2 ·  q p 1 ! A (mod p). ⌘ p 2 · ✓ ◆ Hence, q p 1 a ! A (mod p). (1) ⌘ p 2 · a S ✓ ◆ Y2 pq 1 q 1 p 1 Since we have that 2 = 2 p + 2 and since S contains no multiples of p,weobtain the following: ·

a =(1) (p 1)(p +1) (2p 1)(2p +1) (3p 1) ··· ··· ··· ··· a S Y2 q 3 q 1 q 1 q 1 p 1 p +1 p 1 p +1 p + ··· 2 ··· 2 2 ··· 2 2 ✓ ◆ ✓ ◆✓ ◆ ✓ ◆ q 1 p 1 [(p 1)!] 2 !(modp) ⌘ 2  q 1 p 1 ( 1) 2 !(modp), ⌘ 2  where we have that (p 1)! 1(modp) by Wilson’s Theorem. Thus, ⌘ q 1 p 1 a ( 1) 2 !(modp). (2) ⌘ 2 a S  Y2 It follows from (1) and (2),

q p 1 q 1 p 1 ! A ( 1) 2 !(modp). p 2 · ⌘ 2 ✓ ◆  p 1 p 1 Since 2 !andp are relatively prime, we have that the inverse of 2 ! modulo p exists. p 1 Multiplying the above congruence by the inverse of ! modulo p yields ⇥ ⇤ 2 ⇥ ⇤

q q 1 ⇥ ⇤ A ( 1) 2 (mod p). p · ⌘ ✓ ◆

q q 1 q Multiplying both sides of the congruence by yields A ( 1) 2 (mod p). By a p ⌘ p ✓ ◆ ✓ ◆ p 1 p similar argument, we deduce A ( 1) 2 (mod q). ⌘ q ✓ ◆ 12 q 1 q p 1 p Note that 1=( 1) 2 A (mod p)and 1=( 1) 2 A (mod q). If ± p ⌘ ± q ⌘ ✓ ◆ ✓ ◆ q 1 q p 1 p ( 1) 2 =1=( 1) 2 ,thenwehavethatA 1(modp)andA 1(modq) p q ⌘ ⌘ ✓ ◆ ✓ ◆ q 1 q p 1 p which implies that A 1(modpq). Also, if ( 1) 2 = 1=( 1) 2 ,thenwe ⌘ p q have that A 1(modp)andA 1(modq)whichimpliesthat✓ ◆ A ✓ 1(mod◆ pq). ⌘ ⌘ ⌘ q 1 q p 1 p Therefore, we have that ( 1) 2 =( 1) 2 if and only if A 1or 1(modpq). p q ⌘ We now present the following lemma.✓ ◆ ✓ ◆ Lemma 4. A 1 or 1(modpq) if and only if p q 1(mod4). ⌘ ⌘ ⌘ Proof. Consider the congruence x2 1(modpq). Since p and q are relatively prime, we know that x satisﬁes one of the following⌘ systems of congruences: x 1(modp) x 1(modp) x 1(modp) x 1(modp) x ⌘ 1(modq) x ⌘ 1(modq) x⌘1(modq) x ⌘1(modq) ⌘ ⌘ ⌘ ⌘ By the Chinese Remainder Theorem, we have that each of the systems of congruences has a unique solution modulo pq.Therefore,wehavethatx 1,n, n, 1(modpq) 2 ⌘ are the only solutions to x 1(modpq), where n Zpq such that n 1(modp)and n 1(modq). Now consider⌘ the congruence x2 21(modpq)andsupposethereexists⌘ a solution.⌘ Therefore, we have that pq (x2 +1). Since⌘p and q are relatively prime, we have that p (x2+1) and q (x2+1). It follows| that x2 1(modp)andx2 1(modq). Since x2 |1(modpq)hasasolutionifandonlyif| x2⌘ 1(modp)andx⌘2 1(modq)have a solution⌘ which only occurs when p q 1(mod4),wededucethat⌘ ⌘x2 1(modpq) has a solution if and only if p q ⌘1(mod4).Therefore,if⌘ p q 1(mod⌘ pq), let ` be a solution to x2 1(mod⌘pq).⌘ It follows that `, `, n`, n` are⌘ all⌘ of the solutions to x2 1(modpq). ⌘ ⌘

We will now show that for each a , there exists an a0 suchthataa0 (mod pq) 2 2 ⌘ pq 1 where 1, 1 .Leta . By deﬁnition of , we have that 1 a and 2{ } 2   2 gcd(a, pq)=1.Sincegcd(a, pq)=1,wehavethatthereexistsanonzeroa0 Zpq such 2 that aa0 1(modpq)andgcd(a0,pq)=1.Sincea0 Zpq and a0 is nonzero, we have that ⌘ pq 1 pq+1 2 pq 1 either 1 a0 2 or 2 a0 pq 1. If 1 a 2 ,wehavethata0 . If pq+1     pq+1  2 2 a0 pq 1, we have that 1 pq a0 2 .Itfollowsthat a0 b (mod pq)   pq 1   ⌘ for some integer b with 1 b . Since gcd(a0,pq)=1andsince a0 b (mod pq), we   2 ⌘ have that gcd(b, pq)=1.Therefore,b . Note that ab a( a0) aa0 1(modpq). 2 ⌘ ⌘ ⌘ Thus, we have that for every a , there exists a0 suchthataa0 (mod pq)where 1, 1 . 2 2 ⌘ 2{ }

Let denote the set containing each a suchthata = a0,thatis,let 2 = a2 1(modpq) a . ⌘± | 2 Note that for each a thereexists a0 witha = a0 such that aa0 1(modpq). 2 2 6 ⌘± However, for each a theredoesnotexista0 witha = a0 such that aa0 1(modpq) 2 2 6 ⌘± 13 since a = a0 for all a . It follows that 2 A a a (mod pq). ⌘ ⌘± a a Y2 Y2 Note that each a isasolutiontoeitherx2 1(modpq)orx2 1(modpq). Suppose p q 1(mod4).Itfollowsthat2 ⌘ ⌘ ⌘ ⌘ A a a (1 n ` n`) (n2 `2) 1(modpq). ⌘ ⌘± ⌘± · · · ⌘± · ⌘± a a Y2 Y2 Therefore, A 1(modpq)whenp q 1 (mod 4). Now suppose either p 3(mod4) or q 3(mod4).Itfollowsthat⌘± ⌘ ⌘ ⌘ ⌘ A a a (1 n) n 1(modpq). ⌘ ⌘± ⌘± · ⌘± 6⌘ ± a a Y2 Y2 Therefore, A 1(modpq)whenp 3(mod4)orq 3(mod4).Consequently,we have that A 6⌘ ±1(modpq)ifandonlyif⌘ p q 1(mod4).⌘ ⌘± ⌘ ⌘ q 1 q p 1 p Combining Lemma 3 and Lemma 4 we obtain that ( 1) 2 =( 1) 2 if and p q ✓ ◆ ✓ ◆ q 1 q p 1 p only if p q 1(mod4).Itfollowsthat( 1) 2 = ( 1) 2 if and only if ⌘ ⌘ p q ✓p+1 q◆+1 ✓ ◆ p 3(mod4)orq 3 (mod 4). Note that ( 1) 2 2 =1ifp q 1(mod4)but ⌘ p+1 q+1 ⌘ ⌘ ⌘ ( 1) 2 2 = 1ifp 3(mod4)orq 3(mod4).Therefore,wededucethat ⌘ ⌘ q 1 q p+1 q+1 p 1 p ( 1) 2 = ( 1) 2 2 ( 1) 2 p q ✓ ◆ ✓ ◆ or equivalently, p q p+1 q+1 p 1 q 1 = ( 1) 2 2 ( 1) 2 ( 1) 2 . q p ✓ ◆✓ ◆ Note the following: p 1 q 1 pq p q +1 = 2 2 4 pq + p + q +1 p 1+q 1+2 = 4 2 p +1q +1 p 1 q 1 = + +1 2 2 2 2 p +1q +1 ⇣p 1 q 1 ⌘ + + +1 (mod2). ⌘ 2 2 2 2 It follows that

p+1 q+1 p 1 q 1 p+1 q+1 p 1 q 1 p 1 q 1 1+ + + ( 1) 2 2 ( 1) 2 ( 1) 2 =( 1) 2 2 2 2 =( 1) 2 2 . Therefore, we have p q p 1 q 1 =( 1) 2 2 . q p ✓ ◆✓ ◆

14 4.3 Barnard Virgil Barnard proves the Quadratic Reciprocity Law by using Euler’s Criterion, primitive roots, and properties of the floor function. Upon presenting a modified version of Gauss’ Lemma, he uses an argument similar to counting lattice points which we showed in the first proof we analyzed.

Let p be an odd prime and let a be an integer such that gcd(a, p)=1.Wehaveby a p 1 Euler’s Criterion that a 2 (mod p). Since p is an odd prime, we know that there p ⌘ exists an integer b which✓ ◆ is a primitive root modulo p;thatis,thereisanintegerb such 1 2 p 1 that b ,b ,...,b are each distinct modulo p. It follows that there exists a positive integer j with 0 j0. Then we have the following results: ⌘{ }

1. If a b (mod p), then a b (mod p). ⌘ { }p ⌘{ }p 2. If p - c,then c p = p c p. { } { } a c a c ac 3. { }p = { }p { }p . p p p ⌫

a c p a c p 4. If p - c and p - a,then { } + { } = a 1. p p ⌫ ⌫ 5. For any function f we have the equality:

j k k j f(n)+ f(j + n)= f(n)+ f(k + n). n=1 n=1 n=1 n=1 X X X X These results are given without proof in Barnard’s paper, so we will provide a proof of each result. Proof. (1) Let a, b be integers and let p be a positive integer. Suppose a b (mod p). By the Division Algorithm, we know that a = a + pk and b = b + pk where⌘ 0 a ,b

0 be an integer and let c be an integer such that p - c.Itfollowsthatc =0.By 6 the Division Algorithm, we have that c = c0 + pk0 and c = c1 + pk1 where k0,k1 Z and 0

15 know that 0

(4) Let p>0beanintegerandleta, c be an integer such that p - a and p - c.From(3) a c a c ac a c a c ac we know that { }p = { }p { }p and { }p = { }p { }p . Note the p p p p p p following: ⌫ ⌫

a c a c a c ac a c ac { }p + { }p = { }p { }p + { }p { }p p p p p p p ⌫ ⌫ a c ac a(p c ) p ac = { }p { }p + { }p { }p p p p p a c ac ap a c p ac = { }p { }p + { }p + { }p p p p p p p = a 1 a c a c Therefore, { }p + { }p = a 1. p p ⌫ ⌫ (5) Let f be a function. Note that

j k f(n)+ f(j + n)=f(1) + f(2) + + f(j)+f(j +1)+f(j +2)+ + f(j + k) ··· ··· n=1 n=1 X X = f(1) + f(2) + + f(k)+f(k +1)+f(k +2)+ + f(k + j) ··· ··· k j = f(n)+ f(k + n). n=1 n=1 X X j k k j Thus, n=1 f(n)+ n=1 f(j + n)= n=1 f(n)+ n=1 f(k + n) WeP will now beginP to prove the QuadraticP ReciprocityP Theorem by ﬁrst proving the following proposition.

16 Proposition 1. Let p be an odd prime, a be an integer such that p - a, and let ✓ denote a primitive root modulo p.Ifa ✓j (mod p) where j is an integer such that 1 j

p 1 p 1 p 1 p 1 j j 2w 2 2 2 a ✓n 2 2 2 a ✓n j = ✓n + ✓n { }p = w ✓n + ✓n { }p . (3) p { }p p { }p p p { }p { }p p n=1 n=1 n=1 n=1 n=1 n=1 X X X ⌫ ✓ X X ◆ X ⌫ p 1 p 1 2 n 2 n j n a ✓ p Let x = w n=1 ✓ p + ✓ p.Sincej Z and { } Z,wededucethat { } n=1{ } 2 p 2 n=1 ⌫ 2x P P X 2x Z.Sincep is odd and p - 2, we have that p x.Therefore, is an even integer. p 2 | p Considering equation (3) modulo 2 yields the following:

p 1 p 1 2 a ✓n 2 a ✓n j { }p { }p (mod 2). ⌘ p ⌘ p n=1 n=1 X ⌫ X ⌫ Therefore, n p 1 a ✓ p 2 a n=1 { } =( 1)j =( 1) p . p P ⌫ ✓ ◆

The following proposition is a version of Gauss’ Lemma.

Proposition 3 Let p be an odd prime, a be an odd integer with p - a and ✓ be a primitive root modulo p. Then p 1 an 2 a n=1 =( 1) p . p P ⌫ ✓ ◆ Proof. Let p be an odd prime, a be an odd integer with p - a and ✓ be a primitive root modulo p.Since✓ is a primitive root modulo p,weknowthatp - ✓.Therefore,by(4)

18 a ✓n a ✓n we obtain { }p + { }p = a 1. Since a is odd, we know that a 1 is even. p p ⌫ ⌫ a ✓n a ✓n Therefore, we have that { }p { }p (mod 2). It follows that p ⌘ p ⌫ ⌫

p 1 p 1 2 a ✓n 2 a { }p min( ✓n , ✓n ) (mod 2). p ⌘ p { }p { }p n=1 n=1 X ⌫ X ⌫ n n p 1 From (2) we have that ✓ p = p ✓ p.Therefore,wededuce1 min( ✓ p, ✓ p) 2 . Since ✓ is a primitive root{ } modulo{p,weknowthatthesets}  { } { } 

p 1 p+1 p+3 1 2 p 1 ✓ , ✓ ,..., ✓ 2 and ✓ 2 , ✓ 2 ,..., ✓ { }p { }p { }p { }p { }p { }p ⇢ ⇢ p 1 n +n are disjoint. By (1) from above and since ✓ ✓ 2 (mod p), we know that ⌘

p 1 p+1 p+3 1 2 p 1 ✓ , ✓ ,..., ✓ 2 = ✓ 2 , ✓ 2 ,..., ✓ . { }p { }p { }p { }p { }p { }p ⇢ ⇢ p 1 n n p 1 n n 2 p 1 Since 1 min( ✓ p, ✓ p) 2 ,wededucethat min( ✓ p, ✓ p) n=1 contains 2  { } { }  p 1 { } { } distinct integers that range in value between 1 and .Therefore, 2 p 1 p 1 min( ✓n , ✓n ) 2 = 1, 2,..., . { }p { }p n=1 { 2 }

p 1 p 1 2 a 2 an It follows that min( ✓n , ✓n ) = . Therefore, by Proposition 2, we p { }p { }p p n=1 ⌫ n=1 ⌫ deduce X X p 1 n p 1 a ✓ p a 2 { } 2 an =( 1) n=1 p =( 1) n=1 p . p P P ✓ ◆ ⌅ ⇧ ⌅ ⇧

From here, Barnard uses the same argument as is used in the ﬁrst proof we analyzed where we considered lattice points to complete the proof of the Quadratic Reciprocity Law.

5 Conclusion

The Quadratic Reciprocity Law continues to be a valuable theorem in number theory today. As has been shown previously, the theorem is useful in determining whether an integer a is quadratic residue modulo p where p is an odd prime and p - a.TheQuadratic Reciprocity Law has been implemented in several modern applications. The theorem is ex- tremely useful in the Goldwasser-Micali public key cryptosystem, because the key chosen for this cryptosystem must be a quadratic nonresidue modulo p and q where p, q are very large distinct primes. The Quadratic Reciprocity Law has also been applied in several primality

19 tests, such as Euler’s test.

The three proofs we have analyzed demonstrate the wide variety of techniques that have been implemented over the past 300 years to prove the Quadratic Reciprocity Law. There have been proofs published in many branch of mathematics. Gauss states that the reason he provided multiple proofs was to be able to prove the reciprocity law in general for powers higher than quadratic, for example, the cubic, quartic, and octic reciprocity laws. Gauss was successful in ﬁnding such a generalization in his sixth published proof of the theorem.

After 300 years, why are mathematicians still attempting to provide new proofs of this important theorem today? Mathematician Yuri Manin provides an answer to this question in [9] by saying

“... one proof is suciently convincing. The point is, that proving is the way we are discovering new territories, new features of the mathematical landscape.”

Mathematicians also continue to provide proofs of this theorem because many believe it is one of the most beautiful theorems in number theory. David Burton says about the Quadratic Reciprocity Law in [3] on page 169,

“For those who consider the theory of number ‘The Queen of Mathematics,’ this (QRL) is one of the jewels in her crown.”

References

[1] Barnard, Virgil, “A Proof of Quadratic Reciprocity.” The American Mathematical Monthly 122.6 (2015): 588-592. Web.

[2] Baumgart, Oswald, The Quadratic Reciprocity Law: A Collection of Classical Proofs, Springer International Publishing, [c2015].

[3] Burton, David M., Elementary Number Theory 6th ed. New York, NY: McGraw-Hill, 2007.

[4] Gauss, C. F., Disquisitiones Arithmeticae (trans A. A. Clarke), Yale University Press, New Haven, 1966.

[5] Kim, Sey Y., “An Elementary Proof of the Quadratic Reciprocity Law.” The American Mathematical Monthly 111.1 (2004): 48-50. Web.

[6] Koshy, Thomas, Elementary Number Theory with Applications 2nd ed. Academic Press, 2007.

[7] Legendre, A.M., Essai sur la th´eorie des nombres,Dprat,Paris(1798);cf.p.

[8] Lemmermeyer, F., Reciprocity Laws, Springer-Verlag, Berlin Heidelberg [c2000].

20 [9] Manin, Y. I., Good proofs are proofs that make us wiser, DMV-Mitteilungen 2 (1998), 40-44.

[10] Rosen, Kenneth H., Elementary Number Theory and its Applications 3rd ed. Murray Hill, NJ: AT&T Bell Laboratories, 1993.