QUADRATIC RECIPROCITY

QUADRATIC RECIPROCIT POOJA PATEL Abstract. This aer is an self-contained exosition of the law of uadratic recirocity. We will give two roofs of the Chinese remainder theorem and a roof of uadratic recirocity. These roofs will mainly involve concets from basic algebra and elementary number theory. Contents 1. Introduction 1. Fermat's two suares theorem 1 3. Fermat's Little Theorem, Euler's Theorem, and Wilson's Theorem 3 4. Chinese Remainder Theorem 6 5. Proof of uadratic recirocity 10 References 13 1. Introduction There are numerous aroaches to roving uadratic recirocity, and here we resent a very elegant one discovered by George Rousseau [3]. This aer uses the language of basic abstract algebra and we will see its interlay with elementary number theory throughout. This exosition can also be used as a sulement material for learning basic algebra for the rst time. First we resent a roof [1] of an old theorem due to Fermat: rime numbers in the form 4m + 1 can be exressed as the sum of two suares. Later on, we will see that this gives us a secial case (Lemma 5.4) of Euler's criterion (Theorem 5.) Next, we will resent Fermat's Little Theorem and Euler's Theorem. In the language of elementary grou theory, both of these theorems will become easy corollaries of Lagrange's Theorem. Also we will rove Wilson's Theorem and its corollary. After these, we will rove the Chinese Remainder Theorem in two aroaches: one using only elementary number theory, the other reformulated and roven in the language of rings and ideals. Finally we will resent a roof of the Theorem of Quadratic Recirocity.. Fermat's two suares theorem The main result of this section is the following theorem. Theorem.1. A rime number can be written as a sum of two suares if and only if it is of the form 4m + 1 for some natural number m. Date: August 4, 017. 1

Denition.. An euivalence relation on a set S is a binary relation that satises the following roerties: (1) Reexivity: for all a S; we have a a. () Symmetry: for all a; b S; if a b; then b a: (3) Transitivity: for all a; b; c S; if a b and b c; then a c: Given a set S and an euivalence relation on S. Let a S. We dene the euivalence class of a S to be [a] : fb S j b ag: Then a is called a reresentative of [a]. We denote the collection of all euivalence classes by S : f[a] j a Sg. Here is an examle. Let be a rime number. Let ZZ Z, where is dened as the following: for all a; b Z; a b if and only if j(a b), i.e. there exists k Z; such that k a b. We can denote ZZ by f0; 1; ; 1g. Then we dene (ZZ) to be the collection of nonzero elements f1; ; 1g. Lemma.3. For rimes of the form 4m + 1, there exists s Z such that s 1 mod : Proof. Let 4m + 1 for some m N. Then for all x (ZZ), we consider the set fx; x; x; xg where x is the additive inverse of x and x is the multilicative inverse of x. First we notice that x 6 x because is an odd rime number. For the same reason, x 6 x. Thus jfx; x; x; xgj or 4. Now let's consider the two cases for which jfx; x; x; xgj. Case 1: x x. This imlies x 1, so (x 1)(x + 1) 0 in ZZ. This means x could only be 1 or 1. Let A 0 : f1; 1g: Case : x x. This imlies x 1: Now it suces to show that Case haens for some x (ZZ). Let A f fx; x; x; xg j x (ZZ) g; B f A A j jaj g; C A B f A A j jaj 4g: Since 1 j(zz) j j S AA Aj P AA jaj jbj + 4jC j is divisible by 4, we have jbj is even. Also we already know that A 0 B, so jbj, hence there exists A 1 B with A 1 6 A 0, say A 1 fs; sg for some s (ZZ). Then A 1 falls into Case, so s 1. Proosition.4. Any rime number in the form 4m + 1 is a sum of two suares. Proof. We know there are (b c + 1) airs of integers (x 0 ; y 0 ) with 0 x 0 ; y 0. By the igeon hole rincile, for every s Z, there are two airs (x 0 ; y 0 ); (x 00 ; y 00 ) such that x 0 sy 0 x 00 sy 00 mod : Hence x sy mod, where x jx 0 x 00 j; y jy 0 y 00 j: By Lemma.3, we can take s to be such that s 1 mod. Then x s y y mod. Therefore we have found (x; y) Z with 0 < x + y < and x + y 0 mod. Since is the only rime between 0 and divisible by, we conclude that x + y.

Proosition.5. No number in the form n 4m + 3 is a sum of two suares. Proof. Let n 4m + 3 for some m Z. Assume n x + y for some x; y Z. Then by a simle calculation we see that x 0 or 1 mod 4: Similarly, y 0 or 1 mod 4. Hence n 6 3 mod, so we have a contradiction. Thus n is not a sum of two suares. 3. Fermat's Little Theorem, Euler's Theorem, and Wilson's Theorem In this section we will see the interlay between basic algebra and elementary number theory. We will see that Fermat's Little Theorem follows easily from Lagrange's Theorem. Denition 3.1. A grou is a set G together with a binary oeration \ " : G G! G, satisfying the following three axioms: (1) Associativity: if a; b; c S, then a (b c) (a b) c. () Identity: there exists a uniue e such that for all a G we have a e e a a. (3) Inverse: for all a G there exists a 1 such that a a 1 a 1 a e where e is the identity. A grou is Abelian if it is commutative. That is, for all a; b S, we have a b b a. Examle 3.. (1) (ZnZ, +) () ((ZZ) ; ) Denition 3.3. The order of a grou G is the cardinality of G, in other words it is the number of elements in G. It is denoted by jgj. Denition 3.4. H is a subgrou of a grou G if H is a subset of G and forms a grou under the binary oeration inherited from G. This is denoted by H G. Theorem 3.5 (Lagrange's Theorem). Suose H is a subgrou of a nite grou G. Then jhj divides jgj. Proof. To rove this theorem, we will show that all the cosets of H form a artition of G, and each coset has the same cardinality. Claim: Let k H. Dene kh fkh : h Hg. Then kh H. Proof: For all h H, we have k h H so kh H. We also know that for all h H, h k k 1 h kh; so H kh. Therefore kh H. Next we rove a necessary and sucient condition for two cosets to coincide. Claim: Let ah and bh be any two cosets. Then ah bh if and only if b 1 a H. Proof: ( ) ) Suose ah bh. Let c H. Then there exists d H, ac bd so b 1 a dc 1 H. ( ( ) Let b 1 a H. Then by claim 1, H b 1 ah. So bh b(b 1 ah) ah. Now we will show that if two cosets are not eual then their intersection is emty. Claim: Suose a; b G where ah 6 bh. Then ah \ bh. Proof: Assume ah \ bh 6. So we can take some g ah \ bh. So there exists c; d H such that g ac bd. So b 1 a dc 1 H. Thus by claim, we have ah bh. Furthermore, we show that cosets of a subgrou form a artition of the grou. Claim: If H fch : c Gg, then it forms a artition of G. Proof: It is clear that [ cg ch G, because H, as a subgrou of G, contains e. We also know that ah 6 bh imlies ah \ bh. Thus, all of the distinct ch are disjoint. 3

Now we show that all cosets have the same cardinality. Claim: For all a; b G, jahj jbhj. Proof: It suces to show that for all a G, we have jahj jhj. Consider the function f : H! ah; h 7! a h: We see that it is a bijection because it has a two-sided inverse f 1 : ah! H; g 7! a 1 g: Now our claim follows. From the above claims we know that the left cosets of G form a artition of G, each coset with cardinality jhj. Therefore jhj divides jgj. Theorem 3.6. Let G be a nite grou, with jgj n. Let a G. Then where e is the identity element in G. a n e; Proof. Let H be the cyclic subgrou generated by a, i.e. H fe; a; a ; ; a m 1 g. By Lagrange's theorem, mjn, so there exists some k N such that n m k. Thus a n a m k (a m ) k e k e: Corollary 3.7 (Fermat's little theorem). Let be a rime number, and let a Z. Then a a mod. Proof. We can rove this using Theorem 3.6. Let a Z. It suces to show that a and a reresent the same class in ZZ. If j a, then both a and a will reresent 0 ZZ, and we are done. If - a, then a reresents some [a] (ZZ). Let's also denote it by a. By alying Theorem 3.6 with G (ZZ), we have a 1 1. Thus a a as elements in (ZZ). This roves our statement. Our next goal is to rove Euler's Theorem, a generalization of Fermat's Little Theorem. In order to rove this theorem, we need to dene what a ring is. Denition 3.8. A ring R is an additive Abelian grou that also has a multilicative oeration \ " which is associative and distributive: i.e. For all a; b; c R, we have a (b c) (a b) c a (b + c) a b + a c (b + c) a b a + c a In this aer we always assume a ring has multilicative identity: i.e. There exists a uniue 1 R R; such that for all a R we have 1 R a a 1 R a: A ring is commutative if for all a; b R, we have a b b a. Examle 3.9. (1) (ZnZ; +; ) () Polynomials with coecients in any eld. Denition 3.10. Let R be a ring. Then we say a is a unit in R, if there exists b R such that ab ba 1. We denote the collection of all units in R by R. Examle 3.11. (1) (ZnZ) fa ZnZ j a is co-rime to n:g 4

Lemma 3.1. Let R be a ring, and let U R : Then for all a; b U, we have a b U. Proof. Since a is a unit, there exists c such that a c 1. Similarly there exists d such that b d 1. Then (a b) (d c) a (1) c a c 1, so a b U. Remark 3.13. From the above lemma and the denition of rings, we see that the units of a ring form a multilicative grou. Theorem 3.14 (Euler's theorem). Let m, and let a Z be co-rime to m (i.e. gcd(a; m) 1). Then we have a (m) 1 mod m where (m) is the number of integers in f0; 1; ; ; m m. 1g that are co-rime to Proof. Similar to the roof of Corollary 3.7, we can also rove this by alying Theorem 3.6. Let U m (ZmZ). Then (m) ju m j. Since a is co-rime to m, we have that a reresents some [a] U m. Let's also denote this by a. Then by alying Theorem 3.6 with G U m, we have a (m) 1. This roves our statement. Another imortant result which we will use later is Wilson's Theorem. Theorem 3.15 (Wilson's theorem). If is a rime, then ( 1)! 1 mod. Proof. Each of 1; ; ; 1, when considered as elements in the grou (ZZ), has an multilicative inverse. And among these elements, only 1 and 1 are their own inverses, because if x 1 mod, then x 1 (x + 1)(x 1) 0 mod ; so x 1 mod. Thus all factors in the dening roduct of ( 1)! cancel out with their inverses, excet for 1 and 1. Hence ( 1)! (1) ( 1) 1 mod. Corollary 3.16. If is a rime, then [( 1 )!] ( 1)( 1) 1 mod : Proof. By Wilson's theorem, Therefore, we get 1 ( 1)! 1 ( 1) ( ( 1)) ( )( 1) [(( 1))!] ( 1) 1 mod : ( 1 )! ( 1)( 1) 1 mod : Next we dene what a eld is, a concet that was already alied in section 1. Denition 3.17. A eld F (F; +; ) is a commutative ring (with multilicative identity) such that every non-zero element has a multilicative inverse. This means all non-zero elements are units in a eld. Examle 3.18. (ZZ; +; ). See the following roosition. Proosition 3.19. Let be a rime. Then (ZZ; +; ) is a eld. 5

Before we rove this roosition, we rst rove an imortant lemma which will be alied in later sections as well. Lemma 3.0 (Bezout's Lemma). If gcd(a; b)c, then there exists s; t Z such that as + bt c: Proof. Since gcd(a; b) gcd( a; b), we can assume a and b are ositive. If we were to use the Euclidean algorithm to nd gcd(a; b), we would get the following seuence of uotients and remainders: a b 1 + r 1 b r 1 + r r 1 r 3 + r 3. r n r n 1 n + r n r n 1 r n n+1 + 0 From Euclidean algorithm we know that r n c. We will use strong induction on n to rove that r n is a linear combination of a and b for 1 k n. First, for n 1 we have r 1 a b 1 so we know that P 1 is true. Now for the inductive ste, let ositive integer k < n, we assume that r i is a linear combination of a and b for all 1 i k. Since r k+1 r k 1 r k k+1, and by the inductive hyothesis we know that r k 1 and r k are both linear combinations of a and b, so r k+1 is also a linear combination of a and b. Therefore, by strong induction r n is a linear combination of a and b. Proof of Proosition. We know most of the axioms holds trivially, so it suces to show that every element in (ZZ) has a multilicative inverse. Let a (ZZ), and let a Z be a reresentative of a. Since is a rime, gcd(a; )1. This imlies that there exists s; t Z such that as + t 1: Thus we have ab 1, i.e. a has a multilicative inverse. 4. Chinese Remainder Theorem Theorem 4.1 (Chinese remainder theorem). If 1 t where gcd( i ; j )1 for any i 6 j, and we have some integers b 1 ; b ; ; b t, we can always nd x Z such that x b i mod i for all i. Any two such solutions will dier by a multile of. We will rst begin by roving a lemma. Lemma 4.. If 1 ; ; ; t divide n and gcd( i ; j )1 for i 6 j, then 1 t will divide n. Proof. For t 1 this holds. Let t and suose the lemma holds for t 1 so that 1 t 1 divides n. Let 0 1 t 1 and 0 t. Since t is relatively rime to 1 ; ; ; t, we know that gcd( t ; 0 ) 1: Thus by Lemma 3.0 in the end of revious section, there exists a; b Z such that a t + b 0 1: 6

Thus n na t + nb 0. Since 0 divides n, we get that divides na t. Furthermore, since t divides n, divides nb 0. Thus divides n and this comletes the induction. Now we are ready to rove the Chinese Remainder Theorem. Proof of Chinese Remainder Theorem. Let 1 t, and for all i; let n i i. Then we know that there exists a i ; b i Z such that a i i + b i n i 1: We also know that b i n i 1 mod i, and b i n i 0 mod j for i 6 j: Let tx x 0 r i b i n i : i0 Then for each i, x 0 r i b i n i r i mod i : This imlies that x 0 is a solution. Suose we have another solution x 1. Then x 1 x 0 0 mod i for i 1; ; ; t: Thus 1 t divides (x 1 x 0 ) by lemma 4.. Next we formulate a ring theoretic corollary of the Chinese Remainder Theorem. Denition 4.3. Let R; S be two rings. Then a ma ' : R! S which satises the following three roerties for all a; b R is called a ring homomorhism: (1) '(a + b) '(a) + '(b) () '(a b) '(a) '(b) (3) f(1 R ) 1 S If a homomorhism f is also bijective, then we call it a ring isomorhism. Corollary 4.4. Let ; be two co-rime natural numbers. Then (ZZ) (ZZ) (ZZ) as rings. Proof. Let and Consider the ma : Z! (ZZ) x 7! x mod ; : Z! (ZZ) x 7! x mod ; : Z! (ZZ) x 7! x mod : ' : Z! (ZZ) (ZZ) x 7! (x) ; (x) : We rst show that ' is a ring homomorhism: Let x; y Z. Then '(x + y) (x + y); (x + y) (x)+ (y); (x)+ (y) (x); (x) + (y); (y) '(x) + '(y); '(x y) (x y); (x y) (x) (y); (x) (y) (x); (x) (y); (y) '(x) '(y): '(1) (1); (1) 1 ZZ ; 1 ZZ 1(ZZ) (ZZ) : 7

Now let x; y be two integers with (x) (y). Then divides (x y) evenly, so do and. Thus (x) (y) and (x) (y), hence '(x) '(y). So ' decends to a ring homomorhism e' : (ZZ)! (ZZ) (ZZ): Furthermore, e' is surjective by the Chinese Remainder Theorem. Now observe that j(zz) (ZZ)j j(zz)j < 1; so e' is a surjective ma between two nite sets with the same cardinality. Thus e' is also injective, hence bijective. Therefore e' is an isomorhism. Next we formulate a roof of the Chinese Remainder Theorem that uses rings and ideals. Denition 4.5. A subring Q of ring R is a subset of R which also forms a ring with the oerations inherited from R. A subring also has that 1 Q 1 R. Denition 4.6. An Abelian (additive) subgrou I of R is called a left ideal of R if it absorbs multilication by elements of R, i.e. for all a I, r R, we have r a I. Similarly, if for all a I, r R, we have a r I, we say I is a right ideal of R. If an ideal is both a right and left ideal, then it is called a two-sided ideal. We simly call this an ideal of R. Remark 4.7. Let I and J be two ideals of a ring R. The sum of the two ideals, dened as I + J fx + y; x I; y Jg, is an ideal. Furthermore, the intersection of nitely many ideals is also an ideal. Denition 4.8. Let f : R! S be a ring homomorhism. We dene the kernel of f as ker f fr R; f(r) 0g, and the image of f as imf ff(r) j r Rg. Then we can check that the image of f is a subring of S, and that the kernel of f is a two sided ideal of R. We also notice that f is injective if and only if ker f 0: Denition 4.9. Let I be an ideal of a ring R. We can dene the uotient ring as RI : fr + I; r Rg. The addition and multilication oerations on RI are dened as (r + I) (s + I) r s + I, and (r + I) + (s + I) (r + s) + I, with identity 1 R + I and zero element 0 R + I. Theorem 4.10 (First isomorhism theorem for rings). Let R be a ring. Let ' : R! S be a ring homomorhism. Then the ma is a well dened isomorhism. f : R ker(')! im' r + ker ' 7! '(r) Proof. We rst need to show that the ma f is well dened. Let r; r 0 R with r + ker ' r 0 + ker '. Then r r 0 + s for some s ker f, so '(r) '(r 0 + s) '(r 0 ) + '(s) '(r 0 ) + 0 '(r 0 ): Since ' is a ring homomorhism, we can easily check that f is also a ring homomorhism. Now we show that f is bijective. First it is clear that f is surjective. This is because for all t im'; there exists r R; such that '(r) t; so we have f(r+ker ') t: Furthermore, we can show that f is injective. Let (r+ker ') ker f. Then '(r) f(r + ker ') 0, so r ker '. Thus r + ker ' ker ': This means ker f is trivial, hence f is injective. Therefore, f is an isomorhism as desired. 8

Before we state and rove the ring theoretic version of the Chinese Remainder Theorem, we need the following two denitions. Denition 4.11. Let R 1 ; : : : ; R n be rings. We dene the ring of direct roduct n i1 R i : f(a 1 ; ; a n ) j each a i R i g with comonent-wise addition and multilication. The zero element is (0 R1 ; ; 0 Rn ), and the identity is (1 R1 ; ; 1 Rn ). We can easily check that this forms a ring. Denition 4.1. The ideals I and J of ring R are co-rime if I + J R. Remark 4.13. This terminology is motivated by the examle R Z; I nz; J mz. By Bezout's lemma, I and J are co-rime if and only if n and m are co-rime. Now we can rove the ring theoretic version of the Chinese Remainder Theorem. Theorem 4.14 (Chinese remainder theorem). Let R be a commutative ring with multilicative unit. Let I 1 ; : : : ; I n be airwise co-rime ideals in R, i.e. I i + I j R for all i 6 j. Then we have R \ n i1 I i n i1 Proof. We can dene a ring homomorhism f as f : R! n i1 RI i RI i : a 7! (a + I 1 ; : : : ; a + I n ): First we show that f is surjective. Let a 1 ; ; a n R. Now we look for a R with a a i mod I i for each i. Fix i f1; ; ; ng. We rst construct c i R such that c i 1 mod I i ; and c i 0 mod I j ; for all j 6 i: Let j f1; ; ; ng with j 6 i. Since I i + I j R, there exists b j I i ; d j I j such that b j + d j 1. Let c i : j6i d j : Since for each j 6 i we have b j 0 mod I i ; so c i Q j6i (b j + d j ) 1 mod I i : Also, for each j 6 i we have c i 0 mod I j : This construction works for each i. Now let a a 1 c 1 + + a n c n. This imlies that we have a a i mod I i for each i. Hence f is surjective. We notice that ker f fa R j f(a) (I 1 ; ; I n )g fa R j a I i ; for all i 1; ; ; ng \ n i1 I i: Thus by the rst isomorhism theorem, we have roven the Chinese Remainder Theorem, again. Remark 4.15. Now we see that the ring theoretic corollary of the Chinese Remainder Theorem follows from this version by taking R Z; I 1 Z; I Z: 9

5. Proof of uadratic recirocity We will now give Rousseau's roof for uadratic recirocity. Denition 5.1 (Legendre Symbol). Let a (ZZ). Then we dene ( a 1; if a is a suare in (ZZ) ; : 1; if a is a suare in (ZZ) : And for all a Z not divisible by, we can dene a naturally. Theorem 5. (Euler's criterion). Let be an odd rime. Then, for Z not divisible by, we have 1 mod : Proof. Let x 1 : Then x 1 1 mod, so x 1 mod. Now it suces to show that is a suare in (ZZ) () 1 1 mod : Suose is a suare in (ZZ), i.e. there exists a such that Then by Fermat's theorem, a mod and 1: 1 a 1 1 mod : To show the converse, suose 1 1 mod. Since ZZ is a eld, the olynomial y 1 1 has at most 1 roots in ZZ. In fact, these roots will be 1 ; ; ; ( 1 ). Similar to the roof of the other direction, we can aly Fermat's Little Theorem to nd that all these numbers are roots of the olynomial y 1 1. Since we have 1 of them listed here, it suces to show that these roots are all distinct in ZZ. Let x; y f1; ; ; 1 g. Assume x y mod. Then x y 0 mod, so (x y)(x + y) 0 mod : Since both x; y are less or eual to 1, we have x + y 6 0 mod. Since ZZ is an integral domain, we have x y 0 mod. Also from the range of x and y, we know that 0 jx yj 1. So x y 0, hence x y. Thus all the roots of the olynomial y 1 1 in ZZ are suares in (ZZ). Now since is a root of y 1 1, we have that is a suare in (ZZ). Thus, 1. We will now rove three lemmas that will be used in the roof of uadratic recirocity. Lemma 5.3 (The multilicative roerty of ). Let be a rime number. Let 1 ; Z, both not divisible by. Then we have 1 1 10 1 :

Proof. From Euler's criterion we know Thus 1 1 1 mod ; and 1 1 1 1 mod ( 1 ) 1 mod 1 mod : 1 mod : Lemma 5.4. For all odd rime, we have ( 1 1; if 4n + 1; 1; if 4n + 3: 1 Proof. From Euler's criterion we know that If 4n + 1, then 1 Since is not, we have 1 Lemma 5.5. For an odd rime, we have Proof. Suose 4n + 1. Then we have ( 1) n n (k 1) ( 1) 1 mod. ( 1) n 1 mod : 1. Similarly, if 4n + 3, then 1 (( 1) 1 4 mod ; if 4n + 1; ( 1) +1 4 mod ; if 4n + 3: n n n Q Q n Since (k) n n ( to get: ( 1) n " n (k 1) #" n [ (k 1)] (4n k + ) (n + k) n n [ (k 1)] mod n kn+1 [4n (n + 1 k) + ] k mod : 1. k), we can multily these two euations together k # n " n 11 k #" n kn+1 k # mod

However, we also have " n (k 1) #" n k # " n " n (k 1) k #" n kn+1 #" n k # k : #" n kn+1 So we have ( 1) n n mod, that is 1 ( 1) 1 4 mod : Therefore, by Euler's Criterion 1 ( 1) 4 mod : Similarly for 4n + 3 +1 we can rove ( 1) 4 mod : Thus, is a suare modulo if and only if 8m 1. Theorem 5.6 (The Law of Quadratic Recirocity). Let, be two distinct odd rimes. Then, ( 1) 1 1 : Proof. Let us consider the following set A fx N j 1 x 1 ; where x is relatively rime to both and g: In other words, all elements in A reresent units in (ZZ). We also know for all u (ZZ) ; there exists a uniue x such that either x reresents u or x reresents u. By a straightforward calculation we have (Q Q 1 x k) 1 ( 1 k) [( 1)!] 1 [( 1)]! Q ( 1) 1 1 xa k 1 mod [( 1)]! where we have used the facts ( 1)! 1 by Wilson's theorem and 1 by Euler's criterion. Similarly we have x ( 1) 1 mod : This means (x; x) xa xa ( 1) 1 ; ( 1) 1 k mod (; ): From the Chinese Remainder Theorem, we know that there is a ring isomorhism ' : ZZ! ZZ ZZ. Let B f(a; b) N j 1 a 1; and 1 b ( 1)g: Then, each element in B reresents an element in (ZZ) (ZZ). We can also see that for each x A; exactly one of '(x); '(x) is reresented by an element in B. Thus we have (x; x) (a; b): We also know (a;b)b xa '(x) xa (a; b) (a;b)b [( 1)!] 1 ; [( 1 )!] 1 : 1 #

By a corollary of Wilson's Theorem we know that [( 1 )!] ( 1)( 1) 1 mod : Thus, (a;b)b (a; b) ( 1) 1 ; ( 1) 1 1 ( 1) 1 Q From the calculation above, and the relation xa (x; x) Q (a;b)b have Therefore, ( 1) 1 1 ( 1) : ( 1) 1 ( 1) 1 ( 1) ( 1 )( 1 ) : ( 1) 1 1 : (a; b), we Acknowledgments I would like to thank my mentor, Boming Jia, for teaching me the Chinese Remainder Theorem and for heling me extensively with this aer. I would also like to thank Matthew Emerton for teaching me about uadratic recirocity, and Peter May for organizing the REU and reviewing this aer. References [1] Aigner, M., Ziegler, G. M., & Quarteroni, A. (010). Proofs from the Book (Vol. 74). Berlin: Sringer. [] Dummit, D. S., & Foote, R. M. (004). Abstract algebra (Vol. 3). Hoboken: Wiley. [3] Rousseau, G. (1991). On the uadratic recirocity law. Journal of the Australian Mathematical Society, 51(3), 43-45. [4] Stillwell, J. (00). Elements of number theory. Sringer Science & Business Media. 13