Jacobi sybols efiitio Let be a odd positive iteger If 1, the Jacobi sybol : Z C is the costat fuctio 1 1 If > 1, it has a decopositio ( as ) a product of (ot ecessarily distict) pries p 1 p r The Jacobi sybol : Z C is give by a a a Note: The Jacobi sybol does ot ecessarily distiguish betwee quadratic residues ad oresidues That is, we could have ( a ) 1 just because two of the factors happe to be 1 For istace, ( 1)( 1) 1, 15 3 5 but is ot a square odulo 15 The followig properties of the Jacobi sybol are direct cosequeces of its defiitio Propositio 1 Let, be positive odd itegers ad a, b Z The 1 (i) 1; a (ii) 0 (a, ) > 1; a b (iii) a b (od ) ; ab a b (iv) ; (v) ( a ) ( a ) a ; 1 p 1 p r
(vi) (a, ) 1 a b b Proof Exercise Theore Let, be positive odd itegers The 1 (i) ( 1) 1 ; (ii) ( 1) 1 ; (iii) ( 1) 1 1 Proof The first two forulas are trivially true whe 1 ad so is the third if 1 or 1 or if (, ) > 1 We assue that, > 1 ad (, ) 1 Thus p 1 p r ad q 1 q s for soe pries p i ad q j ad p i q j for all 1 i r, 1 j s The p i (1 + (p i 1)) 1 + (p i 1) + (p i1 1)(p i 1)+ 1 i 1 <i r products of 3, 4 ad so o factors Sice is odd, so are the pries p i Therefore p i 1 0 (od ) ad (p i1 1)(p i 1) 0 (od 4) Therefore all the ters i the above su that are iplicit are also divisible by 4 Hece 1 + (p i 1) (od 4), which is to say 1 (p i 1) (od 4) Sice ad the p i s are odd, it follows that 1 0 (od ) ad p 1 1 0 (od ), 1 i r Thus we ca divide each ter above by ad still get itegers It follows that so Siilarly, 1 ( 1) 1 ( 1) r p i 1 p i 1 ( 1) p i 1 (od ), (1) 1 p i 1
p i ( 1 + (p i 1) ) 1 + (p i 1) + (p i 1 1)(p i 1)+ 1 i 1 <i r products of 3, 4 ad so o factors We use agai the fact that both ad the p i are odd That eas that 1 ( 1)( + 1) is the product of two cosecutive eve itegers, so oe of the is divisible by 4 Thus 1 0 (od ) ad likewise p i 1 0 (od ), 1 i r It follows that the product of two or ore factors i the above suatio is divisible by 64, hece 1 (p i 1) (od 64) Moreover each ter is divisible by, so 1 p i 1 (od ), as itegers It follows that ( 1) 1 ( 1) r p i 1 ( 1) p i 1 p i The last part of the theore, i the case, > 1 ad (, ) 1, is equivalet to ( 1) 1 1 But where t By (1), we have t 1 p i 1 1 ( pi qj q j p i qj 1 ) ( 1) p i 1 q j 1 ( 1) t p i 1 q j 1 (od ) ad the quadratic reciprocity law follows Jacobi sybols have ay applicatios The followig result is a exaple of how they ca be used i the study of certai iophatie equatios 3
Propositio 3 The iophatie equatio y x 3 + k has o solutio if k (4 1) 3 4 ad o prie p 3 (od 4) divides Proof We argue by cotradictio Assue that (x, y) is a solutio Sice k 1 (od 4), it follows that y x 3 1 (od 4) But y 0, 1 (od 4), so x caot be eve ad x 1 (od 4) Therefore x 1 (od 4) Let a 4 1 The a 1 (od 4) ad k a 3 4 We have so y x 3 + k x 3 + a 3 4, y + 4 x 3 + a 3 (x + a)(x ax + a ) () Give that x 1 (od 4) ad a 1 (od 4), we have that the last factor x ax + a 3 (od 4) Thus x ax+a is odd ad it ust have soe prie divisor p 3 (od 4) But () iplies that p y + 4, ie 4 y (od p) so 4 1 p O the other had, sice p 3 (od 4), we have that p ad therefore 4 1 1 (cotradictio!) p p Propositio 4 If, are positive odd itegers ad is a iteger with 0, 1 (od 4) such that (od ), the Proof First we treat the case whe 1 (od 4) If > 0, the ( 1) 1 1 But 1 is eve, hece The arguet holds for ay positive odd iteger, ad it ca therefore be applied just as well to The result follows iediately sice (od ) 4
If < 0, set d The d > 0 ad d 3 (od 4), so d+1 is eve We have d 1 d ( 1) 1 ( 1) 1 d 1 ( 1) 1 d+1 d d Sice the sae holds for, the result follows fro the fact that (od d) Now cosider the other case, 0 (od 4) It follows that a b for soe positive odd iteger b ad a d If > 0, the a b ( 1) 1 a ( 1) 1 b 1 b Siilarly, ( 1) 1 The result would follow if we showed that ad 1 1 b 1 a ( 1) 1 b 1 b a 1 a (od ) (3) 1 b 1 (od ) (4) We have 1 b 1 1 b 1 b 1 ad this is eve sice 4 Thus (4) is proved For the other relatio, we have 1 a 1 a a ( )( + ) a Now + ad a Thus 0 (od 16) whe a 3 ad (3) follows i this case O the other had, if a, the a is agai eve ad we are doe (We used the fact that Z) If < 0, set d The d > 0 ad d 0 (od 4) Fro above it follows that d d We also have d ( 1 )( d ) ( 1) 1 d 5
ad, siilarly, The result follows fro the fact that 1 1 (od ) ( 1) 1 d 4 { (od ) 0 (od 4) Theore 5 Let 0, 1 (od 4) be a ozero iteger The there exists a uique group hooorphis χ : (Z/Z) {±1} such that χ ([p]) (the Legedre sybol odulo p) for all odd pries p p Furtherore, χ ([ 1]) { 1 if > 0; 1 if < 0 Proof First we show existece Let χ : (Z/Z) {±1}, χ([a]) where a (od ) is a odd positive iteger We eed to show that this is a well-defied ap, ad for that we eed to prove the followig two facts Clai 1 For ay (a, ) 1 there exists a positive odd iteger a (od ) Clai If, are positive odd itegers ad (od ), the The secod clai is a iediate cosequece of Propositio 4 The first oe, is also easy There exists soe iteger k for which a + k > 0 If is eve, the a has to be odd ad a + k is odd ad positive If is odd, the either a + k or a + k + is both odd ad positive The ap χ is clearly a group hooorphis sice the Jacobi sybol is copletely ultiplicative The coditio o pries is just as clear Now we have to prove uiqueess Assue that f : (Z/Z) {±1} is a group hooorphis with f([p]) for ay odd prie p Clearly f() 1 Also, for p ay odd iteger > 1, we have p 1 p r for soe odd pries p 1,, p r The f([]) f([p 1 ]) f([p r ]) χ([]) p 1 6 p r
Sice we have show that every class [a] (Z/Z) cotais a positive odd iteger, it follows that f([a]) χ([a]) for all [a] (Z/Z) The proof for the expressio of χ ([ 1]) is left as a exercise 7