Also an integer a is a quadratic resisdue modulo n if

Info icon This preview shows pages 62–64. Sign up to view the full content.

View Full Document Right Arrow Icon
Also, an integer a is a quadratic resisdue modulo n if and only if it is a quadratic residue modulo p i for 1 i r . That completes our investigation of the case where n is an odd positive integer. We shall not investigate the case where n is even, as it is a bit cumbersome, and is not of particular importance. 9.2 The Legendre Symbol For an odd prime p and an integer a with gcd( a, p ) = 1, the Legendre symbol ( a | p ) is defined to be 1 if a is a quadratic residue modulo p , and - 1 otherwise. For completeness, one defines ( a | p ) = 0 if p | a . Theorem 9.4 Let p be an odd prime, and let a, b Z , both not divisible by p . Then 1. ( a | p ) a ( p - 1) / 2 (mod p ) ; in particular, ( - 1 | p ) = ( - 1) ( p - 1) / 2 ; 2. ( a | p )( b | p ) = ( ab | p ) ; 3. a b (mod p ) implies ( a | p ) = ( b | p ) ; 4. (2 | p ) = ( - 1) ( p 2 - 1) / 8 ; 5. if q is an odd prime different from p , then ( p | q )( q | p ) = ( - 1) p - 1 2 q - 1 2 . Part (5) of this theorem is called the Law of Quadratic Reciprocity. Part (1) follows from Theorem 9.1. Part (2) is an immediate cosequence of part (1), and part (3) is clear from the definition. The rest of this section is devoted to a proof of parts (4) and (5) of this theorem. The proof is completely elementary, although a bit technical. The proof we present here is taken almost verbatim from Niven and Zuckerman’s book, An Introduction to the Theory of Numbers . 57
Image of page 62

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Theorem 9.5 (Gauss’ Lemma) Let p be an odd prime and a relatively prime to p . Define α j := ja rem p for 1 j ( p - 1) / 2 , and let n be the number of indices j for which α j > p/ 2 . Then ( a | p ) = ( - 1) n . Proof. Let r 1 , . . . , r n denote the α j ’s exceeding p/ 2, and let s 1 , . . . , s k denote the remaining α j ’s. The r i and s i are all distinct and non-zero. We have 0 < p - r i < p/ 2 for 1 i n , and no p - r i is an s j ; indeed, if p - r i = s j , then s j ≡ - r j (mod p ), and writing s j = k 1 a and r j = k 2 a for 1 k 1 , k 2 ( p - 1) / 2, we have k 1 a ≡ - k 2 a (mod p ), which implies k 1 ≡ - k 2 (mod p ), which is impossible. It follows that the sequence of numbers s 1 , . . . , s k , p - r 1 , . . . , p - r n is just a re-ordering of 1 , . . . , ( p - 1) / 2. Then we have (( p - 1) / 2)! s 1 · · · s k ( - r 1 ) · · · ( - r n ) ( - 1) n s 1 · · · s k r 1 · · · r n ( - 1) n (( p - 1) / 2)! a ( p - 1) / 2 (mod p ) , and cancelling the factor (( p - 1) / 2)!, we obtain a ( p - 1) / 2 ( - 1) n (mod p ), and the result follows from the fact that ( a | p ) a ( p - 1) / 2 (mod p ). 2 Theorem 9.6 If p is an odd prime and gcd( a, 2 p ) = 1 , then ( a | p ) = ( - 1) t where t = ( p - 1) / 2 j =1 b ja/p c . Also, (2 | p ) = ( - 1) ( p 2 - 1) / 8 . Proof. Let a be an integer relatively prime to p (not necessarily odd), and let us adopt the same notation as in the proof of Theorem 9.5. Note that ja = p b ja/p c + α j , for 1 j k , so we have ( p - 1) / 2 X j =1 ja = ( p - 1) / 2 X j =1 p b ja/p c + n X j =1 r j + k X j =1 s j . Also, we saw in the proof of Theorem 9.5 that the integers s 1 , . . . , s k , p - r 1 , . . . , p - p n are a re-ordering of 1 , . . . , ( p - 1) / 2, and hence ( p - 1) / 2 X j =1 j = n X j =1 ( p - r j ) + k X j =1 s j = np - n X j =1 r j + k X j =1 s j . Subtracting, we get ( a - 1) ( p - 1) / 2 X j =1 j = p ( p - 1) / 2 X j =1 b ja/p c - n + 2 n X j =1 r j .
Image of page 63
Image of page 64
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern